GMAT Problem Involving Primes

Enola,

E is the correct answer, however, I think there's a mistake with your method.

h(100) is the product of all even integers from 2 to 100, not the sum of even integers. So, h(100) + 1 would not equal 2551. For example, just look at the product of the last two integers in h(100) -- 100 and 98 -- the product of which is 9800, a number larger than 2551.

I do think, however, that your method has some promise. Let me see if I can get somewhere with it.


Thanks,
Artis

Hello,

I'm having a little trouble with the following problem. I'm looking for the best method to approach it, but have been looking at it for a while and can not come up with anything.

The only explanation I've heard for this problem is that the resulting number is "very large" and so the answer has to be (e), which is not a very satisfying explanation.

Any help would be appreciated.

Thanks,
Artis

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For every positive integer n, 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 2 and 10
- between 10 and 20
- between 20 and 30
- between 30 and 40
- greater than 40

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If you add 1 to a number, the new number can not have any factors greater than 1 in common with the other number. Since you are multiplying every even number from 2 - 100 together, that eliminates any even number from 2 - 100 and all the factors of those numbers. Any prime factor less than 40, multiplied by 2 will equal an even number less than 100 and hence one of the factors of h(100). The prime numbers less the 40 are as follows 2 3 5 7 11 13 17 19 23 29 31 37 41, multiply those by 2 and you have factors of h(100), each of which can be broken down into 2 X that prime number. Therefore, the answer has to be E.

What do you guys think? I am about 3 beers deep and coming off of 6 hours of study, so I could be totally off with this.... but it sounds great to me right now

Yeah sorry I forgot about the +1.....disregard my stupid answer

Thanks, Andrew, that's the best explanation I've heard so far.

Artis

2*4*6*...*100 = (2*1)*(2*2)*...*(2*50) = 2^50 * (1*2*3*...*50)

therefore h(100)+1 = (2^50)*1*2*3*...*50 + 1 so it is not divisible by any number between 2 and 50 (inclusive). this means that its minimal factor must be more than 50, which is more than 40. e is correct.


hope it is better explanation than just "large number".

amit.