Free Math Worksheets

- 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, \(842 - 327 = 515\).

\(515\)

- Begin in the ones column. - Subtract one place-value column at a time. - Regrouping is not needed in these problems.

1. For part a, subtract by place value: \(9 - 6 = 3\), \(8 - 5 = 3\), and \(7 - 2 = 5\). The difference is \(533\). 2. For part b, subtract by place value: \(3 - 1 = 2\), \(4 - 2 = 2\), and \(5 - 1 = 4\). The difference is \(422\). 3. For part c, subtract by place value: \(7 - 5 = 2\), \(6 - 3 = 3\), and \(9 - 4 = 5\). The difference is \(532\).

a) \(533\) b) \(422\) c) \(532\)

- 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, \(852 - 529 = 323\).

a) \(425\) b) \(323\)

- 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 \(1964\).

They were married in \(1964\).

- Subtract the empty mass from the loaded mass for each truck. - Align matching place values in the standard algorithm. - Check each difference by addition.

1. Gravel: Subtract \(12{,}320\,\text{kg}\) from \(21{,}450\,\text{kg}\) using the standard subtraction algorithm. The result is \(9130\,\text{kg}\). 2. Sand: \(35{,}800\,\text{kg} - 24{,}150\,\text{kg} = 11{,}650\,\text{kg}\). 3. Topsoil: Subtract \(10{,}480\,\text{kg}\) from \(18{,}920\,\text{kg}\) using the standard subtraction algorithm. The result is \(8440\,\text{kg}\).

Gravel: \(9130\,\text{kg}\) Sand: \(11{,}650\,\text{kg}\) Topsoil: \(8440\,\text{kg}\)

- The total price is made of two parts. - Which operation finds the value left after one part is removed? - Check your result by adding the cooling unit’s value back.

1. Subtract the cooling unit’s value from the total price: \(\$48{,}200 - \$9350 = \$38{,}850\).

Without the cooling unit, the van costs \(\$38{,}850\).

- 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.

a) \(535{,}418\) b) \(435{,}418\) c) \(335{,}418\) The difference decreases by \(100{,}000\) each time.

- 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, \(1744-1499=245\).

The map from \(1744\) was made later. The maps were made \(245\) years apart.

- 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.

\(632\) years

- 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, \(6305 - 2418 = 3887\).

a) \(3322\) b) \(3887\)

- 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, \(842 - 365 = 477\).

\(477\) books remain.

- 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, \(723 - 456 = 267\).

\(267\) flowerpots remain.

- Align ones, tens, and hundreds. - Begin in the ones column. - Check by adding the difference and the subtrahend.

1. For part a, subtract by place value: \(8 - 4 = 4\), \(9 - 7 = 2\), and \(5 - 2 = 3\). The difference is \(324\). 2. For part b, subtract by place value: \(7 - 1 = 6\), \(6 - 3 = 3\), and \(8 - 5 = 3\). The difference is \(336\). 3. For part c, subtract by place value: \(6 - 3 = 3\), \(5 - 2 = 3\), and \(4 - 1 = 3\). The difference is \(333\).

a) \(324\) b) \(336\) c) \(333\)

- Identify the earlier year and the later year. - To find the number of years between them, subtract the earlier year from the later year. - Line up the place values before subtracting.

1. Subtract the earlier year from the later year using the standard algorithm: \(1886 - 1817 = 69\). 2. Therefore, \(69\) years passed between the inventions.

\(69\) years passed between the two inventions.

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

\(769 - 548 = 221\)

- Check the ones, tens, and hundreds columns one at a time. - Determine which digit must be subtracted to produce the result digit. - Account for regrouping. - You can use the inverse equation: difference plus subtrahend equals minuend.

1. In part a, subtract the difference from the minuend to find the subtrahend: \(745 - 512 = 233\). The missing digit is \(3\). 2. In part b, \(682 - 333 = 349\). The missing digit is \(9\). 3. In part c, \(901 - 445 = 456\). The missing digit is \(4\). 4. In part d, \(527 - 338 = 189\). The missing digit is \(8\).

a) \(3\) b) \(9\) c) \(4\) d) \(8\)

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

1. In the ones column, \(8 - * = 1\), so the missing ones digit is \(7\). 2. In the tens column, \(* - 4 = 2\), so the missing tens digit is \(6\). 3. In the hundreds column, \(9 - * = 5\), 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. - Check with \(377+123\).

