- In written addition, ones must line up with ones. - Look directly above and below the tens position. - Add each place-value column to finish.

1. The tens digits are \(2\) from \(324\) and \(5\) from \(50\). 2. Add by place: \(4+0=4\) ones, \(2+5=7\) tens, and \(3\) hundreds. The sum is \(374\).

The tens digits are \(2\) and \(5\). The sum is \(374\).

- Where should the single digit \(3\) go when you write the numbers vertically? - Look at the ones total before deciding whether to carry. - Check whether adding a few ones changes any larger place.

1. Line up the \(3\) ones under the \(6\) ones. 2. Add the ones: \(6+3=9\). No new ten is made. 3. The tens and hundreds digits stay the same, so the sum is \(239\).

\(236+3=239\). Only the ones place changes.

- Start with the ones column, not the hundreds. - When the ones make more than a group of ten, what happens to that group? - Include the carried group in the next column.

1. Add the ones: \(8+5=13\). Write \(3\) ones and carry \(1\) ten. 2. Add the tens: \(7+1=8\) tens. 3. Keep the \(2\) hundreds. The sum is \(283\).

\(278+5=283\); carry \(1\) ten from the ones column.

- Line up digits that stand for the same place value. - Look for a column whose total cannot fit in one digit. - If a column makes an extra group of ten, where does that group go?

1. The ones column is \(0+0=0\). 2. The tens column is \(6+7=13\) tens. Write \(3\) tens and carry \(1\) hundred. 3. Add the hundreds: \(4+1=5\). The sum is \(530\).

\(460+70=530\). The tens column makes a new hundred.

- Begin in the ones column. - When a column total is \(10\) or more, regroup to the next place value. - Record each regrouped amount above the next column.

1. For \(348 + 475\), \(8 + 5 = 13\), so write \(3\) and regroup \(1\) ten. Then \(4 + 7 + 1 = 12\), so write \(2\) and regroup \(1\) hundred. Finally, \(3 + 4 + 1 = 8\). The sum is \(823\). 2. For \(509 + 291\), \(9 + 1 = 10\), so write \(0\) and regroup \(1\) ten. Then \(0 + 9 + 1 = 10\), so write \(0\) and regroup \(1\) hundred. Finally, \(5 + 2 + 1 = 8\). The sum is \(800\). 3. For \(167 + 633\), \(7 + 3 = 10\), so write \(0\) and regroup \(1\) ten. Then \(6 + 3 + 1 = 10\), so write \(0\) and regroup \(1\) hundred. Finally, \(1 + 6 + 1 = 8\). The sum is \(800\). 4. Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(823\) b) \(800\) c) \(800\) Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

- Align ones with ones, tens with tens, and hundreds with hundreds. - Adding zero does not change a value. - When a column total is \(10\) or more, regroup to the next place value.

1. For \(307 + 452\), add by place value: \(7 + 2 = 9\), \(0 + 5 = 5\), and \(3 + 4 = 7\). The sum is \(759\). 2. For \(560 + 209\), add by place value: \(0 + 9 = 9\), \(6 + 0 = 6\), and \(5 + 2 = 7\). The sum is \(769\). 3. For \(84 + 607\), align the addends by place value. In the ones column, \(4 + 7 = 11\), so write \(1\) and regroup \(1\) ten. In the tens column, \(8 + 0 + 1 = 9\). In the hundreds column, \(0 + 6 = 6\). The sum is \(691\). 4. For \(408 + 302\), \(8 + 2 = 10\), so write \(0\) and regroup \(1\) ten. Then \(0 + 0 + 1 = 1\), and \(4 + 3 = 7\). The sum is \(710\). 5. Carry record: a) none | b) none | c) ones to tens: \(1\) | d) ones to tens: \(1\)

a) \(759\) b) \(769\) c) \(691\) d) \(710\) Carry record: a) none | b) none | c) ones to tens: \(1\) | d) ones to tens: \(1\)

- Decide whether the item weights should be added or subtracted. - Line up the ones, tens, hundreds, and thousands places. - Check that every weight is given in the same unit.

1. Line up the four weights by place value. 2. Use the standard addition algorithm: \(950\,\text{g} + 420\,\text{g} + 380\,\text{g} + 210\,\text{g} = 1960\,\text{g}\). 3. Carry record: tens to hundreds: \(1\); hundreds to thousands: \(1\)

The packed school bag weighs \(1960\,\text{g}\). Carry record: tens to hundreds: \(1\); hundreds to thousands: \(1\)

- Align the ones, tens, and hundreds columns. - Regroup whenever a column sum is \(10\) or greater. - Compare the addends in the two expressions before comparing the sums.

1. For a), add the ones: \(8 + 6 + 5 = 19\); write \(9\) and regroup \(1\) ten. Add the tens: \(4 + 1 + 7 + 1 = 13\); write \(3\) and regroup \(1\) hundred. Add the hundreds: \(3 + 2 + 1 + 1 = 7\). The sum is \(739\). 2. For b), add the ones: \(5 + 8 + 6 = 19\); write \(9\) and regroup \(1\) ten. Add the tens: \(7 + 4 + 1 + 1 = 13\); write \(3\) and regroup \(1\) hundred. Add the hundreds: \(1 + 3 + 2 + 1 = 7\). The sum is \(739\). 3. The results are equal because changing the order of addends does not change their sum. 4. Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(739\) b) \(739\) The results are equal. Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\)

- Right-align numbers with different numbers of digits. - A missing hundreds digit represents \(0\) hundreds. - Include each regrouped value in the next column.

1. For a), the ones sum is \(4 + 9 + 7 = 20\); write \(0\) and regroup \(2\) tens. The tens sum is \(2 + 8 + 6 + 2 = 18\); write \(8\) and regroup \(1\) hundred. The hundreds sum is \(5 + 1 + 1 = 7\). The result is \(780\). 2. For b), the ones sum is \(6 + 4 + 8 = 18\); write \(8\) and regroup \(1\) ten. The tens sum is \(0 + 9 + 5 + 1 = 15\); write \(5\) and regroup \(1\) hundred. The hundreds sum is \(3 + 2 + 1 + 1 = 7\). The result is \(758\). 3. Carry record: a) ones to tens: \(2\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(780\) b) \(758\) Carry record: a) ones to tens: \(2\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\)

- Begin in the ones column. - When a column total is greater than \(9\), regroup to the next place value. - Include each regrouped amount when adding the next column.

1. For part a, \(8 + 4 = 12\), so write \(2\) and regroup \(1\) ten. Then \(5 + 6 + 1 = 12\), so write \(2\) and regroup \(1\) hundred. Finally, \(3 + 2 + 1 = 6\). The sum is \(622\). 2. For part b, \(6 + 5 = 11\), so write \(1\) and regroup \(1\) ten. Then \(7 + 8 + 1 = 16\), so write \(6\) and regroup \(1\) hundred. Finally, \(4 + 1 + 1 = 6\). The sum is \(661\). 3. For part c, \(9 + 3 = 12\), so write \(2\) and regroup \(1\) ten. Then \(2 + 7 + 1 = 10\), so write \(0\) and regroup \(1\) hundred. Finally, \(5 + 3 + 1 = 9\). The sum is \(902\). 4. Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(622\) b) \(661\) c) \(902\) Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

- Align ones with ones, tens with tens, and hundreds with hundreds. - Record a regrouped ten or hundred whenever a column sum is greater than \(9\).

1. For a), add the ones: \(8 + 6 = 14\); write \(4\) and regroup \(1\) ten. Add the tens: \(7 + 5 + 1 = 13\); write \(3\) and regroup \(1\) hundred. Add the hundreds: \(4 + 3 + 1 = 8\). Therefore, \(478 + 356 = 834\). 2. For b), add the ones: \(9 + 4 = 13\); write \(3\) and regroup \(1\) ten. Add the tens: \(8 + 3 + 1 = 12\); write \(2\) and regroup \(1\) hundred. Add the hundreds: \(1 + 6 + 1 = 8\). Therefore, \(189 + 634 = 823\). 3. For c), add the ones: \(7 + 8 = 15\); write \(5\) and regroup \(1\) ten. Add the tens: \(7 + 8 + 1 = 16\); write \(6\) and regroup \(1\) hundred. Add the hundreds: \(5 + 2 + 1 = 8\). Therefore, \(577 + 288 = 865\). 4. Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(834\) b) \(823\) c) \(865\) Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

- Begin in the ones column. - When a column total is \(10\) or more, regroup to the next place value. - Include each regrouped amount when adding the next column.

1. For part a, add the ones: \(4 + 2 = 6\). Add the tens: \(8 + 5 = 13\), so write \(3\) and regroup \(1\) hundred. Add the hundreds: \(3 + 4 + 1 = 8\). The sum is \(836\). 2. For part b, add the ones: \(7 + 8 = 15\), so write \(5\) and regroup \(1\) ten. Add the tens: \(2 + 4 + 1 = 7\). Add the hundreds: \(5 + 2 = 7\). The sum is \(775\). 3. Carry record: a) tens to hundreds: \(1\) | b) ones to tens: \(1\)

a) \(836\) b) \(775\) Carry record: a) tens to hundreds: \(1\) | b) ones to tens: \(1\)

- The rows in the column addition give the amounts for the three sections. - Begin in the ones column and work left. - A column total of ten or more gives something to carry into the next column.

1. Ones: \(0+0+5=5\). 2. Tens: \(5+2+3=10\). Write \(0\) and carry \(1\) hundred. 3. Hundreds: \(4+8+1+1=14\). Write \(4\) and carry \(1\) thousand. 4. Thousands: \(2+1+3+1=7\). 5. Thus, \(2450+1820+3135=7405\). The regrouping is \(1\) hundred from the tens column and \(1\) thousand from the hundreds column.

\(7405\) seedlings. Regrouping: \(1\) hundred and \(1\) thousand.

- Align each pair of addends by place value. - Calculate each sum separately before comparing. - Look for how a decrease in one addend is balanced by an increase in the other.

1. For a), add the ones: \(8 + 1 = 9\). Add the tens: \(3 + 7 = 10\); write \(0\) and regroup \(1\) hundred. Add the hundreds: \(5 + 2 + 1 = 8\). Retain the thousands digit \(4\). Therefore, \(4538 + 271 = 4809\). 2. For b), add the ones: \(8 + 1 = 9\). Add the tens: \(3 + 7 = 10\); write \(0\) and regroup \(1\) hundred. Add the hundreds: \(2 + 5 + 1 = 8\). Retain the thousands digit \(4\). Therefore, \(4238 + 571 = 4809\). 3. For c), add the ones: \(8 + 1 = 9\). Add the tens: \(3 + 7 = 10\); write \(0\) and regroup \(1\) hundred. Add the hundreds: \(7 + 0 + 1 = 8\). Retain the thousands digit \(4\). Therefore, \(4738 + 71 = 4809\). 4. All three sums are equal. 5. Carry record: a) tens to hundreds: \(1\) | b) tens to hundreds: \(1\) | c) tens to hundreds: \(1\)

a) \(4809\) b) \(4809\) c) \(4809\) All three results are equal. Carry record: a) tens to hundreds: \(1\) | b) tens to hundreds: \(1\) | c) tens to hundreds: \(1\)

- Write the two prices with matching place values in the same columns. - Work from right to left and record any regrouped value in the next column. - Check that the total is greater than \(\$78{,}000\).

1. Align \(78{,}000\) and \(4500\) by place value. 2. The ones and tens columns total \(0\). The hundreds column gives \(0+5=5\). 3. In the thousands column, \(8+4=12\). Write \(2\) thousands and regroup \(1\) ten-thousand. 4. In the ten-thousands column, \(7+1=8\). The sum is \(82{,}500\). 5. Carry record: thousands to ten-thousands: \(1\)

The fully equipped boat costs \(\$82{,}500\). Carry record: thousands to ten-thousands: \(1\)

- Align digits by place value, especially when the addends have different numbers of digits. - Begin in the ones column. - Regroup whenever a column sum is \(10\) or greater.

1. For a), add the ones: \(3 + 4 = 7\). Add the tens: \(2 + 9 = 11\); write \(1\) and regroup \(1\) hundred. Add the hundreds: \(8 + 1 + 1 = 10\); write \(0\) and regroup \(1\) thousand. Add the thousands: \(5 + 7 + 1 = 13\); write \(3\) and regroup \(1\) ten-thousand. Add the ten-thousands: \(4 + 2 + 1 = 7\). Therefore, \(45{,}823 + 27{,}194 = 73{,}017\). 2. For b), align \(9347\) by place value. Add the ones: \(6 + 7 = 13\); write \(3\) and regroup \(1\) ten. Add the tens: \(5 + 4 + 1 = 10\); write \(0\) and regroup \(1\) hundred. Add the hundreds: \(0 + 3 + 1 = 4\). Add the thousands: \(2 + 9 = 11\); write \(1\) and regroup \(1\) ten-thousand. Add the ten-thousands: \(8 + 0 + 1 = 9\). Therefore, \(82{,}056 + 9347 = 91{,}403\). 3. Carry record: a) tens to hundreds: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\); thousands to ten-thousands: \(1\)

a) \(73{,}017\) b) \(91{,}403\) Carry record: a) tens to hundreds: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\); thousands to ten-thousands: \(1\)

- No regrouping is needed in this problem. - Compare the digits in each place-value column.

1. In the ones column, \(6 + \square = 9\), so the missing ones digit in the second addend is \(3\). 2. In the tens column, \(5 + 2 = 7\), so the missing tens digit in the result is \(7\). 3. In the hundreds column, \(4 + 3 = 7\). The completed equation is \(456 + 323 = 779\).

The missing digits are \(3\) and \(7\). The completed equation is \(456 + 323 = 779\).

