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25 Jan 2017, 23:01
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
39% (02:32) correct
61%
(02:08)
wrong
based on 161
sessions
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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
(A) 7
(B) 6
(C) 8
(D) 5
(E) 9
30 Jan 2017, 18:01
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saswata4s
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
General Discussion
25 Jan 2017, 23:43
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saswata4s
12600 in prime factors==>2^3×3^2×5^2×7
the highest number from 2^3 , 3^2 , 5^2 & 7 is 5^2. So when counting the nuber of 5==>12. and we require 2 5's each time when dividing 50!.So 12 divide by 2 gives 6.
I thnik the answer is 6.
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