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If $a$ is a positive integer, then $3a^2+19a+30$ and $a^2+6a+9$ are also positive integers. We define the function $f$ such that $f(a)$ is the greatest common divisor of $3a^2+19a+30$ and $a^2+6a+9$. Find the maximum possible value of $f(a)- a$.

Guest Jun 22, 2015

Best Answer 

 #1
avatar+26742 
+10

3a2 + 19a + 30 factors as (3a + 10)(a + 3)

 

a2 + 6a + 9 factors as (a+3)2

 

The greatest common factor is therefore f(a) = (a + 3), which means that f(a) - a = 3 for all a.

Alan  Jun 22, 2015
 #1
avatar+26742 
+10
Best Answer

3a2 + 19a + 30 factors as (3a + 10)(a + 3)

 

a2 + 6a + 9 factors as (a+3)2

 

The greatest common factor is therefore f(a) = (a + 3), which means that f(a) - a = 3 for all a.

Alan  Jun 22, 2015

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