For a natural number m, the product m(m+2)(m+4) is always divisible by:

Concept: Among any three numbers in arithmetic progression with common difference 2 - i.e. m, m+2, m+4 - the remainders on division by 3 are all different from each other precisely when the common difference is coprime to 3. Since gcd(2, 3) = 1, as m ranges over its residue class mod 3, the three terms m, m+2, m+4 cover all three residues 0, 1, 2 in some order, so exactly one of them is always divisible by 3.

Application:

1. Let m leave remainder r on division by 3, so r = 0, 1, or 2.

2. If r = 0, then m itself is divisible by 3.

3. If r = 1, then m + 2 leaves remainder 1 + 2 = 3, i.e. 0, so m + 2 is divisible by 3.

4. If r = 2, then m + 4 leaves remainder 2 + 4 = 6, i.e. 0, so m + 4 is divisible by 3.

5. In every case one of the three factors is divisible by 3, so the product m(m + 2)(m + 4) is always divisible by 3.

Cross-check: For m = 1, the product equals 15 (divisible by 3, but odd and not a multiple of 7). For m = 2, the product equals 48 (divisible by 3, but not a multiple of 5). Both examples confirm divisibility by 3 while ruling out the other options as guaranteed divisors.