Add hexa-decimal numbers (1F5C)₁₆ and (AC2B)₁₆

$\begin{matrix} & \overset{1}{1} & F & \overset{1}{5} & C \ + & A & C & 2 & B \ \hline & \bold{C} & \bold{B} & \bold{8} & \bold{7} \end{matrix}$

Therefore, (1F5C)₁₆ + (AC2B)₁₆ = (CB87)₁₆

Explanation

1. C + B = 12 + 11 = 23. As 23 is greater than 16 so 23 - 16 = 7. Hence, sum = 7 and carry over = 1.
2. 1 + 5 + 2 = 8.
3. F + C = 15 + 12 = 27. As 27 is greater than 16 so 27 - 16 = 11 or B. Hence, sum = B and carry over = 1.
4. 1 + 1 + A = 1 + 1 + 10 = 12 or C.