When two BCD numbers 0x14 and 0x08 are added, what is the binary…

In packed BCD (8421) representation, each 4-bit nibble encodes one decimal digit from 0 to 9; a nibble value of 10 to 15 is invalid. Ordinary binary addition of two digit nibbles can land in that invalid range, so whenever a digit-pair sum exceeds 9 (or produces a binary carry-out), the BCD correction rule adds 0110 (6) to roll that digit over into a valid digit plus a carry into the next digit position.

1. Split each operand into its BCD digit nibbles: 0x14 gives digits (1, 4); 0x08 gives digits (0, 8).

2. Add the lower digits: 4 + 8 = 12, i.e. binary 1100.

3. 1100 (12) exceeds the valid BCD range of 0-9, so add the correction 0110 (6): 1100 + 0110 = 1 0010. This gives a corrected lower digit of 0010 (2) with a carry of 1 into the upper digit.

4. Add the upper digits plus the carry: 1 + 0 + 1 = 2, i.e. binary 0010, which is already within 0-9, so no further correction is needed.

5. Combine the corrected digits in order, upper then lower: 2 and 2, giving the packed-BCD result 0x22, i.e. binary 0010 0010.

Cross-check with plain decimal arithmetic: 0x14 and 0x08 as packed BCD represent the decimal values 14 and 08. Adding them normally gives 14 + 8 = 22, which packs back into the BCD digits 2 and 2, i.e. 0x22, matching the nibble-level correction above.