Which one of the following is the greatest number by which the product of three consecutive even numbers would be exactly divisible?

Correct — C, 48. Write the three consecutive even numbers as 2n, 2n + 2 and 2n + 4, where n is a positive integer. Their product is 2n × 2(n + 1) × 2(n + 2) = 8 × n(n + 1)(n + 2). Now n(n + 1)(n + 2) is a product of three consecutive integers, and among any three consecutive integers one is a multiple of 3 and at least one is even, so that product is always divisible by 6. The whole expression is therefore always divisible by 8 × 6 = 48. To see that nothing larger works, take the smallest case: 2 × 4 × 6 = 48 itself, which no number above 48 can divide. So 48 divides every such product and is the greatest number that does.

• (a)12 — Every such product is divisible by 12, but the question asks for the greatest such number, and 48 also works. Twelve is what you get from 4 × 3, missing one of the three factors of two.
• (b)24 — Also a genuine divisor of every such product, and also not the greatest. Twenty-four comes from 8 × 3 while forgetting that the run of three consecutive integers contributes an extra factor of two as well.
• (d)64 — Too large. The first such product is 2 × 4 × 6 = 48, and 64 does not divide 48. A single counter-example is enough to rule it out.

Two counting facts do all the work. Among any k consecutive integers exactly one is divisible by k, so a product of three consecutive integers always carries a factor of 3 and always carries at least one factor of 2, making it divisible by 6 — in fact by 3! = 6, which is the general result. And pulling a 2 out of each of three consecutive even numbers leaves exactly a run of three consecutive integers behind, contributing 2 × 2 × 2 = 8. The two combine to 48.

The word 'greatest' is the whole difficulty, because three of the four options are true divisors and only one is the largest. The fastest check in the hall is to compute the very first case, 2 × 4 × 6 = 48, since the answer can never exceed it; then confirm that 48 also divides the next case, 4 × 6 × 8 = 192, and the one after, 6 × 8 × 10 = 480. The greatest common divisor of 48, 192 and 480 is 48, which settles it without any algebra at all.

• Three consecutive even numbers can be written 2n, 2n + 2, 2n + 4, whose product is 8 × n(n + 1)(n + 2).
• A product of k consecutive integers is always divisible by k factorial, so three consecutive integers give a factor of 6.
• The product of three consecutive even numbers is therefore always divisible by 8 × 6 = 48.
• The first such product is 2 × 4 × 6 = 48, which caps the answer from above.
• Successive products 48, 192, 480 and 960 have 48 as their greatest common divisor.

• Stopping at 24 because it is a genuine divisor, and missing the word 'greatest'.
• Forgetting the extra factor of two hidden inside the run of consecutive integers.
• Testing only one case and assuming a larger divisor works for all of them.

Asked as a greatest-common-divisor item in disguise, with three true divisors offered alongside the largest one. Testing the first case is the reliable defence.

No directly related past PYQ was found.

• practice — not a real PYQ

  What is the greatest number by which the product of three consecutive natural numbers is always divisible?

  • (a)3
  • (b)6
  • (c)12
  • (d)24

  Answer(b) 6 — three consecutive integers always contain a multiple of 3 and at least one even number; the first product, 1 × 2 × 3, is 6.

• practice — not a real PYQ

  The product of four consecutive even numbers is always divisible by

  • (a)96
  • (b)192
  • (c)384
  • (d)768

  Answer(c) 384 — pulling out four twos gives 16, and four consecutive integers are divisible by 4! = 24, so 16 × 24 = 384; the first product 2 × 4 × 6 × 8 is exactly 384.