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Function/Prime Number Problem

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Function/Prime Number Problem

by Onigbogi Tosin » Tue Sep 15, 2009 6:14 am
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 &10

b. Between 10 & 20

c. Between 20 & 30

d. Between 30 & 40

e. Greater than 40
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by sakurle » Tue Sep 15, 2009 9:37 am
If we plug in values. h(6) = 48
h(6) + 1 = 49, smallest factor = 7.

similarly h(8) = 384
h(8)+1 = 385

again smallest factor is between 2 & 10

Therefor IMO A
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by rohan_vus » Tue Sep 15, 2009 9:47 am
the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive

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

Thus all numbers from 1 to 50 are factors of h(100) as 50! = 1*2*3*4*5*6*..50 .

But when we add 1 to h(100) all factors of h(100) will give remainder as 1 ( except factor 1) . So we can say h(100) + 1 is not divisible by any factor from integer 2 to integer 50 . So any prime factor got to be more than 50 which from he list of options is E . So IMO E
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by Onigbogi Tosin » Wed Sep 16, 2009 5:35 am
Please, I'm not yet clear about the solution to this problem. I still need more clarification.Thanks
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by PussInBoots » Wed Sep 16, 2009 12:09 pm
2*4*6*8*..*100 = 1*2 * 2*2 * 3*2 * 4*2 *... * 50*2 = 50! * 2^50

50! is divisible by any positive integer less or equal to 50, just like 4! is divisible by 1, 2, 3, 4. Hence 50! * 2^50 is divisible by any positive integer less or equal to 50. If you add 1, then the new number is not divisible by any number less or equal to 50. Hence smallest prime number > 40.
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