- Written subtraction keeps ones under ones and tens under tens. - Look at the middle column before subtracting. - No regrouping is needed when the top digit is at least the bottom digit in every column.

1. The tens digits are \(4\) from \(543\) and \(2\) from \(20\). 2. Subtract the ones: \(3-0=3\), the tens: \(4-2=2\), and the hundreds: \(5\). The difference is \(523\).

Tens digits: \(4\) and \(2\). Difference: \(523\).

- A one-digit number belongs beneath the ones column. - Decide whether the ones can be subtracted without trading a ten. - Check the larger places after finishing the ones.

1. Align the \(5\) with the ones digit \(8\). 2. The ones difference is \(8-5=3\). 3. The tens and hundreds stay the same, so the difference is \(463\).

\(468-5=463\). Only the ones place changes.

- Start with the ones column. - When there are too few ones, which nearby place can supply more? - Remember that trading changes the tens column too.

1. There are only \(2\) ones, so trade \(1\) ten for \(10\) ones. 2. There are now \(12\) ones and \(4\) tens. Subtract \(12-8=4\) ones. 3. The tens digit is \(4\) and the hundreds digit is \(3\), giving \(344\).

\(352-8=344\). Trade one ten because \(2\) ones are not enough to subtract \(8\) ones.

- Both numbers end in \(0\), so look at the tens. - A hundred can be exchanged for how many tens? - After a trade, update the hundreds column.

1. The ones difference is \(0-0=0\). 2. There are \(3\) tens, but \(7\) tens must be subtracted. Trade \(1\) hundred for \(10\) tens. 3. Now \(13-7=6\) tens remain, and \(5\) hundreds remain. The difference is \(560\).

\(630-70=560\). Trade one hundred for ten tens.

- Begin in the ones column. - If the top digit is less than the bottom digit, regroup \(1\) unit from the next place value. - Remember that regrouping leaves one fewer unit in the next column.

1. In the ones column, \(2\) is less than \(7\), so regroup \(1\) ten as \(10\) ones. Then \(12 - 7 = 5\). There are \(3\) tens left. 2. In the tens column, \(3 - 2 = 1\). 3. In the hundreds column, \(8 - 3 = 5\). 4. Therefore, \(842 - 327 = 515\). 5. Regrouping record: tens to ones

\(515\) Regrouping record: tens to ones

- Begin with the ones column and work left one place at a time. - Record each column equation, even when no regrouping is needed. - The three column result digits form the final difference.

1. a) Ones: \(9-6=3\). Tens: \(8-5=3\). Hundreds: \(7-2=5\). Therefore, \(789-256=533\). 2. b) Ones: \(3-1=2\). Tens: \(4-2=2\). Hundreds: \(5-1=4\). Therefore, \(543-121=422\). 3. c) Ones: \(7-5=2\). Tens: \(6-3=3\). Hundreds: \(9-4=5\). Therefore, \(967-435=532\).

a) Ones: \(9-6=3\); tens: \(8-5=3\); hundreds: \(7-2=5\); difference: \(533\). b) Ones: \(3-1=2\); tens: \(4-2=2\); hundreds: \(5-1=4\); difference: \(422\). c) Ones: \(7-5=2\); tens: \(6-3=3\); hundreds: \(9-4=5\); difference: \(532\).

- Align the numbers by place value. - Begin in the ones column. - Regroup when the top digit is less than the bottom digit.

1. In the ones column, \(9 - 2 = 7\). 2. In the tens column, \(3 < 6\), so regroup \(1\) hundred as \(10\) tens. Then \(13 - 6 = 7\). 3. In the hundreds column, \(6 - 4 = 2\). 4. Therefore, \(739 - 462 = 277\). Regrouping is needed in the tens column.

\(739 - 462 = 277\) Regrouping is needed in the tens column, from hundreds to tens.

- Begin in the ones column. - Regroup when the top digit is less than the digit below it. - Remember that regrouping leaves one fewer unit in the next column.

1. For part a, regroup one ten in the ones column: \(13 - 8 = 5\). There are \(6\) tens left, so \(6 - 4 = 2\). Then \(6 - 2 = 4\). Thus, \(673 - 248 = 425\). 2. For part b, regroup one ten in the ones column: \(12 - 9 = 3\). There are \(4\) tens left, so \(4 - 2 = 2\). Then \(8 - 5 = 3\). Thus, \(852 - 529 = 323\). 3. Regrouping record: a) tens to ones | b) tens to ones

a) \(425\) b) \(323\) Regrouping record: a) tens to ones | b) tens to ones

- What does a \(60\)th anniversary mean? - To find an earlier year, which operation should you use? - Use the standard subtraction algorithm.

1. A \(60\)th anniversary means that \(60\) years have passed since the wedding. 2. Use the standard subtraction algorithm: \(2024 - 60 = 1964\). 3. Therefore, the wedding year was \(1964\). 4. Regrouping record: thousands to hundreds; hundreds to tens

They were married in \(1964\). Regrouping record: thousands to hundreds; hundreds to tens

- Subtract the empty mass from the loaded mass for each truck. - Align matching place values before subtracting. - Regroup only when the top amount in a column is too small.

1. Gravel: \(21{,}450-12{,}320\). Ones \(0-0=0\), tens \(5-2=3\), hundreds \(4-3=1\). Regroup in the thousands column: \(11-2=9\), leaving \(0\) ten-thousands. The load is \(9130\,\text{kg}\). 2. Sand: \(35{,}800-24{,}150\). Regroup one hundred to make \(10\) tens, so the tens difference is \(5\); then \(7-1=6\) hundreds, \(5-4=1\) thousands, and \(3-2=1\) ten-thousands. The load is \(11{,}650\,\text{kg}\). 3. Topsoil: \(18{,}920-10{,}480\). Regroup one hundred so \(12-8=4\) tens; then \(8-4=4\) hundreds, \(8-0=8\) thousands, and \(1-1=0\) ten-thousands. The load is \(8440\,\text{kg}\). 4. Regrouping record: Gravel ten-thousands to thousands | Sand hundreds to tens | Topsoil hundreds to tens

Gravel: \(9130\,\text{kg}\) Sand: \(11{,}650\,\text{kg}\) Topsoil: \(8440\,\text{kg}\) Regrouping record: Gravel ten-thousands to thousands | Sand hundreds to tens | Topsoil hundreds to tens

- Align the total price and the cooling-unit value by place. - Work from right to left and regroup when the top amount is too small. - Check by adding the cooling-unit value back to your difference.

1. Align \(48{,}200\) and \(9350\) by place value. 2. Ones: \(0-0=0\). Tens: \(0<5\), so regroup one hundred; \(10-5=5\). 3. Hundreds: after the regrouping, \(1<3\), so regroup one thousand; \(11-3=8\). 4. Thousands: after that regrouping, \(7<9\), so regroup one ten-thousand; \(17-9=8\). 5. Ten-thousands: \(3-0=3\). The difference is \(\$38{,}850\). 6. Regrouping record: hundreds to tens; thousands to hundreds; ten-thousands to thousands

Without the cooling unit, the van costs \(\$38{,}850\). Regrouping record: hundreds to tens; thousands to hundreds; ten-thousands to thousands

- Compare the minuends from one line to the next. - The subtrahend stays the same. - Use the first result to predict the next two before calculating.

1. For a), regroup in the ones column: \(12 - 4 = 8\). Then the remaining columns give \(6 - 5 = 1\), \(4 - 0 = 4\), \(6 - 1 = 5\), \(5 - 2 = 3\), and \(8 - 3 = 5\). Thus, \(856{,}472 - 321{,}054 = 535{,}418\). 2. For b), the same lower five columns give \(35{,}418\), and \(7 - 3 = 4\) in the hundred-thousands column. Thus, \(756{,}472 - 321{,}054 = 435{,}418\). 3. For c), the same lower five columns give \(35{,}418\), and \(6 - 3 = 3\) in the hundred-thousands column. Thus, \(656{,}472 - 321{,}054 = 335{,}418\). 4. The minuend decreases by \(100{,}000\) each time while the subtrahend remains fixed, so the difference also decreases by \(100{,}000\) each time. 5. Regrouping record: a) tens to ones | b) tens to ones | c) tens to ones

a) \(535{,}418\) b) \(435{,}418\) c) \(335{,}418\) The difference decreases by \(100{,}000\) each time. Regrouping record: a) tens to ones | b) tens to ones | c) tens to ones

- Compare the two years to decide which one is later. - Use the standard subtraction algorithm to subtract the earlier year from the later year. - Check whether your difference is reasonable by adding it to the earlier year.

1. Since \(1744>1499\), the map from \(1744\) was made later. 2. Use the standard subtraction algorithm. Regroup one ten as \(10\) ones, then regroup one hundred as \(10\) tens. The column differences are \(14-9=5\), \(13-9=4\), \(6-4=2\), and \(1-1=0\). 3. Therefore, \(1744-1499=245\). 4. Regrouping record: tens to ones; hundreds to tens

The map from \(1744\) was made later. The maps were made \(245\) years apart. Regrouping record: tens to ones; hundreds to tens

- Identify the earlier and later years. - Use the standard subtraction algorithm to subtract the earlier year from the later year. - Add your difference to the earlier year to check it.

1. Use the standard subtraction algorithm to subtract the start year from the completion year. In the ones column, regroup one ten as \(10\) ones: \(10-8=2\). 2. Subtract the remaining columns: \(7-4=3\) tens, \(8-2=6\) hundreds, and \(1-1=0\) thousands. 3. Therefore, \(1880-1248=632\), so \(632\) years passed. 4. Regrouping record: tens to ones

\(632\) years Regrouping record: tens to ones

- Identify the total and the number checked out. - Subtract to find the number remaining. - Align the digits by place value and regroup when a top digit is less than the digit below it.

1. Write \(842 - 365\) with matching place values. 2. Regroup \(1\) ten as \(10\) ones: \(12 - 5 = 7\). 3. Regroup \(1\) hundred as \(10\) tens: \(13 - 6 = 7\). 4. Subtract the hundreds: \(7 - 3 = 4\). 5. Therefore, \(842 - 365 = 477\). 6. Regrouping record: tens to ones; hundreds to tens

\(477\) books remain. Regrouping record: tens to ones; hundreds to tens

- Decide whether the number of pots increases or decreases. - Use subtraction because pots are taken away. - Align the digits by place value. - Regroup when a top digit is less than the digit below it.

1. Write \(723 - 456\) with matching place values. 2. Regroup \(1\) ten as \(10\) ones: \(13 - 6 = 7\). 3. Regroup \(1\) hundred as \(10\) tens: \(11 - 5 = 6\). 4. Subtract the hundreds: \(6 - 4 = 2\). 5. Therefore, \(723 - 456 = 267\). 6. Regrouping record: tens to ones; hundreds to tens

\(267\) flowerpots remain. Regrouping record: tens to ones; hundreds to tens

- Align the same place values before subtracting. - Work from right to left and write each column equation. - Use the recorded column results to assemble the difference.

1. a) Ones: \(8-4=4\); tens: \(9-7=2\); hundreds: \(5-2=3\). Difference: \(324\). 2. b) Ones: \(7-1=6\); tens: \(6-3=3\); hundreds: \(8-5=3\). Difference: \(336\). 3. c) Ones: \(6-3=3\); tens: \(5-2=3\); hundreds: \(4-1=3\). Difference: \(333\).

a) Ones: \(8-4=4\); tens: \(9-7=2\); hundreds: \(5-2=3\); \(324\). b) Ones: \(7-1=6\); tens: \(6-3=3\); hundreds: \(8-5=3\); \(336\). c) Ones: \(6-3=3\); tens: \(5-2=3\); hundreds: \(4-1=3\); \(333\).

- Identify the earlier and later year. - Subtract the earlier year from the later year. - Align the place values and regroup where needed.

1. Subtract the earlier year from the later year: \(1886-1817\). 2. Ones: \(6<7\), so regroup one ten; \(16-7=9\). 3. Tens: after regrouping, \(7-1=6\). Hundreds: \(8-8=0\). Thousands: \(1-1=0\). 4. The dates are \(69\) years apart. 5. Regrouping record: tens to ones

\(69\) years Regrouping record: tens to ones

- In each column, determine which bottom digit produces the given result digit. - You can also use the inverse equation: difference plus subtrahend equals minuend.

1. In the ones column, \(9 - \square = 1\), so the missing ones digit is \(8\). 2. In the tens column, \(\square - 4 = 2\), so the missing tens digit is \(6\). 3. In the hundreds column, \(7 - \square = 2\), so the missing hundreds digit is \(5\). 4. The completed equation is \(769 - 548 = 221\).

\(769 - 548 = 221\)

- Work from the ones column to the left. - Account for regrouping before solving the missing digit in a column. - Check each completed subtraction with the displayed result.

1. a) Ones: \(5-3=2\). Tens: \(4-\square=1\), so the missing digit is \(3\). Thus \(745-233=512\). 2. b) Regroup in the ones column: \(12-\square=3\), so the missing digit is \(9\). The remaining columns check: \(7-4=3\) and \(6-3=3\). 3. c) Regroup across the zero in \(901\). This gives \(11-6=5\) ones and \(9-5=4\) tens. In the hundreds column, \(8-\square=4\), so the missing digit is \(4\). 4. d) Regroup in the ones column: \(17-9=8\). Then regroup in the tens column: \(11-\square=3\), so the missing digit is \(8\). The hundreds column gives \(4-1=3\). 5. Regrouping record: a) none | b) tens to ones | c) hundreds to tens; tens to ones | d) tens to ones; hundreds to tens

a) \(3\) b) \(9\) c) \(4\) d) \(8\) Regrouping record: a) none | b) tens to ones | c) hundreds to tens; tens to ones | d) tens to ones; hundreds to tens

- No regrouping is needed. - Subtract one place-value column at a time.

1. In the ones column, \(8 - \square = 1\), so the missing ones digit is \(7\). 2. In the tens column, \(\square - 4 = 2\), so the missing tens digit is \(6\). 3. In the hundreds column, \(9 - \square = 5\), so the missing hundreds digit is \(4\). 4. The completed equation is \(968 - 447 = 521\).

\(968 - 447 = 521\)

- Subtracting from \(500\) requires regrouping across two zeros. - Track how regrouping changes the hundreds and tens places before subtracting. - Use inverse addition only after you have found your corrected difference, as a check.

1. Regroup across both zeros in \(500\). 2. Ones: \(10-3=7\). 3. Tens: \(9-2=7\). 4. Hundreds: \(4-1=3\). 5. The correct difference is \(377\). The displayed result incorrectly keeps \(4\) hundreds after regrouping. 6. Regrouping record: hundreds to tens; tens to ones

The first inconsistent column is the hundreds column. The correct difference is \(377\). Regrouping record: hundreds to tens; tens to ones

- Read the top and bottom rows as the two numbers in a subtraction. - Start in the ones place and compare the available chips with the amount to subtract. - After each exchange, update the next place before continuing.

1. The chart represents \(632-247\). 2. In the ones place, \(2<7\), so the first exchange is from tens to ones. Then \(12-7=5\). 3. The tens count is now \(2\), and \(2<4\), so exchange \(1\) hundred for \(10\) tens. Then \(12-4=8\). 4. Hundreds: \(5-2=3\). The difference is \(385\).

a) The ones place b) \(385\)

- Check the ones column first. - A zero in the tens place does not mean there is no value available in larger places. - Make one exchange at a time, and keep track of the remaining hundreds and tens.

1. There are \(2\) ones, so ones are needed for subtracting \(7\). 2. The tens column is empty. Trade \(1\) hundred into \(10\) tens, then trade \(1\) of those tens into \(10\) ones. 3. Now there are \(3\) hundreds, \(9\) tens, and \(12\) ones. Subtract \(12-7=5\) ones; the difference is \(395\).

\(402-7=395\). Trade one hundred into tens, then one ten into ones.

- Keep the first number as the number you subtract from. - Start with the ones and trade a ten if needed. - Subtract each column only after completing any needed trade.

1. In subtraction, each column must subtract the bottom digit from the top digit. You cannot switch the order just because the bottom digit is greater. 2. Trade one ten for ones: \(12-8=4\) ones. The tens column now has \(4\) tens. 3. Trade one hundred for tens: \(14-7=7\) tens. The hundreds column now has \(5\) hundreds, and \(5-2=3\) hundreds remain. 4. The correct difference is \(374\).

\(652-278=374\). The student switched the subtraction order inside columns instead of regrouping.

