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What is the largest integer k such that 10! is divisible by

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What is the largest integer k such that 10! is divisible by [#permalink]

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What is the largest integer k such that 10! is divisible by \(10^{k}\) ?

A. 1
B. 2
C. 3
D. 4
E. 5

M23-11
[Reveal] Spoiler: OA

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Last edited by arindamsur on 11 Jun 2014, 01:13, edited 2 times in total.
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Re: What is the largest integer k such that 10! is divisible by [#permalink]

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New post 11 Jun 2014, 01:07
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arindamsur wrote:
What is the largest integer k such that 10! is divisible by 10^k ?

A. 1
B. 2
C. 3
D. 4
E. 5

M23-11


Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

Fro example, 125000 has 3 trailing zeros (125000);

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

BACK TO THE ORIGINAL QUESTION:

According to above 10! has \(\frac{10}{5}=2\) trailing zeros. So, the largest integer k such that 10! is divisible by 10^k is 2.

Answer: B.

Check here for theory: everything-about-factorials-on-the-gmat-85592.html

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Hope it helps.
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Re: What is the largest integer k such that 10! is divisible by [#permalink]

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Re: What is the largest integer k such that 10! is divisible by   [#permalink] 29 Mar 2017, 05:32
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