Division With Remainders Practice Problems With Answers

Below are 48 division with remainders problems in three graded levels, each with the quotient and remainder in the answer key. Level 1 uses single-digit divisors, Level 2 moves to two-digit divisors and larger dividends, and Level 3 covers negatives, missing values and problems that ask you to work backwards.

Work through a level, then check every answer against the key. The self-check method at the end catches a wrong answer without looking anything up, so you can mark your own work as you go.

If the method itself needs a refresher first, how to find the remainder covers five ways to get there, and long division with remainders walks through the written layout digit by digit.

How to Check Your Own Answers

Every correct answer satisfies two conditions. Test both before looking at the key.

Condition 1: the identity holds.

Divisor × Quotient + Remainder = Dividend

Condition 2: the remainder is smaller than the divisor.

0 ≤ Remainder < Divisor

Example. For 347 ÷ 12 you might get quotient 28, remainder 11.

12 × 28 + 11 = 336 + 11 = 347     ✓ identity holds
11 < 12                            ✓ remainder is in range

Both pass, so the answer is right.

A remainder equal to or larger than the divisor always means the quotient was too small. If 347 ÷ 12 gave you 27 remainder 23, the 23 is a clear signal: one more 12 fits, so raise the quotient to 28 and drop the remainder by 12.

Level 1: Basic (Problems 1 to 15)

Single-digit divisors. Give the quotient and the remainder for each.

1. 27 ÷ 4
2. 39 ÷ 5
3. 46 ÷ 7
4. 53 ÷ 6
5. 61 ÷ 8
6. 75 ÷ 9
7. 88 ÷ 7
8. 34 ÷ 3
9. 67 ÷ 5
10. 92 ÷ 8
11. 49 ÷ 6
12. 81 ÷ 7
13. 56 ÷ 9
14. 43 ÷ 4
15. 70 ÷ 3

Level 2: Intermediate (Problems 16 to 30)

Two-digit divisors and larger dividends. Long division is the reliable route here.

16. 347 ÷ 12
17. 512 ÷ 15
18. 689 ÷ 23
19. 1,204 ÷ 17
20. 875 ÷ 34
21. 2,360 ÷ 45
22. 999 ÷ 28
23. 1,536 ÷ 19
24. 4,021 ÷ 33
25. 763 ÷ 26
26. 3,105 ÷ 52
27. 1,847 ÷ 41
28. 5,000 ÷ 37
29. 2,748 ÷ 64
30. 918 ÷ 13

Level 3: Challenging (Problems 31 to 42)

Larger numbers, negative dividends, and problems that run the division backwards.

31. 12,345 ÷ 97
32. 100,000 ÷ 333
33. 87,654 ÷ 210
34. Find the quotient and remainder of −47 ÷ 6 using the convention that the remainder is never negative.
35. Find the quotient and remainder of −123 ÷ 11 under the same convention.
36. A division has divisor 14, quotient 23 and remainder 9. Find the dividend.
37. A division has dividend 500, quotient 17 and remainder 24. Find the divisor.
38. Find the remainder when 2¹⁰ is divided by 7.
39. Find the quotient and remainder of 1,000,000 ÷ 13.
40. Find the remainder when 3²⁰ is divided by 8.
41. Find the largest three-digit number that leaves a remainder of 4 when divided by 17.
42. Find the smallest number greater than 500 that is divisible by 23.

Mixed Review (Problems 43 to 48)

43. 156 ÷ 11
44. 2,048 ÷ 16
45. 731 ÷ 29
46. 9,999 ÷ 101
47. 640 ÷ 48
48. 1,111 ÷ 37

Answer Key

Level 1

Level 2

Problem 20 is worth a second look. The quotient and the remainder are both 25, which surprises people, but 34 × 25 + 25 = 850 + 25 = 875 and 25 < 34, so both conditions hold. There is no rule against them matching.

Problem 29 has a remainder of 60, which is large but still below the divisor 64, so it is correct.

Level 3

Mixed Review

Problems 44 and 46 come out exactly, with a remainder of 0. That means 16 is a factor of 2,048 and 101 is a factor of 9,999, which is an exact division.

Worked Solutions to Selected Problems

Problem 24: 4,021 ÷ 33

Long division, one digit at a time.

33 into 40    →  1,  and 33 × 1 = 33,   40 − 33 = 7
bring down 2  →  72
33 into 72    →  2,  and 33 × 2 = 66,   72 − 66 = 6
bring down 1  →  61
33 into 61    →  1,  and 33 × 1 = 33,   61 − 33 = 28

Reading the quotient digits gives 121, and the final leftover is 28.

Answer. 121 remainder 28. Check: 33 × 121 + 28 = 3,993 + 28 = 4,021 ✓ and 28 < 33 ✓

Problem 29: 2,748 ÷ 64

This one is worth doing carefully, because the remainder comes out unusually large.

Estimate first. Round the divisor down to 60. Then 2,748 ÷ 60 is roughly 45, so the true quotient is at most 45 and probably a little less, since 64 is bigger than 60.

Testing 45: 64 × 45 = 2,880, which overshoots 2,748. Come down.

Long division confirms it:

64 into 274   →  4,  and 64 × 4 = 256,   274 − 256 = 18
bring down 8  →  188
64 into 188   →  2,  and 64 × 2 = 128,   188 − 128 = 60

Answer. 42 remainder 60. Check: 64 × 42 + 60 = 2,688 + 60 = 2,748 ✓ and 60 < 64 ✓

A remainder of 60 looks wrong at a glance, but it passes both conditions. It simply means the division fell 4 short of another full copy of 64.