- Examine each column separately. - Because there is no regrouping, each unknown can be found directly from its column sum.

1. No regrouping is needed. In the ones column, \(1 + 2 + \square = 7\), so the missing ones digit is \(4\). 2. In the tens column, \(1 + \square + 3 = 7\), so the missing tens digit is \(3\). 3. In the hundreds column, \(\square + 2 + 3 = 7\), so the missing hundreds digit is \(2\). The completed equation is \(1211 + 2232 + 3334 = 6777\).

\(1211 + 2232 + 3334 = 6777\)

- Read each row as a three-digit addend. - Start adding in the ones column. - Whenever a column reaches \(10\) or more, exchange \(10\) units of that place for \(1\) unit of the next place.

1. The rows represent \(247\) and \(138\). 2. In the ones place, \(7+8=15\), so \(10\) ones are exchanged for \(1\) ten. The ones place is the first place that must be regrouped. 3. Add the tens, including the regrouped ten: \(4+3+1=8\) tens. 4. Add the hundreds: \(2+1=3\) hundreds. 5. The sum is \(385\).

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

- Ask what value the misplaced \(3\) has when it sits in the hundreds column. - For the correct setup, line up the ones digits first. - Compare the values of the second addend in the two layouts.

1. Placing the digits one place too far left makes \(36\) act like \(360\). The wrong sum would be \(427+360=787\). 2. The \(6\) must line up with the \(7\) ones and the \(3\) with the \(2\) tens. 3. Add the ones: \(7+6=13\); write \(3\) ones and carry \(1\) ten. Then \(2+3+1=6\) tens, with \(4\) hundreds, giving \(463\).

a) \(787\). b) \(427+36=463\). The student shifted the second addend one place left, changing its value.

- Line up all three addends by place value. - Start with the ones and carry any full group into the next column. - A carried unit must be included when you add that next column.

1. Ones: \(6+0+4=10\). Write \(0\) and carry \(1\) ten. 2. Tens: \(3+5+7+1=16\) tens. Write \(6\) tens and carry \(1\) hundred. 3. Hundreds: \(1+2+0+1=4\). The sum is \(460\).

\(460\); the ones make a new ten, and the tens make a new hundred.

- Recheck the ones and tens columns in order. - Is there a group of ten to carry in either column? - Check that every carried unit is included in the next column.

1. Ones: \(7+5=12\). Write \(2\) and carry \(1\) ten. 2. Tens: \(2+8+1=11\) tens. Write \(1\) ten and carry \(1\) hundred. 3. Hundreds: \(4+1=5\). The correct sum is \(512\).

\(427+85=512\). The tens column makes \(11\) tens, so one hundred must be carried.

- Read the value shown by each row separately. - Add chips in the same columns. - If a column has at least ten chips, trade a full group to the next column.

1. The top row is \(285\) and the bottom row is \(348\). 2. Adding the ones gives \(5+8=13\) ones. Exchange \(10\) ones for \(1\) ten, leaving \(3\) ones. 3. Adding the tens with the extra ten gives \(8+4+1=13\) tens. Exchange \(10\) tens for \(1\) hundred, leaving \(3\) tens. 4. The hundreds total is \(2+3+1=6\), so the sum is \(633\).

\(633\); trade ten ones for one ten, then ten tens for one hundred.

- Check the ones column before analyzing the tens column. - If the tens total is \(10\) or more, regroup \(10\) tens as \(1\) hundred. - Include any ten regrouped from the ones column. - Line up digits by place value in the standard algorithm.

1. Check the ones first for any regrouping into the tens, and then check the tens column. 2. In \(243 + 526\), the ones sum is \(9\) and the tens sum is \(6\), so no ten is regrouped as a hundred. The sum is \(769\). 3. In \(471 + 382\), the tens sum is \(7 + 8 = 15\), so \(1\) ten is regrouped as \(1\) hundred. The sum is \(853\). 4. In \(615 + 174\), the ones sum is \(9\) and the tens sum is \(8\), so no ten is regrouped as a hundred. The sum is \(789\). 5. In \(354 + 265\), the tens sum is \(5 + 6 = 11\), so \(1\) ten is regrouped as \(1\) hundred. The sum is \(619\).

No regrouping from tens to hundreds: \(243 + 526 = 769\) and \(615 + 174 = 789\). Regrouping from tens to hundreds: \(471 + 382 = 853\) and \(354 + 265 = 619\).

- At what total does a column require regrouping? - First check whether the ones column regroups into the tens. - Test digits from \(0\) through \(9\) in the box.

1. In the ones column, \(8 + 1 = 9\), so there is no regrouping into the tens. 2. To avoid regrouping from tens to hundreds, \(4 + \square\) must be at most \(9\). The possible digits are \(0, 1, 2, 3, 4, 5\). 3. Regrouping from tens to hundreds occurs when \(4 + \square\) is at least \(10\). The possible digits are \(6, 7, 8, 9\). 4. With the smallest digit, \(548 + 201 = 749\). 5. With the largest digit, \(548 + 291 = 839\).

a) \(0, 1, 2, 3, 4, 5\) b) \(6, 7, 8, 9\) c) \(548 + 201 = 749\) and \(548 + 291 = 839\)

- Add the ones before working with the tens. - What happens when a column has \(10\) or more units? - For part c), choose a tens digit that avoids making ten tens.

1. Ones: \(3+4=7\). 2. Tens: \(6+5=11\) tens. Write \(1\) ten and carry \(1\) hundred. 3. Hundreds: \(4+2+1=7\), so \(463+254=717\). 4. To avoid a carry from tens to hundreds, the two tens digits must total no more than \(9\). Since the first tens digit is \(6\), the other can be \(0\), \(1\), \(2\), or \(3\). 5. For example, \(463+234=697\), with no carry from tens to hundreds.

a) \(717\) b) Regroup \(10\) tens as \(1\) hundred in the tens column. c) One possibility: \(463+234=697\). Tens digits \(0\), \(1\), \(2\), or \(3\) work.

- In each column, account for both addend digits and any regrouped amount. - Check whether the previous column created a regrouped \(1\). - Work from the ones column toward the hundreds column.

1. In part a, the ones column must satisfy \(5 + \square = 11\), so the missing ones digit is \(6\), with \(1\) ten regrouped. The tens column must satisfy \(\square + 8 + 1 = 12\), so the missing tens digit is \(3\), with \(1\) hundred regrouped. The completed equation is \(235 + 486 = 721\). 2. In part b, \(7 + 6 = 13\), so write \(3\) and regroup \(1\) ten. The tens column must satisfy \(3 + \square + 1 = 10\), so the missing tens digit is \(6\). The hundreds column must satisfy \(\square + 2 + 1 = 8\), so the missing hundreds digit is \(5\). The completed equation is \(537 + 266 = 803\).

a) \(235 + 486 = 721\) b) \(537 + 266 = 803\)

- Add from right to left using the standard algorithm. - Check whether each place-value sum is \(10\) or greater. - When you regroup, make sure the extra unit is added in the next place.

1. In the ones place, \(6 + 8 = 14\). Lina wrote \(4\) but did not regroup \(1\) ten. 2. Without that regrouped ten, she added \(5 + 7 = 12\) in the tens place. She wrote \(2\) but again did not regroup \(1\) hundred. 3. Correctly applying the algorithm gives: ones, \(6 + 8 = 14\), write \(4\) and regroup \(1\); tens, \(5 + 7 + 1 = 13\), write \(3\) and regroup \(1\); hundreds, \(4 + 2 + 1 = 7\). 4. Therefore, \(456 + 278 = 734\).

Lina forgot to regroup from the ones place and from the tens place. The correct sum is \(734\).

- Begin in the ones column when using the standard addition algorithm. - Record each regrouped amount above the next column. - Regroup whenever a column total is \(10\) or more.

1. For \(368 + 457\), add the ones: \(8 + 7 = 15\). Write \(5\) and regroup \(1\) ten. Add the tens: \(6 + 5 + 1 = 12\). Write \(2\) and regroup \(1\) hundred. Add the hundreds: \(3 + 4 + 1 = 8\). The sum is \(825\). 2. For \(279 + 184\), add the ones: \(9 + 4 = 13\). Write \(3\) and regroup \(1\) ten. Add the tens: \(7 + 8 + 1 = 16\). Write \(6\) and regroup \(1\) hundred. Add the hundreds: \(2 + 1 + 1 = 4\). The sum is \(463\). 3. For \(585 + 238\), add the ones: \(5 + 8 = 13\). Write \(3\) and regroup \(1\) ten. Add the tens: \(8 + 3 + 1 = 12\). Write \(2\) and regroup \(1\) hundred. Add the hundreds: \(5 + 2 + 1 = 8\). The sum is \(823\). 4. For \(496 + 325\), add the ones: \(6 + 5 = 11\). Write \(1\) and regroup \(1\) ten. Add the tens: \(9 + 2 + 1 = 12\). Write \(2\) and regroup \(1\) hundred. Add the hundreds: \(4 + 3 + 1 = 8\). The sum is \(821\). 5. Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\) | d) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(825\) b) \(463\) c) \(823\) d) \(821\) Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\) | d) ones to tens: \(1\); tens to hundreds: \(1\)

- Check the ones column and then the tens column in each problem. - A column requires regrouping when its total is \(10\) or more. - Complete only the problems that require regrouping in both columns.

1. For A, \(6 + 5 = 11\), so regroup from ones to tens. Then \(5 + 3 + 1 = 9\), so there is no regrouping from tens to hundreds. 2. For B, \(7 + 6 = 13\), so regroup from ones to tens. Then \(8 + 5 + 1 = 14\), so regroup from tens to hundreds. The sum is \(387 + 456 = 843\). 3. For C, \(9 + 2 = 11\), so regroup from ones to tens. Then \(2 + 8 + 1 = 11\), so regroup from tens to hundreds. The sum is \(129 + 782 = 911\). 4. For D, \(3 + 1 = 4\), so there is no regrouping from ones to tens. Therefore, D does not meet both conditions.

B) \(387 + 456 = 843\); C) \(129 + 782 = 911\)

- Begin in the ones column. Which digit added to \(8\) gives a sum with a ones digit of \(3\)? - Include the regrouped \(1\) when you work in the tens column. - Check the hundreds column after filling in both digits.

1. In the ones column, \(8 + \square\) must have a ones digit of \(3\) and require regrouping. Therefore, \(8 + 5 = 13\), so the missing ones digit is \(5\), and \(1\) ten is regrouped. 2. In the tens column, \(\square + 7 + 1\) must have a tens digit of \(3\) and require regrouping. Therefore, \(5 + 7 + 1 = 13\), so the missing tens digit is \(5\), and \(1\) hundred is regrouped. 3. Check the hundreds column: \(2 + 4 + 1 = 7\). The completed equation is \(258 + 475 = 733\).

The missing digits are both \(5\). The completed equation is \(258 + 475 = 733\).

- Check which ones, tens, and hundreds digits are aligned. - Ones must be written under ones, and tens under tens. - Moving a digit one place to the left makes its value ten times as great.

1. Mia did not align digits with the same place value. She placed the \(4\) tens under the \(5\) hundreds and the \(3\) ones under the \(2\) tens. Her setup represents \(526 + 430\), not \(526 + 43\). 2. The \(3\) must be aligned under the \(6\) in the ones place, and the \(4\) must be aligned under the \(2\) in the tens place. 3. Add by place value: \(6 + 3 = 9\), \(2 + 4 = 6\), and \(5 + 0 = 5\). Therefore, \(526 + 43 = 569\).

Mia did not align the ones and tens digits by place value. The correct sum is \(569\).

- Recalculate each sum one place at a time. - Check whether any place-value sum requires regrouping. - Compare your intermediate digits with the shown work. - Check that ones are aligned with ones and tens with tens.

1. In Calculation A, \(8 + 6 = 14\). The \(4\) was written in the ones place, but the regrouped \(1\) ten was not added in the tens place. Tip 1 applies. 2. The corrected sum is \(348 + 236 = 584\). 3. In Calculation B, \(42\) was shifted one place to the left, so the work represents \(235 + 420\). Tip 2 applies. 4. The corrected sum is \(235 + 42 = 277\).

Calculation A: Tip 1. The correct sum is \(584\). Calculation B: Tip 2. The correct sum is \(277\).

- Estimate first: \(370 + 260\) is about how much? - Add from right to left and check every sum greater than \(9\). - Count how many times regrouping is needed.

1. In the ones place, \(8 + 7 = 15\). Tim wrote \(5\) but forgot to regroup \(1\) ten. 2. In the tens place, Tim used \(6 + 5 = 11\), wrote \(1\), and again forgot to regroup \(1\) hundred. He also missed the regrouped ten from the ones place. 3. Correctly, the ones sum is \(15\): write \(5\) and regroup \(1\). The tens sum is \(6 + 5 + 1 = 12\): write \(2\) and regroup \(1\). The hundreds sum is \(3 + 2 + 1 = 6\). 4. Therefore, \(368 + 257 = 625\).

Tim failed to regroup from the ones place to the tens place and from the tens place to the hundreds place. The correct sum is \(625\).

- Record each regrouped amount above the next column. - When adding three numbers, the regrouped amount can be \(2\). - In part b, remember that \(82\) has no hundreds digit.

