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by independent » Fri Apr 20, 2012 4:27 am
For every positive even 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 and 10
B: between 10 and 20
C: between 20 and 30
D: between 30 and 40
E: greater than 40


Correct answer: E

How is this problem supposed to be solved?
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by sanju09 » Fri Apr 20, 2012 4:34 am
independent wrote:For every positive even 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 and 10
B: between 10 and 20
C: between 20 and 30
D: between 30 and 40
E: greater than 40


Correct answer: E


How is this problem supposed to be solved?
This has been answered many times before on this forum.

https://www.beatthegmat.com/factor-of-h- ... 08755.html
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by Anurag@Gurome » Fri Apr 20, 2012 6:00 pm
independent wrote:For every positive even 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 and 10
B: between 10 and 20
C: between 20 and 30
D: between 30 and 40
E: greater than 40


Correct answer: E

How is this problem supposed to be solved?

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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