BTGmoderatorDC wrote:If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?

(1) 10^d is a factor of f
(2) d>6

30! = 30*29*28*....*3*2*1.

Statement 1:
To determine the greatest possible value for d, we need to know the number of 10's contained within 30!
Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 30! will yield a 10.
The prime-factorization of 30! is composed of FAR MORE 2'S than 5's.
Thus, the number of 10's depends on the NUMBER OF 5's contained within 30!.

To count the number of 5's, simply divide increasing POWERS OF 5 into 30.

Every multiple of 5 between 1 and 30, inclusive, provides at least one 5:
30/5 = 6 --> 6 5's.
Every multiple of 5² between 1 and 30, inclusive. provides a SECOND 5:
30/5² = 1 --> 1 more 5.
Thus, the total number of 5's contained within 30! = 6+1 = 7.
Implication:
Within the prime-factorization of 30! are a total of 7 10's, with the result that up to 7 10's can divide into 30!.
Thus, d≤7.
Since d can be different values, INSUFFICIENT.

Statement 2:
Since d can be any integer value greater than 6, INSUFFICIENT.

Statements combined:
Since d must be an integer such that 6 < d ≤ 7, d=7.
SUFFICIENT.

The correct answer is C.