Which composite number can divide the sum of the first 12 natural numbers?

Correct — C.

Sum of the first 12 natural numbers = n(n + 1) ⁄ 2
= 12 × 13 ⁄ 2 = 78

Factorise: 78 = 2 × 3 × 13

78 ⁄ 6 = 13, a whole number, and 6 is composite (2 × 3).
78 ⁄ 12 = 6.5, 78 ⁄ 8 = 9.75, 78 ⁄ 4 = 19.5 — none of them whole.
The divisor is 6 → option (c).

Why the others are wrong

• (a)
  12 — 78 ⁄ 12 = 6.5. You reach 12 by forgetting to halve: n(n + 1) = 156 is a multiple of 12, but the sum is 156 ⁄ 2 = 78.
• (b)
  8 — 78 ⁄ 8 = 9.75. 8 is 2³ and needs three factors of 2; 78 = 2 × 3 × 13 supplies exactly one, so an even sum is not enough.
• (d)
  4 — 78 ⁄ 4 = 19.5. Divisibility by 4 is decided by the last two digits, and 78 is not a multiple of 4. The un-halved 156 is, which is where this option comes from.

The sum of the first n natural numbers is n(n + 1) ⁄ 2. For n = 12 that is 12 × 13 ⁄ 2 = 78.

Every option offered is itself composite, so the word composite separates nothing — divisibility does all the work.

Factorise 78 into 2 × 3 × 13 and its complete list of divisors is 1, 2, 3, 6, 13, 26, 39, 78. Reading the options against that list settles the question without a single long division.

Key facts

• The first n natural numbers sum to n(n + 1) ⁄ 2, so the first 12 sum to 78.
• 78 factorises as 2 × 3 × 13, giving the divisors 1, 2, 3, 6, 13, 26, 39 and 78.
• A number is divisible by 4 only when the number formed by its last two digits is a multiple of 4.
• 6 counts as composite because it factors as 2 × 3.

Common traps

• Treating composite as the filter when all four options are composite.
• Assuming an even total must also be divisible by 4 or 8.
• Using n(n + 1) = 156 without halving it, which makes both 12 and 4 look right.

No directly related past PYQ was found.