k>0 and m=40! , find k?
st(1) 10^k is a factor of m is Not Sufficient as many tens are sitting in 40!
st(2) 10^k is a factor of n, where n=9!, n=9*8*7*6*5*4*3*2*1=9*8*7*(6*5) is Sufficient, as k can be only 1

10 can be factored as 2*5, hence 10^k is (2*5)^k in the 40! we have 10, 20, 30 and 40 all contain (2*5)

in 9! we have only one pair (2*5) and 1,3,4,6,7,8,9 <== these can't factored as 2*5

b

GmatKiss wrote:If k is a positive integer and m is the product of the fiÂ…rst 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the fiÂ…rst 9 positive integers.

If k is a positive integer and m is the product of the fiÂ…rst 40 positive integers, what is the value of k ?

m = 1*2*3*4*5*...*10*....15*...20*....40

(1) 10^k is a factor of m. k = 1,2,3,.8 Insufficient
(2) 10^k is a factor of n, where n is the product of the fiÂ…rst 9 positive integers. n = 1*2*3*4*5*6*7*8*9 contains only one '(5*2)', [spoiler]so k = 1 Sufficient[/spoiler]

Answer : B

Agree with above explanations. My working also suggests B. What is the OA?