Long Division with Multi-Digit Divisors — Answer Key
Part A: Fill in the Blank
Write the missing word or number on each line.
1. When you divide 368 by 17, the quotient is 21 with a remainder of 11.
Grade 5 long division: 17 times 21 equals 357, and 368 minus 357 equals 11, so the remainder is 11.
2. In the division 425 divided by 12, the quotient is 35 and the remainder is 5.
Grade 5 long division: 12 times 35 equals 420, and 425 minus 420 equals 5, giving a remainder of 5.
3. The remainder must always be less than the divisor.
Grade 5 rule: the remainder must be less than the divisor; otherwise the quotient could be increased by one more division step.
4. When 514 is divided by 22, the quotient is 23 and the remainder is 8.
Grade 5 long division: 22 times 23 equals 506, and 514 minus 506 equals 8, so the remainder is 8.
5. If you divide 750 by 13, the quotient is 57 with a remainder of 9.
Grade 5 long division: 13 times 57 equals 741, and 750 minus 741 equals 9, giving a remainder of 9.
6. When 600 is divided by 14, the quotient is 42 and the remainder is 12.
Grade 5 long division: 14 times 42 equals 588, and 600 minus 588 equals 12, so the remainder is 12.
7. In the division 289 divided by 16, the quotient is 18 with a remainder of 1.
Grade 5 long division: 16 times 18 equals 288, and 289 minus 288 equals 1, giving a remainder of 1.
8. To check 458 divided by 19 equals 24 R 2, multiply 19 by 24 and add the remainder.
Grade 5 check: 19 times 24 equals 456, and 456 plus 2 equals 458, so the remainder is 2.
9. When 803 is divided by 25, the quotient is 32 and the remainder is 3.
Grade 5 long division: 25 times 32 equals 800, and 803 minus 800 equals 3, so the remainder is 3.
Part B: Matching
Match each item on the left to the correct answer on the right.
1. Match each item to its correct answer.
368 divided by 17
→ 21 R 11
23 R 8
425 divided by 12
→ 35 R 5
21 R 11
514 divided by 22
→ 23 R 8
57 R 9
750 divided by 13
→ 57 R 9
35 R 5
Grade 5 long division produces a quotient with a remainder less than the divisor; verify each match by multiplying back and adding the remainder.