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# For every positive even integer n, the function h(n) is defined to be

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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]
This is a naughty question.

Let's go step by step...

h(100) = (2.4.6.8......100)

There are 50 even numbers between 2 and 100 inclusive
Hence there are 50 "2s"

H(100) = 2^50 * (1*2*3*4*......*50)

and (1*2*3*4*......*50) = 50!

So => H(100) = 50!*2^50

H(100) + 1 and H(100) are consecutive numbers.

Note - two consecutive numbers are co-primes
(i.e. they don't have any common factor other than 1)

We can see that H(100) can have any number smaller than 50 as it's factor

So, H(100) + 1 can't have a factor smaller than 50

Hence choose E
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]