Free Math Worksheets

- 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 \(800\).

a) \(823\) b) \(800\) c) \(800\)

- 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 \(710\).

a) \(759\) b) \(769\) c) \(691\) d) \(710\)

- 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: \(950\,\text{g} + 420\,\text{g} + 380\,\text{g} + 210\,\text{g} = 1960\,\text{g}\).

The packed school bag weighs \(1960\,\text{g}\).

- 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.

a) \(739\) b) \(739\) The results are equal.

- 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 \(758\).

a) \(780\) b) \(758\)

- 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 \(902\).

a) \(622\) b) \(661\) c) \(902\)

- 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, \(577 + 288 = 865\).

a) \(834\) b) \(823\) c) \(865\)

- 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 \(775\).

a) \(836\) b) \(775\)

- Identify the three forest sections and their tree counts. - Align each addend by place value. - Include any regrouped value in the next column.

1. The three sections contain \(2450\), \(1820\), and \(3135\) new trees. 2. Add the ones: \(0 + 0 + 5 = 5\). 3. Add the tens: \(5 + 2 + 3 = 10\); write \(0\) and regroup \(1\) hundred. 4. Add the hundreds: \(4 + 8 + 1 + 1 = 14\); write \(4\) and regroup \(1\) thousand. 5. Add the thousands: \(2 + 1 + 3 + 1 = 7\). Therefore, \(2450 + 1820 + 3135 = 7405\).

The workers plant \(7405\) new trees altogether.

- 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.

a) \(4809\) b) \(4809\) c) \(4809\) All three results are equal.

- Does adding equipment increase or decrease the price? - Line up the place values when adding. - Add the base price and the extra cost.

1. Add the base price and the equipment cost: \(\$78{,}000 + \$4500 = \$82{,}500\).

The fully equipped boat costs \(\$82{,}500\).

- 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, \(82{,}056 + 9347 = 91{,}403\).

a) \(73{,}017\) b) \(91{,}403\)

- Include every leg of the trip. - Align ones, tens, hundreds, and thousands when adding. - Regroup as needed in the standard algorithm.

1. Add the four distances with place values aligned: \(3450 + 3420 + 3650 + 9520\). 2. Add step by step: \(3450 + 3420 = 6870\), \(6870 + 3650 = 10{,}520\), and \(10{,}520 + 9520 = 20{,}040\). 3. The total flight distance is \(20{,}040\) miles.

\(20{,}040\) miles

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

1. In the ones column, \(6 + * = 9\), 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 \(1211 + 2232 + 3334 = 6777\).

\(1211 + 2232 + 3334 = 6777\)

- 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 \(548 + 291 = 839\)

- Work from the ones column toward the hundreds column. - In which column do you record a regrouped \(1\)? - Choose a tens digit that keeps the tens total at \(9\) or less. - There is more than one correct answer in part c.

1. Add the ones: \(3 + 4 = 7\). 2. Add the tens: \(6 + 5 = 11\). Write \(1\) in the tens place and regroup \(1\) ten as \(1\) hundred. 3. Add the hundreds: \(4 + 2 + 1 = 7\). Therefore, \(463 + 254 = 717\). 4. To avoid regrouping from tens to hundreds, the new tens digit \(x\) must satisfy \(6 + x \le 9\), so \(x\) can be \(0, 1, 2\), or \(3\). 5. One possible new problem is \(463 + 234 = 697\).

a) \(463 + 254 = 717\) b) Regrouping occurs in the tens column because \(6 + 5 = 11\). Ten tens are regrouped as \(1\) hundred. c) One possible answer is \(463 + 234 = 697\). Tens digits \(0, 1, 2\), or \(3\) also produce valid new problems.

- 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 \(821\).

a) \(825\) b) \(463\) c) \(823\) d) \(821\)

- 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 \(852\).

a) \(770\) b) \(685\) c) \(852\)

- 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.

Lucas is not correct. The correct sum is \(710\).

- 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 \(168 + 644 = 812\).

1) \(425 + 356 = 781\) 2) \(168 + 644 = 812\)

- 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 \(770\).

a) Correct: \(273 + 154 + 362 = 789\) b) Incorrect: \(416 + 235 + 189 = 840\) c) Incorrect: \(127 + 438 + 205 = 770\)

- 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.

The sum is \(9401\). Check: \(9401-3789=5612\).

- 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 \(600 + 155 = 755\).

a) \(825\) b) \(755\)

- 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 \(247 + 485 = 732\).

- Check an addition equation by subtracting one addend from the stated sum. - The difference should equal the other addend. - Compare the check result with the original equation.

