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07 Oct 2021, 22:02
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Any two consecutive integers will not have any common factor other than 1. We will using this concept to solve this question.
\(h(100)\) =\( 2* 4 * 6 *......*100\) = \(2^{50} (1*2*3*4....*50)\) = \(2 ^ {50} * 50!\)
h(100) will be divisible by all prime numbers below 50.
Since h(100) and h(100)+1 are consecutive integers and they will not have any common factor other than 1.
We can conclude that smallest prime factor of h(100)+1 will be greater than 50.
Option E is the correct answer.
Thanks,
Clifin J Francis,
GMAT SME
\(h(100)\) =\( 2* 4 * 6 *......*100\) = \(2^{50} (1*2*3*4....*50)\) = \(2 ^ {50} * 50!\)
h(100) will be divisible by all prime numbers below 50.
Since h(100) and h(100)+1 are consecutive integers and they will not have any common factor other than 1.
We can conclude that smallest prime factor of h(100)+1 will be greater than 50.
Option E is the correct answer.
Thanks,
Clifin J Francis,
GMAT SME
19 Dec 2021, 23:53
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enigma123
Dear avigutman
Could you validate my approach?
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?
Thus, h(n) = 2*4*6 .... = (2*n+2)!
h(100)+1 = (2*100+2)!+1 = 202!+1
Because 2 consecutive numbers do not share common factors but 1, the answer will be grate then XXX+1 or grater then 40 ?
Thanks beforehand.
avigutman
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Joined: 17 Jul 2019
Last visit: 30 Sep 2025
Posts: 1,285
Given Kudos: 66
Location: Canada
Schools: NYU Stern (WA)
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Expert reply
20 Dec 2021, 05:36
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BLTN
BLTN there's a problem with your logic above.
Just to illustrate the issue with a less abstract situation, consider the following (using your equation with n=4):
h(4) = 2*4 = (2*4+2)! = 10!
But, 2*4 is not equal to 10!
If you want to use factorial for h(n), you'll have to grab a factor of 2 out of every one of the factors of h(n).
For example, if n=10:
h(10) = 2*4*6*8*10 = (2*1) * (2*2) * (2*3) * (2*4) * (2*5) = 2*2*2*2*2 * (1*2*3*4*5) = 2^5 * 5!
