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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 t

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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

#1
+26640
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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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#1
+26640
+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

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