1. For part a, add the ones: \(8 + 7 + 5 = 20\). Write \(0\) and regroup \(2\) tens. Add the tens: \(5 + 6 + 4 + 2 = 17\). Write \(7\) and regroup \(1\) hundred. Add the hundreds: \(2 + 3 + 1 + 1 = 7\). The sum is \(770\). 2. For part b, align \(82\) so that its \(2\) is in the ones column and its \(8\) is in the tens column. Add the ones: \(7 + 2 + 6 = 15\). Write \(5\) and regroup \(1\) ten. Add the tens: \(0 + 8 + 9 + 1 = 18\). Write \(8\) and regroup \(1\) hundred. Add the hundreds: \(4 + 0 + 1 + 1 = 6\). The sum is \(685\). 3. For part c, add the ones: \(4 + 4 + 4 = 12\). Write \(2\) and regroup \(1\) ten. Add the tens: \(8 + 8 + 8 + 1 = 25\). Write \(5\) and regroup \(2\) hundreds. Add the hundreds: \(2 + 2 + 2 + 2 = 8\). The sum is \(852\). 4. Carry record: a) ones to tens: \(2\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(2\)

a) \(770\) b) \(685\) c) \(852\) Carry record: a) ones to tens: \(2\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(2\)

- Add the numbers yourself from right to left. - Keep track of every regrouped ten or hundred. - Compare your result with \(610\) one place at a time.

1. In the ones place, \(6 + 9 + 5 = 20\). Write \(0\) and regroup \(2\) tens. 2. In the tens place, \(4 + 0 + 5 + 2 = 11\). Write \(1\) and regroup \(1\) hundred. 3. In the hundreds place, \(2 + 1 + 3 + 1 = 7\). 4. Therefore, \(246 + 109 + 355 = 710\). Lucas’s result is incorrect. 5. Carry record: ones to tens: \(2\); tens to hundreds: \(1\)

Lucas is not correct. The correct sum is \(710\). Carry record: ones to tens: \(2\); tens to hundreds: \(1\)

- Begin in the ones column. - Include any regrouped amount when you move to the next column. - Use the result digit in each column to determine the missing addend digit. - Check each completed addition with the standard algorithm.

1. In problem 1, the ones column must satisfy \(\square + 6 = 11\), so the missing digit is \(5\), and \(1\) ten is regrouped. The tens column gives \(2 + 5 + 1 = 8\), and the hundreds column gives \(4 + 3 = 7\). The completed equation is \(425 + 356 = 781\). 2. In problem 2, the ones column must satisfy \(8 + \square = 12\), so the missing ones digit is \(4\), and \(1\) ten is regrouped. The tens column must satisfy \(\square + 4 + 1 = 11\), so the missing tens digit is \(6\), and \(1\) hundred is regrouped. The hundreds column gives \(1 + 6 + 1 = 8\). The completed equation is \(168 + 644 = 812\). 3. Carry record: 1) ones to tens: \(1\) | 2) ones to tens: \(1\); tens to hundreds: \(1\)

1) \(425 + 356 = 781\) 2) \(168 + 644 = 812\) Carry record: 1) ones to tens: \(1\) | 2) ones to tens: \(1\); tens to hundreds: \(1\)

- When a column total is greater than \(9\), regroup to the next place value. - Record each regrouped amount above the next column. - Work backward from the result digit to find each missing digit.

1. In problem 1, \(9 + 3 = 12\), so write \(2\) and regroup \(1\) ten. The tens column must satisfy \(4 + \square + 1 = 11\), so the missing tens digit is \(6\), and \(1\) hundred is regrouped. The hundreds column must satisfy \(\square + 2 + 1 = 6\), so the missing hundreds digit is \(3\). The completed equation is \(349 + 263 = 612\). 2. In problem 2, the ones column must satisfy \(\square + 4 = 10\), so the missing ones digit is \(6\), and \(1\) ten is regrouped. In the tens column, \(7 + 8 + 1 = 16\), so \(1\) hundred is regrouped. The hundreds column must satisfy \(5 + \square + 1 = 9\), so the missing hundreds digit is \(3\). The completed equation is \(576 + 384 = 960\).

1) \(349 + 263 = 612\) 2) \(576 + 384 = 960\)

- Add from right to left one place at a time. - Check whether a value must be regrouped into the next place. - Compare each of your place-value sums with the digits in the shown answer.

1. In the ones place, \(6 + 8 + 4 = 18\). Write \(8\) and regroup \(1\) ten. 2. In the shown work, the regrouped ten was not added in the tens place. The correct tens sum is \(5 + 1 + 2 + 1 = 9\), not \(8\). 3. In the hundreds place, \(2 + 3 + 1 = 6\). 4. Therefore, \(256 + 318 + 124 = 698\).

The regrouped ten from the ones place was not added in the tens place. The correct sum is \(698\).

- Recalculate each equation separately with the standard algorithm. - Compare your result with Lina’s result. - When a result differs, check the ones, tens, and hundreds places to locate the error.

1. For part a, the ones sum is \(3 + 4 + 2 = 9\). The tens sum is \(7 + 5 + 6 = 18\), so write \(8\) and regroup \(1\). The hundreds sum is \(2 + 1 + 3 + 1 = 7\). Thus, \(789\) is correct. 2. For part b, the ones sum is \(6 + 5 + 9 = 20\), so write \(0\) and regroup \(2\). The tens sum is \(1 + 3 + 8 + 2 = 14\), so write \(4\) and regroup \(1\). The hundreds sum is \(4 + 2 + 1 + 1 = 8\). The correct sum is \(840\). 3. For part c, the ones sum is \(7 + 8 + 5 = 20\), so write \(0\) and regroup \(2\). The tens sum is \(2 + 3 + 0 + 2 = 7\). The hundreds sum is \(1 + 4 + 2 = 7\). The correct sum is \(770\). 4. Carry record: a) tens to hundreds: \(1\) | b) ones to tens: \(2\); tens to hundreds: \(1\) | c) ones to tens: \(2\)

a) Correct: \(273 + 154 + 362 = 789\) b) Incorrect: \(416 + 235 + 189 = 840\) c) Incorrect: \(127 + 438 + 205 = 770\) Carry record: a) tens to hundreds: \(1\) | b) ones to tens: \(2\); tens to hundreds: \(1\) | c) ones to tens: \(2\)

- Align the addends by place value and regroup as needed. - Use the sum as the minuend in the subtraction check. - The difference should equal the first addend.

1. Add by place value. Ones: \(2+9=11\), so write \(1\) and regroup \(1\) ten. Tens: \(1+8+1=10\), so write \(0\) and regroup \(1\) hundred. Hundreds: \(6+7+1=14\), so write \(4\) and regroup \(1\) thousand. Thousands: \(5+3+1=9\). Therefore, \(5612+3789=9401\). 2. Check with subtraction: \(9401-3789\). Regroup across the zero so the ones calculation is \(11-9=2\) and the tens calculation is \(9-8=1\). Regroup one thousand as \(10\) hundreds, so \(13-7=6\) and \(8-3=5\). Therefore, \(9401-3789=5612\). 3. The difference matches the first addend, so the sum is correct. 4. Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\)

The sum is \(9401\). Check: \(9401-3789=5612\). Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\)

- Add from the ones column through the hundreds column, including any regrouped values. - For the check, look for two addends whose ones digits combine to make \(10\). - Compare the check result with the standard-algorithm result.

1. For a), add the ones: \(2 + 5 + 8 = 15\); write \(5\) and regroup \(1\) ten. Add the tens: \(6 + 2 + 3 + 1 = 12\); write \(2\) and regroup \(1\) hundred. Add the hundreds: \(1 + 4 + 2 + 1 = 8\). The sum is \(825\). 2. Check a) by combining \(162 + 238 = 400\), then \(400 + 425 = 825\). 3. For b), add the ones: \(1 + 5 + 9 = 15\); write \(5\) and regroup \(1\) ten. Add the tens: \(7 + 5 + 2 + 1 = 15\); write \(5\) and regroup \(1\) hundred. Add the hundreds: \(2 + 1 + 3 + 1 = 7\). The sum is \(755\). 4. Check b) by combining \(271 + 329 = 600\), then \(600 + 155 = 755\). 5. Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(825\) b) \(755\) Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\)

- Begin in the ones column and use the result digit to determine the missing addend digit. - Include any regrouped amount when you move to the next column. - Continue from right to left through the place-value columns.

1. In the ones column, \(7 + \square\) must have a ones digit of \(2\) and require regrouping. Therefore, \(7 + 5 = 12\), so the missing ones digit is \(5\), and \(1\) ten is regrouped. 2. In the tens column, \(\square + 8 + 1\) must have a tens digit of \(3\) and require regrouping. Therefore, \(4 + 8 + 1 = 13\), so the missing tens digit is \(4\), and \(1\) hundred is regrouped. 3. Check the hundreds column: \(2 + 4 + 1 = 7\). The completed equation is \(247 + 485 = 732\).

The missing tens digit is \(4\), and the missing ones digit is \(5\). The completed equation is \(247 + 485 = 732\).

- Add from right to left to check each stated sum. - Include a carried value in the next column before adding there. - Explain the carries for the equation whose stated total is incorrect.

1. Ones: \(7+5=12\), tens: \(8+4+1=13\), hundreds: \(3+2+1=6\). The sum is \(632\), so equation 1 is correct. 2. Ones: \(6+8=14\), tens: \(5+7+1=13\), hundreds: \(4+1+1=6\). The sum is \(634\), not \(624\). Carry \(1\) from ones to tens and \(1\) from tens to hundreds. 3. Ones: \(4+7=11\), tens: \(9+6+1=16\), hundreds: \(2+5+1=8\). The sum is \(861\), so equation 3 is correct. 4. Ones: \(8+3=11\), tens: \(6+4+1=11\), hundreds: \(1+7+1=9\). The sum is \(911\), so equation 4 is correct.

1. Correct. 2. Incorrect; \(456+178=634\). Carry \(1\) ten from the ones column and \(1\) hundred from the tens column. 3. Correct. 4. Correct.

- Begin in the ones column. - Record each regrouped amount above the next column. - To check an addition equation, subtract one addend from the sum.

1. Add the ones: \(6 + 9 = 15\). Write \(5\) and regroup \(1\) ten. 2. Add the tens: \(5 + 7 + 1 = 13\). Write \(3\) and regroup \(1\) hundred. 3. Add the hundreds: \(4 + 2 + 1 = 7\). Therefore, \(456 + 279 = 735\). 4. Subtract one addend from the sum to check: \(735 - 279 = 456\). 5. Carry record: ones to tens: \(1\); tens to hundreds: \(1\)

Sum: \(735\) Check: \(735 - 279 = 456\) or \(735 - 456 = 279\) Carry record: ones to tens: \(1\); tens to hundreds: \(1\)

- Begin in the ones column. Which digit added to \(8\) gives a sum with a ones digit of \(3\)? - Include the regrouped \(1\) in the tens column. - Work from right to left through the place-value columns.

1. In the ones column, \(8 + \square\) must have a ones digit of \(3\) and require regrouping. Therefore, \(8 + 5 = 13\), so the missing ones digit is \(5\), and \(1\) ten is regrouped. 2. In the tens column, \(\square + 5 + 1\) must have a tens digit of \(1\) and require regrouping. Therefore, \(5 + 5 + 1 = 11\), so the missing tens digit is \(5\), and \(1\) hundred is regrouped. 3. Check the hundreds column: \(4 + 2 + 1 = 7\). The completed equation is \(458 + 255 = 713\).

The missing digits are \(5\) and \(5\). The completed equation is \(458 + 255 = 713\).

- You can use the standard addition algorithm with more than two addends. - Add all the digits in one column before moving to the next column. - A column total of at least \(10\) requires regrouping; a total of at least \(20\) may require regrouping \(2\).

1. Add the ones: \(3 + 6 + 8 = 17\). Write \(7\) and regroup \(1\) ten. 2. Add the tens: \(4 + 5 + 2 + 1 = 12\). Write \(2\) and regroup \(1\) hundred. 3. Add the hundreds: \(2 + 1 + 3 + 1 = 7\). 4. Therefore, \(243 + 156 + 328 = 727\). 5. Carry record: ones to tens: \(1\); tens to hundreds: \(1\)

\(727\) Carry record: ones to tens: \(1\); tens to hundreds: \(1\)

- List all four categories before calculating. - Align the numbers by place value. - Track each regrouped value when adding several numbers.

1. The four categories contain \(3425\), \(2890\), \(1565\), and \(980\) items. 2. Add the ones: \(5 + 0 + 5 + 0 = 10\); write \(0\) and regroup \(1\) ten. 3. Add the tens: \(2 + 9 + 6 + 8 + 1 = 26\); write \(6\) and regroup \(2\) hundreds. 4. Add the hundreds: \(4 + 8 + 5 + 9 + 2 = 28\); write \(8\) and regroup \(2\) thousands. 5. Add the thousands: \(3 + 2 + 1 + 0 + 2 = 8\). Therefore, \(3425 + 2890 + 1565 + 980 = 8860\). 6. Carry record: ones to tens: \(1\); tens to hundreds: \(2\); hundreds to thousands: \(2\)

The library has \(8860\) items for children altogether. Carry record: ones to tens: \(1\); tens to hundreds: \(2\); hundreds to thousands: \(2\)

- Align the ones digits of all three addends. - Record each regrouped value above the next column. - More than two numbers can be added in the same standard-algorithm setup.

1. Align the three addends by place value. 2. Add the ones: \(9 + 2 + 7 = 18\); write \(8\) and regroup \(1\) ten. 3. Add the tens: \(0 + 8 + 9 + 1 = 18\); write \(8\) and regroup \(1\) hundred. 4. Add the hundreds: \(4 + 6 + 5 + 1 = 16\); write \(6\) and regroup \(1\) thousand. 5. Add the thousands: \(2 + 3 + 1 = 6\), then retain the ten-thousands digit \(1\). The sum is \(16{,}688\). 6. Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\)

\(16{,}688\) Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\)

- Recalculate all three sums using the standard algorithm. - Pay close attention to regrouping in the hundreds place. - Only one equation is incorrect.

