If anyone can explain this question to me it will be greatly appreciated.
Thanks in advance.
For every positive even integer n, the function h(n) is defined to be the product of all of 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 then 40
The answer is apparently E, but I would like to know how they come to that.
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yeah E is the correct because
As the question says what would be the smallest prime factor of 101 and the smallest prime factor of 101 is it self
As the question says what would be the smallest prime factor of 101 and the smallest prime factor of 101 is it self
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I don't follow...johnybravo wrote:yeah E is the correct because
As the question says what would be the smallest prime factor of 101 and the smallest prime factor of 101 is it self
function h(n) is defined to be the product of all of the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1
meaning, h(100)+1 = (2*4*6*8*...98*100)+1
As in, a number much bigger and much different than 101.
More explanation anyone?
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Could you please detail howreachac wrote:Write h(n) = 2*(50!) + 1
ne factor for h(n) has to be then greater than 50. So E
(2*4*6*....100) + 1= 2* 50! +1
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That's not quite right, though it's the right idea. You can take a 2 out of all fifty of the terms in the product:rattanas wrote:just take 2 outside it wil be 2(1*2*3*4*...*49*50)....gives it 2* 50!
2*4*6*...*96*98*100 = (2*1)*(2*2)*(2*3)*...*(2*48)*(2*49)*(2*50)
= (2^50) * 50!
So h(100) + 1 = (2^50)*50! + 1
Now notice that 50! is divisible by every prime less than 50. So h(100) is divisible by every prime less than 50. So h(100) + 1 can't be divisible by any prime less than 50; you'll get a remainder of one.
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