The number 40! is divided by 18, the answer is divided by 18 again, and so on for as long as the result stays a whole number. How many divisions are possible?

Since $18=2\times 3^2$, every division uses up one 2 and TWO 3s. Count the 3s in 40! as ⌊40/3⌋ + ⌊40/9⌋ + ⌊40/27⌋ = 13 + 4 + 1 = 18, then halve them because they are spent in pairs — that allows 9 divisions. The 2s are no obstacle at all: ⌊40/2⌋ + ⌊40/4⌋ + ⌊40/8⌋ + ⌊40/16⌋ + ⌊40/32⌋ = 20 + 10 + 5 + 2 + 1 = 38. The answer is the smaller of the two limits, 9. The trap 18 forgets to halve and treats 18 as though it were 3; 38 reports the supply of 2s; and 19 halves the wrong prime.