1. For equation 1, ones: \(8+2=10\), tens: \(0+9+1=10\), hundreds: \(6+3+1=10\), and thousands: \(5+1=6\). Therefore, \(5608+392=6000\), so equation 1 is correct. 2. For equation 2, ones: \(5+5=10\), tens: \(2+8+1=11\), hundreds: \(7+2+1=10\), and thousands: \(4+1+1=6\). Therefore, \(4725+1285=6010\), so equation 2 is correct. 3. For equation 3, ones: \(9+1=10\), tens: \(1+8+1=10\), hundreds: \(4+6+1=11\), and thousands: \(3+1=4\). Therefore, \(3419+681=4100\), so equation 3 is incorrect. Its stated sum is \(100\) too small. 4. Carry record: 1. ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\) | 2. ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\) | 3. ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\)

Equation 3 is incorrect. The correct sum is \(4100\). Carry record: 1. ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\) | 2. ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\) | 3. ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\)

- Stack the three costs with ones under ones, tens under tens, and so on. - Add one column at a time from right to left. - If a column totals \(10\) or more, regroup to the next column.

1. Align all three costs by place value and add from right to left. 2. Ones: \(0+0+0=0\). Tens: \(0+0+5=5\). 3. Hundreds: \(3+7+4=14\). Write \(4\) hundreds and regroup \(1\) thousand. 4. Thousands: \(4+8+3+1=16\). Write \(6\) thousands and regroup \(1\) ten-thousand. 5. Ten-thousands: \(2+1=3\). The total is \(\$36{,}450\). 6. Carry record: hundreds to thousands: \(1\); thousands to ten-thousands: \(1\)

The town spends \(\$36{,}450\) in total. Carry record: hundreds to thousands: \(1\); thousands to ten-thousands: \(1\)

- Begin in the ones place and work left. - Include each regrouped unit in the next place. - Use the known result digit to determine each missing addend digit.

1. Ones: \(2 + 9 = 11\). Write \(1\) and regroup \(1\) ten. 2. Tens: \(8 + \square + 1 = 14\), so the missing digit is \(5\). Write \(4\) and regroup \(1\) hundred. 3. Hundreds: \(\square + 5 + 1 = 12\), so the missing digit is \(6\). Write \(2\) and regroup \(1\) thousand. 4. Thousands: \(4 + 1 + 1 = 6\), so the missing result digit is \(6\). The completed addition is \(4682 + 1559 = 6241\).

The missing digits, read from top to bottom and left to right, are \(6\), \(5\), and \(6\). The completed addition is \(4682 + 1559 = 6241\).

- Begin with the ones column and work left. - Record every regrouped \(1\). - Check each completed equation with the standard addition algorithm.

1. In part a, the ones column requires \(7+5=12\), giving a regrouped \(1\). The tens column then requires \(6+8+1=15\). The completed equation is \(367+485=852\). 2. In part b, the ones column requires \(9+4=13\). The tens column gives \(2+8+1=11\), and the hundreds column requires \(5+3+1=9\). The completed equation is \(529+384=913\).

a) \(367+485=852\) b) \(529+384=913\)

- Find all three sums first. - Determine the distance from each result to \(800\). - Distance is positive whether the result is above or below \(800\).

1. For a), add by place value: \(7 + 2 = 9\), \(5 + 4 = 9\), and \(4 + 3 = 7\). Therefore, \(457 + 342 = 799\). 2. For b), add the ones: \(9 + 2 = 11\); write \(1\) and regroup \(1\) ten. Add the tens: \(8 + 1 + 1 = 10\); write \(0\) and regroup \(1\) hundred. Add the hundreds: \(2 + 5 + 1 = 8\). Therefore, \(289 + 512 = 801\). 3. For c), add the ones: \(3 + 8 = 11\); write \(1\) and regroup \(1\) ten. Add the tens: \(0 + 9 + 1 = 10\); write \(0\) and regroup \(1\) hundred. Add the hundreds: \(6 + 1 + 1 = 8\). Therefore, \(603 + 198 = 801\). 4. Each result is \(1\) away from \(800\), so all three are equally close. 5. Carry record: a) none | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(799\), which is \(1\) away from \(800\) b) \(801\), which is \(1\) away from \(800\) c) \(801\), which is \(1\) away from \(800\) All three results are equally close to \(800\). Carry record: a) none | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

- Align the addends by place value. - Use a different order for the check. - The commutative property says that changing the order of addends does not change the sum.

1. Add in the given order by place value. Ones: \(2+9+4=15\), so write \(5\) and regroup \(1\) ten. Tens: \(1+8+0+1=10\), so write \(0\) and regroup \(1\) hundred. Hundreds: \(6+3+5+1=15\), so write \(5\) and regroup \(1\) thousand. Thousands: \(5+2+7+1=15\), so write \(5\) and regroup \(1\) ten-thousand. Ten-thousands: \(4+1+1=6\). Therefore, the sum is \(65{,}505\). 2. Check in a different order: \(12{,}389+7504=19{,}893\), and \(19{,}893+45{,}612=65{,}505\). 3. The sums agree because changing the order of addends does not change the sum. 4. Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(1\)

The sum is \(65{,}505\). Changing the order of the addends gives the same sum. Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(1\)

- Look for pairs that make round ten-thousands. - Find each pair’s sum first. - Add the two partial sums and compare with the original total.

1. Add by place value. Ones: \(0+0+0+0=0\). Tens: \(8+2+5+5=20\), so write \(0\) and regroup \(2\) hundreds. Hundreds: \(2+7+1+8+2=20\), so write \(0\) and regroup \(2\) thousands. Thousands: \(5+4+9+0+2=20\), so write \(0\) and regroup \(2\) ten-thousands. Ten-thousands: \(1+3+0+1+2=7\). Therefore, the total is \(70{,}000\). 2. Group convenient pairs: \(15{,}280+34{,}720=50{,}000\) and \(9150+10{,}850=20{,}000\). 3. Add the partial sums: \(50{,}000+20{,}000=70{,}000\). 4. Carry record: tens to hundreds: \(2\); hundreds to thousands: \(2\); thousands to ten-thousands: \(2\)

The total is \(70{,}000\). The grouped check is \(50{,}000+20{,}000=70{,}000\). Carry record: tens to hundreds: \(2\); hundreds to thousands: \(2\); thousands to ten-thousands: \(2\)

- Round the addends to estimate the sum. - Subtract one addend from the stated sum. - The difference should equal the other addend if the sum is correct.

1. Estimate: \(54{,}000+28{,}000=82{,}000\). The stated result is near the estimate but is not confirmed. 2. Inverse check: \(81{,}242-27{,}914=53{,}328\), which does not equal \(54{,}328\). Therefore, the stated sum is incorrect. 3. Add by place value. Ones: \(8+4=12\), so write \(2\) and regroup \(1\) ten. Tens: \(2+1+1=4\). Hundreds: \(3+9=12\), so write \(2\) and regroup \(1\) thousand. Thousands: \(4+7+1=12\), so write \(2\) and regroup \(1\) ten-thousand. Ten-thousands: \(5+2+1=8\). Therefore, \(54{,}328+27{,}914=82{,}242\). 4. Carry record: ones to tens: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(1\)

Ava is incorrect. The correct sum is \(82{,}242\). Carry record: ones to tens: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(1\)

- Check whether the ones digit of each stated sum is possible. - Look for two addends in D that make a round number. - Use the standard addition algorithm for any equation that still seems uncertain.

1. Equation A is correct because \(12{,}456+35{,}123=47{,}579\). 2. Equation B is correct because \(45{,}280+14{,}720=60{,}000\). 3. Equation C is incorrect. The ones digits give \(9+2=11\), so the sum must end in \(1\), not \(5\). Using the standard algorithm, the successive result digits are \(1\), \(9\), \(9\), \(9\), and \(7\), so \(38{,}209+41{,}782=79{,}991\). 4. Equation D is incorrect. Using the standard algorithm, the ones and tens columns each total \(10\), the hundreds column totals \(11\), the thousands column totals \(10\), and the ten-thousands column totals \(4\). Therefore, \(5602+14{,}398+20{,}100=40{,}100\). 5. Carry record: C) ones to tens: \(1\) | D) ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(1\)

C) \(38{,}209 + 41{,}782 = 79{,}991\); D) \(5602 + 14{,}398 + 20{,}100 = 40{,}100\) Carry record: C) ones to tens: \(1\) | D) ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(1\)

- Work one written addition at a time and keep digits aligned by place value. - Record a carry only when a column total is at least \(10\). - After the four additions, compare the two final sums.

1. Method 1, first addition: \(3456+2189=5645\). The ones column carries \(1\) ten, and the tens column carries \(1\) hundred. 2. Method 1, second addition: \(5645+4072=9717\). The tens column carries \(1\) hundred. 3. Method 2, first addition: \(2189+4072=6261\). The ones column carries \(1\) ten, and the tens column carries \(1\) hundred. 4. Method 2, second addition: \(3456+6261=9717\). The tens column carries \(1\) hundred. 5. Both groupings give the same final sum, \(9717\).

1. \(3456+2189=5645\). Carry record: ones \(\to\) tens: \(1\); tens \(\to\) hundreds: \(1\). 2. \(5645+4072=9717\). Carry record: tens \(\to\) hundreds: \(1\). 3. \(2189+4072=6261\). Carry record: ones \(\to\) tens: \(1\); tens \(\to\) hundreds: \(1\). 4. \(3456+6261=9717\). Carry record: tens \(\to\) hundreds: \(1\). Both groupings give \(9717\).

- Align all digits by place value. - Add every digit in a column along with any regrouped value. - Carry each regrouped value into the next column to the left.

1. For a), add the ones: \(2 + 4 + 9 = 15\); write \(5\) and regroup \(1\) ten. Add the tens: \(1 + 9 + 9 + 1 = 20\); write \(0\) and regroup \(2\) hundreds. Add the hundreds: \(6 + 3 + 9 + 2 = 20\); write \(0\) and regroup \(2\) thousands. Add the thousands: \(5 + 8 + 3 + 2 = 18\); write \(8\) and regroup \(1\) ten-thousand. Add the ten-thousands: \(4 + 2 + 1 + 1 = 8\). Therefore, the sum is \(88{,}005\). 2. For b), add the ones: \(0 + 0 + 4 = 4\). Add the tens: \(5 + 0 + 4 = 9\). Add the hundreds: \(4 + 2 + 4 = 10\); write \(0\) and regroup \(1\) thousand. Add the thousands: \(2 + 8 + 4 + 1 = 15\); write \(5\) and regroup \(1\) ten-thousand. Add the ten-thousands: \(1 + 9 + 4 + 1 = 15\); write \(5\) and regroup \(1\) hundred-thousand. Add the hundred-thousands: \(3 + 1 + 4 + 1 = 9\). Therefore, the sum is \(955{,}094\). 3. Carry record: a) ones to tens: \(1\); tens to hundreds: \(2\); hundreds to thousands: \(2\); thousands to ten-thousands: \(1\) | b) hundreds to thousands: \(1\); thousands to ten-thousands: \(1\); ten-thousands to hundred-thousands: \(1\)

a) \(88{,}005\) b) \(955{,}094\) Carry record: a) ones to tens: \(1\); tens to hundreds: \(2\); hundreds to thousands: \(2\); thousands to ten-thousands: \(1\) | b) hundreds to thousands: \(1\); thousands to ten-thousands: \(1\); ten-thousands to hundred-thousands: \(1\)

- Work one column at a time from right to left. - Include any regrouped value from the previous column. - Use the result digit to determine the missing addend digit. - Verify the completed addition.

1. In the ones column, \(5+0+2=7\), so there is no regrouping. 2. In the tens column, \(4+1+\square=9\), so the missing digit is \(4\). 3. In the hundreds column, \(\square+2+3=9\), so the missing digit is \(4\). 4. In the thousands column, \(2+\square+4=12\), so the missing digit is \(6\), with a regrouped \(1\). 5. The completed equation is \(12{,}445+36{,}210+24{,}342=72{,}997\).

The missing digits, from top to bottom, are \(4,6,4\). \(12{,}445+36{,}210+24{,}342=72{,}997\)

- Align digits with the same place value. - Reverse or otherwise change the order of the addends for the check. - Check each place for regrouping.

1. Add by place value. Ones: \(5+0+3+7=15\), so write \(5\) and regroup \(1\) ten. Tens: \(8+2+3+6+1=20\), so write \(0\) and regroup \(2\) hundreds. Hundreds: \(0+7+4+8+2=21\), so write \(1\) and regroup \(2\) thousands. Thousands: \(2+6+9+0+2=19\), so write \(9\) and regroup \(1\) ten-thousand. Ten-thousands: \(1+5+0+0+1=7\). Hundred-thousands: \(3\). Therefore, the sum is \(379{,}105\). 2. Check by changing the order. For example, \(867+9433=10{,}300\). 3. Then \(10{,}300+56{,}720=67{,}020\), and \(67{,}020+312{,}085=379{,}105\). 4. Both orders give the same total. 5. Carry record: ones to tens: \(1\); tens to hundreds: \(2\); hundreds to thousands: \(2\); thousands to ten-thousands: \(1\)

The sum is \(379{,}105\). Carry record: ones to tens: \(1\); tens to hundreds: \(2\); hundreds to thousands: \(2\); thousands to ten-thousands: \(1\)

- Read the three addends from the vertical layout and add one place-value column at a time. - Record each regrouped value in the next column. - Reversing the order of the addends should not change the sum.

