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If 100!/x is not an integer, which of the following could be x?

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If 100!/x is not an integer, which of the following could be x?  [#permalink]

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New post Updated on: 27 Sep 2018, 22:50
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If \(\frac{100!}{x}\) is not an integer, which of the following could be x?


A) \(5^{24}\)

B) \(7^{16}\)

C) \(11^9\)

D) \(13^6\)

E) \(17^6\)

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Originally posted by Abhi077 on 27 Sep 2018, 22:45.
Last edited by Bunuel on 27 Sep 2018, 22:50, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If 100!/x is not an integer, which of the following could be x?  [#permalink]

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New post 27 Sep 2018, 23:21
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Abhi077 wrote:
If \(\frac{100!}{x}\) is not an integer, which of the following could be x?


A) \(5^{24}\)

B) \(7^{16}\)

C) \(11^9\)

D) \(13^6\)

E) \(17^6\)


The question is based on this concept: https://www.veritasprep.com/blog/2011/0 ... actorials/

Looking at the options, we see that we will need to find the highest power of each prime in 100!

Checking each option would be too much work so let's try to make some intelligent guesses. Looking at options (C), (D) and (E), we see that till 100, we will have only a few multiples of 11, 13 and 17.
11*9 < 100 so we will have 9 11s in 100!

Between 13^6 and 17^6, if we have 6 multiples of 17, we will obviously have 6 multiples of 13 too.
17*6 = 102 so till 100, we have only 5 multiples of 17.

Hence 100! will not be divisible by 17^6.

Answer (E)
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Re: If 100!/x is not an integer, which of the following could be x?  [#permalink]

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New post 30 Sep 2018, 22:39
VeritasKarishma wrote:
Abhi077 wrote:
If \(\frac{100!}{x}\) is not an integer, which of the following could be x?


A) \(5^{24}\)

B) \(7^{16}\)

C) \(11^9\)

D) \(13^6\)

E) \(17^6\)


The question is based on this concept: https://www.veritasprep.com/blog/2011/0 ... actorials/

Looking at the options, we see that we will need to find the highest power of each prime in 100!

Checking each option would be too much work so let's try to make some intelligent guesses. Looking at options (C), (D) and (E), we see that till 100, we will have only a few multiples of 11, 13 and 17.
11*9 < 100 so we will have 9 11s in 100!

Between 13^6 and 17^6, if we have 6 multiples of 17, we will obviously have 6 multiples of 13 too.
17*6 = 102 so till 100, we have only 5 multiples of 17.

Hence 100! will not be divisible by 17^6.

Answer (E)



Responding to a pm:
Quote:
Can you please elaborate on answer choice b) also, using "17*6 = 102" concept. Thank you.


100! = 1*2*3*4*5*6*7*8* ... *14* ... *21* ... *49*... *98*99*100

Each 7th humber is a multiple of 7 so it gives us a 7:
7, 14, 21, 28,.. 98
These are 14 numbers because 7*14 = 98
But 49 and 98 gives us two 7s each so we get another two 7s.
In all we get 16 7s so 7^16 divides 100! evenly.

Since 5 and 7 are smaller numbers and have numbers with multiple 5s and 7s (25, 50, 49 etc), I chose to ignore those two options to begin with and focus on the larger numbers. Those don't have squares/cubes etc within 100 numbers (11*11 = 121, 13*13 = 169 etc)

The link I have given above tells you how to quickly calculate the highest power of any number in a factorial.
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Re: If 100!/x is not an integer, which of the following could be x?  [#permalink]

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New post 02 Oct 2018, 02:53
Hi VeritasKarishma

I dont get why youve multiplied the power "11*9 < 100 so we will have 9 11s in 100!" here?
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Re: If 100!/x is not an integer, which of the following could be x?  [#permalink]

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New post 02 Oct 2018, 04:03
hibobotamuss wrote:
Hi VeritasKarishma

I dont get why youve multiplied the power "11*9 < 100 so we will have 9 11s in 100!" here?


100! = 1*2*3*4*5*6*7*8*9*10*11 ... *22* ... *44* ... *77*... *98*99*100

100! has 11, 22, 33, 44, ... 99 as factors. Each one of these 9 numbers provides us with an 11.

So

100! = 1*2*3*4*5*6*7*8*9*10*(11) ... *(2*11)* ... *(4*11)* ... *(7*11)*... *98*(9*11)*100

Hence 100! has 11 multiplied 9 times which gives 11^9
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Re: If 100!/x is not an integer, which of the following could be x? &nbs [#permalink] 02 Oct 2018, 04:03
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