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Current Student D
Joined: 12 Aug 2015
Posts: 2537
Schools: Boston U '20 (M)
GRE 1: Q169 V154 If 22^n is a divisor of 97!+98! ,what is the maximum possible  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 57% (01:48) correct 43% (01:42) wrong based on 94 sessions

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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

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Math Expert V
Joined: 02 Aug 2009
Posts: 8296
Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible  [#permalink]

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5
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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##### General Discussion
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 9895
Location: United States (CA)
Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible  [#permalink]

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2
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.

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CrackVerbal Quant Expert G
Joined: 12 Apr 2019
Posts: 459
Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible  [#permalink]

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In questions related to divisibility, breaking down the dividend will give a lot of information about what the divisor needs to be if it HAS to divide the dividend fully. Therefore, let’s break the dividend down, first.

In 97! + 98!, we can take 97! as common. When we do this, we have 97! + 98! = 97! (1 + 98) = 97! * 99. This helps us understand that $$22^n$$ should be able to divide 97! * 99 fully.

Let’s now look at 22. 22 = 2*11, therefore $$22^n$$ = $$2^n$$ * $$11^n$$.
So, essentially, we are trying to find the highest power of 11 in the dividend since the number of 11s would be far fewer compared to the number of 2s.
Finding out the highest power of any prime number in a given factorial can be done by successive division. In our case, it can be done as shown below in the diagram.

Attachment: 25th Feb 2020 - Reply 3 - 1.jpg [ 73.73 KiB | Viewed 123 times ]

So, the highest power of 11 in 97! is 8. But, 99 also has a 11 in it. Therefore, the highest power of 11 in the numerator (or the dividend) is 9. So, there need to be 9 11s in the denominator as well. As such, the highest power of 22 that can divide 97! + 98! is 9.
The correct answer option is B.

Hope that helps!
_________________ Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible   [#permalink] 24 Feb 2020, 23:55
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