Problem:
Find the three-digit positive integer aβbβcβ whose representation in base nine is bβcβaβnine β, where a,b, and c are (not necessarily distinct) digits.
Solution:
Writing the problem conditions out algebraically gives us
100a+10b+c=81b+9c+a99a=71b+8c
We can take mod 71, which gives us
28aβ‘8c(mod8)7aβ‘2c(mod8)β
We can easily see that (a,c)=(2,7) and b=2.
Thus, aβbβcβ=227β.
The problems on this page are the property of the MAA's American Mathematics Competitions