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(1) f is a factor of 10, but 10^1, 10^2, 10^3.... are also factors of 10
INSUFF
(2) obviously insuff

together
in 30!, there are 3 factors that contain ten: 10:10*1, 20:10*2, 30:10*3, and 3 factors containing 5: 5, 15:5*3, 25:5*5. cobine each of those 5's with one of the many 2's in the factorization for another 4 10's, that makes 7 10's total. if d>6, d must be 7.

I used the same approach as Dahiya, but stuck with 10^6.

When you counted 25 you counted it twice. Why?

Thats because 25 = 5x5.

We compute 5s and 2s in the product to determine the number of zeroes the product would have. Typically since the number of twos are sufficient, we can make our lives easier by counting the number of 5s alone. And 5s are counted by counting in all the factors of 5s (5, 10, 15 etc) but counting extra for powers of 5, like 25, 125 and 625 (which have 2, 3 and 4 5s respectively).