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If n is the product of the integers from1 to 20 inclusive, [#permalink]
 29 Dec 2006, 17:04
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If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2k is a factor of n?
A 10
B 12
C 15
D 18
E 20
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I must be reading this problem wrong or misunderstanding it. There are many integers k larger than 20 for which 2k is a factor of n=20!.
Of the choices given, each of them, multiplied by 2, is a factor of 20!. So, 20 would be the answer since it asks for the greatest value.
However, take k = 50 as an example. 2k = 100. Is 100 a factor of 20!. Yes, since 20! includes 5*2*10. 50 is larger than 20, so 20 can't be the largest integer for which 2k is a factor of 20!.
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sorry~ it was a typo mistake
it should be 2^k
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Senior Manager
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with this correction - 18 is the answer...
there are 10 multiples of 2 in 20!, 5 multiples of 4, 2 multiples of 8 and 1 multiple of 16.
total = 10+5+2+1=18
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Senior Manager
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Getting 18
Just took all the 2s from 1 * 2 * ...20.
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I calculate 19 2's between 1-20 inclusive.
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in factorial 20,this way i counted
2 4 6 8 10 12 14 16 18 20 = 10 factors
2 4 6 8 10 =5 factors
2 4 =2 factors
2 = 1 factor
total = 18 factors
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Iawfy wrote: If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2^k is a factor of n?
A 10
B 12
C 15
D 18
E 20
20! = (product of the odd #s from 1 to 19) * (product of the even #s from 2 to 20) = a * 2*2^2*(2*3)*(2^3)*(2*5)*(2^2*3)*(2*7)*(2^4)*(2*9)*(2^2*5) = a´ * 2^18 => k = 18
(a and a` are abbreviations for the products of the odd numbers that make up 20!).
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Oh shoot, I was counting 19 individual 2s, forgetting that one of them is included in 2^18.
I get D, too.
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