How many terminating zeroes does 200! have?

Hi All,

It looks like everyone who posted in this thread understands the correct answer. For anyone who doesn't quite get why the math works the way that it does, here are some examples to prove the point.

200! = (200)(199)(198)(197)....(2)(1) so we know that it's a gigantic number. The reason why it will end in a "string" of 0s is because of all of the multiples of 5 involved.

When multiplying integers, there are two ways to get a number that ends in a 0:
1) multiply a multiple of 5 by an even number
2) multiply a multiple of 10 by an integer.

(5)(2) = 10 so we get one 0 for every multiple of 5

However, 25 = (5)(5). It has TWO 5s in it, so there will be two 0s.

eg (25)(4) = 100

With 125 = (5)(5)(5), we have THREE 5s, so there will be three 0s.

eg (125)(8) = 1,000

So when we divide 200 by 5.....200/5 = 40 multiples of 5. SOME of those multiples of 5 are multiples of 25 (or 125) though. Each of those special cases has to be accounted for.

25, 50, 75, 100, 125, 150, 175 and 200 are all multiples of 25, so they each include one extra 0 (and 125 includes two extra 0s).

So we have 40 + 8 + 1 = 49 zeroes.

Final Answer:

GMAT assassins aren't born, they're made,
Rich