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How many terminating zeroes does 121! have?

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

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14 Apr 2017, 02:59
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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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Re: How many terminating zeroes does 121! have?  [#permalink]

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14 Apr 2017, 03:06
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2
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, 07:00
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, 07:02
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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21 Aug 2018, 23:52
2
amathews wrote:
How many terminating zeroes does 121! have?

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

Responding to a pm:

Terminating 0s are formed by creation of 10 ( = 2*5)
121! will certainly have more 2s than 5s so 5 will be our limiting factor. We need to find the number of 5s in 121! and that will give us the number of 10s (and hence the number so terminating 0s)

121/5 = 24
24/5 = 4

There will be 24+4 = 28 5s in 121! and hence 28 terminating zeroes.
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Re: How many terminating zeroes does 121! have?  [#permalink]

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22 Aug 2018, 23:19

Quote:
121! will certainly have more 2s than 5s so 5 will be our limiting factor.

To confirm your approach for highlighted text, you did not perform 121! as 121 * 120 * ...1
Instead you found a pattern that just as 10! has more even numbers from 1 to 10
than multiples of 5 from 1 to 10, similarly 121! will have follow same pattern, correct?
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Re: How many terminating zeroes does 121! have?  [#permalink]

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23 Aug 2018, 07:59

Quote:
121! will certainly have more 2s than 5s so 5 will be our limiting factor.

To confirm your approach for highlighted text, you did not perform 121! as 121 * 120 * ...1
Instead you found a pattern that just as 10! has more even numbers from 1 to 10
than multiples of 5 from 1 to 10, similarly 121! will have follow same pattern, correct?

Yes. It is true for all numbers. Every second number has a 2 while only every fifth number has a 5.
Check this post for more:
https://www.veritasprep.com/blog/2011/0 ... actorials/
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Re: How many terminating zeroes does 121! have?  [#permalink]

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26 Aug 2018, 18:14
1
amathews wrote:
How many terminating zeroes does 121! have?

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

To determine the number of trailing zeros in a number, we need to determine the number of factors of 10, which is equal to the number of 5- and-2 pairs within the prime factorization of that number.

Since we know there are fewer 5s in 121! than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 121!, we can use the following shortcut in which we divide 121 by 5, then divide the quotient of 121/5 by 5 and continue this process until we no longer get a nonzero quotient.

121/5 = 24 (we can ignore the remainder)

24/5 = 4 (we can ignore the remainder)

Since 4/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 121!.

Thus, there are 24 +4 = 28 factors of 5 within 120! and so there are 28 5-and-2 pairs, creating 28 trailing zeros.

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

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27 Aug 2018, 04:09
Bunuel Please explain how n5+n52+n53+...+n5kn5+n52+n53+...+n5k , where k must be chosen such that 5^(k+1)>n

I don't get this part "where k must be chosen such that 5^(k+1)>n"
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Re: How many terminating zeroes does 121! have?  [#permalink]

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27 Aug 2018, 05:32
1
onyx12102 wrote:
Bunuel Please explain how n5+n52+n53+...+n5kn5+n52+n53+...+n5k , where k must be chosen such that 5^(k+1)>n

I don't get this part "where k must be chosen such that 5^(k+1)>n"

Hi onyx12102,

IMO, we will carry out the operations(successive division & addition) as long as $${Denominator}\leq{Numerator}$$ Or,$$5^k\leq{n}$$

Here n=121, So, $$5^k$$ is less than 121 when k=2 ($$5^3=125$$, DISCARD)

So, $$\frac{121}{5}+\frac{121}{5^2}$$
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Re: How many terminating zeroes does 121! have? &nbs [#permalink] 27 Aug 2018, 05:32
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