For every positive even integer n, the function (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
I have looked a previous threads that explain this question, but i'm having a hard time grasping the concept.
I understand that if i divide h(100) + 1 by any number 1-50, i will get a remainder of 1, but what is the take-away i need to get from this? the "rule" if you will.
function h(n) number properties
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- Bill@VeritasPrep
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h(100)=2*4*6*8...*100
=2*(2*2)*(2*3)*(2*4)...*(2*50)
We can see that h(100) already has all integers from 1 to 50 as factors. Since we know that consecutive integers have no common factors greater than 1, this means that the smallest possible factor of h(100) + 1 must be greater than 50.
=2*(2*2)*(2*3)*(2*4)...*(2*50)
We can see that h(100) already has all integers from 1 to 50 as factors. Since we know that consecutive integers have no common factors greater than 1, this means that the smallest possible factor of h(100) + 1 must be greater than 50.
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fangtray wrote:For every positive even integer n, the function (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
I have looked a previous threads that explain this question, but i'm having a hard time grasping the concept.
I understand that if i divide h(100) + 1 by any number 1-50, i will get a remainder of 1, but what is the take-away i need to get from this? the "rule" if you will.
h(100) = 2 * 4 * 6 * ... * 100
= (2 * 1) * (2 * 2) * (2 * 3) * ... * (2 * 50)
= 2^(50) * (1 * 2 * 3 ... * 50)
Then h(100) + 1 = 2^(50) * (1 * 2 * 3 ... * 50) + 1
Now, h(100) + 1 cannot have any prime factors 50 or below, because dividing this value by any of these prime numbers will give a remainder of 1.
The correct answer is E.
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what do you mean by "consecutive integers have no common factors greater than 1?"Bill@VeritasPrep wrote:h(100)=2*4*6*8...*100
=2*(2*2)*(2*3)*(2*4)...*(2*50)
We can see that h(100) already has all integers from 1 to 50 as factors. Since we know that consecutive integers have no common factors greater than 1, this means that the smallest possible factor of h(100) + 1 must be greater than 50.
I must be interpretting this incorrectly, because in consecutive integers from 17 - 34, wouldnt 17 be the common factor?
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Here's the takeaway: if h(100) is divisible by every prime from 1 to 50, then h(100)+1 CANNOT be a multiple of any of those numbers.fangtray wrote:For every positive even integer n, the function (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
I have looked a previous threads that explain this question, but i'm having a hard time grasping the concept.
I understand that if i divide h(100) + 1 by any number 1-50, i will get a remainder of 1, but what is the take-away i need to get from this? the "rule" if you will.
What Bill was saying is that the greatest common factor of any TWO consecutive integers is 1. Since h(100) and h(100)+1 are consecutive integers, then any prime factor of h(100) CANNOT also be a prime factor of h(100)+1.
Accordingly, the smallest prime factor of h(100)+1 has to be greater than 50.
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