A is the smallest positive integer which when divided by 9 and 12 leaves remainder 8. B is the smallest positive integer which when divided by 9 and 12 leaves remainder 5. Which one of the following is the value of A – B?

Correct — A, 3. Take A first. A number that leaves remainder 8 on division by 9 and also by 12 must be of the form 36k + 8, because 36 is the lowest common multiple of 9 and 12 and the remainder is common to both. The smallest positive value is at k = 0, which gives A = 8 — and 8 does satisfy the condition, since dividing 8 by 9 leaves 8 and dividing 8 by 12 also leaves 8. By the same reasoning B is of the form 36k + 5 and the smallest positive value is B = 5. So A − B = 8 − 5 = 3. The answer is the same even for a candidate who insists on taking the next value up: 44 − 41 is also 3, because both numbers move by the same lowest common multiple. The difference of the two remainders is what survives, whichever pair is chosen.

• (b)2 — This would need the two remainders to differ by two. They are 8 and 5, and 8 − 5 is 3.
• (c)1 — A single-step slip on the subtraction. Nothing in the problem produces a difference of one between the two smallest qualifying numbers.
• (d)0 — The trap for a candidate who computes only the common part, 36 or the lowest common multiple, and subtracts it from itself while forgetting that the two remainders are different.

When a number leaves the same remainder r on division by several divisors, it takes the form LCM(divisors) × k + r. The remainder must be smaller than every divisor for such a number to exist at all, which is satisfied here since 8 and 5 are both below 9. The structure means that any two such numbers with the same divisors and the same remainder differ by a multiple of the lowest common multiple.

That structure is why this item is robust to the one genuine ambiguity in it. Whether a candidate reads 'smallest positive integer' as allowing a number below the divisors, giving 8 and 5, or takes the first value above them, giving 44 and 41, the difference is 3 either way — the 36 cancels. Noticing that saves the time otherwise spent worrying about the edge case. Also worth remembering: the lowest common multiple of 9 and 12 is 36, not 108, because they share a factor of 3.

• A number leaving remainder r on division by several divisors has the form LCM × k + r.
• The lowest common multiple of 9 and 12 is 36, since 9 = 3² and 12 = 2² × 3 give 2² × 3² = 36.
• The smallest positive integer leaving remainder 8 on division by 9 and 12 is 8 itself.
• The smallest positive integer leaving remainder 5 on both is 5 itself.
• Any two such numbers differ by a multiple of the LCM, so A − B equals the difference of the remainders, 3.

• Taking the lowest common multiple of 9 and 12 as 108 by multiplying them instead of removing the common factor.
• Assuming the smallest such number must be larger than the divisors.
• Computing the two numbers correctly and then subtracting in the wrong order.

Asked as a pair of lowest-common-multiple constructions whose difference is taken, so the multiple itself cancels and only the remainders matter.

No directly related past PYQ was found.

• practice — not a real PYQ

  What is the smallest number which when divided by 6, 8 and 12 leaves a remainder of 5 in each case?

  • (a)5
  • (b)24
  • (c)29
  • (d)53

  Answer(a) 5 — the number has the form 24k + 5, and at k = 0 it is 5, which leaves remainder 5 on division by 6, 8 and 12 alike.

• practice — not a real PYQ

  The lowest common multiple of 9 and 12 is

  • (a)3
  • (b)36
  • (c)72
  • (d)108

  Answer(b) 36 — 9 is 3² and 12 is 2² × 3, so the LCM is 2² × 3² = 36.