In this problem, a and b are positive integers. When a is written in base 9, its last digit is 5. When b is written in base 6, its last two digits are 15. When a-b is written in base 3, what are its last two digits? Assume a-b is positive.
Base 6 number = m(6^2) + 1(6) + 5 = 36m + 11
Let m = 1... = 1156 = 4710
Base 9 number
n(9) + 5
Let n = 5 .... = 559 = 5010
5010 - 4710 = 310 = 103
The last two digits are 1 and 0
