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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # X is the product of integers from 1 to 50, inclusive. If 12^N is a fac

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Math Expert V
Joined: 02 Sep 2009
Posts: 61385
X is the product of integers from 1 to 50, inclusive. If 12^N is a fac  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 58% (01:51) correct 42% (01:46) wrong based on 57 sessions

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X is the product of integers from 1 to 50, inclusive. If 12^N is a factor of X, what is the greatest possible value of N?

A. 4
B. 12
C. 22
D. 28
E. 57

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Joined: 26 Feb 2016
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X is the product of integers from 1 to 50, inclusive. If 12^N is a fac  [#permalink]

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4
The formula for finding the highest power of a prime number in a factorial

If p is a prime number, then the highest power of p in a factorial n is {$$\frac{n}{p}$$} + {$$\frac{n}{p^2}$$} + {$$\frac{n}{p^3}$$}.....
where {$$\frac{a}{b}$$} is the quotient when integer b divides another integer a

Now coming to the problem at hand.

n = $$50$$
p = $$12 = 2^2*3$$

To find the highest power of 12, we need to find the highest power of 3(which is the bigger prime number)
Here, the bigger the prime number the lower will the frequency of that prime be in the factorial.

Therefore, the highest power of 12 in 50 is $${\frac{50}{3}} + {\frac{50}{3^2}} + {\frac{50}{3^3}} = 16 + 5 + 1 = 22$$(Option C)
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X is the product of integers from 1 to 50, inclusive. If 12^N is a fac  [#permalink]

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Bunuel wrote:
X is the product of integers from 1 to 50, inclusive. If 12^N is a factor of X, what is the greatest possible value of N?

A. 4
B. 12
C. 22
D. 28
E. 57

$$50!/{12}^n=50!/(2^23)^n$$
$$quotients(50/[2^1,2^2,2^3…2^x])=25+12+6+3=46$$
$$quotients(50/[3^1,3^2,3^3…3^x])=16+5+1=22$$
$$50!/(12)^n=(2^23)^n=(2^{2n}*3^n)=2^{2(22)}*3^{22}=(12)^{22}$$

Ans (C) X is the product of integers from 1 to 50, inclusive. If 12^N is a fac   [#permalink] 11 Dec 2019, 05:39
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# X is the product of integers from 1 to 50, inclusive. If 12^N is a fac  