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# Factors in Factorials 2

For how many positive integer values of $$k$$ is $${15}^k$$ a factor of $$100!$$?

(A) 22     (B) 23      (C) 24     (D) 25      (E) 26

This problem is a variation on a theme. You should take a look at Factors in Factorials 1 before you jump into this one

## Translation:

This case  is significantly different from Factors in Factorials 1 because 15 is a composite (non-prime) number. In Factors in Factorials 1 we dealt with a prime integer. So the question becomes how do we find factors of 15 which has two factors itself?

## Solution:

To build a 15 we need one 3 and one 5. So, we could find all the factors of 3 and all the factors of 5 and pair them up until we run out of one of the factors, but if we think about this process for a second it should be obvious that we will run out of 5’s before we run out of 3’s because 5 is larger than 3. I call 5 the “limiting factor.”

Now all we have to do is find the number of 5’s – each of these will pair with one three and we’ll have some 3’s left over.

Number of factors divisible by 5: 100/5 = 20

Number of factors divisible by 25: 100/25 = 4

Number of factors divisible by higher powers of 5 = 0

That’s 24 factors of 5 to pair with factors of three (there are 48 factors of three – you should confirm that for yourself), so we can make 24 15’s

## And…

You can use the same thinking to solve trailing zeros problems. For example,

How many consecutive zeros follow the last non-zero digit in the product of the first 1000 positive integers?

A trailing zero represents a factor of 10. Every 10 contains a 5 and a 2, and, since 5 is the limiting fact, we just need to find the number of factors of 5

1000/5 + 1000/25 + 1000/125 + 1000/625 + 1000/3125 = 200 + 40 + 8 + 1 = 249.

Moving from right to left, we’ll count 249 consecutive zeros before we hit a non-zero digit.