1. Add the ones: \(3 + 6 + 1 = 10\). Write \(0\) and regroup \(1\) ten. 2. Add the tens: \(8 + 0 + 1 + 1 = 10\). Write \(0\) and regroup \(1\) hundred. 3. Add the hundreds: \(2 + 9 + 3 + 1 = 15\). Write \(5\) and regroup \(1\) thousand. 4. Add the thousands: \(7 + 5 + 8 + 1 = 21\). Write \(1\) and regroup \(2\) ten-thousands. 5. Add the ten-thousands: \(4 + 1 + 2 + 2 = 9\). The sum is \(91{,}500\). 6. Reversing the addend order gives the same column totals and the same sum, \(91{,}500\). 7. Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(2\)

The sum is \(91{,}500\). Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\); thousands to ten-thousands: \(2\)

- For the standard algorithm, align all four addends by place value. - For the second method, compare the endings of the numbers to find complementary pairs. - Both methods must produce the same total.

1. For a), align all four addends. The ones sum is \(0\). The tens sum is \(0 + 5 + 0 + 5 = 10\); write \(0\) and regroup \(1\) hundred. The hundreds sum is \(6 + 5 + 4 + 4 + 1 = 20\); write \(0\) and regroup \(2\) thousands. The thousands sum is \(2 + 4 + 7 + 5 + 2 = 20\); write \(0\) and regroup \(2\) ten-thousands. The ten-thousands sum is \(1 + 3 + 1 + 2 + 2 = 9\). Therefore, the total is \(90{,}000\). 2. For b), form the pairs \(12{,}600 + 17{,}400 = 30{,}000\) and \(34{,}550 + 25{,}450 = 60{,}000\). 3. Add the partial sums: \(30{,}000 + 60{,}000 = 90{,}000\). 4. Carry record: a) tens to hundreds: \(1\); hundreds to thousands: \(2\); thousands to ten-thousands: \(2\)

a) \(90{,}000\) b) \(12{,}600 + 17{,}400 = 30{,}000\) and \(34{,}550 + 25{,}450 = 60{,}000\), so the total is \(90{,}000\). Carry record: a) tens to hundreds: \(1\); hundreds to thousands: \(2\); thousands to ten-thousands: \(2\)

- Right-align addends that have different numbers of digits. - Add every digit in each column. - Carry each regrouped value into the next column.

1. Align all five addends by place value. 2. Add the ones: \(1 + 4 + 3 + 2 + 0 = 10\); write \(0\) and regroup \(1\) ten. 3. Add the tens: \(8 + 0 + 3 + 6 + 1 + 1 = 19\); write \(9\) and regroup \(1\) hundred. 4. Add the hundreds: \(7 + 3 + 0 + 7 + 4 + 1 = 22\); write \(2\) and regroup \(2\) thousands. 5. Add the thousands: \(6 + 9 + 5 + 0 + 8 + 2 = 30\); write \(0\) and regroup \(3\) ten-thousands. 6. Add the ten-thousands: \(4 + 0 + 2 + 0 + 2 + 3 = 11\); write \(1\) and regroup \(1\) hundred-thousand. 7. Add the hundred-thousands: \(0 + 0 + 1 + 0 + 0 + 1 = 2\). Therefore, the sum is \(210{,}290\). 8. Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(2\); thousands to ten-thousands: \(3\); ten-thousands to hundred-thousands: \(1\)

\(210{,}290\) Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(2\); thousands to ten-thousands: \(3\); ten-thousands to hundred-thousands: \(1\)

- Round each addend to the nearest thousand for the estimate. - To check a sum with three addends, subtract two addends one at a time. - Align digits with the same place value.

1. Add by place value. Ones: \(5+0+5=10\), so write \(0\) and regroup \(1\) ten. Tens: \(0+9+0+1=10\), so write \(0\) and regroup \(1\) hundred. Hundreds: \(4+9+6+1=20\), so write \(0\) and regroup \(2\) thousands. Thousands: \(7+3+8+2=20\), so write \(0\) and regroup \(2\) ten-thousands. Ten-thousands: \(2+1+2=5\). Therefore, the sum is \(50{,}000\). 2. Estimate by rounding to the nearest thousand: \(27{,}000+14{,}000+9000=50{,}000\). 3. Check with inverse operations: \(50{,}000-8605=41{,}395\), and \(41{,}395-13{,}990=27{,}405\). 4. Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(2\); thousands to ten-thousands: \(2\)

The sum is \(50{,}000\). a) Estimate: \(27{,}000+14{,}000+9000=50{,}000\) b) Inverse check: \(50{,}000-8605=41{,}395\) and \(41{,}395-13{,}990=27{,}405\) Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(2\); thousands to ten-thousands: \(2\)

- Include every leg of the trip. - Align ones, tens, hundreds, and thousands before adding. - A column total can require regrouping more than \(1\) into the next place.

1. Stack \(3450\), \(3420\), \(3650\), and \(9520\) with matching place values. 2. Ones: \(0+0+0+0=0\). 3. Tens: \(5+2+5+2=14\). Write \(4\) tens and regroup \(1\) hundred. 4. Hundreds: \(4+4+6+5+1=20\). Write \(0\) hundreds and regroup \(2\) thousands. 5. Thousands: \(3+3+3+9+2=20\). Write \(0\) thousands and regroup \(2\) ten-thousands. 6. The total is \(20{,}040\) miles. 7. Carry record: tens to hundreds: \(1\); hundreds to thousands: \(2\); thousands to ten-thousands: \(2\)

\(20{,}040\) miles Carry record: tens to hundreds: \(1\); hundreds to thousands: \(2\); thousands to ten-thousands: \(2\)

- Begin in the ones column and work from right to left. - Include any regrouped amount in the next column. - When a column total is greater than \(9\), regroup to the next place value. - Substitute the missing digit and check the completed addition.

1. In part a, \(4 + 8 = 12\), so write \(2\) and regroup \(1\) ten. The tens column must satisfy \(2 + \square + 1 = 8\), so \(\square = 5\). 2. In part b, \(7 + 5 = 12\), so write \(2\) and regroup \(1\) ten. The tens column must satisfy \(\square + 6 + 1 = 12\), so \(\square = 5\). 3. In part c, \(9 + 4 = 13\), so write \(3\) and regroup \(1\) ten. Then \(8 + 3 + 1 = 12\), so write \(2\) and regroup \(1\) hundred. The hundreds column must satisfy \(1 + \square + 1 = 6\), so \(\square = 4\). 4. In part d, the ones column must satisfy \(6 + \square = 11\), so \(\square = 5\), and \(1\) ten is regrouped. Then \(7 + 5 + 1 = 13\), and \(4 + 3 + 1 = 8\), which verifies the completed addition. 5. Carry record: a) ones to tens: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\) | d) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(\square = 5\) b) \(\square = 5\) c) \(\square = 4\) d) \(\square = 5\) Carry record: a) ones to tens: \(1\) | b) ones to tens: \(1\); tens to hundreds: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\) | d) ones to tens: \(1\); tens to hundreds: \(1\)

- Add each column from right to left. - Track every regrouped value from one column to the next. - Compare your column sums with the displayed result.

1. Check the columns from right to left. In the ones place, \(5 + 8 + 9 = 22\), so write \(2\) and regroup \(2\) tens. 2. In the tens place, \(4 + 7 + 2 + 2 = 15\), so write \(5\) and regroup \(1\) hundred. 3. In the hundreds place, the shown work used \(2 + 3 + 1 = 6\) and forgot the regrouped hundred. The correct sum is \(2 + 3 + 1 + 1 = 7\). 4. Therefore, the correct result is \(752\), not \(652\).

a) The regrouped \(1\) hundred from the tens place was not added in the hundreds place. b) The correct sum is \(752\).

- Begin in the ones column. - Include any regrouped amount when you move to the next column. - In each column, use the known digits and the result digit to find the missing digit. - Work from right to left.

1. In part a, \(8 + 7 = 15\), so write \(5\) and regroup \(1\) ten. The tens column must satisfy \(5 + \square + 1 = 12\), so \(\square = 6\), and \(1\) hundred is regrouped. The hundreds column gives \(3 + 2 + 1 = 6\). The completed equation is \(358 + 267 = 625\). 2. In part b, the ones column must satisfy \(9 + \square = 12\), so \(\square = 3\), and \(1\) ten is regrouped. The tens column gives \(4 + 8 + 1 = 13\), so the missing result digit is \(3\), and \(1\) hundred is regrouped. The hundreds column must satisfy \(\square + 1 + 1 = 7\), so \(\square = 5\). The completed equation is \(549 + 183 = 732\).

a) \(358 + 267 = 625\) b) \(549 + 183 = 732\)

- Begin in the ones column. - When a column total is \(10\) or more, include the regrouped amount in the next column. - Check your digits by completing the addition from right to left.

1. In the ones column, \(6 + \square = 13\), so the missing ones digit is \(7\), and \(1\) ten is regrouped. 2. In the tens column, \(\square + 7 + 1 = 12\), so the missing tens digit is \(4\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(4 + 2 + 1 = 7\), so the missing result digit is \(7\). 4. The completed equation is \(446 + 277 = 723\).

\(446 + 277 = 723\)

- In the ones column, which digit added to \(8\) gives a total with a ones digit of \(3\)? - Include the regrouped amount in the next column. - Work from right to left through each place-value column.

1. In the ones column, \(\square + 8 = 13\), so the missing ones digit is \(5\), and \(1\) ten is regrouped. 2. In the tens column, \(9 + 0 + 1 = 10\), so the missing tens digit in the result is \(0\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(5 + \square + 1 = 11\), so the missing hundreds digit in the second addend is \(5\), the result digit is \(1\), and \(1\) thousand is regrouped. 4. The regrouped \(1\) becomes the thousands digit. The completed equation is \(595 + 508 = 1103\).

\(595 + 508 = 1103\)

- In each column, add all three digits and any regrouped amount. - Use the result digit to determine how much the missing digit must contribute.

1. In the ones column, \(3 + \square + 5 = 14\), so the missing ones digit is \(6\), and \(1\) ten is regrouped. 2. In the tens column, \(2 + 4 + \square + 1 = 11\), so the missing tens digit is \(4\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(1 + \square + 2 + 1 = 8\), so the missing hundreds digit is \(4\). 4. The completed equation is \(123 + 446 + 245 = 814\).

\(123 + 446 + 245 = 814\)

- Work from the ones place toward the thousands place. - In each column, record the digit you write and any amount you carry to the next place. - The missing hundreds digit must fit both the shown hundreds result and the amount carried onward.

1. Ones: \(6+4+5=15\). Write \(5\) and carry \(1\) ten. 2. Tens: \(8+3+8+1=20\). Write \(0\) and carry \(2\) hundreds. 3. Hundreds: \(7+5+\square+2\) must end in \(1\), so the missing hundreds digit is \(7\). The total is \(21\) hundreds; carry \(2\) thousands. 4. Thousands: \(1+2+2=5\). The completed addition is \(1786+2534+785=5105\).

\(1786+2534+785=5105\). Carrying: \(1\) ten, \(2\) hundreds, and \(2\) thousands.

- Start with the rightmost column that can help you find a missing digit. - Remember to include any carried ten or hundred in the next column. - Part c) asks you to explain the carrying, not merely find the digit.

1. a) The ones column needs \(7+5=12\), so the missing digit is \(7\). Then \(347+215=562\). 2. b) Carry from \(6+8=14\). In the tens column, \(8+3+1=12\), so the missing digit is \(8\). Then \(486+138=624\). 3. c) The ones give \(7+3=10\), carrying \(1\) ten. The tens give \(5+4+1=10\), carrying \(1\) hundred. The missing hundreds digit is \(3\), and \(257+343=600\). 4. d) The ones and tens each carry \(1\). The hundreds column is \(4+3+1=8\), so \(482+319=801\).

a) \(7\) b) \(8\) c) \(3\); carry from ones to tens, then tens to hundreds d) \(4\)

- Work from the ones column toward the left. - Use each visible result digit to determine the missing addend digit. - Verify the digit by adding the completed numbers.

1. For a), the ones column gives \(5 + 7 = 12\), so regroup \(1\). The tens column gives \(4 + 6 + 1 = 11\), so regroup \(1\). In the hundreds column, \(3 + \square + 1 = 11\), so the missing digit is \(7\). 2. For b), the ones column gives \(5 + 1 = 6\). In the tens column, \(4 + \square\) must equal \(12\), so the missing digit is \(8\), and regroup \(1\). The remaining columns verify the result. 3. For c), the ones column gives \(9 + 2 = 11\), so regroup \(1\). In the tens column, \(\square + 3 + 1 = 12\), so the missing digit is \(8\). 4. Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\) | b) tens to hundreds: \(1\); hundreds to thousands: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

a) \(7\) b) \(8\) c) \(8\) Carry record: a) ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\) | b) tens to hundreds: \(1\); hundreds to thousands: \(1\) | c) ones to tens: \(1\); tens to hundreds: \(1\)

- Begin in the ones column and work from right to left. - Decide whether each column requires regrouping. - Use the result digit and the known digits to determine each missing digit.

1. In the ones column, \(5 + \square = 12\), so the missing ones digit in the second addend is \(7\), and \(1\) ten is regrouped. 2. In the tens column, \(\square + 3 + 1 = 10\), so the missing tens digit in the first addend is \(6\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(4 + 2 + 1 = 7\), so the missing result digit is \(7\). 4. The completed equation is \(465 + 237 = 702\).

\(465 + 237 = 702\)

- Keep track of regrouping from one column to the next. - Check your digits by completing the entire addition.