- Read the two rows as two separate numbers. - Which column needs more chips in the top number before subtraction? - A trade changes how many chips are in the next column too.

1. The chart represents \(342-128\). 2. There are \(2\) ones in the top row, but \(8\) ones must be subtracted. Trade \(1\) ten for \(10\) ones, leaving \(3\) tens and \(12\) ones in the top row. 3. Ones: \(12-8=4\); tens: \(3-2=1\); hundreds: \(3-1=2\). The difference is \(214\).

\(214\). Trade one ten for ten ones before subtracting the ones.

- Identify the greatest and least weights first. - Align the place values before subtracting. - When a top digit is too small, regroup from the next place to the left.

1. The heaviest bag is Sami's at \(4210\,\text{g}\), and the lightest is Paul's at \(2890\,\text{g}\). 2. Subtract \(4210-2890\) by place value. Ones: \(0-0=0\). 3. Tens: \(1<9\), so regroup \(1\) hundred as \(10\) tens. Then \(11-9=2\). 4. Hundreds: \(1<8\), so regroup \(1\) thousand as \(10\) hundreds. Then \(11-8=3\). 5. Thousands: \(3-2=1\). The difference is \(1320\,\text{g}\). 6. Regrouping record: hundreds to tens; thousands to hundreds

The difference is \(1320\,\text{g}\). Regrouping record: hundreds to tens; thousands to hundreds

- This problem requires regrouping in two columns. - Work from right to left: ones, tens, then hundreds. - Regroup when the top digit is less than the bottom digit.

1. In the ones column, \(3\) is less than \(5\), so regroup \(1\) ten as \(10\) ones. Then \(13 - 5 = 8\). There are \(0\) tens left. 2. In the tens column, \(0\) is less than \(4\), so regroup \(1\) hundred as \(10\) tens. Then \(10 - 4 = 6\). There are \(5\) hundreds left. 3. In the hundreds column, \(5 - 2 = 3\). 4. Therefore, \(613 - 245 = 368\). 5. Regrouping record: tens to ones; hundreds to tens

\(368\) Regrouping record: tens to ones; hundreds to tens

- In each column, determine which missing digit produces the given result digit. - Account for regrouping when the top digit is less than the bottom digit. - You can check by adding the difference and the subtrahend.

1. In the ones column, the top \(4\) must be regrouped as \(14\) ones. To obtain a difference of \(8\), the missing bottom digit must be \(6\), because \(14 - 6 = 8\). 2. The regrouping leaves one fewer ten. If the original missing tens digit is \(4\), then \(3\) tens remain. Regroup \(1\) hundred to make \(13\) tens, and \(13 - 6 = 7\). Therefore, the missing top tens digit is \(4\). 3. After regrouping \(1\) hundred, \(4 - 2 = 2\), which matches the hundreds digit of the result. 4. The completed equation is \(544 - 266 = 278\).

The missing tens digit in the minuend is \(4\), and the missing ones digit in the subtrahend is \(6\). \(544 - 266 = 278\)

- The number of available seats is the total number of seats minus the tickets sold. - Work from the ones column to the hundreds column. - Include each column equation in your response, not only the final difference.

1. Subtract \(652\) from \(985\). 2. Ones: \(5-2=3\). 3. Tens: \(8-5=3\). 4. Hundreds: \(9-6=3\). 5. Therefore, \(985-652=333\), so \(333\) seats remain.

Ones: \(5-2=3\) Tens: \(8-5=3\) Hundreds: \(9-6=3\) There are \(333\) seats still available.

- Use addition as the inverse operation to find a missing digit. - Treat each place-value column as a separate equation. - Check the completed subtraction from right to left.

1. In the ones column, the missing digit must satisfy \(\square - 3 = 5\), so it is \(8\). 2. In the tens column, the missing digit must satisfy \(6 - \square = 2\), so it is \(4\). 3. In the hundreds column, \(7 - 4 = 3\), so the missing result digit is \(3\). 4. The completed equation is \(768 - 443 = 325\).

\(\begin{array}{r}768\\-\;443\\\hline325\end{array}\)

- Align ones, tens, and hundreds. - In each column, compare the top digit with the bottom digit. - If the top digit is too small, regroup from the next place value. - Check the ones and tens columns separately when classifying each problem.

1. For part a, \(785 - 462 = 323\). Every top digit is at least as large as the digit below it, so no regrouping is needed. 2. For part b, \(691 - 354 = 337\). Since \(1 < 4\) in the ones column, regroup \(1\) ten as \(10\) ones. No further regrouping is needed. 3. For part c, \(528 - 273 = 255\). The ones column does not require regrouping, but \(2 < 7\) in the tens column, so regroup \(1\) hundred as \(10\) tens. 4. For part d, \(813 - 457 = 356\). Since \(3 < 7\), regroup from tens to ones. This leaves \(0\) tens, and \(0 < 5\), so regroup from hundreds to tens as well.

a) \(323\); no regrouping b) \(337\); regrouping only from tens to ones c) \(255\); regrouping only from hundreds to tens d) \(356\); regrouping in both places

- Compare the digits from right to left. - What happens to the top tens digit after regrouping for the ones column? - Compare the remaining number of tens with the bottom tens digit.

1. In the ones column, \(5 < 8\), so Emma must regroup \(1\) ten as \(10\) ones. 2. After that regrouping, only \(1\) ten remains in the top number. Since \(1 < 6\), she must also regroup \(1\) hundred as \(10\) tens. 3. Complete the subtraction: \(15 - 8 = 7\), \(11 - 6 = 5\), and \(6 - 3 = 3\). Therefore, \(725 - 368 = 357\).

Emma sees that \(5 < 8\), so she must regroup from tens to ones. That leaves only \(1\) ten, and \(1 < 6\), so she must also regroup from hundreds to tens. The difference is \(357\).

- Which operation finds the part that is not included? - Align the numbers by place value. - Regroup when a top digit is less than the bottom digit.

1. Subtract the number of participating students from the total: \(514 - 168\). 2. In the ones column, \(4 < 8\), so regroup \(1\) ten. Then \(14 - 8 = 6\). 3. There are now \(0\) tens. Regroup \(1\) hundred as \(10\) tens, and calculate \(10 - 6 = 4\). 4. There are \(4\) hundreds left, and \(4 - 1 = 3\). 5. Therefore, \(514 - 168 = 346\). 6. Regrouping record: tens to ones; hundreds to tens

\(346\) students do not participate in the club. Regrouping record: tens to ones; hundreds to tens

- Complete both subtractions from right to left. - Note each column where the top digit is too small. - Regrouping means exchanging one unit of a greater place value for ten units of the next smaller place value.

1. For Problem A, regroup in the ones column: \(15 - 7 = 8\). This leaves \(1\) ten, so regroup in the tens column: \(11 - 6 = 5\). Then \(7 - 4 = 3\). Therefore, \(825 - 467 = 358\), and regrouping is required in both places. 2. For Problem B, subtract without regrouping: \(9 - 4 = 5\), \(8 - 5 = 3\), and \(7 - 2 = 5\). Therefore, \(789 - 254 = 535\). 3. Only Problem A requires both kinds of regrouping.

Problem A: \(825 - 467 = 358\) Problem B: \(789 - 254 = 535\) Problem A requires regrouping from tens to ones and from hundreds to tens.

- Compare the ones digits in both problems. - In which problem is the top ones digit too small? - Remember that regrouping leaves one fewer ten.

1. Compare the ones digits. In Problem A, \(3 < 7\), so regrouping from tens to ones is required. In Problem B, \(6 > 1\), so it is not required. 2. Solve Problem A: regroup to calculate \(13 - 7 = 6\). There are \(4\) tens left, so \(4 - 2 = 2\). Then \(8 - 4 = 4\). 3. Therefore, \(853 - 427 = 426\).

Problem A requires regrouping from tens to ones. \(853 - 427 = 426\)

- Work through the subtraction from right to left. - Pay attention to regrouping in the tens column. - You can also add the difference and the subtrahend to find the minuend.

1. In the ones column, \(6 - 1 = 5\), which matches the result. 2. In the tens column, \(3 < 5\), so regroup \(1\) hundred. Then \(13 - 5 = 8\), which matches the result. 3. Because one hundred was regrouped, the hundreds column is \(\square - 1 - 2 = 4\). 4. Work backward: \(4 + 2 + 1 = 7\). The missing digit is \(7\), and the completed equation is \(736 - 251 = 485\). 5. Regrouping record: hundreds to tens

The missing digit is \(7\). The completed equation is \(736 - 251 = 485\). Regrouping record: hundreds to tens

- Compare the digits in each place-value column. - Regroup when a top digit is less than the digit below it. - Work from right to left.

1. In Problem A, each top digit is at least as large as the digit below it: \(3 > 1\), \(5 > 2\), and \(9 > 4\). No regrouping is needed. 2. In Problem B, \(5 < 6\) in the tens column, so regroup \(1\) hundred as \(10\) tens. 3. Problem A: \(953 - 421 = 532\). 4. Problem B: \(3 - 1 = 2\). After regrouping in the tens column, \(15 - 6 = 9\), and \(8 - 4 = 4\). Thus, \(953 - 461 = 492\).

Problem B requires regrouping from hundreds to tens. A) \(532\) B) \(492\)

- Begin in the ones column. - When you regroup from a place-value column, that column has one fewer unit. - Check by adding the difference and the subtrahend.

1. In the ones column, regroup \(1\) ten so that \(12 - \square = 5\). The missing ones digit is \(7\). 2. There are now \(2\) tens. Since \(2 < 4\), regroup \(1\) hundred to make \(12\) tens. Then \(12 - 4 = 8\). 3. There are now \(5\) hundreds. The hundreds column must satisfy \(5 - \square = 2\), so the missing hundreds digit is \(3\). 4. The completed equation is \(632 - 347 = 285\). 5. Regrouping record: tens to ones; hundreds to tens

The missing digits are \(7\) in the ones place and \(3\) in the hundreds place. \(632 - 347 = 285\) Regrouping record: tens to ones; hundreds to tens

- First subtract the sheets used for the first project. - Find how many sheets remain after that project. - Use that difference as the starting amount for the second subtraction. - Pay close attention to the zeros in \(800\).

1. Subtract the first amount used: \(800 - 245\). Regroup \(1\) hundred as \(10\) tens, then regroup \(1\) of those tens as \(10\) ones. Calculate \(10 - 5 = 5\), \(9 - 4 = 5\), and \(7 - 2 = 5\). This leaves \(555\) sheets. 2. Subtract the second amount used: \(555 - 138\). Regroup in the ones column: \(15 - 8 = 7\). Then \(4 - 3 = 1\) and \(5 - 1 = 4\). This leaves \(417\) sheets. 3. Regrouping record: step 1 hundreds to tens; tens to ones | step 2 tens to ones

\(417\) sheets of paper remain. Regrouping record: step 1 hundreds to tens; tens to ones | step 2 tens to ones

- Write each subtraction vertically and align place values. - When regrouping across a zero, first regroup from the hundreds place. - Compare the hundreds, tens, and ones of the three differences.

1. For part a, regroup across the zero: \(15 - 8 = 7\), \(9 - 4 = 5\), and \(6 - 3 = 3\). Thus, \(705 - 348 = 357\). 2. For part b, regroup across the zero: \(12 - 4 = 8\), \(9 - 6 = 3\), and \(8 - 5 = 3\). Thus, \(902 - 564 = 338\). 3. For part c, regroup across the zero: \(11 - 9 = 2\), \(9 - 3 = 6\), and \(5 - 2 = 3\). Thus, \(601 - 239 = 362\). 4. Compare the differences: \(338 < 357 < 362\). The least difference is \(338\). 5. Regrouping record: a) hundreds to tens; tens to ones | b) hundreds to tens; tens to ones | c) hundreds to tens; tens to ones

a) \(357\) b) \(338\) c) \(362\) The difference in part b, \(338\), is the least. Regrouping record: a) hundreds to tens; tens to ones | b) hundreds to tens; tens to ones | c) hundreds to tens; tens to ones

- Begin in the ones column. - Regroup whenever the top digit is less than the bottom digit. - When the tens digit is zero, first regroup one hundred as ten tens. - Check with \(576 + \text{the missing number} = 904\).

1. In the ones column, regroup across the zero so that \(14 - \square = 6\). The missing ones digit in the subtrahend is \(8\). 2. Regrouping from the hundreds place creates \(10\) tens, and regrouping one of those tens leaves \(9\) tens. Then \(9 - 2 = 7\), which matches the result. 3. There are \(8\) hundreds left. The hundreds column must satisfy \(8 - \square = 5\), so the missing hundreds digit is \(3\). 4. The missing subtrahend is \(328\), and \(904 - 328 = 576\). 5. Regrouping record: hundreds to tens; tens to ones

\(904 - 328 = 576\) Regrouping record: hundreds to tens; tens to ones

- In “subtract \(245\) from \(602\),” write \(602\) as the minuend. - Regroup across the zero by starting with the hundreds place. - Check a subtraction by adding the difference and the subtrahend.

1. Calculate \(602 - 245\). Regroup \(1\) hundred as \(10\) tens, then regroup \(1\) ten as \(10\) ones. In the ones column, \(12 - 5 = 7\). In the tens column, \(9 - 4 = 5\). In the hundreds column, \(5 - 2 = 3\). The difference is \(357\). 2. Check with addition: \(357 + 245\). Add the ones: \(7 + 5 = 12\). Add the tens: \(5 + 4 + 1 = 10\). Add the hundreds: \(3 + 2 + 1 = 6\). The sum is \(602\). 3. Because the check returns the original minuend, the subtraction is correct. 4. Regrouping record: hundreds to tens; tens to ones

Difference: \(357\) Check: \(357 + 245 = 602\) Regrouping record: hundreds to tens; tens to ones

- Align ones with ones, tens with tens, and hundreds with hundreds. - When the tens digit is zero, regroup from the hundreds place before regrouping to the ones place. - Track each changed digit after regrouping.

1. For part a, regroup across the zero. Calculate \(14 - 8 = 6\), \(9 - 3 = 6\), and \(4 - 2 = 2\). Thus, \(504 - 238 = 266\). 2. For part b, align \(45\) as \(045\), then regroup across the zero. Calculate \(12 - 5 = 7\), \(9 - 4 = 5\), and \(6 - 0 = 6\). Thus, \(702 - 45 = 657\). 3. For part c, regroup across the zero. Calculate \(11 - 7 = 4\), \(9 - 6 = 3\), and \(7 - 3 = 4\). Thus, \(801 - 367 = 434\). 4. Regrouping record: a) hundreds to tens; tens to ones | b) hundreds to tens; tens to ones | c) hundreds to tens; tens to ones

a) \(266\) b) \(657\) c) \(434\) Regrouping record: a) hundreds to tens; tens to ones | b) hundreds to tens; tens to ones | c) hundreds to tens; tens to ones

- Begin in the ones column. - Think about what happens when you need to regroup through a zero. - Track how each regrouping changes the next column.

1. In the ones column, regroup across the zero so that \(13 - 5 = 8\). The missing ones digit in the result is \(8\). 2. Regrouping from the hundreds place creates \(10\) tens, and regrouping one ten to the ones place leaves \(9\) tens. The tens column is \(9 - \square = 4\), so the missing tens digit in the subtrahend is \(5\). 3. There are \(6\) hundreds left. The hundreds column gives \(6 - 2 = 4\), so the missing hundreds digit in the result is \(4\). 4. The completed equation is \(703 - 255 = 448\).

The missing digits are \(5\) in the subtrahend, \(4\) in the hundreds place of the result, and \(8\) in the ones place of the result. \(703 - 255 = 448\)

- Find both differences carefully. - When subtracting from the same number, subtracting a greater amount produces a lesser difference. - Compare the hundreds, tens, and ones of the results.

1. For Problem A, regroup across the zero: \(15 - 8 = 7\), \(9 - 2 = 7\), and \(5 - 3 = 2\). Therefore, \(605 - 328 = 277\). 2. For Problem B, no regrouping is needed in the ones column. Regroup one hundred as ten tens, then calculate \(5 - 2 = 3\), \(10 - 8 = 2\), and \(5 - 3 = 2\). Therefore, \(605 - 382 = 223\). 3. Since \(277 > 223\), Problem A has the greater difference. 4. Regrouping record: A hundreds to tens; tens to ones | B hundreds to tens

Problem A has the greater difference. A) \(277\) B) \(223\) Regrouping record: A hundreds to tens; tens to ones | B hundreds to tens

- Compare the top and bottom digits in the ones, tens, and hundreds columns. - Check whether the ones column requires regrouping. - If it does, remember that the top tens digit decreases by \(1\) before you compare the tens digits.

