Practice Questions
Concept Explanation
A number is divisible by another when division leaves zero remainder. Placement papers often test quick divisibility checks without long division.
Important Divisibility Rules
| Divisor | Rule | Example |
|---|---|---|
| 2 | Last digit is even | 846 |
| 3 | Sum of digits divisible by 3 | 582 → 15 |
| 4 | Last two digits divisible by 4 | 2,316 → 16 |
| 5 | Last digit 0 or 5 | 375 |
| 6 | Divisible by 2 and 3 | 714 |
| 8 | Last three digits divisible by 8 | 4,024 → 024 |
| 9 | Sum of digits divisible by 9 | 891 → 18 |
| 11 | Alternate digit sum difference is 0 or multiple of 11 | 1,331 → 0 |
Composite Divisors
12 needs divisibility by 3 and 4. 15 needs 3 and 5. 18 needs 2 and 9. Break composite tests into prime rules.
Rule for 7
Double the last digit and subtract from the number formed by remaining digits. Repeat until you can judge divisibility by 7.
Example: 343 → 34 − 6 = 28, and 28 ÷ 7 = 4, so 343 is divisible by 7.
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Q1 EasyIf the number 517*324 is completely divisible by 3, then the smallest whole number in the place of * will be:ExplanationSum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = (22 + x), which must be divisible by 3. The smallest x that works is 2.
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Q2 EasyWhich one of the following numbers is exactly divisible by 11?Explanation(4 + 5 + 2) − (1 + 6 + 3) = 1, not divisible by 11. (2 + 6 + 4) − (4 + 5 + 2) = 1, not divisible by 11. (4 + 6 + 1) − (2 + 5 + 3) = 1, not divisible by 11. (4 + 6 + 1) − (2 + 5 + 4) = 0, so 415624 is divisible by 11.
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Q3 EasyWhat smallest number should be added to 4456 so that the sum is completely divisible by 6?Explanation4456 ÷ 6 leaves remainder 4. Required number = (6 − 4) = 2.
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Q4 EasyIf the number 5 * 2 is divisible by 6, then what digit should replace *?Explanation6 = 3 × 2. The number 5x2 is always even. For divisibility by 3, (5 + x + 2) must be divisible by 3, so x = 2.
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Q5 MediumWhich of the following number is divisible by 24?Explanation24 = 3 × 8, where 3 and 8 are co-prime. 35718, 63810 and 537804 fail the divisibility-by-8 test on their last three digits. For 3125736, digit sum = 27 (divisible by 3) and 736 is divisible by 8, so the number is divisible by 24.
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Q6 MediumIf the number 481 * 673 is completely divisible by 9, then the smallest whole number in place of * will be:ExplanationSum of digits = (4 + 8 + 1 + x + 6 + 7 + 3) = (29 + x), which must be divisible by 9. The smallest whole number x is 7.
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Q7 MediumIf the number 97215 * 6 is completely divisible by 11, then the smallest whole number in place of * will be:ExplanationGiven number = 97215x6. (6 + 5 + 2 + 9) − (x + 1 + 7) = (14 − x), which must be divisible by 11. So x = 3.
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Q8 MediumIf the product 4864 × 9P2 is divisible by 12, then what is the value of the digit P?Explanation4864 is divisible by 4, so 9P2 must be divisible by 3. (9 + P + 2) must be divisible by 3, hence P = 1.
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Q9 MediumWhat is the largest four-digit number that is exactly divisible by 88?ExplanationLargest 4-digit number = 9999. On dividing 9999 by 88, remainder = 55. Required number = 9999 − 55 = 9944.
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Q10 MediumWhat least number must be added to 1056, so that the sum is completely divisible by 23?Explanation1056 ÷ 23 leaves remainder 21. Required number = (23 − 21) = 2.
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Q11 MediumWhat is the smallest digit x that should replace the blank in 63x5 so that the number becomes divisible by 15?ExplanationDivisibility by 15 requires divisibility by both 3 and 5. The last digit is 5, so only the divisibility-by-3 test matters: 6 + 3 + x + 5 = 14 + x must be a multiple of 3. The smallest option is x = 1.
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Q12 MediumUsing the standard divisibility rule for 7, which of the following numbers is divisible by 7?ExplanationFor 203: double the last digit and subtract from the remaining number → 20 − 6 = 14, which is divisible by 7. The other options fail this test.
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Q13 Hard476 ** 0 is divisible by both 3 and 11. The non-zero digits in the hundred's and ten's places are respectively:ExplanationLet the number be 476xy0. Then (17 + x + y) must be divisible by 3, and (x − y − 3) must be 0 or a multiple of 11. Solving gives x = 8 and y = 5.
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Q14 HardWhich of the following numbers is divisible by each one of 3, 7, 9 and 11?Explanation639 is not divisible by 7. 2079 passes divisibility tests for 3, 7, 9 and 11. The other options fail at least one test.
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DS1 Easy
You must decide whether an integer x passes the divisibility-by-6 test. Is the integer x divisible by 6?
Statement Ix is divisible by 2.
Statement IIx is divisible by 3.
ExplanationA number is divisible by 6 only when it is divisible by both 2 and 3. Each statement alone is insufficient (e.g. 4 is divisible by 2 but not 6; 9 is divisible by 3 but not 6). Both together guarantee divisibility by 6. -
DS2 Medium
A software firm validates employee codes only when the code is divisible by 12. Is the positive integer N divisible by 12?
Statement IN is divisible by 4.
Statement IIN is divisible by 3.
ExplanationDivisibility by 12 requires both 4 and 3 (since gcd(4,3)=1). Statement I alone allows N = 8 (not divisible by 3). Statement II alone allows N = 6 (not divisible by 4). Together, N must be divisible by lcm(4,3) = 12.
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DS3 Medium
While checking remainders in a number-theory drill, what is the remainder when the positive integer N is divided by 9?
Statement IThe sum of the digits of N is 27.
Statement IIN is a three-digit number less than 500.
ExplanationIf the digit sum is 27, the remainder on division by 9 is 27 mod 9 = 0. Statement I alone fixes the remainder. Statement II only restricts the size of N and does not fix the remainder. -
DS4 Hard
An HR database stores five-digit IDs of the form 63a5b. Is the five-digit number 63a5b divisible by 35?
Statement Ia + b = 8.
Statement IIb = 3.
ExplanationA number divisible by 35 must end in 0 or 5 (divisible by 5). Statement II gives b = 3, so the number cannot be divisible by 5 — hence not divisible by 35. That is enough to answer the question (No). Statement I alone does not fix b to 0 or 5.