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N is the product of the first 5 prime numbers. If 12!/n is divisible b

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N is the product of the first 5 prime numbers. If 12!/n is divisible b  [#permalink]

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New post Updated on: 01 Jun 2016, 07:09
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A
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D
E

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N is the product of the first 5 prime numbers. If 12!/n is divisible by 2^k, what is the greatest value of a positive integer k?
A. 6
B. 7
C. 8
D. 9
E. 10

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Originally posted by MathRevolution on 30 May 2016, 19:46.
Last edited by MathRevolution on 01 Jun 2016, 07:09, edited 1 time in total.
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Re: N is the product of the first 5 prime numbers. If 12!/n is divisible b  [#permalink]

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New post 30 May 2016, 23:29
1
MathRevolution wrote:
N is the product of the first 5 prime numbers. If 12!/n is divisible by 2k, what is the greatest value of a positive integer k?
A. 6
B. 7
C. 8
D. 9
E. 10

*An answer will be posted in 2 days.


N = 2.3.5.7.11.

12! / N = 12. 10. 9. 8. 6. 4

Expressing this as \(2^9\)*\(3^4\)* 5 --> This number is divisible by 2 k.

\(2^8\)*\(3^4\)*5 is divisible by k.

Out of given options 10 is the biggest possible option.

Answer E
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Re: N is the product of the first 5 prime numbers. If 12!/n is divisible b  [#permalink]

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New post 31 May 2016, 02:39
1
N = 2 x 3 x 5 x 7 x 11

12! / N = (12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (2 x 3 x 5 x 7 x 11) = 12 x 10 x 9 x 8 x 6 x 4 x 1

Here we get rid of 2 in the denominator by dividing 12 by 2:
(12! / N) / 2K = (12 x 10 x 9 x 8 x 6 x 4 x 1) / 2K = (6 x 10 x 9 x 8 x 6 x 4 x 1) / K

Hence, largest K from the given options can be 10

Would very much like to hear why the correct answer is D.

Answer: E
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Re: N is the product of the first 5 prime numbers. If 12!/n is divisible b  [#permalink]

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New post 31 May 2016, 08:16
Its 2^k not 2k please correct it.
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Re: N is the product of the first 5 prime numbers. If 12!/n is divisible b  [#permalink]

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New post 31 May 2016, 22:00
HI,

Here given is
1)N is product of first five prime numbers , that is : 2*3*5*7*11
2)12!/N divisible by 2k and that is multiple of 2 and k should be highest value
so , 12!/N = (12*11*10*9*8*7*6*5*4*3*2*1)/(2*3*5*7*11)
and now we have
12!/N = 12*10*9*8*6*4
i.e 2k = 12*10*9*8*6*4 [here except 9 all others are multiples of 2 and can be written in 2 multiples as below]

2k = 2(6*5*4*3*2)*9
2k=2()*9
similar to the given expression as 2k and hence the highest value for K is 9

So D is the option
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Re: N is the product of the first 5 prime numbers. If 12!/n is divisible b  [#permalink]

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New post 01 Jun 2016, 18:03
From 12!/n=12!/(2)(3)(5)(7)(11)=2^k(integer), the value of k is the number in which one of 2 in 12! is taken out. 12! has 2,4,6,8,10,12. In other words, 2^1(2), 2^2(4), 2^1(6), 2^3(8), 2^1(10), 2^2(12). Since it is multiplication, we can add 1+2+1+3+1+2=10. Since we need to take one of 2 out, the number become 10-1=9. Hence, the correct answer is D.
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Re: N is the product of the first 5 prime numbers. If 12!/n is divisible b  [#permalink]

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New post 23 Oct 2017, 14:50
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Re: N is the product of the first 5 prime numbers. If 12!/n is divisible b &nbs [#permalink] 23 Oct 2017, 14:50
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