1. For part a, each top digit is at least as large as the digit below it, so no regrouping is needed. \(485 - 162 = 323\). 2. For part b, \(1 < 8\), so regroup from tens to ones. After regrouping, \(6\) tens remain, and \(6 \ge 3\), so no further regrouping is needed. \(571 - 238 = 333\). 3. For part c, the ones column does not require regrouping, but \(1 < 5\) in the tens column, so regroup from hundreds to tens. \(916 - 453 = 463\). 4. For part d, \(2 < 4\), so regroup from tens to ones. This leaves \(2\) tens, and \(2 < 7\), so regroup from hundreds to tens as well. \(832 - 574 = 258\).

a) No regrouping: \(485 - 162 = 323\) b) Regrouping only from tens to ones: \(571 - 238 = 333\) c) Regrouping only from hundreds to tens: \(916 - 453 = 463\) d) Regrouping in both places: \(832 - 574 = 258\)

- First compare the ones digits in each problem. - Then compare the tens digits, accounting for any regrouping from the ones column. - Three problems require only regrouping from tens to ones; one requires regrouping from hundreds to tens instead. - Solve the problem that does not belong.

1. In A, \(3 < 7\), so regrouping is needed from tens to ones only. 2. In B, \(1 < 5\), so regrouping is needed from tens to ones only. 3. In C, \(2 < 8\), so regrouping is needed from tens to ones only. 4. In D, the ones column does not require regrouping, but \(2 < 5\) in the tens column. Therefore, D requires regrouping from hundreds to tens. 5. The different problem is D, and \(725 - 352 = 373\).

D) is different because it requires regrouping from hundreds to tens, while A, B, and C require regrouping from tens to ones. \(725 - 352 = 373\)

- Align the numbers by place value. - When a needed place-value digit is zero, regroup from the next nonzero place to the left. - One hundred equals ten tens, and one ten equals ten ones. - Work from right to left.

1. For part a, align \(38\) as \(038\) and regroup across the zeros. Calculate \(10 - 8 = 2\), \(9 - 3 = 6\), and \(6 - 0 = 6\). The difference is \(662\). 2. For part b, regroup from the thousands place through the hundreds and tens places. Calculate \(10 - 6 = 4\), \(9 - 5 = 4\), \(9 - 4 = 5\), and \(0 - 0 = 0\). The difference is \(544\). 3. For part c, regroup across the zeros. Calculate \(10 - 7 = 3\), \(9 - 1 = 8\), and \(4 - 2 = 2\). The difference is \(283\). 4. For part d, regroup across the zero. Calculate \(12 - 5 = 7\), \(9 - 3 = 6\), and \(8 - 4 = 4\). The difference is \(467\). 5. Regrouping record: a) hundreds to tens; tens to ones | b) thousands to hundreds; hundreds to tens; tens to ones | c) hundreds to tens; tens to ones | d) hundreds to tens; tens to ones

a) \(662\) b) \(544\) c) \(283\) d) \(467\) Regrouping record: a) hundreds to tens; tens to ones | b) thousands to hundreds; hundreds to tens; tens to ones | c) hundreds to tens; tens to ones | d) hundreds to tens; tens to ones

- Which operation represents sand being unloaded? - Keep the unit attached to the result. - Check subtraction by adding the difference and the subtrahend. - Regroup carefully across both zeros in \(900\).

1. Subtract the amount unloaded: \(900 - 354\). Regroup across the zeros. Calculate \(10 - 4 = 6\), \(9 - 5 = 4\), and \(8 - 3 = 5\). The truck has \(546\,\text{lb}\) of sand left. 2. Check with addition: \(546 + 354\). The ones total is \(10\), the tens total including the regrouped \(1\) is \(10\), and the hundreds total including the regrouped \(1\) is \(9\). Thus, \(546 + 354 = 900\). 3. Regrouping record: hundreds to tens; tens to ones

There are \(546\,\text{lb}\) of sand left on the truck. Regrouping record: hundreds to tens; tens to ones

- Pay close attention to zeros. - Regroup when a top digit is less than the digit below it. - Work from right to left: ones, tens, then hundreds.

1. For part a, regroup from tens to ones: \(10 - 2 = 8\). Then regroup from hundreds to tens: \(12 - 6 = 6\), and \(7 - 4 = 3\). Thus, \(830 - 462 = 368\). 2. For part b, regroup across both zeros: \(10 - 5 = 5\), \(9 - 7 = 2\), and \(5 - 2 = 3\). Thus, \(600 - 275 = 325\). 3. For part c, regroup across the zero: \(12 - 7 = 5\), \(9 - 3 = 6\), and \(8 - 5 = 3\). Thus, \(902 - 537 = 365\). 4. Regrouping record: a) tens to ones; hundreds to tens | b) hundreds to tens; tens to ones | c) hundreds to tens; tens to ones

a) \(368\) b) \(325\) c) \(365\) Regrouping record: a) tens to ones; hundreds to tens | b) hundreds to tens; tens to ones | c) hundreds to tens; tens to ones

- Begin in the ones column. - When the tens digit is zero, regroup first from the hundreds place. - Account for every regrouping when finding the missing digits. - Check by adding the difference and the subtrahend.

1. In the ones column, regroup across the zero so that \(15 - \square = 6\). The missing ones digit in the subtrahend is \(9\). 2. After regrouping across the zero, \(9\) tens remain. Then \(9 - 4 = 5\), so the missing tens digit in the difference is \(5\). 3. There are \(7\) hundreds left. The hundreds column must satisfy \(7 - \square = 5\), so the missing hundreds digit in the subtrahend is \(2\). 4. The completed equation is \(805 - 249 = 556\).

The missing hundreds digit in the subtrahend is \(2\), the missing ones digit in the subtrahend is \(9\), and the missing tens digit in the difference is \(5\). \(805 - 249 = 556\)

- Which operation finds the difference between the total pages and the pages already read? - Align the numbers by place value. - Regroup carefully across the two zeros in \(600\).

1. Subtract the pages already read from the total: \(600 - 342\). 2. Regroup \(600\) as \(5\) hundreds, \(9\) tens, and \(10\) ones. 3. In the ones column, \(10 - 2 = 8\). 4. In the tens column, \(9 - 4 = 5\). 5. In the hundreds column, \(5 - 3 = 2\). 6. Therefore, \(600 - 342 = 258\). 7. Regrouping record: hundreds to tens; tens to ones

Elias still needs to read \(258\) pages. Regrouping record: hundreds to tens; tens to ones

- Check whether the top digit in each place is large enough to subtract the bottom digit. - When regrouping, remember that the value in the next place also changes. - Check each difference by adding it to the subtrahend. - Make sure ones, tens, and hundreds are aligned.

1. For part a, regroup \(1\) ten so that \(12 - 7 = 5\). Then regroup \(1\) hundred so that \(12 - 5 = 7\). Finally, \(5 - 2 = 3\). The difference is \(375\). 2. For part b, regroup across the zero. Rewrite \(504\) as \(4\) hundreds, \(9\) tens, and \(14\) ones. Then \(14 - 6 = 8\), \(9 - 3 = 6\), and \(4 - 1 = 3\). The difference is \(368\). 3. For part c, align \(49\) under the tens and ones places. Regroup \(1\) ten so that \(11 - 9 = 2\). Then \(7 - 4 = 3\) and \(7 - 0 = 7\). The difference is \(732\).

a) \(632 - 257 = 375\) b) \(504 - 136 = 368\) c) \(781 - 49 = 732\)

- Compare the digits in Tim’s answer with the digits in the subtraction problem. - When the top digit is less than the bottom digit, regroup from the next place. - Check whether Tim ignored which digit was on top.

1. Tim appears to have subtracted the smaller digit from the greater digit in each place, regardless of which digit was on top. He used \(6 - 4 = 2\) in the ones place and \(5 - 2 = 3\) in the tens place instead of regrouping. 2. Correctly, regroup \(1\) ten so that \(14 - 6 = 8\). Then regroup \(1\) hundred so that \(11 - 5 = 6\). Finally, \(7 - 3 = 4\). 3. Therefore, \(824 - 356 = 468\).

a) Tim subtracted the smaller digit from the greater digit in the ones and tens places instead of regrouping. b) The correct difference is \(468\).

- Pay special attention when regrouping across a zero. - Make sure ones are aligned with ones and tens with tens. - Recalculate each problem one place at a time. - Remember that one of the shown results may already be correct.

1. For part a, regroup across the zero. Rewrite \(902\) as \(8\) hundreds, \(9\) tens, and \(12\) ones. Then \(12 - 8 = 4\), \(9 - 3 = 6\), and \(8 - 4 = 4\). The correct difference is \(464\). 2. For part b, regroup twice. In the ones place, \(13 - 5 = 8\). In the tens place, \(16 - 8 = 8\). In the hundreds place, \(5 - 0 = 5\). The correct difference is \(588\). 3. For part c, \(456 - 179 = 277\) is correct. After regrouping, \(16 - 9 = 7\), \(14 - 7 = 7\), and \(3 - 1 = 2\).

a) Incorrect. The correct difference is \(464\). b) Incorrect. The correct difference is \(588\). c) Correct. The difference is \(277\).

- Think about what happens when you regroup across zeros in the standard subtraction algorithm. - How far apart are the two numbers on a number line? - When would it be difficult to keep all the subtraction steps organized mentally?

1. For a), the standard algorithm requires regrouping across several place values because the minuend contains two zeros. 2. For b), count up because the numbers are close: from \(4998\) to \(5000\) is \(2\), and from \(5000\) to \(5001\) is \(1\). Therefore, the difference is \(3\). 3. For c), answers vary. One suitable example is \(8342 - 3567 = 4775\), which requires several regrouping steps and is easier to track with the standard algorithm.

a) Regrouping must continue across several place values because of the zeros in \(5001\). b) Count up: \(4998 + 2 + 1 = 5001\), so the difference is \(3\). c) Answers vary; for example, \(8342 - 3567 = 4775\).

- Regroup when the top digit is less than the digit below it. - Addition is the inverse operation of subtraction. - To check, add the difference and the subtrahend.

1. For part a, regroup in the ones and tens columns: \(12 - 6 = 6\), \(10 - 5 = 5\), and \(8 - 4 = 4\). Thus, \(912 - 456 = 456\). Check: \(456 + 456 = 912\). 2. For part b, regroup across the zero: \(13 - 5 = 8\), \(9 - 8 = 1\), and \(6 - 2 = 4\). Thus, \(703 - 285 = 418\). Check: \(418 + 285 = 703\). 3. Regrouping record: a) tens to ones; hundreds to tens | b) hundreds to tens; tens to ones

a) \(456\); check: \(456 + 456 = 912\) b) \(418\); check: \(418 + 285 = 703\) Regrouping record: a) tens to ones; hundreds to tens | b) hundreds to tens; tens to ones

- Begin in the ones column and consider whether a trade is necessary. - Track the change to the top tens digit before subtracting the tens. - Use addition only after completing the subtraction to check it.

1. Ones: \(5\) cannot subtract the missing digit to make \(9\) without regrouping. Trade one ten to get \(15\) ones, so \(15-6=9\). The missing bottom ones digit is \(6\). 2. Tens: after the first trade, the top tens digit is one less than shown. Another trade from hundreds is needed to leave \(7\) tens after subtracting \(4\). The top tens digit must be \(2\), because \(11-4=7\). 3. Hundreds: after that trade, \(7-3=4\), matching the result. The equation is \(825-346=479\). 4. Check: \(479+346=825\).

Top missing digit: \(2\); bottom missing digit: \(6\). \(825-346=479\). Check: \(479+346=825\).

- Find both differences carefully. - Compare the two results after calculating them. - Verify each result by adding the difference and the subtrahend.

1. Problem A: \(621 - 347 = 274\). Check: \(274 + 347 = 621\). 2. Problem B: \(514 - 239 = 275\). Check: \(275 + 239 = 514\). 3. Since \(274 < 275\), the difference in Problem A is less. 4. Regrouping record: A tens to ones; hundreds to tens | B tens to ones; hundreds to tens

Problem A has a difference of \(274\). Problem B has a difference of \(275\). The difference in Problem A is less. Regrouping record: A tens to ones; hundreds to tens | B tens to ones; hundreds to tens

- Examine one place-value column at a time, beginning with the ones column. - Determine whether regrouping affects the next column. - You can use addition to check a completed subtraction.

1. In part a, \(8 - 4 = 4\). In the tens column, \(5 - \square = 2\), so the missing digit is \(3\). The completed equation is \(658 - 334 = 324\). 2. In part b, regroup in the ones column: \(17 - 8 = 9\). This leaves \(1\) ten, so regroup one hundred and calculate \(11 - 5 = 6\). There are \(8\) hundreds left, and \(8 - \square = 3\), so the missing digit is \(5\). The completed equation is \(927 - 558 = 369\). 3. Regrouping record: a) none | b) tens to ones; hundreds to tens

a) The missing digit is \(3\): \(658 - 334 = 324\). b) The missing digit is \(5\): \(927 - 558 = 369\). Regrouping record: a) none | b) tens to ones; hundreds to tens

- Work from the ones column toward the hundreds. - When one column supplies a ten or a hundred, reduce that column's amount before continuing. - Remember that a zero in the ones column can still receive a trade.

1. In a), trade one ten to make \(12\) ones. Since \(12-5=7\), the missing bottom ones digit is \(5\). 2. The top tens are now \(3\). Trade a hundred to make \(13\) tens; \(13-7=6\). After that trade the top hundreds must be \(3\), so the missing top hundreds digit was \(4\). Thus \(442-175=267\). 3. In b), trade one ten to get \(10\) ones; \(10-6=4\). Then the top tens digit has been reduced by one. To get a tens difference of \(8\) after subtracting \(4\), the top tens digit must have been \(3\): after a hundreds trade it becomes \(12\), and \(12-4=8\). 4. The top hundreds are now \(7\), so the missing bottom hundreds digit is \(2\), since \(7-2=5\). Thus \(830-246=584\).

a) \(442-175=267\). b) \(830-246=584\).

- Start with the ones column and work left. - Keep track of each regrouping before moving to the next column. - For the check, add the difference and the subtrahend and compare with the minuend.

1. In the ones column, regroup so that \(14 - \square = 7\). The missing ones digit in the subtrahend is \(7\). This leaves \(3\) tens in the minuend. 2. Since \(3 < 5\), regroup one hundred. Then \(13 - 5 = 8\), so the original missing tens digit in the minuend is \(4\). 3. There are \(6\) hundreds left, and \(6 - 3 = 3\). The missing hundreds digit in the difference is \(3\). 4. The completed equation is \(744 - 357 = 387\). Check: \(387 + 357 = 744\).

The missing tens digit in the minuend is \(4\), the missing ones digit in the subtrahend is \(7\), and the missing hundreds digit in the difference is \(3\). \(744 - 357 = 387\) Check: \(387 + 357 = 744\)

- Subtract the starting reading from the ending reading. - Round each odometer reading to the nearest hundred for the estimate. - Compare each exact distance with its estimate.

1. Trip A: In \(15{,}412-14{,}895\), regroup to calculate \(12-5=7\), \(10-9=1\), and \(13-8=5\); the remaining leading digits subtract to \(0\). The distance is \(517\) miles. Estimate: \(15{,}400-14{,}900=500\) miles. 2. Trip B: In \(16{,}105-15{,}412\), calculate \(5-2=3\), then regroup to calculate \(10-1=9\) and \(10-4=6\); the remaining leading digits subtract to \(0\). The distance is \(693\) miles. Estimate: \(16{,}100-15{,}400=700\) miles. 3. Trip C: In \(17{,}034-16{,}105\), regroup to calculate \(14-5=9\), then calculate \(2-0=2\) and \(10-1=9\); the remaining leading digits subtract to \(0\). The distance is \(929\) miles. Estimate: \(17{,}000-16{,}100=900\) miles. 4. Regrouping record: A tens to ones; hundreds to tens; thousands to hundreds | B hundreds to tens; thousands to hundreds | C tens to ones; thousands to hundreds

Trip A: \(517\) miles; estimate \(500\) miles Trip B: \(693\) miles; estimate \(700\) miles Trip C: \(929\) miles; estimate \(900\) miles Regrouping record: A tens to ones; hundreds to tens; thousands to hundreds | B hundreds to tens; thousands to hundreds | C tens to ones; thousands to hundreds

- Subtract the beginning reading from the ending reading. - To round to the nearest thousand, use the hundreds digit. - Compare the exact result with the estimate.