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.

No. The correct difference is \(377\).

- Which number in the list is greatest, and which is least? - Line up the place values before subtracting.

1. Identify the heaviest bag: Sami's bag weighs \(4210\,\text{g}\). 2. Identify the lightest bag: Paul's bag weighs \(2890\,\text{g}\). 3. Use the standard subtraction algorithm to find the difference: \(4210\,\text{g} - 2890\,\text{g} = 1320\,\text{g}\).

The difference is \(1320\,\text{g}\).

- 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, \(613 - 245 = 368\).

\(368\)

- 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 \(6\). \(544 - 266 = 278\)

- Which operation finds how many seats remain? - Write the larger number above the smaller number and align place values. - State the answer in the context of the problem.

1. Subtract the number of tickets sold from the total number of seats: \(985 - 652\). 2. Subtract by place value: \(5 - 2 = 3\), \(8 - 5 = 3\), and \(9 - 6 = 3\). 3. Therefore, \(985 - 652 = 333\).

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, \(514 - 168 = 346\).

\(346\) students do not participate in the club.

- 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 \(736 - 251 = 485\).

The missing digit is \(7\). The completed equation is \(736 - 251 = 485\).

- 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 \(632 - 347 = 285\).

The missing digits are \(7\) in the ones place and \(3\) in the hundreds place. \(632 - 347 = 285\)

- 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.

\(417\) sheets of paper remain.

- 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 \(338\).

a) \(357\) b) \(338\) c) \(362\) The difference in part b, \(338\), is the least.

- 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 \(904 - 328 = 576\).

\(\begin{array}{r}904\\-\;328\\\hline576\end{array}\)

- 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.

Difference: \(357\) Check: \(357 + 245 = 602\)

- 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, \(801 - 367 = 434\).

a) \(266\) b) \(657\) c) \(434\)

- Begin in the ones column. - When regrouping across a zero, first regroup from the hundreds place. - Check the completed subtraction from right to left.

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 \(9 - * = 4\), 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.

Problem A has the greater difference. A) \(277\) B) \(223\)

- 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 \(467\).

a) \(662\) b) \(544\) c) \(283\) d) \(467\)

- 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, \(546 + 354 = 900\).

There are \(546\,\text{lb}\) of sand left on the truck.

- 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, \(902 - 537 = 365\).

a) \(368\) b) \(325\) c) \(365\)

- 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, \(600 - 342 = 258\).

Elias still needs to read \(258\) pages.

- 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: \(418 + 285 = 703\).

a) \(456\); check: \(456 + 456 = 912\) b) \(418\); check: \(418 + 285 = 703\)

- Begin in the ones column and account for regrouping. - Each regrouping changes the next top digit. - You can also use the related addition equation to find or verify missing digits.

1. In the ones column, regroup so that \(15 - \square = 9\). The missing ones digit in the subtrahend is \(6\). 2. Let \(x\) be the missing tens digit in the minuend. After regrouping one ten to the ones column and one hundred to the tens column, \((x - 1) + 10 - 4 = 7\). Therefore, \(x = 2\). 3. In the hundreds column, \(8 - 1 - 3 = 4\), which matches the result. 4. The completed equation is \(825 - 346 = 479\). Check: \(479 + 346 = 825\).

The missing tens digit in the minuend is \(2\), and the missing ones digit in the subtrahend is \(6\). \(825 - 346 = 479\) 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.

Problem A has a difference of \(274\). Problem B has a difference of \(275\). The difference in Problem A is less.

- 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, \(5 - \text{●} = 2\), 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 \(8 - \text{●} = 3\), so the missing digit is \(5\). The completed equation is \(927 - 558 = 369\).

a) The missing digit is \(3\): \(658 - 334 = 324\). b) The missing digit is \(5\): \(927 - 558 = 369\).

- Regroup when a top digit is less than the digit below it. - Remember that a column has one fewer unit after regrouping from it. - Check each completed subtraction from right to left.

