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

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by Max@Math Revolution » Mon May 30, 2016 6:46 pm
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

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by Max@Math Revolution » Wed Jun 01, 2016 5:04 pm
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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by 800_or_bust » Thu Jun 02, 2016 6:04 am
Max@Math Revolution wrote: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.
You were missing the exponent sign in the original question.
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