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
_________________
5-star rated online GMAT quant
self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.