Question 5426
Let's try base 12.  When you add the last two digits, you add 6+8, which in base 10 you would put down a remainder of 4 and carry a group of 10.  In this base, the result of adding 6+8 is a remainder of 2, so you must be carrying a group of 12.  That makes it appear to be base 12.  Check out the rest of the problem.  In the second digit, you are adding 7+8 plus you had 1 carried over, for a total in base 10 of 16.  However, in base 12 that would be carrying a group of 12, leaving 4 in the second digit of the answer (which is correct!).  Finally add the 5+2 and 1 that was carried for a total of 8, which is also correct.


So this IS base 12.


R^2 at SCC