M7MBA wrote:If n is a positive integer and 10^n is a factor of m, what is the value of n?

(1) m is the product of the first positive 40 integer numbers.
(2) n > 8*m/40!

Since the statements refer to 40!, determine options for n if 10^n is a factor of 40!.

Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 40! will enable 10^n to divide into 40!.
The prime-factorization of 40! includes FAR MORE 2'S than 5's.
Thus, the number of 10's that can divide into 40! depends on the NUMBER OF 5's contained within 40!.

To count the number of 5's, simply divide increasing POWERS OF 5 into 40.

Every multiple of 5 within 40! provides at least one 5:
40/5 = 8 --> eight 5's.
Every multiple of 5² within 40! provides a SECOND 5:
40/5² = 1 --> one more 5.
Thus, the total number of 5's contained within 40! = 8+1 = 9.
Since 40! contains nine 5's,10^n will divide into 40! if n≤9.

Statement 1: m = 40!
Since 10^n is a factor of 40!, n can be any positive integer less than or equal to 9.
Thus, the value of n cannot be determined.
INSUFFICIENT.

Statement 2: n > 8(m/40!)
Here, n can be any positive integer such that n > 8(m/40!).
Thus, the value of n cannot be determined.
INSUFFICIENT.

Statements combined:
Statement 1: n≤9.
Statement 2:
Substituting m=40! into n > 8(m/40!), we get:
n > 8(40!/40!)
n > 8.
Since n≤9 and n>8, only option is possible:
n=9.
SUFFICIENT.

The correct answer is C.