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I must be reading this problem wrong or misunderstanding it. There are many integers k larger than 20 for which 2k is a factor of n=20!.

Of the choices given, each of them, multiplied by 2, is a factor of 20!. So, 20 would be the answer since it asks for the greatest value.

However, take k = 50 as an example. 2k = 100. Is 100 a factor of 20!. Yes, since 20! includes 5*2*10. 50 is larger than 20, so 20 can't be the largest integer for which 2k is a factor of 20!.

If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2^k is a factor of n?

A 10 B 12 C 15 D 18 E 20

20! = (product of the odd #s from 1 to 19) * (product of the even #s from 2 to 20) = a * 2*2^2*(2*3)*(2^3)*(2*5)*(2^2*3)*(2*7)*(2^4)*(2*9)*(2^2*5) = aÂ´ * 2^18 => k = 18

(a and a` are abbreviations for the products of the odd numbers that make up 20!).