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# How many terminating zeroes does 121! have?

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Intern
Joined: 07 Dec 2016
Posts: 43
How many terminating zeroes does 121! have?  [#permalink]

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14 Apr 2017, 03:59
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5% (low)

Question Stats:

78% (00:37) correct 22% (00:56) wrong based on 90 sessions

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How many terminating zeroes does 121! have?

A) 21
B) 24
C) 28
D) 35
E) 41

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Joined: 02 Sep 2009
Posts: 47981
Re: How many terminating zeroes does 121! have?  [#permalink]

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14 Apr 2017, 04:06
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amathews wrote:
How many terminating zeroes does 121! have?

A) 21
B) 24
C) 28
D) 35
E) 41

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 121! has $$\frac{121}{5}+\frac{121}{25}=24+4=28$$ trailing zeros.

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

Hope this helps.
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Re: How many terminating zeroes does 121! have?  [#permalink]

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14 Apr 2017, 08:00
is it really a 700 level qs?

24 + 4 = 28

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Re: How many terminating zeroes does 121! have?  [#permalink]

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14 Apr 2017, 08:02
amathews wrote:
How many terminating zeroes does 121! have?

A) 21
B) 24
C) 28
D) 35
E) 41

$$\frac{121}{5} = 24$$
$$\frac{24}{5} = 4$$

So, Total no of trailing zeroes is 28 , answer will be (C) 28
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Re: How many terminating zeroes does 121! have?  [#permalink]

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14 Apr 2017, 08:39
How do I change it to a 600 level question?
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Re: How many terminating zeroes does 121! have?  [#permalink]

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14 Apr 2017, 08:41
amathews wrote:
How do I change it to a 600 level question?

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Re: How many terminating zeroes does 121! have?  [#permalink]

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15 Jul 2018, 04:52
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: How many terminating zeroes does 121! have? &nbs [#permalink] 15 Jul 2018, 04:52
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