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
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To find number of terminating zero's count the number of 5's

\(\frac{200}{5}\) = 40

\(\frac{200}{25}\) = 8

\(\frac{200}{125}\) = 1


If you add them its 49 zero's

Explanation of this concept on this thread in detail...

question-on-terminating-zero-s-154440.html

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

You have the correct answer.

Please explain how did you go for 5, 25, 125 what is the reasoning for your approach to this soln.

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

number property states

n!/p + n!/p^2 + n!/p^3 +......

where p is the prime number

to find restricting zeros we need to find number of 5's or 2's in 200! as (10=5*2)

200!/2=100 2's
200/2^2=50 2's
200/2^3=25 2's
200!/2^4 =12 2's
200/2^5=6 2's
200/2^6=3 2's
200/2^7=1 2's
total there are

100+50+25+12+6+3+1= 197 2's

number of 5's in 200! are
40+8+1=49

5 being less in number becomes the restriction for number of 10's in the equation.
hence the answer will be C.

To solve this question, it would be good to know the formular: n!/5^x and in this example: 200/5=40 200/25=8 and 200/125=1 and then the sum is 49, Hence C

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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For similiar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.


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Multiplication of 2s and 5s gives a 0.
We need equal number of 2s and 5s.
Since there are less powers of 5 than 2, we'll consider powers of 5 and multiply with equal number of 2s.
5, 25, 125 are the only three powers of 5 below 200.
Now, 200/5 = 40. Hence there are 40 multiples of 5 which when multiplied by multiples of 2 will contribute a 0.
200/25 = 8. Hence there are 8 multiples of 25 which when multiplied by multiples of 2 will contribute a 0.
200/125 = 1 (1 is the quotient). Hence there is 1 multiple of 125 which when multiplied by multiples of 2 will contribute a 0.

Hence there are 40+8+1 = 49 terminating zeros.
Hence option C.

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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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Recall that each 5-and-2 pair (which makes 10 when multiplied) in the factorization of a number results in one trailing zero. Thus, we need to determine the number of 5-and-2 pairs within the prime factorization of that number.

Since we know there are fewer 5s than 2s in 200!, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 200!, we can use the following shortcut in which we divide 200 by 5, then divide the quotient of 200/5 by 5 and continue this process until we no longer get a nonzero quotient.

200/5 = 40

40/5 = 8

8/5 = 1 (we can ignore the remainder)

Since 1/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 200!.

Thus, there are 40 + 8 + 1 = 49 factors of 5 within 200!

Since there are 49 factors of 5 within 200!, there are 49 5-and-2 pairs and thus 49 trailing zeros.

Answer: C
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No. of zeroes at the end of 200! = [200/5]+[200/25]+[200/125] = 40 + 8 + 1 = 49
where [x] denotes greatest integer function less than equal to x

Answer C
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Number of trailing zero's are only possible due to the number 0f pairs of (2*5)
Thus we must count the number of 2's and 5's in 200! But as the number of 5's is less in 200!, it makes sense to count the total number of 5's in it.

Let's count the number of 5's exactly once in each sequence=> Equal to the number of terms in a sequence

1) 5, 10,15,20,25.......200 We can use the direct formula to count no. of terms : [(Last-First)/Difference] +1
=[(200-5)/5]+1
=39+1
=40

2) 25,50, 75,100......200
=[(200-25)/5] + 1
=7+1
=8

3) 125
=1

Thus the total number of 5's in all the above :
40+8+1=49

Thus the total number of trailing zeroes is 49

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