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For every positive even integer n, the function h(n) is [#permalink]
09 Feb 2008, 14:21

1

This post received KUDOS

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 between

A. between 2 and 10 B. between 10 and 20 C. between 20 and 30 D. between 30 and 40 E. greater than 40

So basically you guys aren't using any theorem per say, but just doing some testing/playing around? That's what I did too, but I still don't understand why that works. Argh

So basically you guys aren't using any theorem per say, but just doing some testing/playing around? That's what I did too, but I still don't understand why that works. Argh

only basic principles:

1. N=h(100) contains all prime numbers from 2 to 47. In other words, N is divisible by any prime number from set{2..47}. 2. M=h(100)+1. we can choose any prime number (p) from the set and write: M=p*s+1, where s is an integer. 3. M=p*s+1 means that M has reminder 1 for any prime numbers from the set. 4. Therefore, h(100)+1 is not divisible by prime number less or equal than 47.

So basically you guys aren't using any theorem per say, but just doing some testing/playing around? That's what I did too, but I still don't understand why that works. Argh

only basic principles:

1. N=h(100) contains all prime numbers from 2 to 47. In other words, N is divisible by any prime number from set{2..47}. 2. M=h(100)+1. we can choose any prime number (p) from the set and write: M=p*s+1, where s is an integer. 3. M=p*s+1 means that M has reminder 1 for any prime numbers from the set. 4. Therefore, h(100)+1 is not divisible by prime number less or equal than 47.

Hope this help.

how does N=h(100) contain all prime numbers from 2 to 47 ? h(100) is product of all even numbers from 2 to 100 ... and 2 is the only even prime.

Unfortunately, I actually starting multiplying #s.

I made it 2 --> 10 instead of 100

which came to 3840 + 1. Basically, we need to be able to have a prime divisor of 3840 and 1. The only divisor of 1 is 1, and that is not prime. Hence only 1 and 3841 can be a factor...I believe 3841 is prime. Unfortunately, I needed to actually start multiplying #s in order to see what we needed here.

Is my reasoning right?

Also...if this said h(100) + 2 would the answer for p be 2?

Unfortunately, I actually starting multiplying #s.

I made it 2 --> 10 instead of 100

which came to 3840 + 1. Basically, we need to be able to have a prime divisor of 3840 and 1. The only divisor of 1 is 1, and that is not prime. Hence only 1 and 3841 can be a factor...I believe 3841 is prime. Unfortunately, I needed to actually start multiplying #s in order to see what we needed here.

Is my reasoning right?

I guess it is an incorrect way. You are simply lucky with your answer if you prove that h(10)+1 is a prime number, you cannot say that h(12)+1 is also a prime number.

By the way, 3841 is not a prime number: 3841=23*167.

jimmyjamesdonkey wrote:

Also...if this said h(100) + 2 would the answer for p be 2?

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...