What is the greatest integer m for which the

Basically we need to find the number of trailing zeros in 50!.

50/5 + 50/5^2 = 10 + 2 = 12. (check here: everything-about-factorials-on-the-gmat-85592.html).

Answer: E.

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Question can be asked as , if 10^m is a factorial of 50!, what is the greatest integer?

My version of answer is:
We can write 10^m = 2^m . 5^m
We can only get 10 when we multiply with 5, so if we solve this problem for 5, we get answer.
As we know formula n/x + n/x^1 + n/x^3 ...until x^k <= n

=> 50/5 + 50/5^2 => 10 + 2 => 12


*(if you are curious, you can solve for 2 also, 50/2 + 50/ 2^1 + 50/2^3 => 25+12+6 =>43. This number is greater than 12. And also you can make only 12 10's , that's the reason to consider 5 )

CONCEPT: Power of any Prime Number in any factorial can be calculated by following understanding

Power of prime x in n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] + ... and so on
Where,
[n/x] = No. of Integers that are multiple of x from 1 to n
[n/x^2] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step
[n/x^3] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step
And so on.....

Where [n/x] is greatest Integer value of (n/x) less than or equal to (n/x)
i.e. [100/3] = [33.33] = 33
i.e. [100/9] = [11.11] = 11 etc.

10 = 2*5
But power of 5 in in 50! will always be less than power of 2 because multiple of 5 in expansion of 50! will always be lesser than no. of multiples of 2 in expansion of 50!

Hence, Power of 10 will be equivalent to Power of 5 in 50!

Power of 5 in 50! = [50/5] + [50/5^2] + [50/5^3] + ... = 10 + 2 + 0 + 0 + ... = 12

Answer: Option E

This is a trailing zero question.

By applying trailing zero concept - https://gmatclub.com/forum/everything-a ... 85592.html we can deduce that the answer must be E.

Notice that m is basically the power of 10. let's analyse the character of 10 with exponents.

10^1=10
10^2=100
10^3=1000
10^4=10000

so, one thing is clear that any exponent of 10 increases the no of zeros. ONE important point to be noted that for each zero, one 5 and one 2 are responsible.
50!/10^m will be an integer. to do so we have to eliminate 10^m. 10^m could be 100 , 1000, 1000000 and so on. we will get just zeros in 10^m. As we don't know the value of m it will depend on the no of zero 50! has. it will tell us how many zero we will need to vanish 10^m. Thus we have to find out the no. of zero in 50!.
there is a short-cut to find out zero:

50/5 + 50/(5)^2 = 10+ 2 = 12 . it means will get 12 zero in total in 50!. So, m will be 12.

Now why did we divide 50 by 5 . we have already learnt that each pair of 5 and 2 creates a zero. In 50! we will have no. of 2 as its factor but 5 is limited. So, we found out the no. of 5 in 50!. ultimately, we got 12 fives. thus , 12 pairs of 5 and 2 yield 12 zero.

Thus, the correct answer is E.

m is an integer such that:

50!/10^m = integer

50!/[(2^m)(5^m)] =integer

The expression above is an integer if m is not greater than the total number of 2s in the prime factored form of 50! and if m is not greater than the total number of 5s in the prime factored form of 50!

Since, in the prime factored form of 50!, there are fewer 5s than 2s, we need to concentrate on the total number of 5s, which we can quickly determine with the following technique:

50/5 = 10

10/5 = 2

The total number of 5s is 10 + 2 = 12, so it must be true that:

m ≤ 12

We see that the maximum value of integer m is 12.

Answer: E