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If 22^n is a divisor of 97!+98! ,what is the maximum possible

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If 22^n is a divisor of 97!+98! ,what is the maximum possible [#permalink]

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New post 10 Nov 2016, 10:58
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If 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n?

A)8
B)9
C)10
D)11
E)12
[Reveal] Spoiler: OA

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Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible [#permalink]

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New post 10 Nov 2016, 17:36
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stonecold wrote:
If 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n?

A)8
B)9
C)10
D)11
E)12


22^n means power of 11 as 2 would surely have more power than 11 in a factorial..
97!+98!=97!(1+98)=97!*99....
99 will have ONE 11 and 97! will have 97/11 or 8..
Total 1+8=9

B
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Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible [#permalink]

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stonecold wrote:
If 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n?

A)8
B)9
C)10
D)11
E)12


We are given that 22^n is a divisor of 97!+98!. Thus:

(97!+98!)/22^n = integer

To determine the maximum value of n, we can factor out a 97! in the numerator of the above expression and we have:

97!(1 + 98)/22^n = integer

97!(99)/22^n = integer

Since 22 = 11 x 2, we can rewrite our expression as:

97!(99)/(11^n x 2^n) = integer

In order to determine the maximum value of n such that 22^n divides into 97!(99), we need to determine the maximum number of pairs of factors of 11 and 2 within 97!(99). Since we know there are fewer factors of 11 than factors of 2, we can determine the number of factors of 11 within 97!(99). Let’s start with 97!:

11, 22, 33, 44, 55, 66, 77 and 88 are all factors of 97! Thus, 97! has 8 factors of 11.

We also see that 99 has 1 factor of 11.

Thus, there are 9 factors of 11 within 97!(99), and thus the maximum value of n is 9.

Answer: B
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Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible [#permalink]

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New post 15 Nov 2017, 01:22
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Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible   [#permalink] 15 Nov 2017, 01:22
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