Problem:
A rational number written in base eight is a​b​.c​d​, where all digits are nonzero. The same number in base twelve is b​b​.b​a​. Find the base-ten number a​b​c​.
Solution:
The integer parts of the two representations must match, so ab​eight​=bb​twelve​. This implies 8a+b=12b+b, from which a=23​b. Because both a and b must be positive integers less than 8, the only two possibilities for the ordered pair (b,a) are (2,3) and (4,6). For b=4 and a=6 the fractional part of the number equals 0.46twelve ​=124​+1446​=83​=0.30eight ​, so d would be 0. On the other hand if b=2 and a=3, then the fractional part is 0.23twelve ​=122​+1443​=163​=0.14eight ​, and c=1 and d=4. Indeed, 32.14eight ​=22.23twelve ​. The requested number is 321​.
The problems on this page are the property of the MAA's American Mathematics Competitions