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CrackverbalGMAT
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For every positive even integer n, the function h(n) is defined to be 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
​
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For every positive even integer n, the function h(n) is defined to be 19 Dec 2021, 23:53
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enigma123
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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This is how I am trying to solve this. Please help me if you think I am not right. OA is not provided in the book.

h(100) = 2 * 4 * 6 ****************100

Tn = a1 + (n-1) d-----------------------(1) where Tn is the last term, a1 is the first term and d is the common difference of the evenly spaced set.

100 = 2 + (n-1) 2
n = 50

Product of terms = Average * number of terms

Average = (a1+an)/2
Therefore average = 102/2 = 51
Product of the series = 51*50 = 2550.

H(100) + 1 = 2550+1 = 2551 which is prime. And prime numbers have exactly 2 factors 1 and the number itself. Therefore for me D is the answer i.e. < 10
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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.
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For every positive even integer n, the function h(n) is defined to be 20 Dec 2021, 05:36
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BLTN

h(n) = 2*4*6 .... *n = (2*n+2)!
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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!
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For every positive even integer n, the function h(n) is defined to be 12 Aug 2023, 07:40
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This is a naughty question. :shh: :dazed

Let's go step by step...

h(100) = (2.4.6.8......100)

There are 50 even numbers between 2 and 100 inclusive
Hence there are 50 "2s"

H(100) = 2^50 * (1*2*3*4*......*50)

and (1*2*3*4*......*50) = 50!

So => H(100) = 50!*2^50

H(100) + 1 and H(100) are consecutive numbers.

Note - two consecutive numbers are co-primes
(i.e. they don't have any common factor other than 1)

We can see that H(100) can have any number smaller than 50 as it's factor

So, H(100) + 1 can't have a factor smaller than 50

Hence choose E
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For every positive even integer n, the function h(n) is defined to be 12 May 2024, 04:50
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­The +1 turns a yes into a no. No sweat:

­
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For every positive even integer n, the function h(n) is defined to be 13 Jul 2024, 15:18
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­The most intuitive way to solve it, by reading the answer choice carefully, since range is given in choice, we dont need to get the exact anwer, we need a range only now, coming back to the question:
Since h(100) will have all even number till 100 in the product, it concludes that it will have multiple of all the primes till 50 in the product h(100) ie: 2*4*6*.....*100, it means no prime below 50 divide h(100) +1. hence the answer should be  x>40
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For every positive even integer n, the function h(n) is defined to be 07 Sep 2024, 22:51
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Zinsch123
I'm not taking credit for the following solution. Found it in another forum.

Two consecutive integers cannot be divisible by the same integer greater than 1.

So if we can prove that h(100) is divisible by every number smaller than 50, we proved that h(100)+1 is NOT divisible by any number smaller than 50 (besides 1).

2*4*6*...*98*100=2*(1*2*3*...*50)=2*(50!)
Hence, h(100) is divisible by every number smaller than 51. So the smallest prime factor of h(100)+1 is at least 53.
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­So why is it 53 instead of 51?
Many thanks!
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For every positive even integer n, the function h(n) is defined to be 08 Sep 2024, 02:14
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lnyngayan

Zinsch123
I'm not taking credit for the following solution. Found it in another forum.

Two consecutive integers cannot be divisible by the same integer greater than 1.

So if we can prove that h(100) is divisible by every number smaller than 50, we proved that h(100)+1 is NOT divisible by any number smaller than 50 (besides 1).

2*4*6*...*98*100=2*(1*2*3*...*50)=2*(50!)
Hence, h(100) is divisible by every number smaller than 51. So the smallest prime factor of h(100)+1 is at least 53.
­So why is it 53 instead of 51?
Many thanks!
Show more
­
Check precise soluiton here:

https://gmatclub.com/forum/for-every-po ... l#p1035827
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For every positive even integer n, the function h(n) is defined to be 08 Sep 2025, 07:19
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