1. Subtract by place value. Ones: regroup to calculate \(14-7=7\). Tens: regroup to calculate \(10-6=4\). Hundreds: regroup to calculate \(11-5=6\). Thousands: \(7-4=3\). Therefore, \(38{,}214-34{,}567=3647\), so the yearly use is \(3647\,\text{kWh}\). 2. Round the readings: \(38{,}214\approx38{,}000\) and \(34{,}567\approx35{,}000\). Then \(38{,}000-35{,}000=3000\), so the exact result is reasonable. 3. Regrouping record: tens to ones; hundreds to tens; thousands to hundreds

a) \(3647\,\text{kWh}\) b) Approximately \(3000\,\text{kWh}\) Regrouping record: tens to ones; hundreds to tens; thousands to hundreds

- Subtract each given number from \(1{,}000{,}000\). - Keep matching place values aligned. - Check each result by adding it to the original number.

1. Write each subtraction vertically with matching place values aligned. 2. For a), regroup across the zeros. In the ten-thousands column, \(10-5=5\), and in the hundred-thousands column, \(9-9=0\). Therefore, \(1{,}000{,}000-950{,}000=50{,}000\). 3. For b), regroup across the zeros and subtract by column. This gives \(5\) thousands, \(7\) ten-thousands, and \(2\) hundred-thousands, so \(1{,}000{,}000-725{,}000=275{,}000\). 4. For c), regroup through the tens column. Subtracting by column gives \(1\) ten, \(0\) hundreds, \(0\) thousands, \(0\) ten-thousands, and \(5\) hundred-thousands, so \(1{,}000{,}000-499{,}990=500{,}010\). 5. For d), regroup across the zeros and subtract by column. This gives \(7\) thousands, \(7\) ten-thousands, and \(8\) hundred-thousands, so \(1{,}000{,}000-123{,}000=877{,}000\). 6. Regrouping record: a) millions to hundred-thousands; hundred-thousands to ten-thousands | b) millions to hundred-thousands; hundred-thousands to ten-thousands; ten-thousands to thousands | c) millions to hundred-thousands; hundred-thousands to ten-thousands; ten-thousands to thousands; thousands to hundreds; hundreds to tens | d) millions to hundred-thousands; hundred-thousands to ten-thousands; ten-thousands to thousands

a) \(50{,}000\) b) \(275{,}000\) c) \(500{,}010\) d) \(877{,}000\) Regrouping record: a) millions to hundred-thousands; hundred-thousands to ten-thousands | b) millions to hundred-thousands; hundred-thousands to ten-thousands; ten-thousands to thousands | c) millions to hundred-thousands; hundred-thousands to ten-thousands; ten-thousands to thousands; thousands to hundreds; hundreds to tens | d) millions to hundred-thousands; hundred-thousands to ten-thousands; ten-thousands to thousands

- Work one column at a time from right to left. - Determine which digit makes each result digit correct. - Account for regrouping whenever the top digit is too small.

1. In part a, the ones column gives \(8 - 5 = 3\), so the missing ones digit in the subtrahend is \(5\). Regroup one hundred in the tens column so that \(14 - 6 = 8\), so the missing tens digit in the minuend is \(4\). The completed equation is \(548 - 265 = 283\). 2. In part b, regroup one ten so that \(13 - 4 = 9\), so the missing ones digit in the minuend is \(3\). This leaves \(1\) ten, so regroup one hundred and calculate \(11 - 5 = 6\). There are \(6\) hundreds left, and \(6 - 3 = 3\), so the missing hundreds digit in the subtrahend is \(3\). The completed equation is \(723 - 354 = 369\).

a) The missing digits are \(4\) and \(5\): \(548 - 265 = 283\). b) The missing digits are \(3\) and \(3\): \(723 - 354 = 369\).

- Both problems require regrouping in two columns. - Record each regrouping clearly. - Check by adding the difference and the subtrahend.

1. For part a, regroup in the ones column: \(14 - 7 = 7\). This leaves \(1\) ten, so regroup one hundred and calculate \(11 - 5 = 6\). There are \(8\) hundreds left, and \(8 - 4 = 4\). Thus, \(924 - 457 = 467\). 2. For part b, regroup in the ones column: \(13 - 5 = 8\). This leaves \(0\) tens, so regroup one hundred and calculate \(10 - 6 = 4\). There are \(3\) hundreds left, and \(3 - 2 = 1\). Thus, \(413 - 265 = 148\). 3. Regrouping record: a) tens to ones; hundreds to tens | b) tens to ones; hundreds to tens

a) \(467\) b) \(148\) Regrouping record: a) tens to ones; hundreds to tens | b) tens to ones; hundreds to tens

- Check the statements by subtracting, not by adding back the proposed result. - When a place does not have enough units, trade from the next larger place. - In equation 3, the tens digit is zero: start your trade one place farther left.

1. For \(734-258\), trade one ten to make \(14-8=6\). Then trade a hundred to make \(12-5=7\), leaving \(6-2=4\) hundreds. The difference is \(476\), so equation 1 is correct. 2. For \(612-347\), trade one ten to get \(12-7=5\), then trade one hundred to get \(10-4=6\) tens and \(5-3=2\) hundreds. The difference is \(265\), so equation 2 is incorrect. 3. For \(805-429\), trade one hundred for ten tens; then trade one of those tens for ten ones. This leaves \(7\) hundreds, \(9\) tens, and \(15\) ones. Subtract to get \(376\), so equation 3 is incorrect. 4. For \(541-186\), trade a ten and then a hundred: \(11-6=5\), \(13-8=5\), \(4-1=3\). The difference is \(355\), so equation 4 is correct.

1. Correct. 2. Incorrect; \(612-347=265\). 3. Incorrect; \(805-429=376\). Trade \(1\) hundred for \(10\) tens, then \(1\) ten for \(10\) ones. 4. Correct.

- Use the method named for each part. - In subtraction, start with ones and trade from a greater place value if necessary. - For part c), first turn a hundred into tens before you can trade a ten into ones.

1. In a), trade a ten for ones and a hundred for tens: \(11-3=8\), \(11-4=7\), and \(8-5=3\). Therefore, \(921-543=378\), not \(388\). 2. In b), \(6+5=11\), \(7+8+1=16\), and \(2+4+1=7\). Therefore, \(276+485=761\) is correct. 3. In c), the tens column is zero, so trade one hundred for ten tens, then one ten for ten ones. The remaining units are \(5\) hundreds, \(9\) tens, and \(13\) ones. Subtract: \(13-7=6\), \(9-1=8\), and \(5-2=3\). Therefore, \(603-217=386\) is correct.

a) Incorrect; \(921-543=378\). b) Correct. c) Correct. To subtract through zero, trade \(1\) hundred for \(10\) tens, then \(1\) ten for \(10\) ones.

- Try subtracting hundreds, tens, and ones separately when no regrouping is needed. - When two numbers are close, counting up may be fastest. - Use the standard algorithm when several regrouping steps are needed.

1. For a), use a place-value strategy: \(1000 - 400 - 50 = 550\). 2. For b), use a place-value strategy without regrouping: \(800 - 300 = 500\), \(70 - 20 = 50\), and \(6 - 1 = 5\). The difference is \(555\). 3. For c), use mental math and count up because the numbers are close: from \(498\) to \(502\) is \(4\). 4. For d), use the standard algorithm. Regroup \(1\) ten so \(14 - 6 = 8\). Then regroup \(1\) hundred so \(12 - 5 = 7\). Finally, \(6 - 4 = 2\). Therefore, \(734 - 456 = 278\).

a) \(550\) (place-value strategy) b) \(555\) (place-value strategy) c) \(4\) (mental math) d) \(278\) (standard algorithm)

- How far apart are the two numbers in the first subtraction? - Would regrouping across the zero in \(901\) be more efficient than counting up? - Think about the difference between subtracting directly and finding the distance between two numbers.

1. Lucas is not correct. When two numbers are close, counting up is faster than regrouping across zeros. 2. Count up from \(897\): \(897 + 3 = 900\), and \(900 + 1 = 901\). Therefore, \(901 - 897 = 4\). 3. Use the standard subtraction algorithm for \(823 - 356\). Regroup \(1\) ten so \(13 - 6 = 7\). Then regroup \(1\) hundred so \(11 - 5 = 6\). Finally, \(7 - 3 = 4\). Therefore, \(823 - 356 = 467\).

Lucas is not correct because counting up is faster when the two numbers are close. \(901 - 897 = 4\) \(823 - 356 = 467\)

- Align the numbers by place value. - Regroup when the top digit is less than the digit below it. - After regrouping, remember that the next place-value column has one fewer unit.

1. In the ones column, \(2 < 8\), so regroup one ten. Then \(12 - 8 = 4\). There are \(2\) tens left. 2. In the tens column, \(2 < 5\), so regroup one hundred. Then \(12 - 5 = 7\). There are \(6\) hundreds left. 3. In the hundreds column, \(6 - 4 = 2\). 4. Therefore, \(732 - 458 = 274\). 5. Regrouping record: tens to ones; hundreds to tens

\(274\) Regrouping record: tens to ones; hundreds to tens

- Start with the ones column and check whether a ten must be traded. - The top tens digit may change twice before the tens subtraction is finished. - Check the hundreds column after completing the trades.

1. Ones: trade one ten to get \(14\) ones. Since the result has \(9\) ones, the missing bottom digit is \(5\), because \(14-5=9\). 2. Tens: after that trade, another hundred must be traded for tens. To leave \(5\) tens after subtracting \(7\) tens, the top tens digit must originally have been \(3\). It becomes \(2\) after the first trade and \(12\) after the second, giving \(12-7=5\). 3. Hundreds: \(6\) becomes \(5\) after that trade, and \(5-2=3\). The completed subtraction is \(634-275=359\).

Top missing digit: \(3\); bottom missing digit: \(5\). \(634-275=359\).

- Calculate each difference separately. - Regroup carefully in each subtraction. - Compare the two results.

1. For the left expression, regroup in the ones and tens columns: \(14 - 6 = 8\), \(12 - 5 = 7\), and \(7 - 2 = 5\). Thus, \(834 - 256 = 578\). 2. For the right expression, regroup in the ones and tens columns: \(12 - 4 = 8\), \(10 - 3 = 7\), and \(8 - 3 = 5\). Thus, \(912 - 334 = 578\). 3. Since \(578 = 578\), the correct symbol is \(=\). 4. Regrouping record: left tens to ones; hundreds to tens | right tens to ones; hundreds to tens

\(=\) Regrouping record: left tens to ones; hundreds to tens | right tens to ones; hundreds to tens

- Align the minuend and subtrahend by place value. - Regroup when the lower digit is greater than the upper digit. - Take extra care when regrouping across zeros.

1. For a), in the ones and tens columns, regroup through the \(0\): \(15 - 6 = 9\) and \(9 - 8 = 1\). Then \(3 - 1 = 2\) in the hundreds column. Regroup in the thousands column to get \(12 - 4 = 8\), followed by \(8 - 3 = 5\). Therefore, \(92{,}405 - 34{,}186 = 58{,}219\). 2. For b), regroup from the thousands place through the hundreds, tens, and ones places. This gives \(10 - 5 = 5\), \(9 - 4 = 5\), \(9 - 3 = 6\), and \(4 - 2 = 2\). Then regroup from the hundred-thousands place so \(10 - 1 = 9\) in the ten-thousands column. Therefore, \(105{,}000 - 12{,}345 = 92{,}655\). 3. Regrouping record: a) hundreds to tens; tens to ones; ten-thousands to thousands | b) thousands to hundreds; hundreds to tens; tens to ones; hundred-thousands to ten-thousands

a) \(58{,}219\) b) \(92{,}655\) Regrouping record: a) hundreds to tens; tens to ones; ten-thousands to thousands | b) thousands to hundreds; hundreds to tens; tens to ones; hundred-thousands to ten-thousands

- Evaluate both subtraction expressions first. - Compare their results from the greatest place value. - Subtract the smaller result from the larger result.

1. For Expression 1, subtract by place value: \(3 - 0 = 3\). Regroup to get \(14 - 9 = 5\), then \(4 - 1 = 3\). Regroup again to get \(16 - 8 = 8\), then \(6 - 2 = 4\). Thus, \(76{,}543 - 28{,}190 = 48{,}353\). 2. For Expression 2, \(0 - 0 = 0\). Regroup to get \(10 - 5 = 5\), then regroup again to get \(10 - 7 = 3\). Regroup in the thousands column to get \(11 - 3 = 8\), followed by \(7 - 3 = 4\). Thus, \(82{,}100 - 33{,}750 = 48{,}350\). 3. Expression 1 is greater, and the difference between the results is \(48{,}353 - 48{,}350 = 3\). 4. Regrouping record: 1 hundreds to tens; ten-thousands to thousands | 2 hundreds to tens; thousands to hundreds; ten-thousands to thousands

Expression 1 has the greater result, \(48{,}353\). The difference between the results is \(3\). Regrouping record: 1 hundreds to tens; ten-thousands to thousands | 2 hundreds to tens; thousands to hundreds; ten-thousands to thousands

- Complete the two subtractions in order. - Use the first difference as the minuend in the second subtraction. - Keep all digits aligned by place value.

1. For \(120{,}450 - 34{,}567\), regroup to get \(10 - 7 = 3\) in the ones column and \(14 - 6 = 8\) in the tens column. Regroup across the \(0\) thousands digit to get \(13 - 5 = 8\) in the hundreds column and \(9 - 4 = 5\) in the thousands column. Regroup once more to get \(11 - 3 = 8\) in the ten-thousands column. The intermediate result is \(85{,}883\). 2. For \(85{,}883 - 18{,}902\), the ones and tens digits are \(3 - 2 = 1\) and \(8 - 0 = 8\). Regroup to get \(18 - 9 = 9\) in the hundreds column and \(14 - 8 = 6\) in the thousands column. Finally, \(7 - 1 = 6\). Therefore, the final result is \(66{,}981\). 3. Regrouping record: step 1 tens to ones; hundreds to tens; ten-thousands to thousands; thousands to hundreds; hundred-thousands to ten-thousands | step 2 thousands to hundreds; ten-thousands to thousands

The final result is \(66{,}981\). Regrouping record: step 1 tens to ones; hundreds to tens; ten-thousands to thousands; thousands to hundreds; hundred-thousands to ten-thousands | step 2 thousands to hundreds; ten-thousands to thousands

- Determine how the minuend changes from one problem to the next. - Continue that rule before subtracting. - Align the six-digit numbers carefully, especially in the final problem.

1. Each minuend is \(111{,}111\) less than the previous minuend, so the next two are \(654{,}654\) and \(543{,}543\). 2. Subtracting by place value without regrouping gives \(987{,}987 - 444{,}444 = 543{,}543\). 3. Similarly, \(876{,}876 - 444{,}444 = 432{,}432\). 4. Also, \(765{,}765 - 444{,}444 = 321{,}321\). 5. Continuing the pattern, \(654{,}654 - 444{,}444 = 210{,}210\). 6. For \(543{,}543 - 444{,}444\), regroup in the ones and tens columns to get \(9\) and \(9\), then subtract \(4 - 4 = 0\) in the hundreds column. Regroup again in the thousands and ten-thousands columns to get \(9\) and \(9\), followed by \(4 - 4 = 0\). Thus, the final difference is \(99{,}099\). 7. Regrouping record: 1. none | 2. none | 3. none | 4. none | 5. tens to ones; hundreds to tens; ten-thousands to thousands; hundred-thousands to ten-thousands

1. \(543{,}543\) 2. \(432{,}432\) 3. \(321{,}321\) 4. \(654{,}654 - 444{,}444 = 210{,}210\) 5. \(543{,}543 - 444{,}444 = 99{,}099\) Regrouping record: 1. none | 2. none | 3. none | 4. none | 5. tens to ones; hundreds to tens; ten-thousands to thousands; hundred-thousands to ten-thousands

- Begin in the ones place and decide when regrouping is necessary. - Track how regrouping changes the digit in the next place. - You can check the subtraction by adding the difference and subtrahend.

