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function h(n) number properties

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function h(n) number properties

by fangtray » Sun May 06, 2012 2:45 pm
For every positive even integer n, the function (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 and 10
b. between 10 and 20
c. between 20 and 30
d. between 30 and 40
e. greater than 40

I have looked a previous threads that explain this question, but i'm having a hard time grasping the concept.

I understand that if i divide h(100) + 1 by any number 1-50, i will get a remainder of 1, but what is the take-away i need to get from this? the "rule" if you will.
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by Bill@VeritasPrep » Sun May 06, 2012 3:26 pm
h(100)=2*4*6*8...*100

=2*(2*2)*(2*3)*(2*4)...*(2*50)


We can see that h(100) already has all integers from 1 to 50 as factors. Since we know that consecutive integers have no common factors greater than 1, this means that the smallest possible factor of h(100) + 1 must be greater than 50.
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by Anurag@Gurome » Sun May 06, 2012 6:08 pm
fangtray wrote:For every positive even integer n, the function (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 and 10
b. between 10 and 20
c. between 20 and 30
d. between 30 and 40
e. greater than 40

I have looked a previous threads that explain this question, but i'm having a hard time grasping the concept.

I understand that if i divide h(100) + 1 by any number 1-50, i will get a remainder of 1, but what is the take-away i need to get from this? the "rule" if you will.

h(100) = 2 * 4 * 6 * ... * 100
= (2 * 1) * (2 * 2) * (2 * 3) * ... * (2 * 50)
= 2^(50) * (1 * 2 * 3 ... * 50)
Then h(100) + 1 = 2^(50) * (1 * 2 * 3 ... * 50) + 1
Now, h(100) + 1 cannot have any prime factors 50 or below, because dividing this value by any of these prime numbers will give a remainder of 1.

The correct answer is E.
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by fangtray » Sun May 06, 2012 9:11 pm
Bill@VeritasPrep wrote:h(100)=2*4*6*8...*100

=2*(2*2)*(2*3)*(2*4)...*(2*50)


We can see that h(100) already has all integers from 1 to 50 as factors. Since we know that consecutive integers have no common factors greater than 1, this means that the smallest possible factor of h(100) + 1 must be greater than 50.
what do you mean by "consecutive integers have no common factors greater than 1?"

I must be interpretting this incorrectly, because in consecutive integers from 17 - 34, wouldnt 17 be the common factor?
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by Stuart@KaplanGMAT » Sun May 06, 2012 11:06 pm
fangtray wrote:For every positive even integer n, the function (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 and 10
b. between 10 and 20
c. between 20 and 30
d. between 30 and 40
e. greater than 40

I have looked a previous threads that explain this question, but i'm having a hard time grasping the concept.

I understand that if i divide h(100) + 1 by any number 1-50, i will get a remainder of 1, but what is the take-away i need to get from this? the "rule" if you will.
Here's the takeaway: if h(100) is divisible by every prime from 1 to 50, then h(100)+1 CANNOT be a multiple of any of those numbers.

What Bill was saying is that the greatest common factor of any TWO consecutive integers is 1. Since h(100) and h(100)+1 are consecutive integers, then any prime factor of h(100) CANNOT also be a prime factor of h(100)+1.

Accordingly, the smallest prime factor of h(100)+1 has to be greater than 50.
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