M12-04

Question

How many zeros are there at the end of  100!  ?

A. 20
B. 24
C. 25
D. 30
E. 32

Bunuel · Accepted Answer

Official Solution:

How many zeros are there at the end of  100!  ?

A. 20
B. 24
C. 25
D. 30
E. 32

To find the number of zeros at the end of 100!, we can use this calculation:  \frac{100}{5}+\frac{100}{5^2}=20+4=24 
 
 Some background on trailing zeros: 
 
Trailing zeros refer to a sequence of 0's in a decimal representation of a number, after which no other digits are present.
 
For instance, 125,000 has 3 trailing zeros.
 
The number of trailing zeros in the factorial of a non-negative integer,  n! , can be determined using this formula:
 
 \frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k} , where  k  must be selected such that   5^k \leq n 
 
Let's consider an example:
 
How many zeros are at the end of 32!?
 
 \frac{32}{5}+\frac{32}{5^2}=6+1=7 . Observe that the last denominator ( 5^2 ) must be less than 32. Also, note that we only consider the quotient of the division, that is  \frac{32}{5}=6 . 
 
So, there are 7 zeros at the end of 32!.
 
Another example: how many trailing zeros does 125! have?
 
 \frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31 ,
 
The formula essentially counts the number of factors of 5 in n!, but since there are at least as many factors of 2, this is equivalent to counting the number of factors of 10, each of which contributes one more trailing zero.

Answer: B

dharan · Answer

Well, to find the number of zeros. we can use the following method.

Generally, we need to find the power of 5 that can divide the 100!. i.e 5^x ( we need to find x ).

Do as successive division of just 100 by 5, until you cann't divide further.

100/5 = 20 ---> Quotient
 20/5 = 4 ----> Quotient

Here 4 is the last possible quotient. So the number of zero's are 20 + 4 = 24  .

Similarly, for 125 !
125/5 = 25
 25/5 = 5
 5/5 = 1
 25 + 5 + 1 = 31.

Here finding the number of zero's is an application of finding the highest power of 5 that can divide 100! or 125!.
 i.e this approach is used to find the highest power of any prime number that can divide a factorial without leaving any remainder. 
This is method applicable only if the number whose highest power is to be found is a prime number.

danjbon · Answer

I think this is a  poor-quality  question and I don't agree with the explanation. I am really confused. I am getting that 32! has about 20 trailing zeroes with a calculator.

Bunuel · Answer

100! is a huge number and ordinary calculators round it. BTW 100! has 24 trailing zeros: https://www.wolframalpha.com/input/?i=100%21 For similar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory . Theory on Trailing Zeros: everything-about-factorials-on-the-gmat-85592.html Hope it helps.

dabojula · Answer

I think this is a  high-quality  question and I agree with explanation.

DJ1986 · Answer

I didn't know that there was a formula for trailing zeros so this is how I got it though it took me a while and had I seen this on the gmat I probably would've just considered this question a time sucker and skipped it:

100*99=  bunch of numbers ending with 200 * 98= bunch of numbers ending with 300 *97= you get the point. So you don't actually get another zero until you hit *95.

So every multiple of 5 adds a zero. But then this being the GMAT you know there is a trick and the tricky part is that multiples of 25 give an extra zero. I proved it by just picking 50*20 and 25*4. Since there are 4 multiples of 25 the answer is 24.

lndncalling · Answer

I think this is a  high-quality  question. I think it is a very good question and fantastic explanation.

kelzo123 · Answer

I think this is a  high-quality  question and I agree with explanation.

sushendas · Answer

I think this is a  high-quality  question and I agree with explanation. X>Y means X greater than Y , so please check the signs once again.otherwise explanation is ok.

PriyamRathor · Answer

Hi,
In this the best way to find is to find the no of pairs of 2 and 5.
For this we need to find the number of 2s and 5s.
Now to find the no of 2s
100/2 = 50
50/2 = 25
25/2 =12
12/2 = 6
6/2 = 3 
Which comes to 96. Same method to find 5 
100/5 = 20 
20/5 = 4
hence total 5 s = 24.
Since 2s are 96 and 5s are 24 therefore there are 24 pairs of 5and2.
Hence 24 0s

vchauhan · Answer

This is the problem of trailing zeros.
No. of Trailing zeros = No of 10's = No of 5's.
So, in 100! = Calculate the no of 5's. 
 5|100!____
 5|20
 5|4 => 20+4=24 trailig zeros in 100!

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

Sajjad1994 · Answer

Hi  Bunuel

Can you please post links to some 7-8 similer questions?

Bunuel · Answer

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

Hope this helps.

tmh01 · Answer

I solved it with this --

2*5 = 10, single zero. 4*25 = 100, double zero. We need to find the number of both combinations.

number of 100s : {100} : 2 zeros
number of 10s : {10, ... ,90} : 9 zeros (but this also has 50, with a single 5)

power of 25 : M(25, 50, 75 - {50}) : 25^2, this gives 4 zeros
power of 5 : M(5, 10, 15, 20, ... , 95 - {set of 10s} - {25,75} + {single 5 from 50}) : 5^9, this gives 9 zeros

Using a quick estimate, you find that 4 occurs enough times to result in a zero after multiplying it by 5.

Total zeros = 2 + 9 + 4 + 9 = 24 zeros

Ale300503 · Answer

I like the solution - it’s helpful.

micaelbalioni · Answer

I like the solution - it’s helpful.

Sayan597 · Answer

I like the solution - it’s helpful.

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