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itspinky
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My answer is E: greater than 40. I solved the problem as follows:

h(100) = 2 * 4 * 6 * ... * 98 * 100
h(100) = 2^50 * (50!)

We are trying to determine the smallest prime factor of h(100) + 1 or:

(2^50 * 50!) + 1

For a number, x, to be a valid factor of h(100) + 1,

[(2^50 * 50!) + 1] / x
or
[2^50*50!]/x + 1/x

must be a positive integer.

For x = 1 to 50, the above equation will be equivalent to:

a positive integer + 1/x

since 50! is the product of all numbers from 1 to 50. And 1 is obviously not considered a prime number.

Therefore, the smallest prime factor has to be greater than 50, and therefore the correct answer is E. I think I confused myself trying to explain this, so let me know if you need further clarification.
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shaselai
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so the 1/X represents an integer but doesnt that always results in a fraction? Do i need to add to the other large value to make it into an integer and X has to be greater than 50?
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itspinky
hi all. here is one that I am REALLY stuck on. insight please??


Few discussions:

ps-function-74417.html#p556064
gmatprep-question-need-solution-74516.html#p556411


Similar question: x-is-equal-to-the-product-of-all-even-numbers-83892.html
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This question has already been discussed. Please refer p-is-82314.html



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