Multiply Two 8-bit Numbers

A multiplicand sits at 2000H and a multiplier at 2001H. Multiply them.

There is no MUL on the 8085. Multiplication is something you build, and the simplest construction is the definition itself: 7 * 4 means adding 7 to itself 4 times. That turns a multiplication into a counted loop, which you already know how to write.

Two things make this harder than it sounds, and they are worth separating because they fail in completely different ways:

1. The product does not fit. Two bytes multiplied can reach FE01H. Anything beyond FFH needs sixteen bits, which means the running total cannot live in the accumulator.
2. A multiplier of zero. The natural loop shape on the 8085 decrements a counter and jumps back while it is non-zero, which tests the counter at the bottom. Such a loop always runs at least once — and from zero it runs 256 times, wrapping the whole way round.

So: three steps. Build the loop, widen the total to sixteen bits with the register-pair arithmetic from the previous problem, and then go looking for the input that breaks it.

That last step is in this order for a reason. With an 8-bit product the zero-multiplier bug is invisible: 256 additions wrap to exactly zero, which is the right answer. It only becomes a visible, wrong number once the product is sixteen bits wide — a small lesson in how long a broken program can look correct.