What is number?

$\begin{aligned} \overline{abc} & = \overline{ab} + \overline{ba} + \overline{ac} + \overline{ca} + \overline{bc} + \overline{cb} \\ \Rightarrow 100a + 10b + c & = 22(a+b+c) \quad \quad \color{#3D99F6} {\Rightarrow c \ne 0 \text{ is even }\Rightarrow c =2,4,6,8} \\ 78a & = 12b + 21c \\ 26a & = 4b + 7c \\ 26a - 7c & = 4b \end{aligned}$

The only acceptable positive values of $b$ are when: $\begin{cases} a = 1 \text{ and } c = 2 & \Rightarrow b = \dfrac{26-14}{4} = 3 \\ a = 2 \text{ and } c = 4 & \Rightarrow b = \dfrac{52 -28}{4} = 6 \\ a = 3 \text{ and } c = 6 & \Rightarrow b = \dfrac{78 - 42}{4} = 9 \end{cases}$

For $a > 3$ , $b > 9$ , which is not acceptable. Therefore, the maximum value of $\overline{abc} = \boxed{396}$

Let the maximum value be $x$ . In order to maximize $x$ , we must start by figuring the maximum value of $a$ , since that will be the digit which amplifies $x$ the most.

$100a + 10b + c = 22a + 22b + 22c$

Trying random values of $a$ from 1, it is clear that $a$ cannot cross 3. We set $a = 3$ . This means $234 = 12b + 21c$ . Next, we move to maximizing $b$ . We start with $b =9$ and we find, such a $c$ exists. Hence, by digit jumping, we figure $abc = 396$ .