The max Greatest Common Divisor of 2 digit decimal numbers Assume $a$ and $b$ are different one digit positive integer, both within range $[1,9]$. Then what is the possible maximum Greatest Common Divisor of two digits numbers $\overline{ab} ,\overline{ba}.$ ( $\overline{ab}$ means $a$ is the tens digit, $b$ is the ones digit. So $\overline{ab}=10a+b$).
A iterative way to iterate through $1-9$ of $a,b$ for 36 times can find the answer, I want to know any better mathematical way to solve the question. Thanks.
 A: Let $D=\gcd(\overline{ab},\overline{ba})$.
We have
$$\tag1 D\mid \overline{ab}-\overline{ba}=9(a-b),$$
$$\tag2 D\mid a\cdot \overline{ba}-b\cdot \overline{ab}=a^2-b^2=(a+b)(a-b),$$
$$\tag3 D\mid 10\cdot\overline{ab}-\overline{ba}=99a.$$
$$\tag4 D\mid 10\cdot\overline{ba}-\overline{ab}=99b,$$
$$\tag5 D\mid \overline{ab}+\overline{ba}=11(a+b).$$
From $(1)$, $D$ cannot be difvisible by any prime $>7$ (provided $a\ne b$), hence we can cast out $11$ in $(3)$, $(4)$, $(5)$, and combine these into 
$$\gcd(a,b)\mid D\mid \gcd(9,a+b)\gcd(a,b). $$
Also from $(5)$, $D<18$ and in particular $27\nmid D$.
Then the power of $3$ occuring in $D$ is the same as that in $a+b$ (by the digit sum rules for divisibility by $3$ and $9$).
We conclude 
$$ D=\frac{\gcd(a,b)\gcd(a+b,9)}{\gcd(a,b,9)}.$$
In order to maximize, we can make $a+b$ a multiple of $9$ only by making $a+b=9$. Then $\gcd(a,b)\mid 9$ and $D=\gcd(a+b,9)=9$.
Or we can make $a+b$ a multiple of $3$, but $3\nmid a,b$. Then $D=3\gcd(a,b)<9$.
So the maximal gcd is $9$ and happens with $18$ and $81$, $27$ and $72$, etc.
A: Let a prime number $p \mid \overline{ab}=(10a+b)$  and $ p \mid  \overline{ba}=(10b+a)$.
Then 
$$\tag1 p\mid \overline{ab}-\overline{ba}=9(a-b),$$
So either $p \mid 9$ or $p \mid (a-b)$. And $a-b\leq8$. So $$\tag2 p=3,2,5,7.$$
For $p\neq3$, $p\mid (a-b).$
$$p\mid(a-b)+(10a+b)=11a.$$
And because of $(2)$,
$$\tag3 p\mid a$$
$$\tag4 p\mid a-(a-b)=b$$
Since $a$ and $b$ are distinct numbers and with the range $[1,9]$, It is impossible to get both $p\mid a$ and $p\mid b$ when $p=5$ and $p=7$.
So $$\tag5 p=2,3.$$ 
And $$\tag6 g = \gcd(\overline{ab},\overline{ba}) \textrm{ must be the product of the powers of } 2 \textrm{ and } 3\ .$$ 
$$\tag7 g\mid \overline{ab}+\overline{ba}=11(a+b).$$
Because of $(6)$, 
 $$\tag8 g\mid (a+b)\leq 17$$
Because of $(6)$, the max values of $g$ are $16$,$12$. 
For $16$, 
$$\gcd(97,79)=1 .$$
For $12$, 
$$\gcd(39,93)=3$$
$$\gcd(48,84)=12$$
$$\gcd(57,75)=3$$
So the answer is $$a=8,b=4,\textrm{ which yields } \gcd(48,84)=12.$$
The solution is refined based on the solution here.
