Find a base 7 three-digit number which has its digits reversed when expressed in base 9.

\(\text{let the number be }abc\\ 49a+7b+c = 81c+9b+a\\ 48a-2b=80c\\ c = \dfrac{24a-b}{40}\)

 

I'll let you figure out what the valid numbers are using the equation for c above.

Remember digits can only range from 0-6.

And of course digits have to be whole numbers.