1. In part a, regroup in the ones column so that \(12 - \text{●} = 7\). The missing ones digit is \(5\). There are now \(3\) tens, so regroup one hundred and calculate \(13 - 7 = 6\). In the hundreds column, \(\text{●} - 1 - 1 = 2\), so the missing hundreds digit is \(4\). The completed equation is \(442 - 175 = 267\). 2. In part b, regroup across the zero so that \(10 - 6 = 4\). Let \(x\) be the missing tens digit in the minuend. After regrouping to the ones and then from hundreds to tens, \((x - 1) + 10 - 4 = 8\), so \(x = 3\). In the hundreds column, \(8 - 1 - \text{●} = 5\), so the missing hundreds digit in the subtrahend is \(2\). The completed equation is \(830 - 246 = 584\).

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

- Work from right to left. - Record each regrouping so it is included in the next column. - Check by adding the difference and the subtrahend.

1. In the ones column, regroup so that \(14 - \text{●} = 7\). 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.

Trip A: \(517\) miles; estimate \(500\) miles Trip B: \(693\) miles; estimate \(700\) miles Trip C: \(929\) miles; estimate \(900\) miles

- 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.

a) \(3647\,\text{kWh}\) b) Approximately \(3000\,\text{kWh}\)

- 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 \(1{,}000{,}000-123{,}000=877{,}000\).

a) \(50{,}000\) b) \(275{,}000\) c) \(500{,}010\) d) \(877{,}000\)

- 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, \(413 - 265 = 148\).

a) \(467\) b) \(148\)

- Rewrite each subtraction check as an addition equation. - Add the difference and the subtrahend. - The sum should equal the minuend.

1. Check \(476 + 258 = 734\). Equation 1 is correct. 2. Check \(275 + 347 = 622\), not \(612\). Equation 2 is incorrect. The correct difference is \(612 - 347 = 265\), because \(265 + 347 = 612\). 3. Check \(386 + 429 = 815\), not \(805\). Equation 3 is incorrect. The correct difference is \(805 - 429 = 376\), because \(376 + 429 = 805\). 4. Check \(355 + 186 = 541\). Equation 4 is correct.

1. Correct. 2. Incorrect. \(612 - 347 = 265\) 3. Incorrect. \(805 - 429 = 376\) 4. Correct.

- Use the inverse operation for each equation. - Check subtraction with addition and addition with subtraction. - Regroup carefully when calculating each check.

1. For part a, check \(388 + 543 = 931\), not \(921\). The equation is false. The correct equation is \(921 - 543 = 378\). 2. For part b, check \(761 - 485 = 276\). The equation is correct. 3. For part c, check \(386 + 217 = 603\). The equation is correct.

a) Incorrect. \(921 - 543 = 378\) b) Correct. c) Correct.

- 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, \(732 - 458 = 274\).

\(274\)

- Begin in the ones column. - Regroup when the top amount in a column is too small. - Check the completed subtraction from right to left.

1. In the ones column, regroup so that \(14 - \square = 9\). The missing ones digit in the subtrahend is \(5\). 2. Let \(x\) be the missing tens digit in the minuend. After regrouping one ten to the ones column and one hundred to the tens column, \((x - 1) + 10 - 7 = 5\). Therefore, \(x = 3\). 3. In the hundreds column, \(6 - 1 - 2 = 3\), which matches the result. 4. The completed equation is \(634 - 275 = 359\).

The missing top digit is \(3\), and the missing bottom digit is \(5\). \(634 - 275 = 359\)

- 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 \(=\).

\(=\)

- 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, \(105{,}000 - 12{,}345 = 92{,}655\).

a) \(58{,}219\) b) \(92{,}655\)

- 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 \(48{,}353 - 48{,}350 = 3\).

Expression 1 has the greater result, \(48{,}353\). The difference between the results is \(3\).

- 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 \(66{,}981\).

The final result is \(66{,}981\).

- 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 \(99{,}099\).

1. \(543{,}543\) 2. \(432{,}432\) 3. \(321{,}321\) 4. \(654{,}654 - 444{,}444 = 210{,}210\) 5. \(543{,}543 - 444{,}444 = 99{,}099\)

- 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 \(8334 - 3653 = 4681\).

- 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.

Expression B has the greater result: \(4544 > 4433\).

- Align the numbers by place value. - Regroup whenever the upper digit is smaller than the lower digit. - Work from the ones column toward the left.

