If A = 10! + 12! + 14! + 16! +..........+100!...

LUANDATO wrote:If A = 10! + 12! + 14! + 16! +..........+100!, then the highest power of 2 in A is

A. 7
B. 8
C. 9
D. 10
E. 11

The OA is B.

I don't understand this PS question. Can any expert explain it for me, please? Thanks.

This question isn't terribly well-worded, but the gist seems to ask, if 2^n is a factor of the sum, what is the greatest possible value of n?

First, we know that every term in the series contains 10! for example: 12! = 12*11*10!, and 14! = 14*13*12*11*10!, etc.

So we can re-write the sum as 10!(1 + 12*11 + 14*13*12*11 ... + 100*99*98...) Notice that inside the parentheses, every number will be even, aside from '1.' So we know that this enormous sum will be ODD, as EVEN + 1 = ODD.) Now all we have to do is count the 2's in the prime factorization of 10!

10/2 = 5 terms with at least one 2.
10/(2^2) = 10/4 = 2.5 -->2 terms with at least two 2's.
10/(2^3) = 10/8 = 1.25 --> 1 term with at least three 2's.

5 + 2 + 1 = 8. The answer is B