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N is the product of the first 5 prime numbers. If 12!/n is divisible b
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Updated on: 01 Jun 2016, 06:09
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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 *An answer will be posted in 2 days.
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Re: N is the product of the first 5 prime numbers. If 12!/n is divisible b
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30 May 2016, 22:29
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
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31 May 2016, 01:39
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
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31 May 2016, 07:16
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
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31 May 2016, 21:00
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
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01 Jun 2016, 17:03
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 101=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
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23 Oct 2017, 13:50
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Re: N is the product of the first 5 prime numbers. If 12!/n is divisible b &nbs
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