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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
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
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 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
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
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
Expert
Tutor
Joined: 17 Jul 2019
Last visit: 03 Oct 2024
Posts: 1,296
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
h(n) = 2*4*6 .... *n = (2*n+2)!
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!
