## powerprep #3

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### powerprep #3

by dunkin77 » Tue Apr 24, 2007 8:01 am
Hi,

The answer is E) but I need an explanation -- can anyone help??
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by gabriel » Tue Apr 24, 2007 11:42 am
.... as defined h(100) = 2*3*4*5* .... *100 (that is the product of each consecutive digit form 2 to 100 ) ... so each number between 2 and 100 is a multiple of h(100) ...

... the q asks for the range of the samllest prime number of h(100)+1 ... now consider the samllest prime number 2 .. it is a factor of h(100) (bcoz h(100) is the product of cinsecutive integers from 2 to 100 ) ... but it is not a factor of h(100)+1 (bcoz 2 cannot divide 1) .. nxt consider 3 though it is a factor of h(100) .. it is not a factor of h(100)+1 ( bcoz 3 cannot divide 1)...

...similarly u will see that for each prime number between 2 and 100 ... the prime number will be a factor of h(100) but not a factor h(100)+1 ... so the smallest prime number dividing h(100)+1 will have to be greater than 100 ... so the answer is E ...

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by Cybermusings » Wed Apr 25, 2007 5:36 am
but gabriel the question says...h(n) = product of all consecutive even integers...and u've used product of all integers...

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