When 500! is written as a product of prime factors, how many times does the prime 7 appear?

Count how many multiples of each power of 7 there are up to 500 and add them: ⌊500/7⌋ = 71, ⌊500/49⌋ = 10, ⌊500/343⌋ = 1 and ⌊500/2401⌋ = 0. The exponent is 71 + 10 + 1 = 82. Counting only the multiples of 7 stops at 71 — the standard trap — because it forgets that $49=7^2$ brings a second 7 with it and $343=7^3$ a third. Missing the 343 term alone leaves 81.