Ah, then it's easy: ;* Divide 10 bits by 10. ; --------------------- ; ; Input in (HL) ; ; Quotient in (A), remainder in (L). ; Div10by10: ld de,(10 shl 6) ; divisor ld bc,0701h ; iteration count + 1 for xor xor a ; quotient Dtbt2: xor c ; assume set sbc hl,de ; subtract jr nc,Dbt4 ; if no carryout add hl,de ; add back xor c ; clear the bit Dbt4: rr d ; (carry is clear) rr e ; shift divisor add a,a ; shift quotient djnz dbt2 ; loop rra ; correct quotient ret ; a = quotient, l = remainder That will do it for values of a divident up to 1023. The algorithm can be extended somewhat by shifting the value of 10 left more places and increasing the number of iterations. I haven't tested it out on real hardware, but it should work. I've also got a divide 32-bits by 10 along the same line, if you're interested. It's fairly deterministic in terms of cycles; i.e., there's not a lot of difference in timing between a dividends of 0 and 799. Cheers, Chuck