1. Ones: \(4 - \square = 1\), so the missing ones digit in the subtrahend is \(3\). 2. Tens: Regroup to make \(13 - 5 = 8\), so the missing tens digit in the minuend is \(3\). 3. Hundreds: After the first regrouping, the hundreds digit is \(2\). Regroup from the thousands place: \(12 - 6 = 6\), so the missing hundreds digit in the difference is \(6\). 4. Thousands: The regrouped thousands digit is \(7\). Since \(7 - \square = 4\), the missing thousands digit in the subtrahend is \(3\). The completed subtraction is \(8334 - 3653 = 4681\).

The missing digits, read from top to bottom and left to right, are \(3\), \(3\), \(3\), and \(6\). The completed subtraction is \(8334 - 3653 = 4681\).

- Begin in the ones column. - In b), regroup when the upper digit is smaller than the lower digit. - Keep the numbers aligned by place value.

1. For a), subtract without regrouping: \(2 - 0 = 2\), \(9 - 7 = 2\), \(4 - 1 = 3\), and \(8 - 5 = 3\). Therefore, \(8492 - 5170 = 3322\). 2. For b), regroup through the \(0\) in the tens place. This gives \(15 - 8 = 7\) in the ones column and \(9 - 1 = 8\) in the tens column. Regroup again to get \(12 - 4 = 8\) in the hundreds column, followed by \(5 - 2 = 3\). Therefore, \(6305 - 2418 = 3887\). 3. Regrouping record: a) none | b) hundreds to tens; tens to ones; thousands to hundreds

a) \(3322\) b) \(3887\) Regrouping record: a) none | b) hundreds to tens; tens to ones; thousands to hundreds

- Calculate both differences first. - Regroup carefully across the zeros. - Compare the final results by place value.

1. For Expression A, regroup from the thousands place through all three zeros. Then \(10 - 7 = 3\), \(9 - 6 = 3\), \(9 - 5 = 4\), and \(8 - 4 = 4\). Thus, \(9000 - 4567 = 4433\). 2. For Expression B, regroup from the thousands place through all three zeros. Then \(10 - 6 = 4\), \(9 - 5 = 4\), \(9 - 4 = 5\), and \(7 - 3 = 4\). Thus, \(8000 - 3456 = 4544\). 3. Since \(4544 > 4433\), Expression B has the greater result. 4. Regrouping record: A thousands to hundreds; hundreds to tens; tens to ones | B thousands to hundreds; hundreds to tens; tens to ones

Expression B has the greater result: \(4544 > 4433\). Regrouping record: A thousands to hundreds; hundreds to tens; tens to ones | B thousands to hundreds; hundreds to tens; tens to ones

- Start at the ones; look left if the next column has zero to trade. - Each trade changes the place you take from. - Check by adding the difference to the number being subtracted.

1. The ones place has \(3-5\). Trade \(1\) hundred for \(10\) tens, then trade \(1\) ten for \(10\) ones. Now \(13-5=8\) in the ones and \(9-1=8\) in the tens. 2. The hundreds place has \(4-7\). Trade \(1\) thousand for \(10\) hundreds: \(14-7=7\). 3. In the thousands place, trade \(1\) ten-thousand for \(10\) thousands, giving \(11-6=5\). 4. The ten-thousands place gives \(7-4=3\). Therefore, \(82{,}503-46{,}715=35{,}788\).

\(35{,}788\). Trade from hundreds to tens and then from tens to ones to get past the zero.

- Work through the columns from ones to thousands. - Each trade changes the value remaining in the column it comes from. - Check whether the known answer in a column requires borrowing from the next higher place.

1. Ones: \(5\) trades one ten to make \(15\) ones, and \(15-8=7\), matching the answer. 2. Tens: after that trade, the top tens must be \(6\) so that \(6-4=2\). The missing original tens digit is \(7\). 3. Hundreds: \(2\) hundreds cannot subtract the missing hundreds digit to leave \(4\) hundreds, so trade one thousand. Now \(12\) hundreds minus \(8\) hundreds gives \(4\) hundreds. 4. The remaining thousands give \(5-2=3\), so \(6275-2848=3427\).

Top missing digit: \(7\); bottom missing digit: \(8\). \(6275-2848=3427\).

- Regroup when the top digit is less than the digit below it. - Exchange one unit of the next greater place value for ten units of the current place value. - Keep the numbers aligned by place value.

1. For part a, regroup in the ones column: \(15 - 8 = 7\). Regroup in the tens column: \(11 - 4 = 7\). Then \(6 - 3 = 3\). The difference is \(377\). 2. For part b, regroup in the ones column: \(13 - 6 = 7\). Regroup in the tens column: \(10 - 5 = 5\). Then \(5 - 2 = 3\). The difference is \(357\). 3. For part c, regroup in the ones column: \(11 - 5 = 6\). Regroup in the tens column: \(13 - 7 = 6\). Then \(8 - 4 = 4\). The difference is \(466\). 4. Regrouping record: a) tens to ones; hundreds to tens | b) tens to ones; hundreds to tens | c) tens to ones; hundreds to tens

a) \(377\) b) \(357\) c) \(466\) Regrouping record: a) tens to ones; hundreds to tens | b) tens to ones; hundreds to tens | c) tens to ones; hundreds to tens

- Align ones, tens, and hundreds. - Regroup when the top digit is less than the digit below it. - Remember that regrouping leaves one fewer unit in the next column. - Check by adding the difference and the subtrahend.

1. For part a, regroup in the ones column: \(13 - 8 = 5\). There are \(6\) tens left, so \(6 - 2 = 4\). Then \(6 - 4 = 2\). The difference is \(245\). 2. For part b, \(4 - 1 = 3\). In the tens column, regroup one hundred and calculate \(11 - 5 = 6\). There are \(8\) hundreds left, so \(8 - 3 = 5\). The difference is \(563\). 3. For part c, regroup across the zero: \(15 - 7 = 8\), \(9 - 6 = 3\), and \(7 - 4 = 3\). The difference is \(338\). 4. Regrouping record: a) tens to ones | b) hundreds to tens | c) hundreds to tens; tens to ones

a) \(245\) b) \(563\) c) \(338\) Regrouping record: a) tens to ones | b) hundreds to tens | c) hundreds to tens; tens to ones

- When a place-value digit is zero, regroup from the next nonzero place to the left. - Continue regrouping through any zeros. - Record each regrouped value above its column. - Work from right to left.

1. For part a, regroup across both zeros. Calculate \(10 - 4 = 6\), \(9 - 3 = 6\), and \(4 - 2 = 2\). The difference is \(266\). 2. For part b, regroup across the zero. Calculate \(16 - 9 = 7\), \(9 - 8 = 1\), and \(6 - 4 = 2\). The difference is \(217\). 3. For part c, regroup from the thousands place through the hundreds and tens places. Calculate \(10 - 7 = 3\), \(9 - 5 = 4\), and \(9 - 6 = 3\). The difference is \(343\). 4. Regrouping record: a) hundreds to tens; tens to ones | b) hundreds to tens; tens to ones | c) thousands to hundreds; hundreds to tens; tens to ones

a) \(266\) b) \(217\) c) \(343\) Regrouping record: a) hundreds to tens; tens to ones | b) hundreds to tens; tens to ones | c) thousands to hundreds; hundreds to tens; tens to ones

- Work from right to left. - Regroup when the top digit is less than the digit below it. - With zeros, you may need to regroup through more than one place value. - Align all numbers by place value.

1. For part a, subtract without regrouping: \(8 - 4 = 4\), \(5 - 2 = 3\), and \(7 - 3 = 4\). The difference is \(434\). 2. For part b, regroup in both the ones and tens columns: \(12 - 5 = 7\), \(13 - 8 = 5\), and \(5 - 1 = 4\). The difference is \(457\). 3. For part c, regroup across both zeros: \(10 - 7 = 3\), \(9 - 6 = 3\), and \(8 - 5 = 3\). The difference is \(333\). 4. For part d, align \(42\) as \(042\). Calculate \(3 - 2 = 1\), regroup one hundred to calculate \(11 - 4 = 7\), and then calculate \(7 - 0 = 7\). The difference is \(771\). 5. Regrouping record: a) none | b) tens to ones; hundreds to tens | c) hundreds to tens; tens to ones | d) hundreds to tens

a) \(434\) b) \(457\) c) \(333\) d) \(771\) Regrouping record: a) none | b) tens to ones; hundreds to tens | c) hundreds to tens; tens to ones | d) hundreds to tens

- Treat each place-value column as a small equation. - Account for regrouping when determining a missing digit. - You can use addition as the inverse operation. - Add the difference and the subtrahend to check.

1. In part a, regroup one hundred in the tens column. The original missing tens digit must be \(2\), because after regrouping \(12 - 8 = 4\). The completed equation is \(527 - 284 = 243\). 2. In part b, regroup across the zero. There are \(7\) hundreds left, and \(7 - \square = 1\), so the missing hundreds digit is \(6\). The completed equation is \(804 - 636 = 168\). 3. In part c, regroup in the ones column so that \(16 - \square = 8\). The missing ones digit is \(8\). The completed equation is \(456 - 178 = 278\). 4. Regrouping record: a) hundreds to tens | b) hundreds to tens; tens to ones | c) tens to ones; hundreds to tens

a) \(2\) b) \(6\) c) \(8\) Regrouping record: a) hundreds to tens | b) hundreds to tens; tens to ones | c) tens to ones; hundreds to tens

- Begin in the ones column. - Determine which number minus \(7\) gives a ones digit of \(8\) after regrouping. - Carry the effect of each regrouping into the next column. - Check by adding the difference and the subtrahend.

1. In the ones column, regroup so that \(15 - 7 = 8\). The missing ones digit in the minuend is \(5\). 2. Regrouping leaves \(1\) ten. Since the result has \(4\) tens, regroup one hundred and solve \(11 - ? = 4\). The missing tens digit in the subtrahend is \(7\). 3. There are \(4\) hundreds left, and \(4 - 1 = 3\), which matches the result. 4. The completed equation is \(525 - 177 = 348\).

The missing top ones digit is \(5\), and the missing bottom tens digit is \(7\). \(525 - 177 = 348\)

- Addition is the inverse operation of subtraction. - Regroup carefully when a place contains a zero. - Align digits with the same place value.

1. For part a, regroup to calculate \(11-3=8\), \(14-6=8\), \(3-1=2\), \(12-5=7\), and \(8-4=4\). Therefore, \(92{,}451-45{,}163=47{,}288\). 2. Check part a: \(47{,}288+45{,}163=92{,}451\). 3. For part b, regroup across the three ending zeros. The ones, tens, and hundreds calculations are \(10-2=8\), \(9-8=1\), and \(9-4=5\). Regroup from the hundred-thousands place so the remaining calculations are \(15-7=8\), \(9-3=6\), and \(4-2=2\). Therefore, \(506{,}000-237{,}482=268{,}518\). 4. Check part b: \(268{,}518+237{,}482=506{,}000\). 5. Regrouping record: a) tens to ones; hundreds to tens; ten-thousands to thousands | b) thousands to hundreds; hundreds to tens; tens to ones; hundred-thousands to ten-thousands; ten-thousands to thousands

a) \(47{,}288\); check: \(47{,}288+45{,}163=92{,}451\) b) \(268{,}518\); check: \(268{,}518+237{,}482=506{,}000\) Regrouping record: a) tens to ones; hundreds to tens; ten-thousands to thousands | b) thousands to hundreds; hundreds to tens; tens to ones; hundred-thousands to ten-thousands; ten-thousands to thousands

- The minuend equals the difference plus the subtrahend. - Think of the difference and subtrahend as two parts of the minuend. - In the second check, subtract the difference you found from the original minuend.

1. Regroup in every place from ones through ten-thousands: \(10-5=5\), \(11-3=8\), \(10-4=6\), \(12-8=4\), \(13-5=8\), and \(6-2=4\). Therefore, \(743{,}120-258{,}435=484{,}685\). 2. First check: \(484{,}685+258{,}435=743{,}120\). 3. Second check: \(743{,}120-484{,}685=258{,}435\). 4. Both checks reproduce a number from the original subtraction, so the difference is correct. 5. Regrouping record: tens to ones; hundreds to tens; thousands to hundreds; ten-thousands to thousands; hundred-thousands to ten-thousands

The difference is \(484{,}685\). First check: \(484{,}685+258{,}435=743{,}120\) Second check: \(743{,}120-484{,}685=258{,}435\) Regrouping record: tens to ones; hundreds to tens; thousands to hundreds; ten-thousands to thousands; hundred-thousands to ten-thousands

- Add each stated difference to its subtrahend. - Compare each addition result with the original minuend. - In part c, regroup carefully across the zeros.

1. Check part a: \(44{,}121+14{,}271=58{,}392\), so part a is correct. 2. Check part b: \(55{,}219+37{,}186=92{,}405\), so part b is correct. 3. Check part c: \(45{,}695+24{,}315=70{,}010\), not \(70{,}000\), so part c is incorrect. 4. For the correct subtraction, regroup across the four zeros. The place-value calculations are \(10-5=5\), \(9-1=8\), \(9-3=6\), \(9-4=5\), and \(6-2=4\). Therefore, \(70{,}000-24{,}315=45{,}685\). 5. Regrouping record: c) ten-thousands to thousands; thousands to hundreds; hundreds to tens; tens to ones

c) is incorrect. The correct equation is \(70{,}000-24{,}315=45{,}685\). Regrouping record: c) ten-thousands to thousands; thousands to hundreds; hundreds to tens; tens to ones

- Align each pair by place value. - Regroup carefully, especially across zeros. - Compare the completed results digit by digit.

1. For \(512{,}403 - 234{,}567\), regroup through the \(0\) in the tens place to get \(13 - 7 = 6\) and \(9 - 6 = 3\). Continue regrouping to get \(13 - 5 = 8\), \(11 - 4 = 7\), \(10 - 3 = 7\), and \(4 - 2 = 2\). The difference is \(277{,}836\). 2. For \(800{,}000 - 522{,}164\), regroup from the hundred-thousands place through all five zeros. The columns give \(10 - 4 = 6\), \(9 - 6 = 3\), \(9 - 1 = 8\), \(9 - 2 = 7\), \(9 - 2 = 7\), and \(7 - 5 = 2\). The difference is \(277{,}836\). 3. For \(401{,}020 - 123{,}184\), regroup to get \(10 - 4 = 6\) and \(11 - 8 = 3\). Then \(9 - 1 = 8\). Regroup across the next \(0\) to get \(10 - 3 = 7\), \(9 - 2 = 7\), and \(3 - 1 = 2\). The difference is \(277{,}836\). 4. All three differences are equal. 5. Regrouping record: 1. hundreds to tens; tens to ones; thousands to hundreds; ten-thousands to thousands; hundred-thousands to ten-thousands | 2. hundred-thousands to ten-thousands; ten-thousands to thousands; thousands to hundreds; hundreds to tens; tens to ones | 3. tens to ones; thousands to hundreds; hundreds to tens; hundred-thousands to ten-thousands; ten-thousands to thousands

All three results are equal to \(277{,}836\). Regrouping record: 1. hundreds to tens; tens to ones; thousands to hundreds; ten-thousands to thousands; hundred-thousands to ten-thousands | 2. hundred-thousands to ten-thousands; ten-thousands to thousands; thousands to hundreds; hundreds to tens; tens to ones | 3. tens to ones; thousands to hundreds; hundreds to tens; hundred-thousands to ten-thousands; ten-thousands to thousands

- Align the numbers by place value. - Regroup when the top digit is less than the digit below it. - Begin in the ones column. - Remember that regrouping changes the next top digit.

1. For part a, regroup in the ones and tens columns: \(13 - 5 = 8\), \(13 - 7 = 6\), and \(5 - 2 = 3\). The difference is \(368\). 2. For part b, regroup in the ones and tens columns: \(12 - 6 = 6\), \(10 - 5 = 5\), and \(7 - 4 = 3\). The difference is \(356\). 3. For part c, regroup across the zero: \(14 - 8 = 6\), \(9 - 2 = 7\), and \(4 - 1 = 3\). The difference is \(376\). 4. For part d, align \(83\) as \(083\). Regroup in the ones and tens columns: \(11 - 3 = 8\), \(11 - 8 = 3\), and \(8 - 0 = 8\). The difference is \(838\). 5. Regrouping record: a) tens to ones; hundreds to tens | b) tens to ones; hundreds to tens | c) hundreds to tens; tens to ones | d) tens to ones; hundreds to tens

a) \(368\) b) \(356\) c) \(376\) d) \(838\) Regrouping record: a) tens to ones; hundreds to tens | b) tens to ones; hundreds to tens | c) hundreds to tens; tens to ones | d) tens to ones; hundreds to tens

- When a top digit is zero, regroup from the next nonzero place to the left. - Continue regrouping through consecutive zeros. - Work from right to left. - Check with estimation or addition.

