Again GCD

Relevant wiki: Extended Euclidean Algorithm

By Euclid's Algorithm

$\gcd(2n+3,3n-6)=\gcd(3n-6-2n-3,2n+3)=\gcd(n-9,2n+3)=\gcd(n-9,n+12)=\gcd(n-9,21) .$

So, the gcd can be $\leq 21$ .Possible values of gcd 's are $=1,3,7,21$

Case(1): When $\gcd=1$

$\gcd(2n+3,3n-6)=\gcd(2n+3),3(n-2))$ The $2n+3$ is always odd if $n-2$ is even and $n \neq 3k$ then their gcd will be $1$ .

Case(2) When gcd will be $3$

if $n=3k$ then both numbers $2n+3$ and $3n-6$ will always have gcd $3$

Case(3) When gcd will be $7$

So $7 | 2n+3$ and $7| 3n-6$ .From simple observation $n=2,9,16...$ will be solutions So $n=7k-5$ then gcd will be always $7$

Case(4) When gcd will be $21$

So, $21 | 2n+3$ and $21 | 3n-6$ Again from observation $n=9,30,51,.....$ then $n=21k-12$ .Here gcd will be $21$

Therefore, possiblle gcd will be $1,3,7,21$ Sum is $32$

Using the Euclidean algorithm we get: gcd(2n+3, 3n-6)=gcd(2n+3, n-9)= gcd(21, n-9). We notice that the maximum value of this expression is 21, so the gcd must divide 21. The list of numbers that divide 21 is: 1, 3, 7, 21. Adding all of these numbers we get the answer, which is 32

Let $d$ be a common divisor of $(2n+3)$ and $(3n-6)$ .

Then, $d \mid (2n+3) \implies d\mid 3(2n+3) \implies d\mid (6n+9)$ .

$d \mid (3n-6) \implies d\mid 2(3n-6) \implies d\mid (6n-12)$ .

As $d\mid (6n+9)$ and $d\mid (6n-12)$ , then $d \mid (6n+9)-(6n-12) \implies d\mid 21$ .

Now, we show that every divisor of $21$ is possible as a value of $gcd(2n+3,3n-6)$ .

For $n=9$ , $gcd(2n+3,3n-6)=gcd(21,21)=21$ .

For $n=16$ , $gcd(2n+3,3n-6)=gcd(35,47)=7$ .

For $n=12$ , $gcd(2n+3,3n-6)=gcd(27,30)=3$ .

For $n=10$ , $gcd(2n+3,3n-6)=gcd(23,24)=1$ .

So, $1+3+7+21=\boxed{32}$ is the answer.