The addition operation failed and the result is invalid

The 16-bit result fits within 16 bits (no 17-bit overflow); if C=1, the true result would be 17 bits (0x1xxxx)

The low byte addition produced no carry, so the high byte operation was unnecessary

The subtraction borrow from the previous operation is still set

0xC5, unchanged because SWAP requires two operands

0xE7, by performing arithmetic on the nibbles

0x3A, by inverting bits during the swap operation

0x5C, because SWAP exchanges the upper nibble (C) with the lower nibble (5), resulting in 0101 1100

Use repeated addition (add the number to itself 15 times), which avoids multiplication entirely

Use MUL AB instruction with B equals 16, which takes 48 clocks but is more reliable than shifts

Use software multiplication algorithm from Practical 26, which is more accurate than hardware

Perform four consecutive left shifts using RL A (4 machine cycles total), since each left shift multiplies by 2 and 2 to the power of 4 equals 16

0x0A, because ANL performs bitwise AND, preserving only the lower nibble (9AH = 1001 1010, 0FH = 0000 1111, AND = 0000 1010 = 0AH)

0x9A, unchanged because AND doesn't modify the accumulator

0x90, by clearing the lower nibble to all 0s

0x9F, by setting the lower nibble bits to all 1s

Because ADD handles unsigned numbers while ADDC handles signed numbers

Because ADDC requires the carry flag to be cleared first, which ADD does automatically

Because ADD clears and ignores the carry flag, while ADDC includes the carry from the previous byte addition, enabling the carry chain necessary for multi-precision arithmetic

Because ADD is faster than ADDC, so we use it for the first byte to optimize performance

Software uses floating-point arithmetic which is slower, while hardware uses integer division

Software must check for division by zero on every iteration, while hardware checks only once

Software executes one iteration per unit of quotient (255÷1 takes approximately 255 iterations), while hardware uses a restoring/non-restoring algorithm with fixed iterations

Hardware can perform multiple subtractions simultaneously using parallel circuits

The value of the current bit of the multiplier—if the bit is 1, add the multiplicand; if 0, skip the addition

Whether the multiplicand is larger than the current partial product

The value of the current bit of the multiplicand—if the bit is 1, add to the product

Whether the partial product is even or odd—add if even, skip if odd

A=17 (remainder), B=13 (quotient), reversing the storage locations

A=13, B=0 (quotient only, with remainder discarded)

A=13 (quotient: 251÷18=13), B=17 (remainder: 251 mod 18=17), with OV=0 indicating successful division

A=251, B=18 (operands unchanged because division failed)

A=0x01C2 (complete 16-bit result stored in A register)

A=25, B=18 (operands unchanged because MUL requires separate instruction)

A=0xC2 (low byte of product), B=0x01 (high byte of product), because 25×18=450=0x01C2

A=0x01 (high byte), B=0xC2 (low byte), reversing the byte order

C=1 (unsigned overflow), OV=0 (no signed overflow)

C=1 (unsigned overflow: 0x90+0x80=0x110, result > 0xFF), OV=1 (signed overflow: -112+(-128) should be -240, but result appears as +16)

C=0, OV=0 (no overflow in either unsigned or signed arithmetic)

C=0 (no unsigned overflow), OV=1 (signed overflow occurred)