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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