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Manhattan GMAT Challenge Problem

Expert replies
by jrbrown2 » Mon Jun 25, 2007 8:45 pm
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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Source: — Problem Solving |

by jayhawk2001 » Mon Jun 25, 2007 9:11 pm
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.
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by jrbrown2 » Mon Jun 25, 2007 9:16 pm
wow ...i didn't realize a manhattan gmat challenge prob could be done in so few steps
Last edited by jrbrown2 on Thu Jun 28, 2007 4:17 pm, edited 1 time in total.
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by jrbrown2 » Mon Jun 25, 2007 9:23 pm
no prob jay...any suggestions that help shorten the amount of work required is gladly accepted
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