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Given that a is an odd multiple of 7767, find the greatest common divisor of 6a^2+49a+108 and 2a+9.

Guest Jun 21, 2018
 #1
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Odd Multiples of 7767 are 1, 3, 9, 863, 2589, 7767

\(6a^2+49a+108\) and \(2a+9\)

 

Lets do 1: 

\(6(1)^2+49(1)+108\)=163

\(2(1)+9\)=11

So GCF is 1

 

Lets do 3:

\(6(3)^2+49(3)+108\)=309

\(2(3)+9\)=15

So GCF is 3

 

Lets do 9:

\(6(9)^2+49(9)+108\)=1035

\(2(9)+9\)=27

So GCF is 9

 

I can go on with the remaining multiples, but I feel like I am not understanding the question.

apostos1  Jun 21, 2018

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