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For every positive even integer n, the function h(n) is defi [#permalink]

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12 Aug 2009, 23:30

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

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: 1) between 2 and 10 2) between 10 and 20 3) between 20 and 30 4) between 30 and 40 5) greater than 40

We can thus conclude that h(100) is divisible by every number from 1 to 50. Thus all the prime numbers between 1 and 50 are prime factors of h(100).

But when 1 is added [h(100) + 1] is no more divisible by any of these prime numbers. Thus the smallest prime factor of [h(100) + 1] would have to be a number greater than 50.

found this somewhere , but not sure how far this is true, n!+1 cannot be divisible by any number less than or equal to n , may be helpful..

Is it true? Coz if it is.. I will be nothing less than happy! Do you remember where you got this from? What is the source?

Oh yes, it IS true! Isn't it the same property we have used in the explanation above!

If you write down a few numbers on paper, the property becomes quite simple to understand and remember.....

Take 420 = 2^2 * 3 * 5 * 7 Lets take some of the factors of 420 => 2, 3, 4, 5, 6, 7, 10

420 is divisible by all the nos above. But if I add 1 to 420, the new number 421 is not divisible by any of the nos above or any of the actors of 420. REASON? If 420 is divisible by 2, then what is the next smallest number divisible by 2? - Its 420 + 2

|||ly, If 420 is divisible by 3, then the next smallest number divisible by 3 is 420 + 3 If 420 is divisible by 4, then the next smallest number divisible by 4 is 420 + 4

=> 420 + 1 will not be divisible by 2, 3, 4, 5, 6, 7, 10, etc.

Now lets consider n!, n! is divisible by all the number from 1 to n inclusive. Thus n! + 1 will not be divisible by any of the factors of n, which includes the numbers from 1 to n inclusive!!

found this somewhere , but not sure how far this is true, n!+1 cannot be divisible by any number less than or equal to n , may be helpful..

Is it true? Coz if it is.. I will be nothing less than happy! Do you remember where you got this from? What is the source?

Oh yes, it IS true! Isn't it the same property we have used in the explanation above!

If you write down a few numbers on paper, the property becomes quite simple to understand and remember.....

Take 420 = 2^2 * 3 * 5 * 7 Lets take some of the factors of 420 => 2, 3, 4, 5, 6, 7, 10

420 is divisible by all the nos above. But if I add 1 to 420, the new number 421 is not divisible by any of the nos above or any of the actors of 420. REASON? If 420 is divisible by 2, then what is the next smallest number divisible by 2? - Its 420 + 2

|||ly, If 420 is divisible by 3, then the next smallest number divisible by 3 is 420 + 3 If 420 is divisible by 4, then the next smallest number divisible by 4 is 420 + 4

=> 420 + 1 will not be divisible by 2, 3, 4, 5, 6, 7, 10, etc.

Now lets consider n!, n! is divisible by all the number from 1 to n inclusive. Thus n! + 1 will not be divisible by any of the factors of n, which includes the numbers from 1 to n inclusive!!

Hope this helps....

Great thing, samrus98! I have tried on like 10 different numbers and it worked on all of them. That convincing enough for me. However, I would say, that I have read several of your posts and man, you know loads of properties that I don't even know exist. Like that of "Square of any odd number will always have a remainder of 1, when divided by 8". I mean where do you get all that from? Whats the secret, mate?! _________________

GMAT offended me. Now, its my turn! Will do anything for Kudos! Please feel free to give one.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Now, two numbers \(h(100)=2^{50}*50!\) and \(h(100)+1=2^{50}*50!+1\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

As \(h(100)=2^{50}*50!\) has all prime numbers from 1 to 50 as its factors, according to above \(h(100)+1=2^{50}*50!+1\) won't have ANY prime factor from 1 to 50. Hence \(p\) (\(>1\)), the smallest prime factor of \(h(100)+1\) will be more than 50.

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