The sequence f(n) = (2n)! ÷ n! is defined for all positive integer values of n. If x is defined as the product of the first 10 ten terms of this sequence, which of the following is the greatest factor of x?
(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55
I think the answer is E. Look at the image below to see my reasoning (MS Word makes the calculations look alot neater). Let me know of any flaws. Thanks in advance.
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Manhattan GMAT Challenge Problem
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jayhawk2001
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jrbrown2, your approach looks good.
Just posting a slightly different version if you don't want to deal with
the various factorials.
For n=1, we have 2!/1! = 2 = 2^1
For n=2, we have 4!/2! = 4*3 - factor of 2^2
For n=3, we have 6!/3! = 6*5*4 - factor of 2^3
...
For n=10, result will be a factor of 2^10
Product of the above will hence be a factor of 2^(1 + 2 + 3 + ... + 10)
Using arithmetic progression, sum of 1 + 2 + ... + 10 = 10*11/2 = 55
So, 2^55 will be a factor of the product.
Just posting a slightly different version if you don't want to deal with
the various factorials.
For n=1, we have 2!/1! = 2 = 2^1
For n=2, we have 4!/2! = 4*3 - factor of 2^2
For n=3, we have 6!/3! = 6*5*4 - factor of 2^3
...
For n=10, result will be a factor of 2^10
Product of the above will hence be a factor of 2^(1 + 2 + 3 + ... + 10)
Using arithmetic progression, sum of 1 + 2 + ... + 10 = 10*11/2 = 55
So, 2^55 will be a factor of the product.