1. For part a, regroup from the thousands place through the hundreds and tens places. Calculate \(10 - 2 = 8\), \(9 - 5 = 4\), and \(9 - 4 = 5\). The difference is \(548\). 2. For part b, regroup across both zeros. Calculate \(10 - 8 = 2\), \(9 - 2 = 7\), and \(6 - 3 = 3\). The difference is \(372\). 3. For part c, align \(67\) as \(0067\), then regroup through the zeros. Calculate \(10 - 7 = 3\), \(9 - 6 = 3\), and \(9 - 0 = 9\). The difference is \(933\). 4. For part d, regroup across both zeros. Calculate \(10 - 1 = 9\), \(9 - 9 = 0\), and \(4 - 2 = 2\). The difference is \(209\). 5. Regrouping record: a) thousands to hundreds; hundreds to tens; tens to ones | b) hundreds to tens; tens to ones | c) thousands to hundreds; hundreds to tens; tens to ones | d) hundreds to tens; tens to ones

a) \(548\) b) \(372\) c) \(933\) d) \(209\) Regrouping record: a) thousands to hundreds; hundreds to tens; tens to ones | b) hundreds to tens; tens to ones | c) thousands to hundreds; hundreds to tens; tens to ones | d) hundreds to tens; tens to ones

- Addition is the inverse operation of subtraction. - Add the difference and subtrahend to recover the minuend. - Regroup carefully in each subtraction.

1. Part a: regroup to calculate \(12-8=4\), \(14-7=7\), \(2-1=1\), and \(6-2=4\). Therefore, \(6352-2178=4174\). Check: \(4174+2178=6352\). 2. Part b: regroup across the zero to calculate \(12-5=7\), \(9-2=7\), \(17-9=8\), \(4-3=1\), and \(4-1=3\). Therefore, \(45{,}802-13{,}925=31{,}877\). Check: \(31{,}877+13{,}925=45{,}802\). 3. Part c: regroup to calculate \(10-3=7\), \(16-8=8\), and \(13-5=8\). After regrouping across the zero in the thousands place, calculate \(9-4=5\), \(10-2=8\), and \(2-1=1\). Therefore, \(310{,}470-124{,}583=185{,}887\). Check: \(185{,}887+124{,}583=310{,}470\). 4. Regrouping record: a) tens to ones; hundreds to tens | b) hundreds to tens; tens to ones; thousands to hundreds | c) tens to ones; hundreds to tens; ten-thousands to thousands; thousands to hundreds; hundred-thousands to ten-thousands

a) \(4174\); check: \(4174+2178=6352\) b) \(31{,}877\); check: \(31{,}877+13{,}925=45{,}802\) c) \(185{,}887\); check: \(185{,}887+124{,}583=310{,}470\) Regrouping record: a) tens to ones; hundreds to tens | b) hundreds to tens; tens to ones; thousands to hundreds | c) tens to ones; hundreds to tens; ten-thousands to thousands; thousands to hundreds; hundred-thousands to ten-thousands

- Read each row by counting the chips in every place-value column. - An empty column represents a digit of \(0\). - Subtract the lesser number from the greater number.

1. The first row has \(4\) ten-thousands, \(3\) thousands, \(0\) hundreds, \(7\) tens, and \(5\) ones, so it represents \(43{,}075\). 2. The second row has \(2\) ten-thousands, \(8\) thousands, \(5\) hundreds, \(1\) ten, and \(9\) ones, so it represents \(28{,}519\). 3. Use the standard subtraction algorithm for \(43{,}075 - 28{,}519\). Regroup \(1\) ten as \(10\) ones, giving \(15 - 9 = 6\) in the ones place and \(6 - 1 = 5\) in the tens place. 4. Regroup \(1\) thousand as \(10\) hundreds, giving \(10 - 5 = 5\) in the hundreds place. Then regroup \(1\) ten-thousand as \(10\) thousands, giving \(12 - 8 = 4\) in the thousands place and \(3 - 2 = 1\) in the ten-thousands place. 5. Therefore, \(43{,}075 - 28{,}519 = 14{,}556\). 6. Regrouping record: b) tens to ones; thousands to hundreds; ten-thousands to thousands

a) The numbers are \(43{,}075\) and \(28{,}519\). b) The difference is \(14{,}556\). Regrouping record: b) tens to ones; thousands to hundreds; ten-thousands to thousands

- Start at the ones column for each subtraction. - Record every place that lends units to another column. - Use the result digit to decide which hidden digit can work after the trades.

1. In a), ones: \(12-5=7\) after trading one ten. Tens: trade a hundred to make \(13\) tens; \(13-8=5\). So the missing digit is \(8\). 2. In b), ones: \(13-8=5\). The top tens become \(3\) after that trade. Trade one hundred for ten tens, so \(13-4=9\). The missing top tens digit was \(4\). 3. In c), ones: \(15-8=7\). Then \(11-7=4\) tens after a hundreds trade. The hundreds column is now one less, so the missing top hundreds digit is \(8\), making \(7-4=3\) hundreds. 4. In d), trade through the zero tens digit to reach \(14\) ones. Since \(14-6=8\), the missing bottom ones digit is \(6\). The remaining digits give \(9-2=7\) tens and \(8-5=3\) hundreds. 5. Transfers: a) tens to ones, hundreds to tens; b) tens to ones, hundreds to tens; c) tens to ones, hundreds to tens; d) hundreds to tens, tens to ones.

a) \(8\); tens to ones, hundreds to tens. b) \(4\); tens to ones, hundreds to tens. c) \(8\); tens to ones, hundreds to tens. d) \(6\); hundreds to tens, tens to ones.

- Start in the ones column and track each trade. - When the tens give a ten away, update their amount before subtracting tens. - The hundreds result must also account for any hundred traded away.

1. Ones: trade a ten to make \(12\) ones. To leave \(4\) ones, the missing subtrahend digit is \(8\), since \(12-8=4\). 2. Tens: the top tens have already given away one ten. Another hundred must be traded. To leave \(5\) tens after subtracting \(7\), the original tens digit must have been \(3\): it becomes \(2\), then \(12\), and \(12-7=5\). 3. Hundreds: after the trade, \(5\) hundreds become \(4\), and \(4-2=2\). Thus the missing result digit is \(2\). 4. The completed equation is \(532-278=254\).

\(532-278=254\).

- When a zero in the minuend cannot supply the needed amount, trace the regrouping leftward until you reach a nonzero place. - After completing the missing digits, check the subtraction with the related addition without using the check to discover the digits.

1. Regroup from the thousands place through the hundreds, tens, and ones places. In the ones column, \(10 - \square = 7\), so the missing ones digit is \(3\). 2. In the tens column, \(9 - 4 = 5\). In the hundreds column, \(9 - \square = 6\), so the missing hundreds digit is \(3\). The completed equation is \(1000 - 343 = 657\). 3. Regrouping record: thousands to hundreds; hundreds to tens; tens to ones

\(1000 - 343 = 657\) Regrouping record: thousands to hundreds; hundreds to tens; tens to ones

- Focus on the tens column of the first subtraction before using the second one. - The same digit must make both tens-column equations true. - Check the completed equations column by column.

1. In the first subtraction, the tens column is \(9-\square=6\), so the missing digit is \(3\). 2. This gives \(999-133=866\). 3. In the second subtraction, the tens column is \(6-\square=3\), again giving \(\square=3\). 4. This gives \(866-235=631\). 5. The same missing digit, \(3\), satisfies both standard-subtraction columns.

First tens column: \(9-3=6\). Second tens column: \(6-3=3\). The common missing digit is \(3\): \(999-133=866\) and \(866-235=631\).

- Work from right to left in the displayed subtraction. - Regroup before using the visible result digit to solve an unknown. - When a zero blocks regrouping, exchange from the next nonzero place to the left.

1. a) Regroup in the ones column: \(12-4=8\). Then \(7-3=4\). In the hundreds column, \(5-\square=3\), so the missing digit is \(2\). 2. b) Regroup in the ones column: \(13-7=6\). To subtract \(6\) tens, regroup through the zero hundreds digit, giving \(13-6=7\) tens and \(9-5=4\) hundreds. In the thousands column, \(11-\square=9\), so the missing digit is \(2\). 3. c) Ones: \(5-0=5\). Tens: \(\square-1=4\), so the missing digit is \(5\). The remaining columns give \(2,4,6\), matching the result. 4. Regrouping record: a) tens to ones | b) tens to ones; thousands to hundreds; hundreds to tens; ten-thousands to thousands | c) none

a) \(2\) b) \(2\) c) \(5\) Regrouping record: a) tens to ones | b) tens to ones; thousands to hundreds; hundreds to tens; ten-thousands to thousands | c) none

- Work from ones to hundreds and track the effect of each trade. - Find how many tens are needed to leave the shown tens result. - Check the completed digits against the hundreds result too.

1. Ones: trade one ten to make \(12\) ones. Since \(12-5=7\), the missing bottom ones digit is \(5\). 2. Tens: after the ones trade, the top tens digit is one less than before. To leave \(8\) tens after subtracting \(4\), the column needs \(12\) tens. One hundred is traded into \(10\) tens, so \(2\) tens had remained after the first trade. The missing original tens digit is \(3\). 3. Hundreds: \(8\) becomes \(7\) hundreds after the trade, and \(7-4=3\). Thus \(832-445=387\).

\(832-445=387\).

- Start in the ones column and decide whether regrouping is required. - Each regrouping changes the top digit in the next column; keep track of that change. - Use the visible result digit in each column to determine the missing digit.

1. In the ones column, \(\square-8=5\) requires regrouping. Therefore \(13-8=5\), so the missing ones digit in the minuend is \(3\). 2. After regrouping, the tens column is \(10-\square=6\), so the missing tens digit in the subtrahend is \(4\). One hundred is regrouped. 3. In the hundreds column, \(\square-1-2=1\), so the missing hundreds digit in the minuend is \(4\). 4. The completed equation is \(413-248=165\).

\(413-248=165\)

- Decide whether you must regroup from the next place value. - You can also work upward with addition: the difference plus the subtrahend equals the minuend.

1. In the ones column, \(\square - 7 = 4\) requires regrouping. Therefore, \(11 - 7 = 4\), so the missing ones digit in the minuend is \(1\). 2. After regrouping, \(5\) tens remain. The tens column is \(5 - \square = 1\), so the missing tens digit in the subtrahend is \(4\). 3. In the hundreds column, \(5 - 2 = 3\), so the missing hundreds digit in the result is \(3\). 4. The completed equation is \(561 - 247 = 314\).

The missing digits are \(1\), \(4\), and \(3\). The completed equation is \(561 - 247 = 314\).

- When the tens digit is zero, regroup from the hundreds place before regrouping to the ones place. - Check by adding the difference and the subtrahend.

1. In the ones column, regroup across the zero so that \(13 - 8 = 5\). The missing ones digit in the result is \(5\). 2. Regrouping across the zero leaves \(9\) tens. The tens column is \(9 - \square = 6\), so the missing tens digit in the subtrahend is \(3\). 3. There are \(5\) hundreds left, and \(5 - 2 = 3\), which matches the result. 4. The completed equation is \(603 - 238 = 365\).

The missing digits are \(3\) and \(5\). The completed equation is \(603 - 238 = 365\).

- Begin with the missing trade needed in the ones column. - Decide how many tens the top number must have after regrouping to get the shown result. - A hundreds trade may be necessary to supply those tens.

1. Ones: the top ones digit is \(0\), so trade a ten to make \(10\) ones. Then \(10-5=5\). 2. Tens: the missing top tens digit has already lent one ten. Borrow one hundred to make \(12\) tens; \(12-6=6\), so the original top tens digit was \(3\). 3. After the hundreds trade, \(0\) hundreds remain. The completed equation is \(130-65=65\). 4. Transfers: tens to ones, then hundreds to tens.

Missing digit: \(3\). \(130-65=65\). Transfers: tens to ones, hundreds to tens.

- Work from the ones column to the hundreds. - Remember that borrowing changes the top digit in the next column. - Check the completed number by adding the difference and the subtrahend.

1. Ones: trade one ten to make \(14\) ones. The missing bottom ones digit is \(8\), because \(14-8=6\). 2. Tens: the top tens have lent one ten. Trade a hundred for ten tens; to leave \(4\) tens after subtracting \(7\), the top tens must originally have been \(2\). After the first trade there is \(1\) ten, then \(11-7=4\). 3. Hundreds: \(5\) has become \(4\) after that trade, and \(4-2=2\). The equation is \(524-278=246\).

Missing top tens digit: \(2\); missing bottom ones digit: \(8\). \(524-278=246\).

- Work from the ones column and keep track of how each regrouping changes the next column. - Use the visible result digits as column constraints rather than reconstructing the whole minuend or subtrahend at once.

1. In the ones column, \(2 - 8\) requires regrouping, so \(12 - 8 = 4\). 2. After that regrouping, \(0\) tens remain. Regroup one hundred and solve \(10 - \square = 3\), so the missing tens digit in the subtrahend is \(7\). 3. In the hundreds column, \(\square - 1 - 4 = 3\), so the missing hundreds digit in the minuend is \(8\). 4. The completed equation is \(812 - 478 = 334\).

The missing digits are \(8\) and \(7\). The completed equation is \(812 - 478 = 334\).

- Check that each column is calculated as the top digit minus the bottom digit. - Regroup from the next place when the top digit is too small.

1. The shown work appears to subtract the smaller digit from the greater digit in each column, regardless of which digit is on top. 2. In the ones place, regroup \(1\) ten and calculate \(14 - 8 = 6\). 3. One ten remains, so regroup \(1\) hundred and calculate \(11 - 7 = 4\). 4. In the hundreds place, \(4 - 2 = 2\). 5. Therefore, \(524 - 278 = 246\).

The smaller digit was subtracted from the greater digit in each column instead of regrouping when needed. The correct difference is \(246\).

- Check what must happen when subtracting \(7\) from \(5\). - Remember that regrouping in the ones place affects the next place.

1. The first error is in the ones place. Since \(5<7\), regroup one ten and calculate \(15-7=8\). The shown work appears to use \(7-5=2\) instead. 2. After regrouping, \(1\) ten remains. Regroup one hundred, then calculate \(11-7=4\) in the tens place. 3. After that regrouping, \(1\) hundred remains. Regroup one thousand, then calculate \(11-8=3\) in the hundreds place. 4. Six thousands remain, and \(6-3=3\). Therefore, \(7225-3877=3348\).

The ones place was calculated as \(7-5=2\) instead of regrouping to calculate \(15-7=8\). The correct difference is \(3348\).

- Start in the ones column and decide whether regrouping is required. - Remember that regrouping changes the top digit in the tens column. - Use the visible result digit to determine the missing subtrahend digit.

1. In the ones column, \(\square-8=4\) requires regrouping. Therefore \(12-8=4\), so the missing ones digit in the minuend is \(2\). 2. After regrouping, \(4\) tens remain. Since that is too small to produce the visible tens digit, regroup one hundred and solve \(14-\square=6\). The missing tens digit in the subtrahend is \(8\). 3. In the hundreds column, \(7-1-3=3\), which matches the result. 4. The completed equation is \(752-388=364\).

\(752-388=364\)

- You cannot take the ones in the lower number from the upper zero without trading. - Find the nearest nonzero place you can trade from. - Check the adjusted tens digit before deciding the hundreds digit.

1. The top number ends in two zeros. To subtract in the ones place, trade \(1\) hundred for \(10\) tens, then \(1\) ten for \(10\) ones. 2. The ones column must be \(10-4=6\), so the missing ones digit below is \(4\). 3. There are now \(9\) tens. Since \(9-4=5\), the tens column checks. 4. In the hundreds column, after trading, \(3-1=2\), so the original hundreds digit was \(4\). The completed subtraction is \(400-144=256\).

\(400-144=256\). Trade one hundred into ten tens, then one ten into ten ones.

- Regrouping across zeros begins at the next nonzero place to the left. - Each place used for regrouping decreases by \(1\).