1. In the ones column, \(3 + 9 = 12\), so write \(2\) and regroup \(1\) ten. 2. In the tens column, \(8 + \square + 1 = 12\), so the missing tens digit in the second addend is \(3\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(\square + 3 + 1 = 5\), so the missing hundreds digit in the first addend is \(1\). 4. The completed equation is \(183 + 339 = 522\).

\(183 + 339 = 522\)

- Begin in the ones column. - Regroup when a column total is greater than \(9\). - Use each result digit to determine the missing digit in that column.

1. In the ones column, \(\square + 4\) must have a ones digit of \(0\). Therefore, \(6 + 4 = 10\), so the missing ones digit is \(6\), and \(1\) ten is regrouped. 2. In the tens column, \(7 + 5 + 1 = 13\), so the missing tens digit in the result is \(3\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(0 + \square + 1 = 9\), so the missing hundreds digit in the second addend is \(8\). 4. The completed equation is \(76 + 854 = 930\).

\(76 + 854 = 930\)

- Work from right to left. - When a result digit is \(0\), check whether the column total must be \(10\). - Carry each regrouped value into the next column.

1. The ones and tens columns each total \(10\), so each writes \(0\) and regroups \(1\). 2. In the hundreds column, \(0 + 9 + 1 = 10\). The completed equation is \(5034 + 966 = 6000\). 3. Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\)

\(5034 + 966 = 6000\) Carry record: ones to tens: \(1\); tens to hundreds: \(1\); hundreds to thousands: \(1\)

- Notice that the result has more digits than either addend. - Include the regrouped value as you move left.

1. In the ones column, \(\square + 8\) must end in \(2\). Therefore, the missing ones digit is \(4\), because \(4 + 8 = 12\). Write \(2\) and regroup \(1\) ten. 2. In the tens column, \(4 + 7 + 1 = 12\), so the missing result digit is \(2\). Write \(2\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 1\) must equal \(10\), so the missing hundreds digit is \(9\). Write \(0\) and regroup \(1\) thousand. The completed equation is \(44 + 978 = 1022\).

\(44 + 978 = 1022\)

- Which digit added to \(9\) gives a sum with a ones digit of \(3\)? - Include the regrouped \(1\) in the tens column.

1. In the ones column, \(\square + 9 = 13\), so the missing ones digit is \(4\), and \(1\) ten is regrouped. 2. In the tens column, \(\square + 0 + 1 = 10\), so the missing tens digit is \(9\), and \(1\) hundred is regrouped. 3. The regrouped \(1\) becomes the hundreds digit. The completed equation is \(94 + 9 = 103\). 4. Carry record: ones to tens: \(1\); tens to hundreds: \(1\)

\(94 + 9 = 103\) Carry record: ones to tens: \(1\); tens to hundreds: \(1\)

- Begin in the ones column. - Record the regrouped amount above the tens column.

1. In the ones column, \(\square + 7 = 11\), so the missing ones digit in the first addend is \(4\), and \(1\) ten is regrouped. 2. In the tens column, \(8 + \square + 1 = 16\), so the missing tens digit in the second addend is \(7\), and \(1\) hundred is regrouped. 3. The completed equation is \(84 + 77 = 161\).

\(84 + 77 = 161\)

- Work systematically from right to left. - Include each regrouped amount in the next column.

1. In the ones column, \(\square + 5 = 10\), so the missing ones digit is \(5\), and \(1\) ten is regrouped. 2. In the tens column, \(6 + 3 + 1 = 10\), so the missing tens digit in the result is \(0\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(0 + \square + 1 = 7\), so the missing hundreds digit in the second addend is \(6\). 4. The completed equation is \(65 + 635 = 700\).

\(65 + 635 = 700\)

- The ones result is \(0\). What must happen when the two ones digits are added? - Include what was carried into the tens column. - Use the shown thousands digit to check the regrouping from hundreds.

1. Ones: the missing digit plus \(7\) must end in \(0\). It is \(3\), with \(1\) ten carried. 2. Tens: \(3+6+1=10\). The missing tens digit in the result is \(0\), with \(1\) hundred carried. 3. Hundreds: the missing hundreds digit plus the carried \(1\) must make \(10\) hundreds, so the digit is \(9\). 4. The completed addition is \(33+967=1000\).

\(33+967=1000\). The result's missing tens digit is \(0\); the missing addend digits are \(3\) and \(9\).

- Start with the ones column and use the visible result digit. - Decide whether the ones column must regroup before you work in the tens column. - Include any regrouped value when solving for the tens digit.

1. In the ones column, the missing digit plus \(6\) must produce a ones digit of \(2\) and a regrouped ten, so the missing ones digit is \(6\). 2. In the tens column, \(4+\square+1\) must produce a tens digit of \(0\) and regroup \(1\) hundred, so the missing tens digit is \(5\). 3. The completed equation is \(46+56=102\).

\(46+56=102\)

- Use the visible ones digit of the result to constrain the missing addend digit. - Decide whether the ones column must regroup. - Carry that regrouped amount into the tens column.

1. In the ones column, the missing digit plus \(8\) must end in \(6\) and regroup \(1\) ten, so the missing ones digit is \(8\). 2. In the tens column, \(9+0+1=10\), so the missing tens digit in the result is \(0\), and \(1\) hundred is regrouped. 3. The regrouped \(1\) becomes the hundreds digit. The completed equation is \(98+8=106\).

\(98+8=106\)

- Use the zeros in the result to reason about each column total. - Keep track of whether a regrouped \(1\) enters the tens column. - Check the completed equation with the standard algorithm.

1. In the ones column, the missing digit plus \(3\) must end in \(0\) and regroup \(1\) ten, so the missing ones digit is \(7\). 2. In the tens column, \(3+\square+1\) must end in \(0\) and regroup \(1\) hundred, so the missing tens digit is \(6\). 3. The completed equation is \(37+63=100\).

\(37+63=100\)

- Begin in the ones column and work from right to left. - Include any regrouped amount in the next column. - Use the result digit to determine the missing digit in each column.

1. In the ones column, \(7 + \square = 12\), so the missing ones digit is \(5\), and \(1\) ten is regrouped. 2. In the tens column, \(\square + 8 + 1 = 12\), so the missing tens digit is \(3\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(4 + 2 + 1 = 7\). The completed equation is \(437 + 285 = 722\).

The missing digits are \(3\) and \(5\). The completed equation is \(437 + 285 = 722\).

- Check one column at a time, beginning with the ones column. - Include any regrouped amount when you move to the next column.

1. In the ones column, \(\square + 8 = 12\), so the missing ones digit in the first addend is \(4\), and \(1\) ten is regrouped. 2. In the tens column, \(5 + \square + 1 = 11\), so the missing tens digit in the second addend is \(5\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(\square + 4 + 1 = 8\), so the missing hundreds digit in the first addend is \(3\). 4. The completed equation is \(354 + 458 = 812\).

The missing digits are \(3\), \(4\), and \(5\). The completed equation is \(354 + 458 = 812\).

- Add the known digits in each column first. - Determine how much is needed to reach the visible result digit. - With three addends, remember to include the regrouped value.

1. In the ones column, \(2 + 5 + 3 = 10\), so write \(0\) and regroup \(1\) ten. 2. In the tens column, \(2 + 4 + \square + 1 = 10\), so the missing tens digit is \(3\). Write \(0\) and regroup \(1\) hundred. 3. In the hundreds column, \(1 + \square + 2 + 1 = 8\), so the missing hundreds digit is \(4\). The completed equation is \(122 + 445 + 233 = 800\).

The missing digits are \(4\) and \(3\). The completed equation is \(122 + 445 + 233 = 800\).

- Solve one column at a time from right to left. - Record each regrouped value before moving left.

1. In the ones column, \(1 + \square = 7\), so the missing ones digit is \(6\). 2. In the tens column, \(4 + 8 = 12\), so write \(2\) and regroup \(1\) hundred. 3. In the hundreds column, \(5 + \square + 1\) must end in \(4\), so the missing hundreds digit is \(8\). Write \(4\) and regroup \(1\) thousand. 4. In the thousands column, \(1 + 3 + 1 = 5\), so the missing result digit is \(5\). The completed equation is \(1541 + 3886 = 5427\).

\(1541 + 3886 = 5427\)

- Track the chain of regrouping through the columns. - Use the visible result digits to determine each unknown.

1. In the ones column, \(6 + 3 = 9\), so the ones digit is \(9\). 2. In the tens column, \(7 + 4 = 11\), so the missing result digit is \(1\), and regroup \(1\) hundred. 3. In the hundreds column, \(5 + 5 + 1 = 11\), so write \(1\) and regroup \(1\) thousand. 4. In the thousands column, \(4 + \square + 1\) must end in \(0\), so the missing digit is \(5\). Write \(0\) and regroup \(1\) ten-thousand. The completed equation is \(34{,}576 + 45{,}543 = 80{,}119\).

\(34{,}576 + 45{,}543 = 80{,}119\)

- Move from right to left one column at a time. - When a column sum is \(10\) or more, carry the regrouped value into the next column.

1. In the ones column, \(5 + 2 = 7\), so the ones digit is \(7\). 2. In the tens column, \(4 + \square\) must end in \(0\), so the missing tens digit is \(6\). Write \(0\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 7 + 1\) must end in \(1\), so the missing hundreds digit is \(3\). Write \(1\) and regroup \(1\) thousand. 4. In the thousands column, \(3 + \square + 1 = 8\), so the missing thousands digit is \(4\). The completed equation is \(3345 + 4762 = 8107\).

\(3345 + 4762 = 8107\)

- The missing ones digit must combine with \(8\) to produce a result ending in \(1\). - Check for regrouping after every column.

1. In the ones column, \(\square + 8\) must end in \(1\), so the missing ones digit is \(3\). Write \(1\) and regroup \(1\) ten. 2. In the tens column, \(2 + \square + 1\) must end in \(1\), so the missing tens digit is \(8\). Write \(1\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 4 + 1\) must end in \(2\), so the missing hundreds digit is \(7\). Write \(2\) and regroup \(1\) thousand. 4. In the thousands column, \(6 + 2 + 1 = 9\). The completed equation is \(6723 + 2488 = 9211\).

\(6723 + 2488 = 9211\)

- With three addends, add the known digits first and determine the remaining amount. - Keep any regrouped value when moving to the next column.

1. In the ones column, \(2 + \square + 5\) must end in \(1\), so the missing ones digit is \(4\). Write \(1\) and regroup \(1\) ten. 2. In the tens column, \(\square + 4 + 0 + 1\) must end in \(3\), so the missing tens digit is \(8\). Write \(3\) and regroup \(1\) hundred. 3. In the hundreds column, \(1 + 2 + \square + 1 = 7\), so the missing hundreds digit is \(3\). The completed equation is \(182 + 244 + 305 = 731\).

\(182 + 244 + 305 = 731\)

- Add all known digits in a column first. - Determine how much is needed to reach the visible result digit or the next ten.

1. In the ones column, \(5 + \square + 4\) must end in \(6\), so the missing ones digit is \(7\). Write \(6\) and regroup \(1\) ten. 2. In the tens column, \(\square + 1 + 2 + 1\) must end in \(2\), so the missing tens digit is \(8\). Write \(2\) and regroup \(1\) hundred. 3. In the hundreds column, \(3 + 1 + 2 + 1 = 7\). The completed equation is \(385 + 117 + 224 = 726\).

\(385 + 117 + 224 = 726\)

- Recalculate the addition from right to left. - Compare your regrouped values with the displayed work.

1. In the ones place, \(7 + 5 = 12\). Write \(2\) and regroup \(1\) ten. 2. In the tens place, \(1 + 5 + 5 = 11\). Write \(1\) and regroup \(1\) hundred. 3. In the hundreds place, the shown work used \(4 + 2 = 6\) and forgot the regrouped hundred. The correct hundreds sum is \(1 + 4 + 2 = 7\). 4. Therefore, \(457 + 255 = 712\).

The regrouped \(1\) hundred from the tens place was not added in the hundreds place. The correct sum is \(712\).

- Check every regrouped value carefully. - Verify the addition one place at a time.

1. Ones: \(4+8=12\). Write \(2\) and regroup \(1\) ten. 2. Tens: \(3+7+1=11\). Write \(1\) and regroup \(1\) hundred. 3. Hundreds: \(2+6+1=9\). The shown sum has \(8\) in this place because the regrouped hundred was omitted. 4. Thousands: \(1+5=6\). The correct sum is \(6912\).

The regrouped \(1\) was omitted in the hundreds place. The correct sum is \(6912\).

- Begin in the ones column. - Regroup whenever a column total is \(10\) or more. - Check the completed addition from right to left.

1. In the ones column, \(\square + 5 = 12\), so the missing ones digit in the first addend is \(7\), and \(1\) ten is regrouped. 2. In the tens column, \(7 + \square + 1 = 13\), so the missing tens digit in the second addend is \(5\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(2 + 1 + 1 = 4\), which matches the result. 4. The completed equation is \(277 + 155 = 432\).

\(277 + 155 = 432\)

- Work from the ones column to the hundreds column. - Record and include each regrouped amount.

1. In the ones column, \(4 + \square = 12\), so the missing ones digit is \(8\), and \(1\) ten is regrouped. 2. In the tens column, \(3 + 7 + 1 = 11\), so write \(1\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 2 + 1 = 7\), so the missing hundreds digit is \(4\). 4. The completed equation is \(434 + 278 = 712\).

\(434 + 278 = 712\)

- Calculate carefully one column at a time. - Notice whether each empty box is in an addend or in the result.

