Find a base 7 three-digit number which has its digits reversed when expressed in base 9.
Let the digits of the number be a,b,c where
a is between 1 and 6 inclusive and
b is between 0 and 6 inclusive
and
c is between 1 and 6 inclusive
a>c
\(49a+7b+c=81c+9b+a\\ 0=80c+2b-48a\\ 0=40c+b-24a\\ 24a-b=40c\\ \)
24a could be 24, 48, 72, 96, 120, 144
40c could be 40, 80,
I do not think I will finish it since it is obviously a challenge question for you. You need to examine all the possibilities in orger to find triplets that work. It is not as tiresome as it may seem.
When you finish it, it would be nice for you to display the rest, (or the alternate) logic that you use.
That is another challenge for you.
Please let Kitten continue this. Do not finish it for her. Thanks
LaTex:
49a+7b+c=81c+9b+a\\
0=80c+2b-48a\\
0=40c+b-24a