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Number theory

by priyank.hirani » Mon Apr 16, 2012 9:03 am
h(n) is product of all even integers 2 to n, inclusive. if P is smallest prime factor of h(100) + 1 , the n P is":

between 2 to 10
between 10 to 20
between 20 to 30
between 30 to 40
greater than 40
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by bryan88 » Mon Apr 16, 2012 9:12 am
https://www.manhattangmat.com/forums/for ... t1152.html
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by shubham_k » Mon Apr 16, 2012 9:17 am
What is h(100)?

h(100) = 2*4*6*...*96*98*100 = (2*1)*(2*2)*(2*3)*...*(2*48)*(2*49)*(2*50) = (2^50)*50!

So we need to know about the smallest prime factor of (2^50)*50! + 1.

Notice that 50! is divisible by every prime less than 50. That ensures that (2^50)*50! is divisible by every prime less than 50, which ensures that (2^50)*50! + 1 will be divisible by no prime less than 50: the remainder will be 1 each time. The smallest prime factor of h(100) + 1 must therefore be larger than 50 (and therefore certainly larger than 40). E.
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by Anurag@Gurome » Mon Apr 16, 2012 7:18 pm
priyank.hirani wrote:h(n) is product of all even integers 2 to n, inclusive. if P is smallest prime factor of h(100) + 1 , the n P is":

between 2 to 10
between 10 to 20
between 20 to 30
between 30 to 40
greater than 40
h(100) = 2 * 4 * 6 * ... * 100
= (2 * 1) * (2 * 2) * (2 * 3) * ... * (2 * 50)
= 2^(50) * (1 * 2 * 3 ... * 50)
Then h(100) + 1 = 2^(50) * (1 * 2 * 3 ... * 50) + 1
Now, h(100) + 1 cannot have any prime factors 50 or below, because dividing this value by any of these prime numbers will give a remainder of 1.

The correct answer is E.
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by amit28it » Mon Apr 16, 2012 9:33 pm
I am completely agree with Anurag@Gurome the correct answer is E I have just solved it and found the same answer given by him,I always watch your explanation and its very good.
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