1. Regroup through the \(0\) in the tens place. In the ones column, \(13 - 5 = 8\), and in the tens column, \(9 - 1 = 8\). 2. Regroup in the hundreds column to get \(14 - 7 = 7\). Regroup again in the thousands column to get \(11 - 6 = 5\), followed by \(7 - 4 = 3\) in the ten-thousands column. Therefore, \(82{,}503 - 46{,}715 = 35{,}788\).

\(35{,}788\)

- Work from right to left. - Track how regrouping changes the next digit. - Use the result digit to determine each blank. - Check with addition.

1. In the ones column, regroup to get \(15-8=7\). 2. Let the missing tens digit in the minuend be \(z\). After regrouping, \((z-1)-4=2\), so \(z=7\). 3. In the hundreds column, regroup from the thousands place. Then \(12-\square=4\), so the missing subtrahend digit is \(8\). 4. The completed equation is \(6275-2848=3427\).

The missing digits are \(7\) in the minuend and \(8\) in the subtrahend. \(6275-2848=3427\)

- 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 \(466\).

a) \(377\) b) \(357\) c) \(466\)

- 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 \(338\).

a) \(245\) b) \(563\) c) \(338\)

- 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 \(343\).

a) \(266\) b) \(217\) c) \(343\)

- 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 \(771\).

a) \(434\) b) \(457\) c) \(333\) d) \(771\)

- 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 \(456 - 178 = 278\).

a) \(2\) b) \(6\) c) \(8\)

- 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: \(268{,}518+237{,}482=506{,}000\).

a) \(47{,}288\); check: \(47{,}288+45{,}163=92{,}451\) b) \(268{,}518\); check: \(268{,}518+237{,}482=506{,}000\)

- 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.

The difference is \(484{,}685\). First check: \(484{,}685+258{,}435=743{,}120\) Second check: \(743{,}120-484{,}685=258{,}435\)

- 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, \(70{,}000-24{,}315=45{,}685\).

c) is incorrect. The correct equation is \(70{,}000-24{,}315=45{,}685\).

- 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.

All three results are equal to \(277{,}836\).

- 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 \(838\).

a) \(368\) b) \(356\) c) \(376\) d) \(838\)

- 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 \(209\).

a) \(548\) b) \(372\) c) \(933\) d) \(209\)

- 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: \(185{,}887+124{,}583=310{,}470\).

a) \(4174\); check: \(4174+2178=6352\) b) \(31{,}877\); check: \(31{,}877+13{,}925=45{,}802\) c) \(185{,}887\); check: \(185{,}887+124{,}583=310{,}470\)

- 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, \(43{,}075 - 28{,}519 = 14{,}556\).

a) The numbers are \(43{,}075\) and \(28{,}519\). b) The difference is \(14{,}556\).

- Begin in the ones column. - Track how regrouping changes the next top digit. - If the top digit is too small, regroup from the next place value. - Substitute the missing digit and check the complete subtraction.

1. In part a, regroup in the ones column: \(12 - 5 = 7\). This leaves \(3\) tens, so regroup one hundred and solve \(13 - * = 5\). Thus, \(* = 8\). 2. In part b, regroup in the ones column: \(13 - 8 = 5\). Let the missing tens digit be \(x\). After regrouping one ten and one hundred, \(10 + x - 1 - 4 = 9\), so \(x = 4\). 3. In part c, regroup in the ones column: \(15 - 8 = 7\). Then regroup in the tens column: \(11 - 7 = 4\). In the hundreds column, \((* - 1) - 4 = 3\), so \(* = 8\). 4. In part d, regroup across the zero so that \(14 - * = 8\). Thus, \(* = 6\). The remaining columns verify the result: \(9 - 2 = 7\) and \(8 - 5 = 3\).

a) \(* = 8\) b) \(* = 4\) c) \(* = 8\) d) \(* = 6\)

- Regroup when the top digit is less than the digit below it. - Record each regrouped unit so that the next column is adjusted correctly.

