Numberbases

This equation can be written in Base 10 as such: $49a+7b+c=81c+9b+a$ Which can be simplified to: $24a=40c+b$ Now, whatever the value of c is, b will always = 24a mod 10 because 40c does not affect the units.

Therefore we can create a table of values as such: $\begin{matrix} a & b \\ 1 & 4 \\ 2 & 8 \\ 3 & 2 \\ 4 & 6 \\ 5 & 0 \\ 6 & 4 \\ 7 & 8 \\ 8 & 2 \\ 9 & 6 \end{matrix}$ From this we can add the values of c using the equation above: $\begin{matrix} a & b & c \\ 1 & 4 & (0.5) \\ 2 & 8 & 1 \\ 3 & 2 & (1.75) \\ 4 & 6 & (2.25) \\ 5 & 0 & 3 \\ 6 & 4 & (3.5) \\ 7 & 8 & 4 \\ 8 & 2 & (4.75) \\ 9 & 6 & (5.25) \end{matrix}$ From this we can see that there are only 3 solutions. (Bracketed values for c are decimals and invalid for the situations): $\begin{matrix} { abc }_{ 7 } & 281 & 503 & 784 \\ { abc }_{ 9 } & 182 & 305 & 487 \\ { abc }_{ 10 } & 155 & 248 & 403 \end{matrix}$ To achieve these values you need to convert between bases.(See here: Base Converter )

However, we're not done yet, as two answers can't be possible $(281)$ and $(784)$ , this is because base 7 has a maximum digital value of 6, both of these contain an 8, which is impossible.

Therefore the only solution is: $\boxed {248_{10}} = 503_7 = 305_9$

The equation in decimal system

49a+7b+c=81c+9b+a

ie 48a-80c=2b

or 24a-40c=b

therefore b is clearly a multiple of 8 which gives b=0 or 8

since b cannot be greater than 7 therefore b=0;

now the above equation can be written as 24a-40c=0;

ie, 3a=5c; which gives a=5 and c=3 only solution (in the range 0to7)

therefore a=5,b=0,c=3;

converteng the number in decimal system we get 248 as right answer;

Write equation as 49a + 7b + c = 81c + 9b + a.

Therefore, 24a - b - 40c = 0, where 1<=a,c<=6 and 0<=b<=6

Try a=1,2,3,4,5,6. It should be pretty quick if you can see that a=1 is impossible, and if a=2,3,4,6, b will be larger than 6.

So, a = 5, b = 0, c=3.

503 in base 7 is 248 in decimal.