1. In the ones column, regroup across the zeros so that \(10 - \square = 7\). The missing ones digit in the subtrahend is \(3\). 2. Regrouping leaves \(9\) tens, and \(9 - 7 = 2\). 3. There are \(4\) hundreds left. The hundreds column is \(4 - \square = 1\), so the missing hundreds digit in the subtrahend is \(3\). 4. The completed equation is \(500 - 373 = 127\). 5. Regrouping record: hundreds to tens; tens to ones

\(500 - 373 = 127\) Regrouping record: hundreds to tens; tens to ones

- Pay special attention when regrouping across a zero. - Check by adding the difference and subtrahend.

1. Ones: Regroup to calculate \(12-5=7\). 2. Tens: Regroup across the zero, leaving \(9\) tens. Then \(9-4=5\). 3. Hundreds: After regrouping, \(7\) hundreds remain. Calculate \(7-3=4\), not \(5\). 4. The correct difference is \(802-345=457\).

The error is in the hundreds place because the regrouping was not included. The correct difference is \(457\).

- Look at the tens column before trying to regroup the ones. - If a place has no units available to exchange, move left until you find a nonzero place. - Track what remains in each column after both exchanges.

1. The chart represents \(504-278\). 2. There are only \(4\) ones, so an exchange is needed, but there are \(0\) tens. 3. Exchange \(1\) hundred for \(10\) tens. Then exchange \(1\) of those tens for \(10\) ones. This leaves \(4\) hundreds, \(9\) tens, and \(14\) ones. 4. Subtract: \(14-8=6\), \(9-7=2\), and \(4-2=2\). 5. The difference is \(226\).

Regrouping must pass from hundreds through tens because the tens place starts at \(0\). The difference is \(226\).

- Read the chart as one complete minuend and one complete subtrahend. - In the ones place, ask whether the minuend has enough ones to subtract \(8\). - Regroup across the zero before subtracting each column.

1. The chart represents \(702-358\). 2. Subtraction is not performed by choosing the larger digit in each column; the place-value amounts belong to the whole minuend and subtrahend. 3. Regroup through the zero: exchange \(1\) hundred for \(10\) tens, then exchange \(1\) ten for \(10\) ones. 4. Compute \(12-8=4\), \(9-5=4\), and \(6-3=3\). 5. The correct difference is \(344\).

Sam's method ignores the direction of subtraction and regrouping. The correct difference is \(344\).

- Start at the ones column and find a larger place that can supply units. - A trade can pass through more than one empty column. - Update every column touched by the trade before subtracting.

1. The ones column has \(2\), so trade from the thousands: the \(1\) thousand becomes \(10\) hundreds. 2. Trade one hundred for ten tens, then one ten for ten ones. The top number is now \(9\) hundreds, \(9\) tens, and \(12\) ones, with no thousands left. 3. Subtract: \(12-5=7\) ones, \(9-8=1\) ten, and \(9-3=6\) hundreds. 4. The difference is \(617\).

\(1002-385=617\). Trade the thousand into hundreds, a hundred into tens, and a ten into ones.

- Choose digits according to whether each place-value column should require regrouping. - To force regrouping from tens to ones, the bottom ones digit must be greater than the top ones digit. - Remember that regrouping in one column changes the available value in the next column. - Each part has many correct answers.

1. For part a, choose each digit of the subtrahend so it is no greater than the corresponding digit of \(674\). One example is \(674-123=551\). 2. For part b, choose a ones digit greater than \(4\) so that one ten must be regrouped. Then choose the tens digit so no second regrouping is needed. One example is \(674-125=549\). 3. For part c, choose a ones digit that forces regrouping and then a tens digit that also forces regrouping after the first exchange. One example is \(674-285=389\).

Possible answers: a) \(674-123=551\) b) \(674-125=549\) c) \(674-285=389\)

- Begin in the ones column and work from right to left. - Regroup from the next place value when the top digit is too small. - Track how each regrouping changes the next top digit. - Check by adding the difference and the subtrahend.

1. In the ones column, regroup \(1\) ten so that \(11 - 6 = 5\). The missing ones digit in the result is \(5\). 2. The regrouping leaves \(1\) ten in the top number. Since the result has \(6\) tens, regroup \(1\) hundred to make \(11\) tens. Then \(11 - \square = 6\), so the missing tens digit in the subtrahend is \(5\). 3. After regrouping, the top hundreds digit is one less than the original missing digit. It must satisfy \((\square - 1) - 1 = 2\), so the original hundreds digit is \(4\). 4. The completed equation is \(421 - 156 = 265\).

The missing hundreds digit in the minuend is \(4\), the missing tens digit in the subtrahend is \(5\), and the missing ones digit in the result is \(5\). \(421 - 156 = 265\)

- Why is regrouping from tens to ones guaranteed? - After regrouping one ten, how many tens remain in \(652\)? - Choose a missing digit greater than the number of tens that remain.

1. In the ones column, \(2 < 8\), so regrouping from tens to ones is always required. 2. After regrouping, the top number has \(4\) tens. To require regrouping from hundreds to tens, the missing tens digit must be greater than \(4\). 3. Therefore, the possible digits are \(5\), \(6\), \(7\), \(8\), and \(9\). 4. For \(\square = 5\), \(652 - 358 = 294\). 5. For \(\square = 6\), \(652 - 368 = 284\). 6. For \(\square = 7\), \(652 - 378 = 274\). 7. For \(\square = 8\), \(652 - 388 = 264\). 8. For \(\square = 9\), \(652 - 398 = 254\).

Possible digits: \(5\), \(6\), \(7\), \(8\), and \(9\) \(\square = 5\): \(652 - 358 = 294\) \(\square = 6\): \(652 - 368 = 284\) \(\square = 7\): \(652 - 378 = 274\) \(\square = 8\): \(652 - 388 = 264\) \(\square = 9\): \(652 - 398 = 254\)

- First calculate the original difference. - Apply each stated change to the correct number. - Consider how decreasing the minuend and increasing the subtrahend both reduce the difference.

1. For a), regroup in the ones column to get \(15 - 6 = 9\), then regroup in the tens column to get \(14 - 5 = 9\). The remaining columns give \(4 - 4 = 0\), \(5 - 3 = 2\), \(5 - 2 = 3\), and \(5 - 1 = 4\). Therefore, \(555{,}555 - 123{,}456 = 432{,}099\). 2. The new minuend is \(555{,}555 - 1111 = 554{,}444\), and the new subtrahend is \(123{,}456 + 1111 = 124{,}567\). The new problem is \(554{,}444 - 124{,}567\). 3. For the new problem, successive regrouping gives \(14 - 7 = 7\), \(13 - 6 = 7\), \(13 - 5 = 8\), and \(13 - 4 = 9\). The remaining columns give \(4 - 2 = 2\) and \(5 - 1 = 4\), so the new difference is \(429{,}877\). 4. The new difference is \(432{,}099 - 429{,}877 = 2222\) less. This matches the combined effect of decreasing the minuend by \(1111\) and increasing the subtrahend by \(1111\). 5. Regrouping record: original tens to ones; hundreds to tens | new tens to ones; hundreds to tens; thousands to hundreds; ten-thousands to thousands

a) \(432{,}099\) b) \(554{,}444 - 124{,}567\) c) The new difference is \(429{,}877\), which is \(2222\) less than the original difference. Regrouping record: original tens to ones; hundreds to tens | new tens to ones; hundreds to tens; thousands to hundreds; ten-thousands to thousands

- Work systematically from right to left. - Record each regrouping before moving to the next place. - Use the known difference digit to determine each missing digit.

1. Ones: Regroup to make \(12 - \square = 7\), so the missing ones digit in the subtrahend is \(5\). 2. Tens: After regrouping, \(11 - 8 = 3\), so the missing tens digit in the minuend is \(2\). 3. Hundreds: Regroup to make \(13 - \square = 8\), so the missing hundreds digit in the subtrahend is \(5\). 4. Thousands: After regrouping, \(11 - 7 = 4\), so the missing thousands digit in the minuend is \(2\). 5. Ten-thousands: After the final regrouping, \(6 - 1 = 5\), so the missing ten-thousands digit in the minuend is \(7\). The completed subtraction is \(72{,}422 - 17{,}585 = 54{,}837\).

\(\begin{array}{r} 72{,}422 \\ - 17{,}585 \\ \hline 54{,}837 \end{array}\)

- Begin at the ones place. - Regroup across the string of zeros carefully. - Track the changed digits after regrouping. - Check the completed subtraction with addition.

1. Regroup across the zeros. In the ones column, \(10-6=4\), so the missing ones digit in the minuend is \(0\). 2. The regrouped tens, hundreds, and thousands digits are each \(9\), giving \(9-5=4\), \(9-4=5\), and \(9-3=6\). 3. In the ten-thousands column, \(9-\square=7\), so the missing subtrahend digit is \(2\). 4. The hundred-thousands digit becomes \(2\), and \(2-1=1\), so the missing result digit is \(1\). 5. The completed equation is \(300{,}000-123{,}456=176{,}544\).

The missing digits are \(0,2,1\). \(300{,}000-123{,}456=176{,}544\)

- Find all three exact differences first. - Then find each result’s distance from the target number. - Regroup carefully across the zeros in B.

1. For A, regroup through the two zeros to obtain \(13 - 4 = 9\), \(11 - 5 = 6\), \(9 - 6 = 3\), and \(9 - 7 = 2\). Regroup once more to get \(14 - 8 = 6\), followed by \(3 - 1 = 2\). Thus, A equals \(262{,}369\). 2. For B, regroup from the hundred-thousands place through all five zeros. The columns give \(10 - 1 = 9\), \(9 - 3 = 6\), \(9 - 6 = 3\), \(9 - 7 = 2\), \(9 - 3 = 6\), and \(5 - 3 = 2\). Thus, B also equals \(262{,}369\). 3. For C, successive regrouping gives \(10 - 3 = 7\), \(13 - 7 = 6\), \(12 - 9 = 3\), \(11 - 9 = 2\), \(10 - 4 = 6\), and \(4 - 2 = 2\). Thus, C equals \(262{,}367\). 4. The distances from \(260{,}000\) are \(2369\), \(2369\), and \(2367\), respectively. Therefore, C is closest. 5. Regrouping record: A tens to ones; ten-thousands to thousands; thousands to hundreds; hundreds to tens; hundred-thousands to ten-thousands | B hundred-thousands to ten-thousands; ten-thousands to thousands; thousands to hundreds; hundreds to tens; tens to ones | C tens to ones; hundreds to tens; thousands to hundreds; ten-thousands to thousands; hundred-thousands to ten-thousands

A) \(262{,}369\) B) \(262{,}369\) C) \(262{,}367\) C is closest to \(260{,}000\). Regrouping record: A tens to ones; ten-thousands to thousands; thousands to hundreds; hundreds to tens; hundred-thousands to ten-thousands | B hundred-thousands to ten-thousands; ten-thousands to thousands; thousands to hundreds; hundreds to tens; tens to ones | C tens to ones; hundreds to tens; thousands to hundreds; ten-thousands to thousands; hundred-thousands to ten-thousands

- Rewrite each missing-number statement as a subtraction you can calculate. - In part b, first find the entire subtrahend. - Compare the calculated number with the digit pattern. - Check each answer with addition.

1. For part a, subtract the result from the minuend: \(625{,}000-187{,}654\). Regroup across the three zeros. Ones: \(10-4=6\). Tens: \(9-5=4\). Hundreds: \(9-6=3\). Thousands: regroup and calculate \(14-7=7\). Ten-thousands: regroup and calculate \(11-8=3\). Hundred-thousands: \(5-1=4\). Therefore, the missing subtrahend is \(437{,}346\). 2. For part b, find the complete subtrahend with \(403{,}506-215{,}212\). Ones: \(6-2=4\). Tens: regroup and calculate \(10-1=9\). Hundreds: \(4-2=2\). Thousands: regroup across the zero and calculate \(13-5=8\). Ten-thousands: \(9-1=8\). Hundred-thousands: \(3-2=1\). Thus, the subtrahend is \(188{,}294\). 3. Comparing \(188{,}294\) with \(1\square8{,}2\square4\), the missing digits are \(8\) and \(9\).

a) \(437{,}346\) b) The missing digits are \(8\) and \(9\), giving \(403{,}506-188{,}294=215{,}212\).

- Begin in the ones column and move left. - Record each regrouping before solving the next column. - Check your completed subtraction by adding the difference and subtrahend.

1. For a), the ones column requires regrouping, so \(11 - 3 = 8\); the missing ones digit in the minuend is \(1\). In the tens column, regroup again to get \(13 - 8 = 5\), so the missing result digit is \(5\). In the hundreds column, \(11 - \square = 6\), so the missing subtrahend digit is \(5\). The thousands column gives \(12 - 7 = 5\), and the ten-thousands column gives \(5 - 2 = 3\), so the missing leading result digit is \(3\). The completed equation is \(63{,}241 - 27{,}583 = 35{,}658\). 2. For b), the ones column requires \(15 - \square = 7\), so the missing ones digit in the subtrahend is \(8\). Regroup through the \(0\) hundreds digit to get \(12 - 7 = 5\) in the tens column and \(9 - 3 = 6\) in the hundreds column, so the missing result digit is \(6\). In the thousands column, \(11 - 5 = 6\), and in the ten-thousands column, \(7 - \square = 3\), so the missing leading subtrahend digit is \(4\). The missing thousands digit in the minuend is \(2\). The completed equation is \(82{,}035 - 45{,}378 = 36{,}657\).

a) \(63{,}241 - 27{,}583 = 35{,}658\) b) \(82{,}035 - 45{,}378 = 36{,}657\)

- Begin in the ones column and move left. - Track every regrouping carefully. - Check each completed subtraction by adding the difference and subtrahend.

1. For a), regroup in the ones column: \(13 - 7 = 6\), so the missing result digit is \(6\), and the minuend’s tens digit becomes \(7\). Then \(7 - 1 = 6\). In the hundreds column, \(4 - \square = 0\), so the missing subtrahend digit is \(4\). In the thousands column, \(\square - 1 = 8\), so the missing minuend digit is \(9\). In the ten-thousands column, \(4 - \square = 1\), so the missing subtrahend digit is \(3\). The completed equation is \(49{,}483 - 31{,}417 = 18{,}066\). 2. For b), regroup through the \(0\) in the tens place. The ones column gives \(16 - 7 = 9\), so the missing result digit is \(9\). The tens column gives \(9 - 8 = 1\). In the hundreds column, \(6 - \square = 4\), so the missing subtrahend digit is \(2\). In the thousands column, \(\square - 4 = 3\), so the missing minuend digit is \(7\). In the ten-thousands column, \(8 - \square = 7\), so the missing subtrahend digit is \(1\). The completed equation is \(87{,}706 - 14{,}287 = 73{,}419\).

a) \(49{,}483 - 31{,}417 = 18{,}066\) b) \(87{,}706 - 14{,}287 = 73{,}419\)

- Regroup carefully one column at a time. - A zero in the minuend may require regrouping from a place farther to the left.

1. In the ones column, regroup to get \(12 - \square = 3\), so the missing ones digit in the subtrahend is \(9\). The minuend’s tens digit is therefore \(1\) before regrouping. 2. Regroup through the \(0\) hundreds digit. The tens column gives \(10 - 1 = 9\), and the hundreds column gives \(9 - 8 = 1\), so the missing result digit is \(1\). 3. In the thousands column, regroup to get \(14 - \square = 7\), so the missing thousands digit in the subtrahend is \(7\). The ten-thousands column then gives \(3 - 0 = 3\). The completed equation is \(45{,}012 - 7819 = 37{,}193\).

\(45{,}012 - 7819 = 37{,}193\)

- Note every place where regrouping is needed. - Check the completed subtraction by adding the difference and the completed subtrahend to recover the minuend.

1. In the ones column, regroup to get \(\square + 10 - 7 = 5\), so the missing ones digit in the minuend is \(2\). 2. After the regrouping, the tens digit is \(3\). Since \(3 - \square = 2\), the missing tens digit in the subtrahend is \(1\). 3. In the hundreds column, regroup from the thousands place. Then \(\square + 10 - 5 = 8\), so the missing hundreds digit in the minuend is \(3\). 4. The thousands digit has become \(7\). Since \(7 - \square = 4\), the missing thousands digit in the subtrahend is \(3\). The completed equation is \(8342 - 3517 = 4825\).

\(8342 - 3517 = 4825\)

- When subtracting from a number with several zeros, regroup from the first nonzero place. - After regrouping, most intermediate zero digits become \(9\).