1. Check \(632 - 245 = 387\). Equation 1 is correct. 2. Check \(624 - 178 = 446\), not \(456\). Equation 2 is incorrect. The correct sum is \(456 + 178 = 634\). 3. Check \(861 - 567 = 294\). Equation 3 is correct. 4. Check \(911 - 743 = 168\). Equation 4 is correct.

1. Correct. 2. Incorrect. \(456 + 178 = 634\) 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: \(735 - 279 = 456\).

Sum: \(735\) Check: \(735 - 279 = 456\) or \(735 - 456 = 279\)

- 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, \(243 + 156 + 328 = 727\).

\(727\)

- 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, \(3425 + 2890 + 1565 + 980 = 8860\).

The library has \(8860\) items for children altogether.

- 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 \(16{,}688\).

\(16{,}688\)

- 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.

Equation 3 is incorrect. The correct sum is \(4100\).

- Add all three costs. - Watch the regrouping when using the standard addition algorithm. - Combining the round thousands first may help.

1. Add the first two costs: \(\$24{,}300 + \$8700 = \$33{,}000\). 2. Add the fencing cost: \(\$33{,}000 + \$3450 = \$36{,}450\).

The town spends \(\$36{,}450\) in total.

- 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.

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 \(800\).

- 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.

The sum is \(65{,}505\). Changing the order of the addends gives the same sum.

- 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: \(50{,}000+20{,}000=70{,}000\).

The total is \(70{,}000\). The grouped check is \(50{,}000+20{,}000=70{,}000\).

- 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, \(54{,}328+27{,}914=82{,}242\).

Ava is incorrect. The correct sum is \(82{,}242\).

- 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, \(5602+14{,}398+20{,}100=40{,}100\).

C) \(79{,}991\) D) \(40{,}100\)

- Align digits with the same place value when you add. - Watch for regrouping in each partial sum. - Think about whether parentheses change a sum when the order of the addends stays the same.

1. First, \(3456+2189=5645\): regroup in the ones and tens columns. Then \(5645+4072=9717\): regroup in the tens column. 2. First, \(2189+4072=6261\): regroup in the ones and tens columns. Then \(3456+6261=9717\): regroup in the tens column. 3. Both groupings give the same sum. Changing how addends are grouped does not change their sum.

Both methods give \(9717\). The sum does not change when the addends are grouped differently.

- 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 \(955{,}094\).

a) \(88{,}005\) b) \(955{,}094\)

- 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 \(4,6,4\). \(12{,}445+36{,}210+24{,}342=72{,}997\)

- Round each addend to a convenient ten-thousand or hundred-thousand for the estimate. - Align all exact addends by place value. - Compare the size of the exact sum with the estimate.

1. For a), align the addends. The column sums from ones through hundred-thousands are \(7\), \(7\), \(8\), \(9\), \(8\), and \(4\), so the exact sum is \(489{,}877\). One estimate is \(120{,}000 + 50{,}000 + 320{,}000 = 490{,}000\), which is close to the exact result. 2. For b), align all five addends. The column sums from ones through hundred-thousands are \(9\), \(8\), \(9\), \(8\), \(9\), and \(6\), so the exact sum is \(698{,}989\). One estimate is \(230{,}000 + 10{,}000 + 100{,}000 + 50{,}000 + 300{,}000 = 690{,}000\), which supports the reasonableness of the exact result.

a) Estimate: \(490{,}000\); exact sum: \(489{,}877\) b) Estimate: \(690{,}000\); exact sum: \(698{,}989\)

- 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.

The sum is \(379{,}105\).

- Align all digits by place value. - Record each regrouped value in the next column. - Reversing the order of the addends provides a useful check.

1. Add from top to bottom. First, \(47{,}283 + 15{,}906\): the ones sum is \(9\), the tens sum is \(8\), the hundreds sum is \(11\), the thousands sum including the regrouped \(1\) is \(13\), and the ten-thousands sum including the regrouped \(1\) is \(6\). This gives \(63{,}189\). Then add \(28{,}311\): regroup from the ones and tens columns, giving \(91{,}500\). 2. Check from bottom to top. First, \(28{,}311 + 15{,}906 = 44{,}217\), with regrouping from the hundreds and thousands columns. Then \(44{,}217 + 47{,}283 = 91{,}500\), with regrouping from the ones, tens, and thousands columns. 3. Both orders give the same sum.

The sum is \(91{,}500\).

- 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: \(30{,}000 + 60{,}000 = 90{,}000\).

a) \(90{,}000\) b) \(12{,}600 + 17{,}400 = 30{,}000\) and \(34{,}550 + 25{,}450 = 60{,}000\), so the total is \(90{,}000\).