1. In the ones column, \(5 + 8 = 13\), so write \(3\) and regroup \(1\) ten. 2. In the tens column, \(4 + \square + 1\) must end in \(2\), so the missing tens digit is \(7\). Write \(2\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 7 + 1\) must end in \(1\), so the missing hundreds digit is \(3\). Write \(1\) and regroup \(1\) thousand. 4. In the thousands column, \(2 + 8 + 1 = 11\), so the missing result digit is \(1\), and regroup \(1\) ten-thousand. The completed equation is \(12{,}345 + 38{,}778 = 51{,}123\).

\(12{,}345 + 38{,}778 = 51{,}123\)

- A result containing several zeros often means consecutive column totals of \(10\). - Track regrouping consistently from right to left.

1. In the ones column, \(\square + 3 = 10\), so the missing ones digit is \(7\). Write \(0\) and regroup \(1\) ten. 2. In the tens column, \(9 + \square + 1 = 10\), so the missing tens digit is \(0\). Write \(0\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 2 + 1 = 10\), so the missing hundreds digit is \(7\). Write \(0\) and regroup \(1\) thousand. 4. In the thousands column, \(7 + 1 + 1 = 9\). The completed equation is \(7797 + 1203 = 9000\).

\(7797 + 1203 = 9000\)

- Add all three digits in each column. - Regroup whenever the column total reaches \(10\) or more.

1. In the ones column, \(\square + 7 + 2 = 9\), so the missing ones digit is \(0\). No regrouping is needed. 2. In the tens column, \(5 + 6 + \square\) must end in \(9\), so the missing tens digit is \(8\), because \(5 + 6 + 8 = 19\). Write \(9\) and regroup \(1\) hundred. 3. In the hundreds column, \(4 + 3 + 1 + 1 = 9\). The completed equation is \(450 + 367 + 182 = 999\).

\(450 + 367 + 182 = 999\)

- Add the known digits in each column first. - Then determine the missing digit needed to match the result digit or reach the next multiple of \(10\).

1. In the ones column, \(4 + \square + 3 = 10\), so the missing ones digit is \(3\), and \(1\) ten is regrouped. 2. In the tens column, \(\square + 7 + 2 + 1 = 10\), so the missing tens digit is \(0\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(2 + \square + 1 + 1 = 7\), so the missing hundreds digit is \(3\). 4. The completed equation is \(204 + 373 + 123 = 700\).

\(204 + 373 + 123 = 700\)

- Remember that regrouping changes the total in the next column. - Use the known tens digit in the second addend and work from right to left.

1. In the ones column, \(3 + 9 = 12\), so write \(2\) and regroup \(1\) ten. 2. In the tens column, \(\square + 5 + 1 = 12\), so the missing tens digit is \(6\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(1 + 4 + 1 = 6\), which matches the result. 4. The completed equation is \(163 + 459 = 622\). 5. Carry record: ones to tens: \(1\); tens to hundreds: \(1\)

\(163 + 459 = 622\) Carry record: ones to tens: \(1\); tens to hundreds: \(1\)

- Add each place again carefully. - Check that every regrouped value is included.

1. Ones: \(3+5+8=16\). Write \(6\) and regroup \(1\) ten. 2. Tens: \(2+4+7+1=14\). Write \(4\), not \(3\), and regroup \(1\) hundred. 3. Hundreds: \(1+0+6+1=8\). 4. The correct sum is \(846\).

The error is in the tens place. The correct sum is \(846\).

- Read both rows from the chart before adding. - Work from right to left and include each regrouped unit in the next column. - A column needs an exchange whenever its total is \(10\) or more.

1. The rows represent \(3685\) and \(2479\). 2. Ones: \(5+9=14\), so exchange \(10\) ones for \(1\) ten. 3. Tens: \(8+7+1=16\), so exchange \(10\) tens for \(1\) hundred. 4. Hundreds: \(6+4+1=11\), so exchange \(10\) hundreds for \(1\) thousand. 5. Thousands: \(3+2+1=6\). The sum is \(6164\).

a) Ones, tens, and hundreds b) \(6164\)

- Read the two addends from the chart. - Remember that a regrouped ten changes the next column's total. - Check the tens column only after accounting for the regrouped ten from the ones.

1. The rows represent \(456\) and \(378\). 2. Ones: \(6+8=14\), so regroup \(1\) ten. 3. Tens: \(5+7+1=13\), so the tens column also needs regrouping. 4. Hundreds: \(4+3+1=8\). 5. Maya is not correct. Both the ones and tens columns require regrouping, and the sum is \(834\).

Maya is not correct. The ones and tens columns both need regrouping, and the sum is \(834\).

- Begin in the ones column, even when there are three addends. - The regrouped amount can be \(2\) when a column total is \(20\) or more. - Add the known digits in a column, then determine what is needed to match the result digit.

1. In problem 1, the ones column must satisfy \(4 + \square + 8 = 19\), so the missing ones digit is \(7\), and \(1\) ten is regrouped. The tens column must satisfy \(\square + 2 + 5 + 1 = 11\), so the missing tens digit is \(3\), and \(1\) hundred is regrouped. The hundreds column gives \(1 + 2 + 2 + 1 = 6\). The completed equation is \(134 + 227 + 258 = 619\). 2. In problem 2, the ones column must satisfy \(2 + 5 + \square = 10\), so the missing ones digit is \(3\), and \(1\) ten is regrouped. The tens column must satisfy \(4 + \square + 7 + 1 = 20\), so the missing tens digit is \(8\), and \(2\) hundreds are regrouped. The hundreds column gives \(3 + 1 + 2 + 2 = 8\). The completed equation is \(342 + 185 + 273 = 800\).

1) \(134 + 227 + 258 = 619\) 2) \(342 + 185 + 273 = 800\)

- Work from the ones place to the left. - Record every regrouped unit. - When a result digit is \(0\), consider whether the column total is \(10\).

1. Ones: \(7 + \square = 15\), so the missing digit is \(8\). Write \(5\) and regroup \(1\) ten. 2. Tens: \(\square + 8 + 1 = 10\), so the missing digit is \(1\). Write \(0\) and regroup \(1\) hundred. 3. Hundreds: \(4 + \square + 1 = 12\), so the missing digit is \(7\). Write \(2\) and regroup \(1\) thousand. 4. Thousands: \(\square + 3 + 1 = 10\), so the missing digit is \(6\). Write \(0\) and regroup \(1\) ten-thousand. 5. Ten-thousands: \(2 + 1 + 1 = 4\), so the missing result digit is \(4\). The completed addition is \(26{,}417 + 13{,}788 = 40{,}205\).

The missing digits, read from top to bottom and left to right, are \(6\), \(1\), \(7\), \(8\), and \(4\). The completed addition is \(26{,}417 + 13{,}788 = 40{,}205\).

- Add the digits in each column, including any regrouped amount. - A column total may be greater than \(19\), so check the regrouping value. - Begin with the ones place.

1. Ones: \(\square + 7 + 2 = 14\), so the missing digit is \(5\). Regroup \(1\) ten. 2. Tens: \(1 + 4 + \square + 3 = 12\), so the missing digit is \(4\). Regroup \(1\) hundred. 3. Hundreds: \(1 + 8 + 9 + \square = 19\), so the missing digit is \(1\). Regroup \(1\) thousand. 4. Thousands: \(1 + \square + 4 = 9\), so the missing digit is \(4\). The completed addition is \(4845 + 4947 + 132 = 9924\).

\(\begin{array}{r} 4845 \\ + 4947 \\ + 132 \\ \hline 9924 \end{array}\)

- Add one aligned place-value column at a time, starting with the ones. - Record the full column total before writing its ones digit and carrying the remaining tens to the next column. - Include each incoming carry in the next column total. - Use rounded addends only for the final reasonableness check.

1. a) Ones: \(2+9+5=16\), write \(6\) and carry \(1\). Tens: \(0+8+5+1=14\), write \(4\) and carry \(1\). Hundreds: \(3+7+4+1=15\), write \(5\) and carry \(1\). Thousands: \(4+5+0+1=10\), write \(0\) and carry \(1\). Ten-thousands: \(2+4+2+1=9\), with no carry. Hundred-thousands: \(1+0+3=4\). The exact sum is \(490{,}546\). A reasonable estimate is \(120{,}000+50{,}000+320{,}000=490{,}000\). 2. b) Ones: \(5+0+9+1+1=16\), carry \(1\). Tens: \(0+4+8+0+2+1=15\), carry \(1\). Hundreds: \(4+3+7+9+1+1=25\), carry \(2\). Thousands: \(1+2+4+1+0+2=10\), carry \(1\). Ten-thousands: \(3+1+0+5+0+1=10\), carry \(1\). Hundred-thousands: \(2+0+1+0+3+1=7\). The exact sum is \(700{,}556\). A reasonable estimate is \(230{,}000+10{,}000+100{,}000+50{,}000+300{,}000=690{,}000\).

a) Pre-regrouping column totals, ones to hundred-thousands: \(16,14,15,10,9,4\). Carries: \(1,1,1,1,0\). Exact sum: \(490{,}546\). Estimate: \(490{,}000\). b) Pre-regrouping column totals, ones to hundred-thousands: \(16,15,25,10,10,7\). Carries: \(1,1,2,1,1\). Exact sum: \(700{,}556\). Estimate: \(690{,}000\).

- Work one column at a time from right to left. - Include any regrouped value from the previous column. - Use the visible result digit to determine each unknown digit, then verify the completed addition.

1. For a), start in the ones column. Since \(7 + 5 = 12\), the missing ones digit is \(5\), the result digit is \(2\), and \(1\) is regrouped. In the tens column, \(2 + 8 + 1 = 11\), so the missing result tens digit is \(1\) and \(1\) is regrouped. In the hundreds column, \(6 + 5 + 1 = 12\), so the missing digit in the first addend is \(6\) and \(1\) is regrouped. In the thousands column, \(4 + 8 + 1 = 13\), so the missing digit in the second addend is \(8\) and \(1\) is regrouped. In the ten-thousands column, \(3 + 1 + 1 = 5\), so the leading result digit is \(5\). The completed equation is \(34{,}627 + 18{,}585 = 53{,}212\). 2. For b), the ones column gives \(5 + 3 + 9 = 17\), so the missing ones digit is \(3\) and \(1\) is regrouped. The tens column gives \(4 + 8 + 5 + 1 = 18\), so the missing tens digit is \(5\) and \(1\) is regrouped. The hundreds column gives \(5 + 3 + 2 + 1 = 11\), so the missing hundreds digit is \(5\) and \(1\) is regrouped. The thousands column gives \(2 + 4 + 1 + 1 = 8\), so the leading result digit is \(8\). The completed equation is \(2545 + 4383 + 1259 = 8187\).

a) \(34{,}627 + 18{,}585 = 53{,}212\) b) \(2545 + 4383 + 1259 = 8187\)

- Begin in the ones column. - Include any regrouped value in the next column. - Verify all missing digits by recalculating the completed addition.

1. For a), the ones column gives \(7 + 5 = 12\), so the missing result digit is \(2\) and \(1\) is regrouped. The tens column gives \(7 + 9 + 1 = 17\), so the missing tens digit in the first addend is \(7\) and \(1\) is regrouped. The hundreds column gives \(4 + 4 + 1 = 9\), so the missing hundreds digit in the second addend is \(4\). The thousands column gives \(4 + 3 = 7\), so the missing thousands digit in the first addend is \(4\). The ten-thousands column gives \(3 + 3 = 6\), so the missing leading digit in the second addend is \(3\). The completed equation is \(34{,}477 + 33{,}495 = 67{,}972\). 2. For b), the ones column gives \(8 + 7 = 15\), so the missing result digit is \(5\) and \(1\) is regrouped. The tens column gives \(0 + 0 + 1 = 1\), so the missing tens digit in the first addend is \(0\). The hundreds column gives \(3 + 2 = 5\), so the missing hundreds digit in the second addend is \(2\). The thousands column gives \(4 + 7 = 11\), so the missing thousands digit in the first addend is \(4\) and \(1\) is regrouped. The ten-thousands column gives \(4 + 2 + 1 = 7\), so the missing leading digit in the second addend is \(2\). The completed equation is \(44{,}308 + 27{,}207 = 71{,}515\).

a) \(34{,}477 + 33{,}495 = 67{,}972\) b) \(44{,}308 + 27{,}207 = 71{,}515\)

- Record each regrouped value above the next column. - In each column, determine the digit needed to produce the visible result digit.

1. In the ones column, \(2 + 9 = 11\), so write \(1\) and regroup \(1\) ten. In the tens column, \(1 + 8 + 1 = 10\), so the missing tens digit is \(8\), write \(0\), and regroup \(1\) hundred. In the hundreds column, \(7 + 8 + 1 = 16\), so the missing hundreds digit is \(7\), write \(6\), and regroup \(1\) thousand. In the thousands column, \(5 + 8 + 1 = 14\), so the missing thousands digit is \(8\), write \(4\), and regroup \(1\) ten-thousand. In the ten-thousands column, \(4 + 3 + 1 = 8\), so the missing result digit is \(8\). 2. The completed equation is \(45{,}712 + 38{,}889 = 84{,}601\).

\(45{,}712 + 38{,}889 = 84{,}601\)

- Use the visible digits in the result to constrain each unknown digit. - Include the regrouped value from the previous column before solving the next one.

1. In the ones column, \(8 + 4 = 12\), so write \(2\) and regroup \(1\) ten. In the tens column, \(1 + 9 + 1 = 11\), so the missing tens digit is \(9\), write \(1\), and regroup \(1\) hundred. In the hundreds column, \(2 + 9 + 1 = 12\), so write \(2\) and regroup \(1\) thousand. In the thousands column, \(4 + 8 + 1 = 13\), so the missing thousands digit is \(8\), write \(3\), and regroup \(1\) ten-thousand. In the ten-thousands column, \(6 + 5 + 1 = 12\), so the missing ten-thousands digit is \(6\), write \(2\), and regroup \(1\) hundred-thousand. In the hundred-thousands column, \(6 + 1 + 1 = 8\), so the missing result digit is \(8\). 2. The completed equation is \(664{,}218 + 158{,}994 = 823{,}212\).

