saswata4s wrote:

Find the maximum value of n such that 50! is perfectly divisible by 12600^n.

(A) 7

(B) 6

(C) 8

(D) 5

(E) 9

Let’s break 12,600 into prime factors.

12,600 = 126 x 100 = 9 x 7 x 2 x 25 x 4 = 3 x 3 x 7 x 2 x 5 x 5 x 2 x 2 = 2^3 x 3^2 x 5^2 x 7^1

Since 12,600 = 2^3 x 3^2 x 5^2 x 7^1, we really need to determine how many of 5^2 divide into 50!

Let's first determine the number of 5s within 50! by using the following shortcut in which we divide 50 by 5 then divide the quotient of 50/5 by 5, and continue this process until we no longer get a nonzero quotient.

50/5 = 10

10/5 = 2

Since 2/5 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 50!.

Thus, there are 10 + 2 = 12 factors of 5 within 50!.

Since there are 12 factors of 5 in 50!, there are 6 factors of 5^2 = 25 in 50!. Recall that the maximum value of n such that 50! is divisible by 12600^n is determined by the number of 5^2 in 50!. Therefore, n = 6.

Thus, the maximum value of n such that 50!/[(2^3 x 3^2 x 5^2 x 7^1)^n] = integer is 6.

Answer: B

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