Problem 32: 100,000 ÷ 333

Estimate first. 333 × 300 = 99,900, which is just under 100,000.

100,000 − 99,900 = 100

Since 100 < 333, no further copies fit.

Answer. 300 remainder 100. Check: 333 × 300 + 100 = 100,000 ✓

Problem 34: −47 ÷ 6

Under the convention that the remainder is never negative, the quotient rounds down rather than toward zero.

-47 / 6 = -7.833…
round down (toward negative infinity)  →  quotient = -8

Then the remainder follows from the identity:

-47 = 6 × (-8) + r
-47 = -48 + r
  r = 1

Answer. −8 remainder 1. Both conditions hold: the identity checks out and 0 ≤ 1 < 6.

A different convention, used by C, Java and JavaScript, rounds the quotient toward zero instead and gives −7 remainder −5. That also satisfies the identity, which is why both answers appear in the wild. The two are compared in the remainder of negative numbers.

Problem 37: Find the divisor

Start from the identity and substitute what is known:

Dividend = Divisor × Quotient + Remainder

    500  = d × 17 + 24
    476  = 17d
      d  = 28

Answer. The divisor is 28. Check: 28 × 17 + 24 = 476 + 24 = 500 ✓ and 24 < 28 ✓

That last check matters. If the arithmetic had produced a divisor smaller than 24, the problem would have had no valid answer.

Problem 38: The remainder of 2¹⁰ ÷ 7

Two routes reach the same place.

Direct. 2¹⁰ = 1,024, and 7 × 146 = 1,022, so the remainder is 2.

By pattern. The remainders of the powers of 2 divided by 7 cycle every three steps:

2¹ ≡ 2,  2² ≡ 4,  2³ ≡ 1,  2⁴ ≡ 2,  2⁵ ≡ 4,  2⁶ ≡ 1, …

Since 10 = 3 × 3 + 1, the exponent 10 sits at the same position in the cycle as exponent 1, so the remainder is 2.

The second route is the one that still works when the power is far too large to write out, which is the subject of modular exponentiation.

Problem 41: Largest three-digit number leaving remainder 4 on division by 17

Start from the largest three-digit number and work down.

999 ÷ 17 = 58 remainder 13

A remainder of 13 is 9 more than the 4 we want, so subtract 9:

999 − 9 = 990
990 ÷ 17 = 58 remainder 4     ✓

Answer. 990. Adding 17 would give 1,007, which has four digits, so 990 is the largest.

Estimating the Quotient With a Two-Digit Divisor

Level 2 gets much easier once you stop guessing digits and start estimating.

Round the divisor to a friendly number, work out a rough quotient, then adjust.

Notice the direction of the error. Rounding the divisor up makes the estimate too small, so the real quotient is higher. Rounding it down makes the estimate too large, so the real quotient is lower. Knowing which way to adjust turns two or three trial multiplications into one.

Multiples of the divisor are also worth writing down once before you start. For 1,204 ÷ 17, listing 17, 34, 51, 68, 85 and so on makes every subtraction in the long division a lookup instead of a calculation.

Common Errors to Watch For

• A remainder equal to or larger than the divisor. The quotient is one too small. Raise it by 1 and reduce the remainder by the divisor.
• Rounding the quotient up. 46 ÷ 7 is 6 remainder 4, not 7. The quotient counts only complete copies.
• Reporting the decimal part as the remainder. A calculator shows 46 ÷ 7 as 6.571…, and 0.571 is not the remainder 4. Recovering it takes an extra step, covered in finding the remainder on a calculator.
• Losing a digit in long division. Every digit of the dividend must be brought down, including zeros. Missing one in 1,000,000 ÷ 13 changes the answer completely.
• Skipping the check. Multiplying back takes ten seconds and catches nearly everything.

What to Practise Next

Once the arithmetic is automatic, the harder skill is deciding what a remainder means in a real situation, since the same numbers can give three different final answers. That is covered in division with remainders word problems, which works through 30 problems sorted by interpretation.

To express a leftover as part of a share rather than as a whole number, see remainders as fractions and remainders as decimals.

The Remainder Calculator checks any of these divisions instantly, and the Long Division Calculator shows the full written working for the Level 2 and Level 3 problems.

Division Practice FAQ

How do I check a division with a remainder without an answer key?

Multiply the divisor by your quotient, add your remainder, and see whether you get back the dividend. Then confirm the remainder is smaller than the divisor. If both hold, the answer is correct, and no key is needed.

Can the remainder ever equal the quotient?

Yes. Problem 20 in this set gives 875 ÷ 34 = 25 remainder 25. Nothing forbids it. The only rule is that the remainder stays below the divisor.

What is the largest remainder possible?

One less than the divisor. Dividing by 12 can leave at most 11, and dividing by 7 can leave at most 6. A remainder equal to the divisor would mean another whole copy still fits.

Why is the remainder of a negative number sometimes positive?

Because two conventions exist for rounding the quotient. Rounding down keeps the remainder in the range 0 to divisor − 1, so −47 ÷ 6 gives −8 remainder 1. Rounding toward zero gives −7 remainder −5. Both satisfy the identity, and mathematics normally prefers the first.

How can I find a remainder without doing long division?

For divisors like 2, 5, 9, 10 and 11 there are digit-based shortcuts that give the answer in one step, collected in remainder shortcuts and tricks. For large powers, look for a repeating cycle rather than expanding the number.

How many practice problems should I do?

Enough that the two checks become automatic. Working Level 1 without errors, then Level 2 with the multiply-back check every time, is usually the point at which Level 3 stops feeling different.