BCD addition rule Compute the BCD result of adding 0100 (BCD 4) and 0110 (BCD 6).

Introduction / Context:Binary-Coded Decimal (BCD) encodes each decimal digit in four bits. Arithmetic in BCD requires digit-wise operations and correction when a nibble exceeds 9. This question tests basic BCD addition and correction logic.

Given Data / Assumptions:

• BCD digits: 0100 (4) and 0110 (6).
• Single-digit BCD addition; no incoming carry.
• We express the final answer as an 8-bit BCD (two nibbles) with the proper correction.

Concept / Approach:Add the two nibbles as binary. If the sum is greater than 1001 (9) or a carry occurs, add 0110 (decimal 6) to correct, producing a carry into the next decimal digit. For 4 + 6, the raw sum is 10 (1010), which requires BCD correction.

Step-by-Step Solution:Raw binary sum: 0100 + 0110 = 1010 (decimal 10).BCD correction: since 1010 > 1001, add 0110 → 1010 + 0110 = 1 0000 (carry into tens place).Final BCD: 0001 0000 (representing decimal 10).

Verification / Alternative check:Decimal check: 4 + 6 = 10. In BCD, 10 is encoded as two digits: 1 and 0 → 0001 0000.

Why Other Options Are Wrong:00010111BCD (17): incorrect; 4 + 6 ≠ 17.00001011BCD (11): incorrect; not the sum of 4 and 6.00010011BCD (13): incorrect result.00000110BCD (6): this is just digit 6, not the sum.

Common Pitfalls:Forgetting the 0110 correction or misplacing the carry into the tens place. Always check whether the nibble exceeds 9 or produced a carry.

Final Answer:00010000BCD