A person starts multiplying consecutive positive integers from 20. How many numbers should he multiply before the will have result that will end with 3 zeroes?

To get 3 trailing zeros, we need three factors of 5×10 = at least three 5s and three 2s in the prime factorization. Counting from 20: 20 contributes one 5, 25 contributes two 5s. By the 6th number (20×21×22×23×24×25), we have three 5s (one from 20, two from 25) and many 2s, giving exactly 3 trailing zeros. Option A (11) would give 4+ trailing zeros.