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prime number question, who can solve this?

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prime number question, who can solve this?

by gian-28 » Sat Dec 01, 2007 4:24 am
If g(n) represented the product of every even integer from 2 to n, then g(80) + 1 is divisible by the lowest prime number p. P is:

A. Between 1 and 10
B. Between 11 and 20
C. Between 21 and 30
D. Between 31 and 40
E. Greater than 40

what I found out is that there's gonna be quite a few zeros at the end ....0000001, so I'd probably go for answer E. However, how would I be able to check if such a number was divisible by 3? pleaaaaase help
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thinking out loud back

by gian-28 » Tue Dec 04, 2007 10:54 am
Thinking out loud back to ya:

"...So I would go for E on the basis that the lowest prime that it could be a multiple of would have to be more than 80."

I think the lowest prime that it could be a mutliple of would have to be more than 37, no? (74 is the the biggest even number between 2 and 78, which is itself a product of a prime factor (2x37)). so if all prime numbers up to 37 are used up, the next biggest prime number would be 41... which would still leave us with choice E. no?

I hope this is not too confusing.
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by abhi75 » Tue Dec 04, 2007 2:41 pm
Here is the explanation from the previous discussion on this board. I then tried to convey the concept through another example with small number.

h(100) = 2* 4* 6* ..... *98*100

=> 2*(2*2)*(2*3)....*(2*49)*(2*50)

=> (2^50)(1*2*3....*49*50)

Means - all integers from 1 to 50 are factors of h(100). So, none of them will be factor of h(100)+1

So, smallest prime factor of h(100)+1 will be greater than 50

To understand this problem, try h(10) instead of h(100)

H(10) = 2 * (2*2) * (2*3) * (2*4) * (2*5) [this value is 3840]
 2^5 (1 * 2 * 3 * 4 *5) [this is also 3840]
So the smalles prime factor of h(10) will be greater than 5.

The answer is 3840 ( 2, 3 and 5 are factors of 3840)
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