Question

What is the result of adding A3₁₆ and 4B₁₆?

Answer 1

To add the hexadecimal numbers \( A3_{16} \) and \( 4B_{16} \), we can follow these steps:

1. **Convert each hexadecimal number to decimal.**

- For \( A3_{16} \):
- \( A = 10 \) (in decimal),
- \( 3 = 3 \) (in decimal),
- So, \( A3_{16} = (10 \times 16^1) + (3 \times 16^0) = 160 + 3 = 163_{10} \).

- For \( 4B_{16} \):
- \( 4 = 4 \) (in decimal),
- \( B = 11 \) (in decimal),
- So, \( 4B_{16} = (4 \times 16^1) + (11 \times 16^0) = 64 + 11 = 75_{10} \).

2. **Add the decimal conversions together:**
\[
163 + 75 = 238_{10}
\]

3. **Convert the result back to hexadecimal.**

To convert \( 238_{10} \) to hexadecimal:
- Divide by 16:
- \( 238 \div 16 = 14 \) with a remainder of \( 14 \).
- The quotient \( 14 \) corresponds to \( E_{16} \).
- The remainder \( 14 \) corresponds to \( E_{16} \).

Thus, reading from the least significant digit to the most significant digit, we get:
\[
238_{10} = EE_{16}
\]

Therefore, the result of adding \( A3_{16} \) and \( 4B_{16} \) is \( \boxed{EE_{16}} \).