\(664{,}218 + 158{,}994 = 823{,}212\)

- Align addends with different numbers of digits by place value. - Use the result digit and any regrouped value to determine each unknown.

1. In the ones column, \(4 + 5 + 1 = 10\), so the missing ones digit is \(1\), write \(0\), and regroup \(1\) ten. In the tens column, \(6 + 5 + 1 + 1 = 13\), so the missing tens digit in the second addend is \(5\), write \(3\), and regroup \(1\) hundred. In the hundreds column, \(5 + 8 + 5 + 1 = 19\), so the missing hundreds digit in the first addend is \(5\), write \(9\), and regroup \(1\) thousand. The thousands column gives \(2 + 1 + 0 + 1 = 4\), and the ten-thousands digit is \(1\). 2. The completed equation is \(12{,}564 + 1855 + 511 = 14{,}930\).

\(12{,}564 + 1855 + 511 = 14{,}930\)

- Begin in the ones column. - Pay close attention to regrouping in the tens and hundreds columns.

1. In the ones column, \(5 + 2 + 0 = 7\), so the missing ones digit is \(0\). In the tens column, \(4 + 4 + 4 = 12\), so write \(2\) and regroup \(1\) hundred. In the hundreds column, \(3 + 8 + 4 + 1 = 16\), so the missing hundreds digit in the first addend is \(3\), write \(6\), and regroup \(1\) thousand. In the thousands column, \(2 + 1 + 0 + 1 = 4\), so the missing leading result digit is \(4\). 2. The completed equation is \(2345 + 1842 + 440 = 4627\).

\(2345 + 1842 + 440 = 4627\)

- Begin with the ones column. - Record each regrouped value before solving the next column. - Ask which digit is needed to produce the visible result digit.

1. In the ones column, \(6 + 4 = 10\), so write \(0\) and regroup \(1\) ten. In the tens column, \(5 + 3 + 1 = 9\), so the missing tens digit in the second addend is \(3\). In the hundreds column, \(8 + 3 = 11\), so the missing hundreds digit in the first addend is \(8\), write \(1\), and regroup \(1\) thousand. In the thousands column, \(2 + 7 + 1 = 10\), so the missing result thousands digit is \(0\) and \(1\) ten-thousand is regrouped. In the ten-thousands column, \(4 + 3 + 1 = 8\). 2. The completed equation is \(42{,}856 + 37{,}334 = 80{,}190\).

\(42{,}856 + 37{,}334 = 80{,}190\)

- In this calculation, nearly every column produces regrouping. - When a column totals exactly \(10\), write \(0\) and regroup \(1\).

1. In the ones column, \(7 + 3 = 10\), so write \(0\) and regroup \(1\) ten. The tens and hundreds columns also each total \(10\), so write \(0\) and regroup \(1\) each time. In the thousands column, \(4 + 5 + 1 = 10\), so the missing thousands digit in the second addend is \(5\), write \(0\), and regroup \(1\) ten-thousand. In the ten-thousands column, \(3 + 6 + 1 = 10\), so the missing ten-thousands digit in the first addend is \(3\), write \(0\), and regroup \(1\) hundred-thousand. In the hundred-thousands column, \(3 + 4 + 1 = 8\). 2. The completed equation is \(334{,}157 + 465{,}843 = 800{,}000\).

\(334{,}157 + 465{,}843 = 800{,}000\)

- Add all four digits in each column. - Regrouping with four addends can produce \(2\) or more. - Use the visible result digit to determine each unknown.

1. In the ones column, \(4 + 5 + 6 + 8 = 23\), so write \(3\) and regroup \(2\) tens. 2. In the tens column, \(8 + 2 + 5 + 6 + 2 = 23\), so the missing result digit is \(3\), and regroup \(2\) hundreds. 3. In the hundreds column, the missing digit must satisfy \(\square + 9 + 1 + 2 + 2 = 17\). Therefore, the missing digit is \(3\), and regroup \(1\) thousand. 4. In the thousands column, \(0 + 7 + 4 + 6 + 1 = 18\), so write \(8\) and regroup \(1\) ten-thousand. 5. In the ten-thousands column, \(6 + 3 + 4 + 1 = 14\), so write \(4\) and regroup \(1\) hundred-thousand. The completed equation is \(60{,}384 + 37{,}925 + 44{,}156 + 6268 = 148{,}733\).

\(60{,}384 + 37{,}925 + 44{,}156 + 6268 = 148{,}733\)

- The same standard-algorithm reasoning applies to six-digit numbers. - Keep every addend aligned by place value. - The final regrouping helps determine the leading result digit.

1. In the ones column, \(6 + 1 + 7 = 14\), so write \(4\) and regroup \(1\) ten. 2. In the tens column, \(5 + 0 + 6 + 1 = 12\), so write \(2\) and regroup \(1\) hundred. 3. In the hundreds column, \(4 + 9 + \square + 1\) must end in \(9\). Therefore, the missing digit is \(5\), because \(4 + 9 + 5 + 1 = 19\). Write \(9\) and regroup \(1\) thousand. 4. In the thousands column, \(3 + 8 + 4 + 1 = 16\), so the missing result digit is \(6\), and regroup \(1\) ten-thousand. 5. In the ten-thousands column, \(2 + 7 + 3 + 1 = 13\), so write \(3\) and regroup \(1\) hundred-thousand. Then \(1 + 2 + 1 = 4\) in the hundred-thousands column. The completed equation is \(123{,}456 + 78{,}901 + 234{,}567 = 436{,}924\).

\(123{,}456 + 78{,}901 + 234{,}567 = 436{,}924\)

- Add the known digits in each column first. - Determine how much is needed to reach the visible result digit after accounting for any regrouped value. - Check each completed equation from ones through hundreds.

1. For a), the ones column gives \(3 + 5 + 7 = 15\), so write \(5\) and regroup \(1\). In the tens column, \(2 + \square + 6 + 1 = 15\), so the missing tens digit is \(6\), and regroup \(1\). In the hundreds column, \(1 + 4 + \square + 1 = 9\), so the missing hundreds digit is \(3\). Thus, \(123 + 465 + 367 = 955\). 2. For b), the ones column gives \(5 + 4 + 1 = 10\), so write \(0\) and regroup \(1\). In the tens column, \(2 + \square + 1 + 1 = 10\), so the missing tens digit is \(6\), and regroup \(1\). In the hundreds column, \(\square + 3 + 1 + 1 = 9\), so the missing hundreds digit is \(4\). Thus, \(425 + 364 + 111 = 900\).

a) \(123 + 465 + 367 = 955\) b) \(425 + 364 + 111 = 900\)

- Begin in the ones column. - Write each regrouped value before moving to the next column.

1. In the ones column, \(1 + 8 = 9\), so the missing result digit is \(9\). 2. In the tens column, \(\square + 6\) must equal \(10\), so the missing digit is \(4\). Write \(0\) and regroup \(1\) hundred. 3. In the hundreds column, \(3 + 7 + 1 = 11\), so write \(1\) and regroup \(1\) thousand. 4. In the thousands column, \(4 + \square + 1\) must equal \(12\), so the missing digit is \(7\). Write \(2\) and regroup \(1\) ten-thousand. The completed equation is \(14{,}341 + 37{,}768 = 52{,}109\).

\(14{,}341 + 37{,}768 = 52{,}109\)

- When a column sum is greater than \(9\), regroup into the next column. - Use each visible result digit to solve for the unknown.

1. In the ones column, \(\square + 5 = 7\), so the missing ones digit is \(2\). 2. In the tens column, \(8 + 3 = 11\), so write \(1\) and regroup \(1\) hundred. 3. In the hundreds column, \(7 + \square + 1\) must end in \(6\). Therefore, the missing hundreds digit is \(8\), because \(7 + 8 + 1 = 16\). Write \(6\) and regroup \(1\) thousand. 4. The thousands column gives \(8 + 3 + 1 = 12\), and the ten-thousands column gives \(4 + 2 + 1 = 7\). The completed equation is \(48{,}782 + 23{,}835 = 72{,}617\).

\(48{,}782 + 23{,}835 = 72{,}617\)

- With three addends, a regrouped value can be \(2\) or greater. - Solve the columns from right to left and include every regrouped value.

1. In the ones column, \(6 + \square + 5\) must equal \(20\), so the missing ones digit is \(9\). Write \(0\) and regroup \(2\) tens. 2. In the tens column, \(3 + 8 + 7 + 2 = 20\), so write \(0\) and regroup \(2\) hundreds. 3. In the hundreds column, \(\square + 6 + 1 + 2\) must equal \(10\), so the missing hundreds digit is \(1\). Write \(0\) and regroup \(1\) thousand. 4. In the thousands column, \(3 + 4 + 2 + 1 = 10\), so write \(0\) and regroup \(1\) ten-thousand. The completed equation is \(3136 + 4689 + 2175 = 10{,}000\).

\(3136 + 4689 + 2175 = 10{,}000\)

- Start in the ones column and determine which digit makes that column end in \(0\). - Each time a column creates a regrouped \(1\), include it in the next column before solving the next missing digit.

1. In the ones column, \(\square + 6 = 10\), so the missing ones digit is \(4\). Write \(0\) and regroup \(1\) ten. 2. In the tens column, \(2 + 7 + 1 = 10\), so write \(0\) and regroup \(1\) hundred. 3. In the hundreds column, \(5 + \square + 1 = 10\), so the missing hundreds digit is \(4\). Write \(0\) and regroup \(1\) thousand. 4. In the thousands column, \(\square + 2 + 1 = 10\), so the missing thousands digit is \(7\). Write \(0\) and regroup \(1\) ten-thousand. 5. In the ten-thousands column, \(8 + 1 + 1 = 10\), producing the leading \(1\). The completed equation is \(87{,}524 + 12{,}476 = 100{,}000\).

\(87{,}524 + 12{,}476 = 100{,}000\)

- A result digit of \(0\) often indicates a column total of \(10\). - Check every column from right to left and retain each regrouped value.

1. In the ones column, \(1 + \square = 10\), so the missing ones digit is \(9\). Write \(0\) and regroup \(1\) ten. 2. In the tens column, \(5 + 4 + 1 = 10\), so write \(0\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 7 + 1 = 10\), so the missing hundreds digit is \(2\). Write \(0\) and regroup \(1\) thousand. 4. In the thousands column, \(4 + \square + 1 = 9\), so the missing thousands digit is \(4\). The completed equation is \(4251 + 4749 = 9000\).

\(4251 + 4749 = 9000\)

- Work backward from the visible hundreds digit: what must have happened in the tens column? - Consider the greatest total two single digits can make in the tens column. - Use the carry into the tens column to constrain the two ones digits.

1. The hundreds digit of the sum is \(1\), so the tens column must regroup exactly \(1\) hundred. 2. Therefore the two tens digits, together with any carry from the ones column, must total \(19\). Two digits can total at most \(18\), so there must be a carry of \(1\) from the ones column and both tens digits must be \(9\). 3. A carry of \(1\) from the ones column means the two ones digits total at least \(10\). Because the visible ones digit of the sum is \(8\), their total must be \(18\), forcing both ones digits to be \(9\). 4. The completed equation is \(99+99=198\).

The four missing digits are all \(9\), so \(99+99=198\). The tens column forces a regrouped hundred, and the resulting carry pattern forces both ones digits to be \(9\) as well.

- Work from right to left. - Record each regrouped value before solving the next unknown digit.

1. In the ones column, \(\square + 7\) must end in \(5\), so the missing ones digit is \(8\). Write \(5\) and regroup \(1\) ten. 2. In the tens column, \(8 + \square + 1\) must end in \(4\), so the missing tens digit is \(5\). Write \(4\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 2 + 1\) must end in \(0\), so the missing hundreds digit is \(7\). Write \(0\) and regroup \(1\) thousand. 4. In the thousands column, \(3 + \square + 1 = 8\), so the missing thousands digit is \(4\). The completed equation is \(3788 + 4257 = 8045\).

\(3788 + 4257 = 8045\)

- A round result such as \(6000\) often requires regrouping through several columns. - Verify every completed column and the final total.

1. In the ones column, \(6 + \square = 10\), so the missing ones digit is \(4\). Write \(0\) and regroup \(1\) ten. 2. In the tens column, \(7 + 2 + 1 = 10\), so write \(0\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 3 + 1 = 10\), so the missing hundreds digit is \(6\). Write \(0\) and regroup \(1\) thousand. 4. In the thousands column, \(2 + \square + 1 = 6\), so the missing thousands digit is \(3\). The completed equation is \(2676 + 3324 = 6000\).

\(2676 + 3324 = 6000\)

- Some unknown digits depend on regrouping created by earlier columns. - Test a possible digit only if it remains consistent with all columns to its left.

1. In the ones column, \(\square + 4\) must end in \(1\), so the missing ones digit is \(7\). Write \(1\) and regroup \(1\) ten. 2. In the tens column, \(6 + \square + 1\) must end in \(0\), so the missing tens digit is \(3\). Write \(0\) and regroup \(1\) hundred. 3. In the hundreds column, \(\square + 8 + 1\) must end in \(4\), so the missing hundreds digit is \(5\). Write \(4\) and regroup \(1\) thousand. 4. In the thousands column, \(4 + \square + 1 = 7\), so the missing thousands digit is \(2\). The completed equation is \(4567 + 2834 = 7401\).

\(4567 + 2834 = 7401\)