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Products of consecutive intervals

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Products of consecutive intervals

by saracuse » Thu Sep 17, 2009 6:38 pm
Function h(n) = product of all even numbers from 2 up to n.

if h(100) ... and if p is the small prime factor of h(100) + 1.... what is p?

a) between 10 and 20
b) btwn 20 and 30
c) btwn 30 and 40
d) greater than 40

D is the answer

I took this to mean: h(100) + 1 = [ 2 x 4 x 6 x 8.. 100 ] + 1. Not sure what to do next.

Thanks for your help!
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by aa2kash » Fri Sep 18, 2009 3:51 am
There is a simple rule that states that n!+1 cannot be divisible by any number less than or equal to n.

modulo arithmetic:
n! mod x (where x <= n) = 0
n! + 1 mod x = 1, therefore not divisible by x.
In this particular question the function can be written as:
h(100) + 1 = (2^50)*50! + 1
therefore it's primes are > 50

correct answer is D
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by viju9162 » Fri Sep 18, 2009 4:20 am
Hi aa2kash,

I didn't understand the explanation :( . is this a standar formula ?
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by aa2kash » Fri Sep 18, 2009 2:30 pm
Hi Viju
For better understanding of this try to analyse it by putting different values of n. for eg n=5
n!+1 --> 5! +1 =121 which is not divisible by 2,3,4,5 i.e. any number less than or equal to 5 . it is divisible by 11 .

Its a general rule therefore just remember this.
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by viju9162 » Sun Sep 20, 2009 12:01 am
Thank you :)
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