- 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 \(210{,}290\).

\(210{,}290\)

- 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 \(41{,}395-13{,}990=27{,}405\).

The sum is \(50{,}000\). a) Estimate: \(27{,}000+14{,}000+9000=50{,}000\) b) Inverse check: \(50{,}000-8605=41{,}395\) and \(41{,}395-13{,}990=27{,}405\)

- 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 \(2 + * + 1 = 8\), so \(* = 5\). 2. In part b, \(7 + 5 = 12\), so write \(2\) and regroup \(1\) ten. The tens column must satisfy \(* + 6 + 1 = 12\), so \(* = 5\). 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 \(1 + * + 1 = 6\), so \(* = 4\). 4. In part d, the ones column must satisfy \(6 + * = 11\), so \(* = 5\), and \(1\) ten is regrouped. Then \(7 + 5 + 1 = 13\), and \(4 + 3 + 1 = 8\), which verifies the completed addition.

a) \(* = 5\) b) \(* = 5\) c) \(* = 4\) d) \(* = 5\)

- 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 \(5 + * + 1 = 12\), so \(* = 6\), 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 \(9 + * = 12\), so \(* = 3\), 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 \(* + 1 + 1 = 7\), so \(* = 5\). 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, \(6 + * = 13\), so the missing ones digit is \(7\), and \(1\) ten is regrouped. 2. In the tens column, \(* + 7 + 1 = 12\), 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, \(* + 8 = 13\), 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, \(5 + * + 1 = 11\), 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, \(3 + * + 5 = 14\), so the missing ones digit is \(6\), and \(1\) ten is regrouped. 2. In the tens column, \(2 + 4 + * + 1 = 11\), so the missing tens digit is \(4\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(1 + * + 2 + 1 = 8\), so the missing hundreds digit is \(4\). 4. The completed equation is \(123 + 446 + 245 = 814\).

\(123 + 446 + 245 = 814\)

- Work through each addition from the ones column to the hundreds column. - Include any regrouped amount in the next column. - You can also subtract the known addend from the sum to find the missing addend.

1. For part a, use the related subtraction equation: \(562 - 215 = 347\). The missing digit is \(7\). 2. For part b, \(624 - 138 = 486\). The missing digit is \(8\). 3. For part c, \(600 - 257 = 343\). The missing digit is \(3\). 4. For part d, \(801 - 319 = 482\). The missing digit is \(4\).

a) \(7\) b) \(8\) c) \(3\) 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 \(8\).

a) \(7\) b) \(8\) c) \(8\)

- 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, \(5 + * = 12\), so the missing ones digit in the second addend is \(7\), and \(1\) ten is regrouped. 2. In the tens column, \(* + 3 + 1 = 10\), 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, \(8 + * + 1 = 12\), so the missing tens digit in the second addend is \(3\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(* + 3 + 1 = 5\), 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, \(* + 4\) 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, \(0 + * + 1 = 9\), 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 \(5034 + 966 = 6000\).

\(5034 + 966 = 6000\)

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

\(94 + 9 = 103\)

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

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

\(84 + 77 = 161\)

- The ones-column total must have a ones digit of \(8\), so consider \(8\) and \(18\). - If the ones digits totaled \(8\), could two tens digits produce the required tens digit and hundreds digit? - Remember that two digits can total at most \(18\).

1. In the ones column, the two missing digits must add to a number with ones digit \(8\), so their sum is either \(8\) or \(18\). 2. If their sum were \(8\), there would be no regrouped ten. The two tens digits would then have to total \(19\) to produce a tens digit of \(9\) and regroup \(1\) hundred, but two digits can total at most \(18\). Therefore, the ones-column sum must be \(18\), so both ones digits are \(9\), and \(1\) ten is regrouped. 3. In the tens column, the two missing digits plus the regrouped \(1\) must total \(19\). Therefore, the two tens digits total \(18\), so both are \(9\). 4. The completed equation is \(99 + 99 = 198\).

\(99 + 99 = 198\)

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

1. In the ones column, \(* + 5 = 10\), 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, \(0 + * + 1 = 7\), so the missing hundreds digit in the second addend is \(6\). 4. The completed equation is \(65 + 635 = 700\).

\(65 + 635 = 700\)

- A result of \(1000\) requires regrouping in each place-value column. - Begin by finding the digit that combines with \(7\) to make \(10\).

