fizzbuzz, by hand

The classic interview question, without the safety net of a remainder operator: print the numbers 1 through 15, except say fizz for multiples of 3, buzz for multiples of 5, and fizzbuzz for multiples of both.

AArch64 has division but no remainder instruction. The course idiom is udivUnsigned divide; divide-by-zero writes 0. then msubRd = Ra - Rn * Rm.: divide, multiply the quotient back, subtract, and what is left is the remainder.

what is checked

• all fifteen lines, exactly as fizzbuzz demands
• the program exits cleanly
• the remainders come from msubRd = Ra - Rn * Rm., not from a lookup