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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

The answer is (C)

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.