1. Regroup from the ten-thousands place through all four zeros. The ones column becomes \(10 - \square = 7\), so the missing ones digit is \(3\). 2. In the tens column, \(9 - 4 = 5\). In the hundreds column, \(9 - \square = 1\), so the missing hundreds digit is \(8\). 3. In the thousands column, \(9 - 2 = 7\). In the ten-thousands column, \(4 - \square = 2\), so the missing ten-thousands digit is \(2\). The completed equation is \(50{,}000 - 22{,}843 = 27{,}157\). 4. Regrouping record: ten-thousands to thousands; thousands to hundreds; hundreds to tens; tens to ones

\(50{,}000 - 22{,}843 = 27{,}157\) Regrouping record: ten-thousands to thousands; thousands to hundreds; hundreds to tens; tens to ones

- Move from right to left one column at a time. - Decide in each column whether regrouping from the next place is required.

1. In the ones column, regroup to get \(12 - \square = 7\), so the missing ones digit in the subtrahend is \(5\). 2. The minuend’s tens digit is now \(0\). Regroup from the hundreds place, so \(10 - 5 = 5\); the missing tens digit in the result is \(5\). 3. The minuend’s hundreds digit is now \(3\). Regroup from the thousands place, so \(13 - \square = 6\); the missing hundreds digit in the subtrahend is \(7\). 4. The minuend’s thousands digit is now \(6\). Since \(6 - 3 = 3\), the missing thousands digit in the result is \(3\). The completed equation is \(7412 - 3755 = 3657\).

\(7412 - 3755 = 3657\)

- When an unknown is in the result, subtract normally after regrouping. - When an unknown is in an operand, use the inverse relationship between addition and subtraction.

1. In the ones column, regroup to get \(\square + 10 - 9 = 5\), so the missing ones digit in the minuend is \(4\). 2. The minuend’s tens digit is now \(6\). Since \(6 - 5 = 1\), the missing tens digit in the result is \(1\). 3. In the hundreds column, regroup from the thousands place. Then \(10 - \square = 2\), so the missing hundreds digit in the subtrahend is \(8\). 4. The minuend’s thousands digit is now \(0\), so regroup from the ten-thousands place. Then \(10 - 2 = 8\), and \(7 - 3 = 4\) in the ten-thousands column. The missing thousands digit in the minuend is \(1\). The completed equation is \(81{,}074 - 32{,}859 = 48{,}215\).

\(81{,}074 - 32{,}859 = 48{,}215\)

- Work from right to left and use each visible result digit. - Regroup before solving a missing digit when the top amount is too small. - Check each completed subtraction with the standard algorithm.

1. a) Ones: regroup to get \(15-\square=7\), so the ones digit is \(8\). Tens: after the regrouping, regroup again so \(10-\square=7\), giving \(3\). Hundreds: \(7-6=1\). Thus the subtrahend is \(638\). 2. b) Ones: \(9-\square=1\), so the ones digit is \(8\). Tens: regroup so \(16-\square=7\), giving \(9\). Hundreds: \(6-\square=5\), giving \(1\). Thus the subtrahend is \(198\). 3. c) Ones: regroup so \(12-3=9\). Tens: \(2-1=1\). Hundreds: \(\square-5=2\), so the missing digit is \(7\). The minuend is \(732\). 4. d) Ones: \(5-3=2\). Tens: \(4-\square=0\), so the subtrahend tens digit is \(4\). Hundreds: \(\square-2=5\), so the minuend hundreds digit is \(7\). The equation is \(745-243=502\). 5. Regrouping record: a) tens to ones; hundreds to tens | b) hundreds to tens | c) tens to ones | d) none

a) \(3\) and \(8\) b) \(1\), \(9\), and \(8\) c) \(7\) d) \(7\) and \(4\) Regrouping record: a) tens to ones; hundreds to tens | b) hundreds to tens | c) tens to ones | d) none

- Move from ones toward ten-thousands, one column at a time. - The amount left after lending affects the next subtraction. - Check each hidden digit against the known answer digit in its column.

1. Ones: \(3\) is too small to subtract \(6\), so trade a ten and subtract \(13-6=7\). 2. Tens: after that trade, the unknown top tens digit must be \(1\). It becomes \(0\), then borrows a hundred to become \(10\), giving \(10-4=6\). 3. Hundreds: \(2\) is now \(1\), so borrow a thousand to make \(11\) hundreds. Then \(11-3=8\). 4. Thousands: \(1\) becomes \(0\) after lending to hundreds. Borrow one ten-thousand to make \(10\) thousands; \(10-7=3\), so the missing bottom thousands digit is \(7\). 5. Ten-thousands: \(5\) is now \(4\), and \(4-2=2\). Thus \(51{,}213-27{,}346=23{,}867\).

\(51{,}213-27{,}346=23{,}867\).

- The zero in the tens place may need a trade. - Track the reduced hundreds digit before deciding the missing digit beneath it. - Account for the thousand traded into the hundreds column.

1. Ones: \(7-5=2\), so no trade is needed in that column. 2. Tens: \(0\) cannot subtract \(8\), so trade a hundred to get \(10\) tens. Then \(10-8=2\). 3. Hundreds: \(3\) has become \(2\). Trade a thousand to make \(12\) hundreds. Since \(12-5=7\), the missing bottom hundreds digit is \(5\). 4. Thousands: after the trade, the top thousands digit has been reduced by one. The difference of thousands must be \(2\) after subtracting \(5\), so the original top thousands digit was \(8\): \(8-1-5=2\). 5. Ten-thousands: \(4-2=2\). The completed subtraction is \(48{,}307-25{,}585=22{,}722\).

\(48{,}307-25{,}585=22{,}722\).

- The ones difference requires a trade from the tens. - Follow any chain of regrouping all the way to the left. - Use the shown result digits to work out hidden digits after each trade.

1. Ones: trade a ten to get \(15-6=9\), so the missing result ones digit is \(9\). 2. Tens: after that trade the top tens digit is one less. Trade a hundred to make \(13\) tens; \(13-6=7\), so the original top tens digit was \(4\). 3. Hundreds: the \(3\) hundreds are now \(2\). Trade a thousand to make \(12\) hundreds; \(12-8=4\), so the missing bottom hundreds digit is \(8\). 4. Thousands: after lending to hundreds, \(2\) becomes \(1\). Trade a ten-thousand to make \(11\) thousands; \(11-5=6\). 5. Ten-thousands: \(9\) becomes \(8\), and \(8-4=4\). Thus \(92{,}345-45{,}866=46{,}479\).

\(92{,}345-45{,}866=46{,}479\).

- Begin with the ones and consider every trade needed. - Update the tens before determining the missing bottom tens digit. - The hundreds trade also changes the thousands digit.

1. Ones: the top missing digit must be \(3\), because \(13-7=6\) after borrowing one ten. 2. Tens: \(2\) tens become \(1\) after that trade. Trade a hundred to make \(11\) tens; since \(11-6=5\), the missing bottom tens digit is \(6\). 3. Hundreds: the top hundreds have lent one hundred, leaving \(0\). To leave \(4\) hundreds after subtracting \(6\), trade a thousand to make \(10\) hundreds. Thus the missing original hundreds digit was \(1\). 4. Thousands: after the trade, \(5\) thousands become \(4\), and \(4-2=2\), so the missing bottom thousands digit is \(2\). 5. The completed subtraction is \(5123-2667=2456\).

\(5123-2667=2456\).

- Locate the nearest nonzero place that can supply units to the ones. - Move the trade through hundreds and tens in order. - Finish each column after updating the amount it has lent.

1. Ones: \(2\) cannot subtract \(7\), so trade through the two zero columns. One thousand becomes \(10\) hundreds; one hundred becomes \(10\) tens; one ten becomes \(10\) ones. Then \(12-7=5\). 2. Tens: after lending to ones, \(9-4=5\). Therefore the missing bottom tens digit is \(4\). 3. Hundreds: after lending to tens, \(9-4=5\). The missing original top hundreds digit was \(0\). 4. Thousands: the top thousands are now \(8\), so \(8-3=5\). The missing bottom thousands digit is \(3\). 5. The completed equation is \(9002-3447=5555\).

\(9002-3447=5555\).

- The ones digit can be found without a trade. - To get the shown tens result, think about borrowing from hundreds. - Follow that borrowing into the thousands column if necessary.

1. Ones: \(5-3=2\), so the missing top ones digit is \(5\). 2. Tens: \(1\) ten cannot subtract the missing bottom tens digit to leave \(9\) tens. Trade a hundred to get \(11\) tens; \(11-2=9\), so that digit is \(2\). 3. Hundreds: \(1\) hundred has been traded, leaving none. Trade one thousand into \(10\) hundreds. Since \(10-2=8\), the missing top hundreds digit was \(1\). 4. Thousands: \(6\) becomes \(5\) after the trade. Then \(5-2=3\), so the missing bottom thousands digit is \(2\). 5. The completed equation is \(6115-2223=3892\).

\(6115-2223=3892\).

- Work from the ones place and keep track of each trade. - The zero hundreds digit makes the tens trade take two steps. - Check whether the thousands column needs to borrow from ten-thousands afterward.

1. Ones: \(5-4=1\), so the missing answer ones digit is \(1\). 2. Tens: \(1\) ten cannot subtract \(7\) tens. The hundreds digit is \(0\), so trade from the thousands to make \(10\) hundreds, then one hundred to make \(11\) tens. Thus \(11-7=4\), and the missing top tens digit was \(1\). 3. Hundreds: after that trade, \(9-3=6\). 4. Thousands: after lending to hundreds, the top thousands digit is \(1\). Trade one ten-thousand to make \(11\) thousands, and \(11-8=3\), so the missing bottom thousands digit is \(8\). 5. Ten-thousands: \(4\) becomes \(3\), and \(3-1=2\). The completed subtraction is \(42{,}015-18{,}374=23{,}641\).

\(42{,}015-18{,}374=23{,}641\).

- Begin with the ones digit and decide what must be borrowed. - Each new trade reduces the top amount in the next place. - Use the shown answer digit to find each hidden digit after regrouping.

1. Ones: trade one ten to make \(15\) ones, so \(15-7=8\); the missing result ones digit is \(8\). 2. Tens: after that trade there is \(1\) ten left. Trade one hundred to get \(11\) tens. Since \(11-7=4\), the missing bottom tens digit is \(7\). 3. Hundreds: there is now \(1\) hundred. Trade a thousand to make \(11\) hundreds. Since \(11-8=3\), the missing original top hundreds digit is \(2\). 4. Thousands: \(7\) becomes \(6\), and \(6-3=3\). Therefore \(7225-3877=3348\).

Missing top hundreds digit: \(2\); bottom tens digit: \(7\); result ones digit: \(8\). \(7225-3877=3348\).

- Start with the ones and watch the change to the tens. - A zero in the hundreds place may need a trade from thousands. - Use the visible answer digits to check the hidden digits.

1. Ones: borrow a ten to make \(11\) ones, then \(11-8=3\). 2. Tens: after lending one ten, the missing top tens digit must be \(4\), leaving \(3\) tens. Then \(3-3=0\). 3. Hundreds: the top has \(0\) hundreds, so trade one thousand into \(10\) hundreds. Since \(10-2=8\), the missing bottom hundreds digit is \(2\). 4. Thousands: \(5\) becomes \(4\), and \(4-2=2\), so the missing result thousands digit is \(2\). 5. The completed equation is \(5041-2238=2803\).

\(5041-2238=2803\).

- When the minuend has several zeros, regroup across all of them from the first nonzero digit to the left. - A zero that passes a regrouped unit to the next place becomes \(9\).

1. Ones place: Regroup across the zeros to get \(10 - 4 = 6\). 2. Tens place: After regrouping, \(9 - 3 = 6\), so the missing result digit is \(6\). 3. Hundreds place: The missing digit in the minuend is \(0\). After regrouping across that place, \(9 - 2 = 7\). 4. Thousands place: After regrouping, \(8 - 1 = 7\). The completed subtraction is \(9000 - 1234 = 7766\).

The completed subtraction is \(9000 - 1234 = 7766\).

- Work left from the ones only after finishing each trade. - Follow the chain of lending through hundreds and thousands. - The correct hidden digits must match the visible result in every column.

1. Ones: trade one ten to get \(11\) ones, so \(11-6=5\). 2. Tens: the missing top tens digit must originally have been \(2\). After lending to ones it becomes \(1\); after borrowing from hundreds it becomes \(11\), and \(11-8=3\). 3. Hundreds: after lending a hundred to tens, there is \(1\) hundred. Borrow a thousand to get \(11\) hundreds; \(11-7=4\), so the missing result hundreds digit is \(4\). 4. Thousands: after lending to hundreds, \(4\) thousands become \(3\). Borrow one ten-thousand to make \(13\) thousands. Since \(13-7=6\), the missing bottom thousands digit is \(7\). 5. Ten-thousands: \(5\) becomes \(4\). The completed subtraction is \(54{,}221-7786=46{,}435\).

\(54{,}221-7786=46{,}435\).

- Start from the right and find a higher place with a nonzero value to trade. - Track each place the trade passes through. - Use the result digits to find the remaining hidden digits after the trades.

1. The top number ends in four zeros. To subtract at the ones place, trade one ten-thousand into thousands, then trade successively through hundreds and tens until the ones column has \(10\) ones. 2. The top now has \(7\) ten-thousands, \(9\) thousands, \(9\) hundreds, \(9\) tens, and \(10\) ones. 3. Ones: \(10-5=5\); tens: \(9-3=6\), so the missing bottom tens digit is \(3\). Hundreds: \(9-3=6\), so the missing result hundreds digit is \(6\). 4. Thousands: \(9-7=2\), so the missing bottom thousands digit is \(7\). Ten-thousands: \(7-1=6\), so the missing original top digit is \(8\). 5. The completed subtraction is \(80{,}000-17{,}335=62{,}665\).

\(80{,}000-17{,}335=62{,}665\).

- Work from the ones place toward the left. - After a trade, remember that the digit to the left has decreased by one. - Check the completed subtraction by addition.

1. Trade one ten for ones: \(15-6=9\). 2. The tens digit has lost one ten. Trade one hundred for ten tens. The tens calculation must give \(9\), so the original tens digit is \(4\): \(13-4=9\). 3. The hundreds digit is now \(2\). Trade one thousand for ten hundreds: \(12-4=8\). 4. The thousands digit is now \(1\). Trade one ten-thousand for ten thousands: \(11-2=9\). 5. The ten-thousands digit is now \(0\). The completed subtraction is \(12{,}345-2446=9899\).

\(12{,}345-2446=9899\)

- Start at the ones place and note each trade before moving left. - A missing digit can be found from the result in that column after regrouping. - Add the difference to the bottom number to check the top number.

1. Trade one ten for ones: \(11-9=2\). The missing ones digit in the answer is \(2\). 2. The tens digit is now \(1\). Trade one hundred for ten tens. To leave \(3\) tens, subtract \(8\) from \(11\); the missing tens digit in the bottom number is \(8\). 3. The hundreds digit is now \(1\). Trade one thousand for ten hundreds: \(11-8=3\). The missing hundreds digit in the answer is \(3\). 4. After that trade, the thousands digit must be \(6\). Trade one ten-thousand for ten thousands to get \(16-8=8\). The missing thousands digit in the top number is \(7\). 5. The ten-thousands digit is now \(6\): \(6-3=3\). The completed subtraction is \(77{,}221-38{,}889=38{,}332\).

\(77{,}221-38{,}889=38{,}332\)

- The standard subtraction process is unchanged for longer numbers. - Verify each column before moving to the next one.

1. In the ones column, regroup to get \(11 - 8 = 3\), so the missing result digit is \(3\). The missing tens digit in the minuend is \(1\) before regrouping. 2. In the tens column, the minuend digit is now \(0\). Regroup from the hundreds place to get \(10 - 5 = 5\). 3. In the hundreds column, the minuend digit is now \(2\). Regroup from the thousands place so \(12 - \square = 6\); the missing hundreds digit in the subtrahend is \(6\), and the missing thousands digit in the minuend is \(3\) before regrouping. 4. In the thousands column, the minuend digit is now \(2\). Regroup from the ten-thousands place to get \(12 - 5 = 7\). 5. In the ten-thousands column, the minuend digit is now \(3\). Since \(3 - \square = 1\), the missing ten-thousands digit in the subtrahend is \(2\). Finally, \(7 - 2 = 5\) in the hundred-thousands column. The completed equation is \(743{,}311 - 225{,}658 = 517{,}653\).

\(743{,}311 - 225{,}658 = 517{,}653\)