1. In the ones column, \(* + 7 = 10\), so the missing ones digit is \(3\), and \(1\) ten is regrouped. 2. In the tens column, \(3 + 6 + 1 = 10\), so the missing tens digit in the result is \(0\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(0 + * + 1 = 10\), so the missing hundreds digit in the second addend is \(9\), and \(1\) thousand is regrouped. 4. The regrouped \(1\) becomes the thousands digit. The completed equation is \(33 + 967 = 1000\).

\(33 + 967 = 1000\)

- Which digit added to \(6\) gives \(12\)? - Include the regrouped \(1\) in the tens column.

1. In the ones column, \(* + 6 = 12\), so the missing ones digit in the first addend is \(6\), and \(1\) ten is regrouped. 2. In the tens column, \(4 + * + 1 = 10\), so the missing tens digit in the second addend is \(5\), and \(1\) hundred is regrouped. 3. The completed equation is \(46 + 56 = 102\).

\(46 + 56 = 102\)

- Which digit added to \(8\) gives \(16\)? - Keep track of the regrouped amount.

1. In the ones column, \(* + 8 = 16\), so the missing ones digit is \(8\), 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. The regrouped \(1\) becomes the hundreds digit. The completed equation is \(98 + 8 = 106\).

\(98 + 8 = 106\)

- First complete the ones column to make \(10\). - Include the regrouped \(1\) when you work in the tens column.

1. In the ones column, \(* + 3 = 10\), so the missing ones digit in the first addend is \(7\), and \(1\) ten is regrouped. 2. In the tens column, \(3 + * + 1 = 10\), so the missing tens digit in the second addend is \(6\), and \(1\) hundred is regrouped. 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, \(7 + * = 12\), so the missing ones digit is \(5\), and \(1\) ten is regrouped. 2. In the tens column, \(* + 8 + 1 = 12\), 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, \(* + 8 = 12\), so the missing ones digit in the first addend is \(4\), and \(1\) ten is regrouped. 2. In the tens column, \(5 + * + 1 = 11\), so the missing tens digit in the second addend is \(5\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(* + 4 + 1 = 8\), 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 \(354 + 458 = 812\).

- 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, \(* + 5 = 12\), so the missing ones digit in the first addend is \(7\), and \(1\) ten is regrouped. 2. In the tens column, \(7 + * + 1 = 13\), 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, \(4 + * = 12\), 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, \(* + 2 + 1 = 7\), so the missing hundreds digit is \(4\). 4. The completed equation is \(434 + 278 = 712\).

\(434 + 278 = 712\)

- 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, \(4 + * + 3 = 10\), so the missing ones digit is \(3\), and \(1\) ten is regrouped. 2. In the tens column, \(* + 7 + 2 + 1 = 10\), so the missing tens digit is \(0\), and \(1\) hundred is regrouped. 3. In the hundreds column, \(2 + * + 1 + 1 = 7\), 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, \(* + 5 + 1 = 12\), 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 \(163 + 459 = 622\).

\(163 + 459 = 622\)

- 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 \(846\).

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

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

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

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

- First add the numbers that are already complete. - Use the visible digits in the result to determine the missing addend and final digit.

1. In the ones column, \(1 + 2 + 3 + 4 = 10\), so the missing result digit is \(0\) and \(1\) ten is regrouped. In the tens column, \(1 + 2 + 3 + 4 + 1 = 11\), so write \(1\) and regroup \(1\) hundred. In the hundreds column, \(1 + 2 + 3 + 4 + 1 = 11\), so the missing hundreds digit in the fourth addend is \(4\), write \(1\), and regroup \(1\) thousand. In the thousands column, \(1 + 2 + 3 + 0 + 1 = 7\). 2. The completed equation is \(1111 + 2222 + 3333 + 444 = 7110\).

\(1111 + 2222 + 3333 + 444 = 7110\)

- 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. - With three addends, a regrouped value may be \(2\).

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) Missing digits: \(6\) and \(3\) b) Missing digits: \(4\) and \(6\)

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

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

- The result is \(100{,}000\). Consider what each column must total. - Track the regrouped \(1\) through every column.

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

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

- 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 \(44 + 978 = 1022\).

\(44 + 978 = 1022\)

- 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 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 \(2676 + 3324 = 6000\).

\(2676 + 3324 = 6000\)

- 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 \(385 + 117 + 224 = 726\).

\(385 + 117 + 224 = 726\)

- 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 \(4567 + 2834 = 7401\).

\(4567 + 2834 = 7401\)

- Calculate carefully one column at a time. - Notice whether each star 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 \(450 + 367 + 182 = 999\).

\(450 + 367 + 182 = 999\)