What is the smallest whole number n for which n! is a multiple of $2^{10}$?

Count the 2s inside $n$! by adding ⌊$n/2$⌋ + ⌊$n/4$⌋ + ⌊$n/8$⌋ + … . For $n=11$ that gives 5 + 2 + 1 = 8 — two short. For $n=12$ it gives 6 + 3 + 1 = 10, exactly enough, because $12=2^2\times 3$ brings two fresh 2s at once. So the answer is 12. The trap 10 matches the index 10 to the number 10, but 10! carries only 5 + 2 + 1 = 8 twos. Answering 11 fails because 11 is odd and contributes no 2 at all, and 16 overshoots by insisting the answer must itself be a power of 2.