1. In the ones column, \(2 - *\) cannot equal \(4\) without regrouping. Regroup one ten and solve \(12 - * = 4\). The missing ones digit is \(8\). 2. Let the missing tens digit be \(x\). After regrouping one ten to the ones column, and then regrouping one hundred to the tens column, \(10 + x - 1 - 7 = 5\). Thus, \(x = 3\). 3. In the hundreds column, \(5 - 1 - 2 = 2\), so the missing result digit is \(2\). 4. The completed equation is \(532 - 278 = 254\).

\(532 - 278 = 254\)

- Subtracting from a number with several zeros requires regrouping across place values. - Check the result by adding \(657\) to the completed subtrahend.

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 \(1000 - 343 = 657\).

\(1000 - 343 = 657\)

- Work one column at a time. - Because the same digit appears in both subtrahends, represent it with one variable and use the visible result digit to form an equation.

1. In the ones column, \(9 - 3 - 5 = 1\), so no regrouping is needed. 2. Let the repeated missing digit be \(x\). In the tens column, \(9 - x - x = 3\), so \(2x = 6\) and \(x = 3\). 3. In the hundreds column, \(9 - 1 - 2 = 6\), so the missing result digit is \(6\). The completed equation is \(999 - 133 - 235 = 631\).

\(999 - 133 - 235 = 631\)

- Work systematically from the ones place to the left. - The inverse operation, addition, can help recover an unknown subtrahend. - Check whether regrouping changes the next column.

1. For a), the missing subtrahend is \(4582 - 3348 = 1234\), so the missing digit is \(2\). 2. For b), the missing subtrahend is \(12{,}043 - 9476 = 2567\), so the missing digit is \(2\). 3. For c), \(87{,}655 - 23{,}410 = 64{,}245\), so the missing digit is \(5\).

a) \(2\) b) \(2\) c) \(5\)

- Regroup from the next greater place value when needed. - You can also use addition as the inverse operation.

1. In the ones column, regroup so that \(12 - * = 7\). The missing ones digit in the subtrahend is \(5\). 2. Let the missing tens digit in the minuend be \(x\). After regrouping one ten to the ones column and one hundred to the tens column, \((x - 1) + 10 - 4 = 8\). Thus, \(x = 3\). 3. In the hundreds column, \(8 - 1 - 4 = 3\), which matches the result. 4. The completed equation is \(832 - 445 = 387\).

\(832 - 445 = 387\)

- When the top digit is less than the bottom digit, regroup from the next place value.

1. In the ones column, \(* - 8 = 5\) requires regrouping. Therefore, \(13 - 8 = 5\), so the missing ones digit in the minuend is \(3\). 2. After regrouping, the tens column is \(10 - * = 6\), so the missing tens digit in the subtrahend is \(4\). One hundred is regrouped. 3. In the hundreds column, \(* - 1 - 2 = 1\), 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, \(* - 7 = 4\) 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 \(5 - * = 1\), 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 \(9 - * = 6\), 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 \(603 - 238 = 365\).

- Work backward from the result. - What tens digit leaves \(6\) tens after subtracting \(6\) and accounting for regrouping?

1. In the ones column, regroup one ten so that \(10 - 5 = 5\). 2. Let the missing tens digit be \(x\). After regrouping one ten to the ones column, the tens column still requires regrouping from the hundreds place. Thus, \((x - 1) + 10 - 6 = 6\), so \(x = 3\). 3. After regrouping one hundred, no hundreds remain. The completed equation is \(130 - 65 = 65\).

The missing digit is \(3\). The completed equation is \(130 - 65 = 65\).

- Pay close attention to regrouping. - When the top digit is less than the digit below it, regroup from the next place value.

1. In the ones column, \(4 - * = 6\) requires regrouping. Therefore, \(14 - 8 = 6\), so the missing ones digit in the subtrahend is \(8\). 2. Let the missing tens digit in the minuend be \(x\). After regrouping one ten to the ones column and one hundred to the tens column, \((x - 1) + 10 - 7 = 4\). Thus, \(x = 2\). 3. In the hundreds column, \(5\) becomes \(4\) after regrouping, and \(4 - 2 = 2\), which matches the result. 4. The completed equation is \(524 - 278 = 246\).

The missing digits are \(2\) in the minuend and \(8\) in the subtrahend. The completed equation is \(524 - 278 = 246\).

- You can use the related addition equation \(334 + 4\square8 = \square12\). - Account for each regrouping.

