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Help with a question

by Zitch33 » Fri Jul 11, 2008 1:03 pm
If anyone can explain this question to me it will be greatly appreciated.

Thanks in advance.

For every positive even integer n, the function h(n) is defined to be the product of all of 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 and 10

B. between 10 and 20

C. between 20 and 30

D. between 30 and 40

E. greater then 40

The answer is apparently E, but I would like to know how they come to that.
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by johnybravo » Fri Jul 11, 2008 11:21 pm
yeah E is the correct because
As the question says what would be the smallest prime factor of 101 and the smallest prime factor of 101 is it self
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by dipika upadhyay » Sat Jul 12, 2008 8:01 am
yes plz anybody tell me how we get the ans as E..........plz tell me the reason???????
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by tzink » Sat Jul 12, 2008 1:06 pm
johnybravo wrote:yeah E is the correct because
As the question says what would be the smallest prime factor of 101 and the smallest prime factor of 101 is it self
I don't follow...
function h(n) is defined to be the product of all of the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1

meaning, h(100)+1 = (2*4*6*8*...98*100)+1
As in, a number much bigger and much different than 101.

More explanation anyone?
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by reachac » Sat Jul 12, 2008 11:58 pm
Write h(n) = 2*(50!) + 1

ne factor for h(n) has to be then greater than 50. So E
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by asigheartau » Mon Jul 14, 2008 6:57 am
reachac wrote:Write h(n) = 2*(50!) + 1

ne factor for h(n) has to be then greater than 50. So E
Could you please detail how

(2*4*6*....100) + 1= 2* 50! +1

Thanks,
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by rattanas » Mon Jul 14, 2008 7:10 am
just take 2 outside it wil be 2(1*2*3*4*...*49*50)....gives it 2* 50!
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by Ian Stewart » Mon Jul 14, 2008 7:36 am
rattanas wrote:just take 2 outside it wil be 2(1*2*3*4*...*49*50)....gives it 2* 50!
That's not quite right, though it's the right idea. You can take a 2 out of all fifty of the terms in the product:

2*4*6*...*96*98*100 = (2*1)*(2*2)*(2*3)*...*(2*48)*(2*49)*(2*50)

= (2^50) * 50!

So h(100) + 1 = (2^50)*50! + 1

Now notice that 50! is divisible by every prime less than 50. So h(100) is divisible by every prime less than 50. So h(100) + 1 can't be divisible by any prime less than 50; you'll get a remainder of one.

Many other solutions to this problem can be found on the forum- do a search for a more detailed explanation.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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