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Product & prime factors
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poonam1279
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DavoodBeater
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the answer is more than 40, but it is a hard question. you can find the solution is a recently posted topic, with 3-4 various approach to solve. search for than.
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vivek.kapoor83
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it hs been dicussed 1000 times. I ll search and ll let u know the link. But its a tuf ques. no doubt abt it
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logitech
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Ian Stewart wrote: You are not free to pick a value of n here; the question asks specifically about h(100) + 1. That is, n = 100.
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.
LGTCH
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