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function, factors

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by giaxou » Wed Feb 04, 2009 7:52 am
Can someone pls explain steps in solving this? Thanks!

For every positive integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is
A) between 2 - 10
B) between 10 - 20
C) between 20 - 30
D) between 30 - 40
D) > 40
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Source: — Problem Solving |

by DeepakR » Wed Feb 04, 2009 8:58 am
I am not too sure about this. May be someone can correct me if am wrong.
h(100)=(2*4...100)=2(1*2*3...*50)=2(50*51)/2=2550.

Now prime factors for h(100)=2*17*5*3 which is >40. 1 is neither prime nor composite so i am leaving it. Hence Ans.) D ?

- Deepak
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by Hopper39 » Wed Feb 04, 2009 12:02 pm
Good explanation here:

https://www.manhattangmat.com/forums/for ... t1152.html

That said, Deepak, I think the formula that you used below (1*2*3*...50)=(50*51)/2 only works for an additive formula, not a product formula.

(1+2+3+4+5) = (5*6)/2 = 15
BUT (1*2*3*4*5) does not = (5*6)/2 = 15, actual is 120.
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