Functions: h(n) is product of even integers from 2 to n....

mcheytan

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For every positive integer n, h(n) is the product of all the even integers from 2 to n (inclusive). P is the smallest prime factor of h(100)+1. What is P=?

Since we have only even numbers multiplied and then we add a 1, we should get odd number. Shouldn't the smallest prime factor be between 2 nd 10? The answers are:

a) between 2 and 10
b)between 10 and 20
c)between 20 and 30
d)between 30 and 40
e)over 40

I was thinking that if it is the SMALLEST PRIME FACTOR it will be a 2 or 3 or 5 or 7....
 

stapel

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mcheytan said:
For every positive integer n, h(n) is the product of all the even integers from 2 to n (inclusive). P is the smallest prime factor of h(100)+1. What is P=?
Since n = 100, then the product is:

. . . . .h(100) = 2 * 4 * 6 * ... * 100

Take out the factors of 2:

. . . . .h(100) = 2[sup:157bmry7]50[/sup:157bmry7](1 * 2 * 3 * ... * 50)

Now add the 1:

. . . . .h(100) + 1 = 2[sup:157bmry7]50[/sup:157bmry7](1 * 2 * 3 * ... * 50) + 1

If one number is a factor of another number, then dividing the latter by the former will leave a remainder of zero. Look at the above. If you divide by any number (prime or otherwise) from 1 to 50, what will be the remainder? :wink:

Eliz.
 

mcheytan

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but why should the SMALLEST factor be over 40....such a huge number can surely be divisible by 2.... I have a hard time understanding it..
 

stapel

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mcheytan said:
such a huge number can surely be divisible by 2
Really? You've determined that the odd number, h(100) + 1, is even? :wink:

Eliz.
 

mcheytan

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Oh yeas, I think that tha number will be odd, and divisible at least by 3...but why such a huge number....40 and over....???tha t is where i am stuck, how do I define what is the range of numbers...
how do you know that h(100)=2^50 (.....)
I should not be using a calculator that is why I don't know how to estimate the smallest prime factor.
and it seems to me that whatever the number is, it will be divisible by something smaller than 40
 

Subhotosh Khan

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mcheytan said:
Oh yeas, I think that tha number will be odd, and divisible at least by 3...but why such a huge number....40 and over....???tha t is where i am stuck, how do I define what is the range of numbers...
how do you know that h(100)=2^50 (.....)
I should not be using a calculator that is why I don't know how to estimate the smallest prime factor.
and it seems to me that whatever the number is, it will be divisible by something smaller than 40
No!

The number will not be divisible by any number - less than or equal to 50.

Lets do a small example

Take 2*4*6*8*10 + 1 = 3841

Now this number is NOT divisible by any number less than 5.

Take 2*4*6*8*10*12 + 1 = 46081

Now this number is NOT divisible by any number less than 6.

Similar logic was used by Euler - hundreds of years ago - to prove that there are infinite numbers of prime number.
 

mcheytan

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Oh yes, ok, ..I think I start to understand it!...But h(100)+1 will be odd right?
 

mcheytan

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Thank you very much guys!!!!
 
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