11. In a box, there are twenty balls numbered from 1 to 20. Each number is present on one and only one of these balls. How many different balls do we need to draw at a minimum to be sure that the product of their numbers is a multiple of 12?
(A) 7
(B) 11
(C) 12
(D) 15
(E) 18.
11. The answer is (D). The multiples of 3 between 1 and 20 are 6, so there are 14 numbers that are not multiples of 3. If we just draw those 14 numbers, their product would not be a multiple of 3, let alone 12, so the minimum number \( n \) of draws to ensure that the product is a multiple of 12 is greater than 14. If we draw 15 numbers, however, we will definitely have at least one multiple of 3. Since the even numbers between 1 and 20 are 10, and the odd numbers are 10, with 15 draws we ensure at least 5 even numbers. Thus, the product will be a multiple of 3 and a multiple of \( 2^5 = 32 \). In particular, it will be a multiple of 3 and 4, and thus of 12. It follows that \( n \) is indeed 15.