Divisibility

1. Meaning and relevance in CSAT

If 84 objects can be arranged in groups of 7 without any object remaining, 84 is divisible by 7. Here, 7 is a divisor or factor, and 84 is a multiple of 7. The notation 7 | 84 means that 7 divides 84. Divisibility describes exact integer division, whereas ordinary division may produce a fraction or a non-zero remainder.

Every positive integer is divisible by 1 and itself. Zero is divisible by every non-zero integer because 0 = d × 0. However, division by zero is undefined, so zero must not be used as a divisor. Negative signs do not affect whether division is exact: divisibility can generally be checked using absolute values.

In CSAT, divisibility commonly appears within questions on missing digits, number formation, equal grouping, fractions, HCF and LCM, and remainders. Rather than carrying out long division, aspirants should recognise the structure of the divisor and apply the shortest valid test. These skills also support rapid elimination of answer options.

• Factor: a number that divides a given integer exactly.
• Multiple: the result of multiplying an integer by another integer.
• HCF: the greatest positive integer dividing each of the given integers.
• LCM: the smallest positive integer divisible by each of the given positive integers.

2. Essential divisibility tests

Decimal divisibility tests arise from place value. Since 10, 100 and higher powers of 10 are multiples of 2 and 5, the units digit decides divisibility by 2 or 5. Similarly, because 100 is divisible by 4, only the last two digits matter for 4; because 1,000 is divisible by 8, only the last three digits matter for 8.

For 3 and 9, add all digits and test the sum. For example, the digit sum of 57,834 is 27, so the number is divisible by both 3 and 9. This works because every power of 10 leaves remainder 1 when divided by 3 or 9. The number and its digit sum therefore have the same remainder for these divisors.

For 11, find the difference between the sums of digits in alternate positions. If the difference is zero or a positive or negative multiple of 11, the number is divisible by 11. For 918,082, the two sums are 17 and 11; their difference is 6, so the number is not divisible by 11.

For 7, remove the units digit, double it and subtract the result from the remaining number. Repeat if useful. Thus, 2,401 gives 240 − 2 = 238, then 23 − 16 = 7; hence 2,401 is divisible by 7. This transformation preserves divisibility, but not necessarily the original remainder.

• 2: the last digit is 0, 2, 4, 6 or 8.
• 5: the last digit is 0 or 5; 10: the last digit is 0.
• 4: the number formed by the last two digits is divisible by 4.
• 8: the number formed by the last three digits is divisible by 8.
• 25: the last two digits are 00, 25, 50 or 75.
• 125: the last three digits form a multiple of 125, including 000.

3. Composite divisors and prime-power reasoning

Prime factorisation gives the general method for divisibility. If d = pᵃqᵇ, where p and q are distinct primes, a number is divisible by d precisely when its prime factorisation contains at least a copies of p and b copies of q. For example, 72 = 2³ × 3², so divisibility by 72 requires divisibility by both 8 and 9.

A number divisible by a and b is always divisible by their LCM. It is necessarily divisible by their product only when a and b are coprime, meaning their HCF is 1. Thus, testing 3 and 4 establishes divisibility by 12. Testing 2 and 6 does not: 6 passes both tests but is not divisible by 12.

For a positive integer written as a product of prime powers, the number of positive divisors is found by adding 1 to each exponent and multiplying. Since 360 = 2³ × 3² × 5, it has (3 + 1)(2 + 1)(1 + 1) = 24 positive divisors. Each divisor is formed by independently selecting an allowed exponent for every prime.

• 6: test divisibility by 2 and 3.
• 12: test divisibility by 3 and 4.
• 15: test divisibility by 3 and 5.
• 18: test divisibility by 2 and 9.
• 24: test divisibility by 3 and 8.
• 36: test divisibility by 4 and 9.

4. Missing digits, remainders and counting multiples

For missing-digit questions, translate each divisibility requirement into a constraint. Suppose 4a2b is divisible by 45. Since 45 = 5 × 9, b must be 0 or 5, and 6 + a + b must be divisible by 9. If b = 0, a = 3; if b = 5, a = 7. The possible numbers are therefore 4,320 and 4,725.

The division algorithm states that N = dq + r, where d is positive and 0 ≤ r < d. The smallest non-negative amount to subtract from N to obtain a multiple of d is r. The smallest non-negative amount to add is d − r when r is non-zero, and zero when r = 0. If the question asks for a strictly positive addition and N is already divisible, the answer is d.

The number of positive multiples of d not exceeding N is floor(N/d), where floor means the greatest integer not exceeding the value. In the inclusive positive-integer interval L to U, the count is floor(U/d) − floor((L − 1)/d). Thus, multiples of 7 from 50 to 100 number 14 − 7 = 7.

For numbers divisible by a or b, add the separate counts and subtract the count divisible by LCM(a, b), since those numbers were counted twice. From 1 to 100, numbers divisible by 4 or 6 total 25 + 16 − 8 = 33. The overlap consists of multiples of 12.

• A digit must be an integer from 0 to 9; the leading digit of a multi-digit number cannot be zero.
• For divisibility by both divisors, count multiples of their LCM.
• For divisibility by exactly one of two divisors, subtract twice the overlap from the sum of the separate counts.

5. Efficient exam strategy and common errors

Start by factorising the divisor and inspecting the answer options. Conditions on the last digit or last two digits are usually quickest, followed by digit-sum and alternating-sum tests. In multiple-condition questions, apply the most restrictive condition first. Finally, verify surviving candidates against every requirement, including digit restrictions and interval endpoints.

Divisibility also behaves predictably under addition and subtraction. If d divides both a and b, it divides a + b and a − b, and more generally every integer combination ma + nb. Conversely, if d divides a and a + b, it must divide b. These properties can replace lengthy calculations with a short structural argument.

Do not cancel a factor from a divisibility claim without checking the conditions. If d divides ab and HCF(a, d) = 1, then d divides b. Without coprimality, this inference may fail: 6 divides 2 × 3, but 6 does not divide 3. Likewise, divisibility by 3 does not imply divisibility by 9, and an even digit sum does not imply an even number.

• Distinguish a remainder-preserving method from a test that preserves only divisibility.
• Check whether endpoints are included before counting multiples.
• Use prime powers, not merely distinct prime factors: divisibility by 2 and 3 is insufficient for divisibility by 12.
• Prefer a short valid test over memorising many unfamiliar rules.

Previous year questions

No UPSC question has been asked directly on this micro-topic yet. Use the practice questions below.

Practice questions