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X is the product of integers from 1 to 50

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X is the product of integers from 1 to 50  [#permalink]

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New post 08 Apr 2017, 07:52
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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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X is the product of integers from 1 to 50  [#permalink]

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New post 08 Apr 2017, 10:28
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ashikaverma13 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


\(12^n = 2^(2n)*3^n\)

We have many 2's in the product of 1 to 50. thus we need to find number of 3's in product 1-50

1-30 = 10 + 1(9) + 1(18) +2(27)
31- 50 = 6(33- 48) + 1(36) + 1(45) = 22
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Re: X is the product of integers from 1 to 50  [#permalink]

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New post 08 Apr 2017, 10:33
RMD007 wrote:
ashikaverma13 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


\(12^n = 2^(2n)*3^n\)

We have many 2's in the product of 1 to 50. thus we need to find number of 3's in product 1-50

1-30 = 10 + 1(9) + 1(18) +2(27)
31- 50 = 6(33- 48) + 1(36) + 1(45) = 22


Can you explain a bit more? Why is only finding the number of 3s enough?
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Re: X is the product of integers from 1 to 50  [#permalink]

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New post 08 Apr 2017, 11:27
1
ashikaverma13 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


X is the product of integers from 1 to 50, inclusive = 50!

\(12 = 2^2*3\)

Highest power of 2 that will divide 50! is 47

\(\frac{50}{2} = 25\)
\(\frac{25}{2} = 12\)
\(\frac{12}{2} = 6\)
\(\frac{6}{2} = 3\)
\(\frac{3}{2} = 1\)

Highest power of 3 that will divide 50! is 22

\(\frac{50}{3} = 16\)
\(\frac{16}{3} = 5\)
\(\frac{5}{3} = 1\)

Since the highest power of 3 < 2 , the highest power of \(12^n\) that will divide 50! will be 22 , hence, answer will be (C) 22.
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Re: X is the product of integers from 1 to 50  [#permalink]

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New post 08 Apr 2017, 23:45
ashikaverma13 wrote:
RMD007 wrote:
We have many 2's in the product of 1 to 50. thus we need to find number of 3's in product 1-50

1-30 = 10 + 1(9) + 1(18) +2(27)
31- 50 = 6(33- 48) + 1(36) + 1(45) = 22


Can you explain a bit more? Why is only finding the number of 3s enough?


Sure. We have a product of numbers from 1 to 50 and the question says it is divisible by \(12^n\) = \(2^(2n)*3^n\)

We have two prime factors 2 and 3 and we need to find such maximum value of n for which \(2^(2n)\)and \(3^n\) both will be part of this product.

Now, as we glance through the product of 1 to 50, we know that number of 2's in the product will be greater than number of 3's present . So we if find the number of 3's, such number will obviously be less than 2's and hence it will be sufficient to say that \(12^n\) is divisible by product of 1-50.

Hope this helps..
Re: X is the product of integers from 1 to 50 &nbs [#permalink] 08 Apr 2017, 23:45
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