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# For every positive even integer n, the function h(n) is

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

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09 Feb 2008, 14:21
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
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09 Feb 2008, 14:28
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Expert's post
E

h(100) contains all prime numbers between 2 and 47 inclusive. Obviously, these prime numbers cannot be factors of h(100)+1
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11 Feb 2008, 04:49
I got this one today. E.

Right answer, but I'm not sure either.

What rule is +1 makes it the next prime.... some sieve algorithm?
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11 Feb 2008, 08:58
yeah, what is the fundamental rule and concept we are looking for here ?
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11 Feb 2008, 09:19
pmenon wrote:
yeah, what is the fundamental rule and concept we are looking for here ?

n! + 1 doesn't have 2, 3, ... n as its factors.

generalize m*(n!) + 1 doesn't have 2, 3, ... n as its factors.

h(100) + 1 = 2^50*(50!)+ 1 doesn't have 2,3 ... 50 as its factors -> E
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11 Feb 2008, 16:19
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
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11 Feb 2008, 22:41
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.
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12 Feb 2008, 17:47
walker wrote:
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.
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12 Feb 2008, 18:22
I got E as well...here is my thinking...

Unfortunately, I actually starting multiplying #s.

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?
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12 Feb 2008, 19:25
pmenon wrote:
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.

N=h(100)=2*4*6*8....*98*100=2^50*1*2*3*4*5....*47*48*49*50.
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12 Feb 2008, 19:48
walker, can you dumb it down one more step for me please ?

N= h(100) = 2*4*6*8*...8*100 = 2*2^2*6*2^3 ...

where do you go from here ?
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12 Feb 2008, 19:59
pmenon wrote:
walker, can you dumb it down one more step for me please ?

N= h(100) = 2*4*6*8*...8*100 = 2*2^2*6*2^3 ...

where do you go from here ?

h(100) = 2*4*6*8*10*12...96*98*100
h(100) = 2*(2*2)*(2*3)*(2*4)*(2*5)*(2*6)...(2*48)*(2*49)*(2*50)
h(100) = 2^50*(2*3*4*5*6...48*49*50)
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12 Feb 2008, 20:11
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Expert's post
jimmyjamesdonkey wrote:
I got E as well...here is my thinking...

Unfortunately, I actually starting multiplying #s.

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?

Agree.
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13 Feb 2008, 06:37
walker wrote:
pmenon wrote:
walker, can you dumb it down one more step for me please ?

N= h(100) = 2*4*6*8*...8*100 = 2*2^2*6*2^3 ...

where do you go from here ?

h(100) = 2*4*6*8*10*12...96*98*100
h(100) = 2*(2*2)*(2*3)*(2*4)*(2*5)*(2*6)...(2*48)*(2*49)*(2*50)
h(100) = 2^50*(2*3*4*5*6...48*49*50)

That's some incredible number properties skills there. nice job.
Re: Prime numbers   [#permalink] 13 Feb 2008, 06:37
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