How it works
- 2: last digit is even. 5: last digit is 0 or 5. 10: last digit is 0.
- 4: last two digits form a multiple of 4. 8: last three digits form a multiple of 8.
- 3: digit sum is a multiple of 3. 9: digit sum is a multiple of 9.
- 11: alternating sum of digits (+ − + − …) is a multiple of 11.
For a number like 6, 12 or 72, split it into factors that share nothing (6 = 2 × 3, 12 = 3 × 4, 72 = 8 × 9) and check both rules. Splitting 12 as 2 × 6 does not work, because 2 and 6 overlap. To count multiples in a range, use (largest multiple − smallest multiple) ÷ k + 1.
Divisibility Rules practice problems with solutions
These problems come from Mathyessy's question bank. Try each one before opening the solution.
What is the smallest digit d so that 234d is divisible by 3?
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Answer: 0
The digit sum is 2 + 3 + 4 + d = 9 + d. This is a multiple of 3 when d is 0, 3, 6 or 9, so the smallest is 0.
What is the smallest 4 digit number divisible by 2 and 3 and 5?
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Answer: 1020
Divisible by 2, 3 and 5 means divisible by 30. 30 × 33 = 990 is too small and 30 × 34 = 1020.
How many integers between 100 and 200 are divisible by 3 but not by 9?
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Answer: 22
Multiples of 3 from 102 to 198: (198 − 102) ÷ 3 + 1 = 33. Multiples of 9 from 108 to 198: (198 − 108) ÷ 9 + 1 = 11. Every multiple of 9 is a multiple of 3, so 33 − 11 = 22.
Six digit number 5a312b is divisible by 72. What is the maximum possible value of a plus b?
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Answer: 16
72 = 8 × 9. For 8, the last three digits 12b must be a multiple of 8, so b = 0 or 8. For 9, the digit sum 11 + a + b must be a multiple of 9. With b = 0, a = 7 (sum 7); with b = 8, a = 8 (sum 16). The maximum is 16.
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Related topics
Related practice sets in the app
All Number Theory topics · Contest math practice by topic · MATHCOUNTS practice · AMC 8 practice