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 \(10 - * = 3\), so the missing tens digit in the subtrahend is \(7\). 3. In the hundreds column, \(* - 1 - 4 = 3\), 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 \(3348\).

- Regroup when the top digit is less than the digit below it. - You can also work upward with addition: \(4 + 8 = 12\).

1. In the ones column, \(* - 8 = 4\) requires regrouping. Therefore, \(12 - 8 = 4\), so the missing ones digit in the minuend is \(2\). 2. After regrouping, \(4\) tens remain. Since \(4 - * = 6\) is not possible, regroup one hundred and solve \(14 - * = 6\). 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 \(752 - 388 = 364\).

\(752 - 388 = 364\)

- When a place-value digit is zero, regroup from the next nonzero place to the left. - Track each regrouped unit carefully.

1. In the ones column, regroup across the zeros so that \(10 - * = 6\). The missing ones digit in the subtrahend is \(4\). 2. Regrouping from the hundreds place creates \(10\) tens, and regrouping one ten to the ones column leaves \(9\) tens. Then \(9 - 4 = 5\). 3. There are one fewer hundred after regrouping. The hundreds column must satisfy \(* - 1 - 1 = 2\), so the missing hundreds digit in the minuend is \(4\). 4. The completed equation is \(400 - 144 = 256\).

\(400 - 144 = 256\)

- 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 \(10 - * = 7\). 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 \(4 - * = 1\), so the missing hundreds digit in the subtrahend is \(3\). 4. The completed equation is \(500 - 373 = 127\).

\(500 - 373 = 127\)

- 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 \(457\).

- Choose digits according to whether each place-value column should require regrouping. - To force regrouping from tens to ones, choose a bottom ones digit greater than \(4\). - Regrouping in the ones column reduces the top tens digit from \(7\) to \(6\). - 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. After regrouping, \(6\) tens remain, so choose a tens digit no greater than \(6\). One example is \(674 - 125 = 549\). 3. For part c, choose a ones digit greater than \(4\), and then choose a tens digit greater than the \(6\) tens that remain. One example is \(674 - 285\). Regrouping gives \(14 - 5 = 9\), \(16 - 8 = 8\), and \(5 - 2 = 3\), so the difference is \(389\).

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 \(1111\).

a) \(432{,}099\) b) \(554{,}444 - 124{,}567\) c) The new difference is \(429{,}877\), which is \(2222\) less than the original difference.

- 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.

A) \(262{,}369\) B) \(262{,}369\) C) \(262{,}367\) C is closest to \(260{,}000\).

- 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 \(50{,}000 - 22{,}843 = 27{,}157\).

\(50{,}000 - 22{,}843 = 27{,}157\)

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

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

- When several digits are missing, use the related addition equation. - Work from right to left and record regrouping. - Check each completed subtraction.

1. In part a, use the minuend and difference to find the subtrahend: \(815 - 177 = 638\). The missing digits are \(3\) and \(8\). 2. In part b, \(769 - 571 = 198\). The missing digits are \(1\), \(9\), and \(8\). 3. In part c, use addition to find the minuend: \(219 + 513 = 732\). The missing digit is \(7\). 4. In part d, \(502 + 243 = 745\). The missing digits are \(7\) in the minuend and \(4\) in the subtrahend.

a) \(3\) and \(8\) b) \(1\), \(9\), and \(8\) c) \(7\) d) \(7\) and \(4\)

- Subtract from right to left, and keep track of each regrouping.

1. Ones place: Regroup to get \(13 - 6 = 7\). 2. Tens place: Let the missing digit be \(x\). After the regrouping, \((x-1)+10-4=6\), so \(x=1\). Regroup again. 3. Hundreds place: \((2-1)+10-3=8\). Regroup again. 4. Thousands place: \((1-1)+10-y=3\), so the missing digit \(y=7\). Regroup again. 5. Ten-thousands place: \(5-1-2=2\). The completed subtraction is \(51{,}213 - 27{,}346 = 23{,}867\).

The completed subtraction is \(51{,}213 - 27{,}346 = 23{,}867\).

- Work one place at a time and record each regrouping.

