For every positive even integer n, the function h(n) is defined to be the product of all 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
This is a question from GMATPrep Exam. Please explain.
Thankyou
[spoiler]OA:e[/spoiler]
Problem solving
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h(100) = 2 * 4 * 6 * ... * 100sakshis wrote:For every positive even integer n, the function h(n) is defined to be the product of all 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
This is a question from GMATPrep Exam. Please explain.
Thankyou
[spoiler]OA:e[/spoiler]
= (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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