The largest power of 12 that divides 100! exactly is $12^k$. What is $k$?

Since $12=2^2\times 3$, count both primes inside 100!. Twos: ⌊100/2⌋ + ⌊100/4⌋ + ⌊100/8⌋ + ⌊100/16⌋ + ⌊100/32⌋ + ⌊100/64⌋ = 50 + 25 + 12 + 6 + 3 + 1 = 97. Threes: ⌊100/3⌋ + ⌊100/9⌋ + ⌊100/27⌋ + ⌊100/81⌋ = 33 + 11 + 3 + 1 = 48. Each 12 swallows TWO twos and one three, so k is the smaller of ⌊97/2⌋ = 48 and 48, i.e. 48. Reporting 97 forgets that 12 needs two 2s each time.