1. Ones place: \(7 - 5 = 2\). 2. Tens place: Regroup to get \(10 - 8 = 2\). 3. Hundreds place: After regrouping, \(12 - x = 7\), so the missing digit in the subtrahend is \(x=5\). Regroup again. 4. Thousands place: \(y-1-5=2\), so the missing digit in the minuend is \(y=8\). 5. Ten-thousands place: \(4 - 2 = 2\). The completed subtraction is \(48{,}307 - 25{,}585 = 22{,}722\).

The completed subtraction is \(48{,}307 - 25{,}585 = 22{,}722\).

- Several places require regrouping, so keep careful track as you move from right to left.

1. Ones place: Regroup to get \(15 - 6 = 9\), so the missing result digit is \(9\). 2. Tens place: Let the missing minuend digit be \(x\). After regrouping, \((x-1)+10-6=7\), so \(x=4\). Regroup again. 3. Hundreds place: \((3-1)+10-y=4\), so the missing subtrahend digit is \(y=8\). Regroup again. 4. Thousands place: \((2-1)+10-5=6\). Regroup again. 5. Ten-thousands place: \(9-1-4=4\). The completed subtraction is \(92{,}345 - 45{,}866 = 46{,}479\).

The completed subtraction is \(92{,}345 - 45{,}866 = 46{,}479\).

- Work from right to left and regroup when the top digit is too small. - Check your answer with the inverse operation: add the difference and the subtrahend.

1. Ones place: The top digit must make \(13 - 7 = 6\), so the missing ones digit is \(3\). Regroup one ten. 2. Tens place: After regrouping, regroup again to get \(11 - x = 5\), so the missing subtrahend digit is \(x=6\). 3. Hundreds place: Let the missing minuend digit be \(y\). Then \((y-1)+10-6=4\), so \(y=1\). Regroup from the thousands place. 4. Thousands place: \(5-1-z=2\), so the missing subtrahend digit is \(z=2\). The completed subtraction is \(5123 - 2667 = 2456\).

The completed subtraction is \(5123 - 2667 = 2456\).

- When you need to regroup from a place containing zero, continue left until you reach a nonzero digit. - Check your result by adding the subtrahend and the difference.

1. Ones place: Regroup to get \(12 - 7 = 5\). 2. Tens place: Regroup across the zero. Then \(9 - x = 5\), so the missing subtrahend digit is \(x=4\). 3. Hundreds place: Let the missing minuend digit be \(y\). After regrouping, \((y-1)+10-4=5\), so \(y=0\). 4. Thousands place: \(9-1-z=5\), so the missing subtrahend digit is \(z=3\). The completed subtraction is \(9002 - 3447 = 5555\).

The completed subtraction is \(9002 - 3447 = 5555\).

- Start with the ones place. - Keep track of the regrouping in the tens and hundreds places.

1. Ones place: \(x - 3 = 2\), so the missing digit is \(x=5\). 2. Tens place: Regroup to get \(11-y=9\), so the missing subtrahend digit is \(y=2\). 3. Hundreds place: Let the missing minuend digit be \(z\). After regrouping, \((z-1)+10-2=8\), so \(z=1\). Regroup from the thousands place. 4. Thousands place: \(6-1-w=3\), so the missing subtrahend digit is \(w=2\). The completed subtraction is \(6115 - 2223 = 3892\).

The completed subtraction is \(6115 - 2223 = 3892\).

- You will need to regroup across more than one place. - Be especially careful when regrouping through a zero in the minuend.

1. Ones place: \(5 - 4 = 1\), so the missing result digit is \(1\). 2. Tens place: Regroup across the zero to get \(11 - 7 = 4\), so the missing tens digit in the minuend is \(1\). 3. Hundreds place: After the regrouping, \(9 - 3 = 6\). 4. Thousands place: After regrouping, \(11-x=3\), so the missing subtrahend digit is \(x=8\). Regroup again. 5. Ten-thousands place: \((4-1)-1=2\). The completed subtraction is \(42{,}015 - 18{,}374 = 23{,}641\).

The completed subtraction is \(42{,}015 - 18{,}374 = 23{,}641\).

- Begin at the ones place. - For a missing digit in the minuend or subtrahend, use the other two digits in that place to determine what fits.

