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Product & prime factors

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Source: — Problem Solving |
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by DavoodBeater » Sun Jan 04, 2009 1:27 pm
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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by vivek.kapoor83 » Sun Jan 04, 2009 10:59 pm
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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by logitech » Mon Jan 05, 2009 2:26 am
And it is the 3rd question in the test!! NICE!
LGTCH
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by logitech » Mon Jan 05, 2009 2:32 am
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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