+1164 Find A + B(in base 10), given the following addition problem
\(\[ \begin{array}{c@{}c@{\;}c@{}c@{}c@{}c}& & & 4 & A & B_{6}\\ &+& & & 4 & 1_{6}\\ & & & 5 & 3 & A_{6}\\ \end{array} \]\)
We can solve this problem by converting the numbers to base 10 and then adding them.
The number (4AB)_6 in base 10 is 46^2 + A6^1 + B*6^0 = 144 + 6A + B.
The number (41)_6 in base 10 is 46^1 + 16^0 = 24 + 1.
The number (53A)_6 in base 10 is 56^2 + 36^1 + A*6^0 = 180 + 18 + A.
We are given that (4AB)_6 + (41)_6 = (53A)_6, so we have the equation:
144 + 6A + B + 24 + 1 = 180 + 18 + A
This simplifies to 6A + B = 37.
Since A and B are digits in base 6, they must be between 0 and 5. So, the only possible values for A and B are A = 4 and B = 3.
Therefore, A+B = 4+3 = 7.