1. Ones place: Regroup to get \(15 - 7 = 8\), so the missing result digit is \(8\). 2. Tens place: After the first regrouping, regroup again to get \(11-x=4\), so the missing subtrahend digit is \(x=7\). 3. Hundreds place: Let the missing minuend digit be \(y\). After regrouping, \((y-1)+10-8=3\), so \(y=2\). 4. Thousands place: \(7-1-3=3\). The completed subtraction is \(7225 - 3877 = 3348\).

The missing digits are \(2\), \(7\), and \(8\). The completed subtraction is \(7225 - 3877 = 3348\).

- A zero in the minuend may require regrouping from a place farther to the left. - Check your work by adding the subtrahend and the difference.

1. Ones place: Regroup to get \(11 - 8 = 3\). 2. Tens place: After regrouping, \((x-1)-3=0\), so the missing tens digit in the minuend is \(x=4\). 3. Hundreds place: Regroup to get \(10-y=8\), so the missing hundreds digit in the subtrahend is \(y=2\). 4. Thousands place: After regrouping, \(4-2=2\), so the missing result digit is \(2\). The completed subtraction is \(5041 - 2238 = 2803\).

The completed subtraction is \(5041 - 2238 = 2803\).

- 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 \(9000 - 1234 = 7766\).

- When the top digit is smaller than the bottom digit, regroup from the place to its left. - Check by adding the difference and the subtrahend.

1. Ones place: Regroup to get \(11 - 6 = 5\). 2. Tens place: Let the missing minuend digit be \(x\). After regrouping, \((x-1)+10-8=3\), so \(x=2\). Regroup again. 3. Hundreds place: \((2-1)+10-7=4\), so the missing result digit is \(4\). Regroup again. 4. Thousands place: \((4-1)+10-y=6\), so the missing subtrahend digit is \(y=7\). Regroup again. 5. Ten-thousands place: \(5-1=4\). The completed subtraction is \(54{,}221 - 7786 = 46{,}435\).

The completed subtraction is \(54{,}221 - 7786 = 46{,}435\).

- Subtracting from a number with several zeros requires regrouping across multiple places. - After regrouping across the zeros, the intermediate zeros become \(9\), while the ones place receives \(10\).

1. Ones place: Regroup across the zeros to get \(10 - 5 = 5\). 2. Tens place: After regrouping, \(9-x=6\), so the missing subtrahend digit is \(x=3\). 3. Hundreds place: \(9-3=6\), so the missing result digit is \(6\). 4. Thousands place: \(9-y=2\), so the missing subtrahend digit is \(y=7\). 5. Ten-thousands place: After regrouping, \(z-1-1=6\), so the missing minuend digit is \(z=8\). The completed subtraction is \(80{,}000 - 17{,}335 = 62{,}665\).

The completed subtraction is \(80{,}000 - 17{,}335 = 62{,}665\).

- Regrouping changes the digit in the next place to the left. - Check any uncertain digit by completing the subtraction and using addition to verify it.

1. Ones place: Regroup to get \(15 - 6 = 9\). 2. Tens place: Let the missing minuend digit be \(x\). Then \((x-1)+10-4=9\), so \(x=4\). Regroup again. 3. Hundreds place: After regrouping, \((3-1)+10-4=8\), so the missing result digit is \(8\). Regroup again. 4. Thousands place: \((2-1)+10-2=9\). Regroup again. 5. Ten-thousands place: \(1-1=0\). The completed subtraction is \(12{,}345 - 2446 = 9899\).

The completed subtraction is \(12{,}345 - 2446 = 9899\).

- Record each regrouping clearly. - Check each place before moving to the next one.

1. Ones place: Regroup to get \(11 - 9 = 2\), so the missing result digit is \(2\). 2. Tens place: After regrouping, regroup again to get \(11-x=3\), so the missing subtrahend digit is \(x=8\). 3. Hundreds place: After regrouping, \(11-8=3\), so the missing result digit is \(3\). Regroup again. 4. Thousands place: Let the missing minuend digit be \(y\). After regrouping, \((y-1)+10-8=8\), so \(y=7\). Regroup again. 5. Ten-thousands place: \(7-1-3=3\). The completed subtraction is \(77{,}221 - 38{,}889 = 38{,}332\).

The completed subtraction is \(77{,}221 - 38{,}889 = 38{,}332\).

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

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