Hey guys,
Here's the problem:
"For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+ 1, then p is:
(A) between 2 and 10
(B) between 10 and 20
(C) between 20 and 30
(D) between 30 and 40
(E) greater than 40
Correct answer: E
Believe it or not, it had to guess on this one... Would really appreciate your time-saving approaches to this problem.
Cheers,
Tobi
Smallest prime factor of product of even numbers
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This question has been discussed many times in the forum.gander123 wrote:For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+ 1, then p is:
h(100) + 1 = (2*4*6*8*10*...*100) + 1 = (2^50)*(1*2*3*4*5*...*50) + 1
Thus when [h(100) + 1] is divided by any integer (including all the primes) less than or equal to 50, it'll leave a remainder of 1. Thus p must be greater than 50 in turn p > 40.
The correct answer is E.
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Each term of (2*4*6*8*10*...*100) is multiple of 2. I'm just taking one 2 from each term out of the bracket. Hence, I'm taking fifty 2s out of the bracket.gander123 wrote:How do you get there?! Whats the intermediate step?(2^50)*(1*2*3*4*5*...*50) + 1
For example, (2*4*6*8) can be written as [(2*1)*(2*2)*(2*3)*(2*4)] = (2*2*2*2)*(1*2*3*4) = (2^4)*(1*2*3*4)
Hope that helps.
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