A Pre-RMO question! -23

Given that

$\begin{aligned} \overline{abc}_9 & = \overline{cba}_7 \\ 9^2a + 9b + c & = 7^2c + 7b + a & \small \blue{\text{Rearranging}} \\ 80a + 2b & = 48c \\ 40a + b & = 24c \end{aligned}$

Since the right-hand side $24c$ is divisible by $4$ , the left-hand side must be divisible by $4$ . Since $40a$ is divisible by $4$ , $b$ must be divisible by $4$ . Since $0 \le b \le 6$ , $b$ can only be $0$ or $4$ .

When $\begin{cases} \begin{array} {llll} b = 0, & \implies 40a + 0 = 24c & \implies 5a = 3c & \implies a = 3, c = 5 \\ b = 4, & \implies 40a + 4 = 24c & \implies 10a + 1 = 6c & \implies \red{\text{Odd } \ne \text{ even, no solution}} \end{array} \end{cases}$ .

Therefore the number is $9^2(3)+9(0) + 5 = 248_{10}$ with a sum of digits of $2+4+8=\boxed{14}$ .

It is equivalent to $81a+9b+c=49c+7b+a$ , $40a+b=24c$ , hence $b$ is a multiple of $8$ .

However $b<7$ since it appears in the base 7 representation, so $b=0, 5a=3c,a=3,c=5$

The number is $305_9=503_7=248$