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Integer Question

by Baldini » Fri Mar 06, 2009 3:17 am
For every positive 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.

OA is e.
Could someone please explain the answer and reasoning?
Thanks
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by praxis » Fri Mar 06, 2009 6:29 am
All the factors of h(n) are even, so h(n) is even and h(n)+1 is odd. The smallest prime factor is just the smallest factor, not including 1.

Examples:

n=2: h(n) + 1 = 3: 3
n=4: h(n) + 1 = 9: 3
n=6: h(n) + 1 = 49: 7
n=8: h(n) + 1 = 385: 5

Clearly we cannot calculate h(100), and the smallest prime factor is not trivial. However, we can say that it is even, and therefore divisible by 2. Also, every even number between 2 and 100 is included in the factors, and
so is every prime number less than 50 because 100 = 2*50. But none of these prime numbers can be a factor of h(n)+1 because they are factors of h(n) and the are larger than 1.

Therefore the smallest factor must be larger than the largest prime less than 50 (which is 43), and the smallest factor must be greater than 40. (e).

Generally, the smallest prime factor of h(n)+1 is larger than n/2.

Want more questions about divisibility and primes? Try https://www.testsandtutors.com/course/qu ... and-Primes
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https://www.testsandtutors.com/course/view.php/GMAT
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by Baldini » Fri Mar 06, 2009 8:58 am
thank you very much. Now i understand the question!
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by Ian Stewart » Fri Mar 06, 2009 1:26 pm
praxis wrote: Therefore the smallest factor must be larger than the largest prime less than 50 (which is 43)
The largest prime less than 50 is 47, not 43.
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by keeyu2 » Mon Mar 09, 2009 12:11 pm
praxis wrote:All the factors of h(n) are even, so h(n) is even and h(n)+1 is odd. The smallest prime factor is just the smallest factor, not including 1.

Examples:

n=2: h(n) + 1 = 3: 3
n=4: h(n) + 1 = 9: 3
n=6: h(n) + 1 = 49: 7
n=8: h(n) + 1 = 385: 5

Clearly we cannot calculate h(100), and the smallest prime factor is not trivial. However, we can say that it is even, and therefore divisible by 2. Also, every even number between 2 and 100 is included in the factors, and
so is every prime number less than 50 because 100 = 2*50. But none of these prime numbers can be a factor of h(n)+1 because they are factors of h(n) and the are larger than 1.

Therefore the smallest factor must be larger than the largest prime less than 50 (which is 43), and the smallest factor must be greater than 40. (e).

Generally, the smallest prime factor of h(n)+1 is larger than n/2.

Want more questions about divisibility and primes? Try https://www.testsandtutors.com/course/qu ... and-Primes

Why couldn't the prime factors be under 100 since all even number include 2 to 100 as factors?
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