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factor of h (100) + 1

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factor of h (100) + 1

by sanju09 » Thu Mar 22, 2012 12:57 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
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by killer1387 » Thu Mar 22, 2012 1:10 am
sanju09 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
E
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by shanconnoisseur » Thu Mar 22, 2012 1:27 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
===================================

This is tricky.
h(100)= 2*4*6*.........*100
take out 2 from each even number
h(100) becomes 2^50 * (1*2*3*......*50).
h(100) is divisible by all the nos. from 1 to 50 which include prime nos. as well.
so h(100)+1 will not be divisible by any of the number from 1 to 50.
Therefore, ans. prime no. greater than 50.
Option E is correct.
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by sanju09 » Thu Mar 22, 2012 1:29 am
shanconnoisseur 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
===================================

This is tricky.
h(100)= 2*4*6*.........*100
take out 2 from each even number
h(100) becomes 2^50 * (1*2*3*......*50).
h(100) is divisible by all the nos. from 1 to 50 which include prime nos. as well.
so h(100)+1 will not be divisible by any of the number from 1 to 50.
Therefore, ans. prime no. greater than 50.
Option E is correct.
beautiful
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by Anurag@Gurome » Thu Mar 22, 2012 2:12 am
sanju09 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

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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