Division using only addition and multiplication?

Question

I need to perform a division operation using only addition and multiplication. I can't use substraction. Is it somehow possible to do that with only these two operations?

Answer 1

A long division involves multiplication and subtraction at each stage, so you only need to consider converting an addition into a subtraction. You can do this (in the decimal number system) by taking the 10's complement of the subtrahend, which you then add and ignore any carry-out (similar to the subtraction algorithm for binary numbers).

Example for $23 - 15 = 08$: The $9$'s complement of $15$ is $(9-1)(9-5) = 84$, hence the $10$'s complement is $84+1=85$. Thus, $23+85=108$. Ignoring the carry-out leading $1$ gives $08$.

NB: it's debatable whether taking the complement is a subtraction operation or not. As there are only $10$ digits, it can be done using a look-up table instead of performing an actual mathematical operation.

Answer 2

If you can do comparisons, then you can manage this to an arbitrary precision.

Let the two numbers be $a$ and $b$, and you are trying to compute $b \div a$. If $a = 0$, the division is illegal, and you are done. If $b = 0$, the quotient is $0$, and you are done.

Otherwise, count up how many times you can add $a$ to itself before reaching or surpassing $b$. Make sure you keep track of the running sums. If you reach $b$ exactly, this is the integer quotient $q$ exactly, and you are done.

Otherwise, one less than the count is $q$, the integer portion of the quotient, and the next-to-last sum was was $qa$. Now multiply both it and $b$ by $10$, so you now are comparing $10qa$ and $10b$. Again, add $a$ to the former until you reach or surpass $10b$. If you reach $10b$ exactly, then the count is $r_1$, the first and last decimal digit of the quotient, and you are done.

Otherwise, $r_1$ is one less than the count. If you kept track of your running sums, the previous sum will be $(10q+r_1)a$. Multiply both this and $10b$ by $10$, so now you are comparing $(100q+10r_1)a$ and $100b$. Again, add $a$ to the former until you reach or surpass $100b$, which will allow you to determine $r_2$, the second digit of the decimal portion of the quotient. I trust you can take it from here.

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Example. Let $a = 4$ and $b = 19$. We can add $a = 4$ to itself $5$ times to surpass $b = 19$, so one less than that is $q = 4$, the integer portion of the quotient.

The previous sum was $qa = 16$. Multiply both this and $b$ by $10$ to get $10qa = 160$ and $10b = 190$. We can add $a = 4$ to $10qa = 160$ a total of $8$ times before surpassing $10b = 190$, so one less than that is $r_1 = 7$, the first digit of the decimal portion of the quotient.

The previous sum was $10qa+r_1a = 188$, so we multiply both this and $190$ by $10$ to get $1880$ and $1900$. We can add $a = 4$ a total of $5$ times to the $1880$ to reach $1900$ exactly, so $r_2 = 5$ is the second and final digit of the decimal portion of the quotient.