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# What is the greatest integer m for which the

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What is the greatest integer m for which the [#permalink]

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28 Nov 2013, 07:24
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What is the greatest integer m for which the number 50!/10^m is an integer?

(A) 5
(B) 8
(C) 10
(D) 11
(E) 12
[Reveal] Spoiler: OA
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Re: What is the greatest integer m for which the [#permalink]

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29 Nov 2013, 02:51
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Expert's post
mohnish104 wrote:
What is the greatest integer m for which the number 50!/10^m is an integer?

(A) 5
(B) 8
(C) 10
(D) 11
(E) 12

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).

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Re: What is the greatest integer m for which the [#permalink]

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28 Nov 2013, 23:51
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Expert's post
mohnish104 wrote:
What is the greatest integer m for which the number 50!/10^m is an integer?

(A) 5
(B) 8
(C) 10
(D) 11
(E) 12

Here is a conceptual discussion on this question: http://www.veritasprep.com/blog/2011/06 ... actorials/

As for 50!/10^m, the greatest value of m will be obtained by finding the maximum number of 5s in 50!

The quick process for that (discussed in the link above) is
50/5 = 10
10/5 = 2
Total number of 5s is 12. So maximum value of m will also be 12.
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Re: What is the greatest integer m for which the [#permalink]

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29 Nov 2013, 01:12
mohnish104 wrote:
What is the greatest integer m for which the number 50!/10^m is an integer?

(A) 5
(B) 8
(C) 10
(D) 11
(E) 12

10 = 5 x 2; 2s are in abundance so all we need is to count the number of 5s in 50! and get to the answer.

50! = 1 x 2 x 3 x ....50

5, 10, 15, 20, 25, 25, 30, 35, 40, 45, 50, 50

25 and 50 come twice because they have 2 fives.

There are 12 fives hence the highest 10 power is 12.

The answer is E.

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Re: What is the greatest integer m for which the [#permalink]

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29 Nov 2013, 02:51
Bunuel wrote:
mohnish104 wrote:
What is the greatest integer m for which the number 50!/10^m is an integer?

(A) 5
(B) 8
(C) 10
(D) 11
(E) 12

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).

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Hope it helps.
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Re: What is the greatest integer m for which the [#permalink]

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26 Feb 2014, 14:32
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 )
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Re: What is the greatest integer m for which the [#permalink]

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Re: What is the greatest integer m for which the [#permalink]

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04 Nov 2015, 10:43
mohnish104 wrote:
What is the greatest integer m for which the number 50!/10^m is an integer?

(A) 5
(B) 8
(C) 10
(D) 11
(E) 12

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

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Re: What is the greatest integer m for which the   [#permalink] 04 Nov 2015, 10:43
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