How many terminating zeroes does 200! have?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64

You have 40 multiples of 5, 8 of 25 and 1 of 125. This will give 49 zeros

in order to determine the zeros you need to know how many 2 and 5 200! contains. Apparently, 2 are much more than 5 since every second number is a multiple of 2. That is why when you find the multiples of 5 you will get the number of zeros. Now every 5-th number is a 5 multiple, where every 25-th has two fives, and 125 has 3 fives. So you have 40 times 5, 8 times 5x5 and 1 time 5x5x5. But note that 25 and 125 already contain one five from the fives multiples, and you will have 40+8+1 times 5 which is 49.
Hope that helps

Nice question. Answer is C.

To answer a question about trailing zeros in a factorial, we must realize that the trailing zeros depend on the number of 10s (i.e. 10s or its multiples) in the factorial. 10=5*2, so we are looking for 5s and 2s. By virtue of being the more frequent number, we don't need to count the number of 2s, because there will always be equal or more 2s than 5s in a factorial. So, essentially we are looking for 5s and its exponents.

Start with finding 5s. 200/5= 40. So we have 40 fives (and 40 twos) which means we'll have 40 zeros. If the answer was a decimal, such as 201/5, round down.

Now look for two 5s, that is 25. So, 200/25= 8, so we have eight more zeros.

Lastly, 5^3= 125, and we have one such number in 200!, which gives us one more trailing zero. Therefore, we have 40+8+1=49 trailing zeros.

Hope this helps.
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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:

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Rich
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1) To paraphrase this question, we need to find the number of prime factor pairs 2*5 in the product 200!
2) In a range of consecutive integers, there will always be more prime factors 2, than prime factors 5, so we only need to calculate the number of prime factors of 5.
3) 200/5=40; 200/25=8; 200/125=1
4) 40+8+1=49

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