title: Division and remainder

text:
# Question
Title: Division and remainder
This challenge, while probably trivial in most "standard" languages, is addressed to those languages which are so esoteric, low-level, and/or difficult to use that are very rarely seen on this site. It should provide an interesting problem to solve, so this is your occasion to try that weird language you've read about!

### The task

Take two natural numbers `a` and `b` as input, and output two other numbers: the result of the integer division `a/b`, and the remainder of such division (`a%b`).

This is : shortest answer (in bytes), for each language, wins!

### Input/Output

* 0\<=`a`\<=255, 1\<=`b`\<=255. Each of your inputs (and outputs too) will fit in a single byte.
* You may choose any format you like for both input and output, as long as the two numbers are clearly distinguishable (e.g. no printing the two results together without a delimiter)

### Examples

```
a,b->division,remainder
5,7->0,5
5,1->5,0
18,4->4,2
255,25->10,5

```

**Note:** Builtins that return both the result of the division and the remainder are *forbidden*. At least show us how your language deals with applying two functions to the same arguments.

**Note 2:** As always, an explanation of how your code works is very welcome, even if it looks readable to you it may not be so for someone else!

---

## The Catalogue

The Stack Snippet at the bottom of this post generates the catalogue from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

```
## Language Name, N bytes

```

where `N` is the size of your submission. If you improve your score, you *can* keep old scores in the headline, by striking them through. For instance:

```
## Ruby, <s>104</s> <s>101</s> 96 bytes

```

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the *last* number in the header:

```
## Perl, 43 + 2 (-p flag) = 45 bytes

```

You can also make the language name a link which will then show up in the snippet:

```
## [><>](https://esolangs.org/wiki/Fish), 121 bytes

```

```
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        link: a.share_link,
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```
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  <table class="language-list">
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```

# Answer
# BitCycle, ~~146~~ ~~79~~ 64 bytes

*Just realized a whole section of my original code was unneccessary. Huge reduction!*

```
v  <>!
A\B^^=
? D^>^<
 >\v^~ D@
 >/ C/
  > C ^
A/B v
^   <
?  D^

```

The program takes input in unary from the command line, with the divisor first. It outputs the quotient and remainder in unary, separated by a `0`. For example, here's `a=11`, `b=4`, `a/b=2`, `a%b=3`:

```
C:\>python bitcycle.py divmod.btc 1111 11111111111
110111

```

### Ungolfed, in action

Here's my ungolfed version computing `a=3`, `b=5` with animation turned on (sorry about the glitchiness):

### Attempt at an explanation

*The explanation applies to the ungolfed version. Before you tackle it, I highly recommend you read the Esolangs page to get a feel for how the language works.*

The algorithm goes like this:

* Run an outer loop until the program is terminated.
    + Run an inner loop over the bits of the divisor, pairing them off with bits from the dividend.
        - If all bits of the divisor have matching dividend bits, output a single bit.
        - If not all bits of the divisor have matching dividend bits, output the separator `0` followed by what dividend bits there were, then terminate.

The heart of the code is the relationships among the collectors (the uppercase letters). Since there are multiple separate collectors with each letter, let's refer to them as `A1`, `A2`, `B1`, `B2`, etc., numbering from top to bottom.

* `A1` and `A2` hold the divisor and dividend, respectively, at the beginning of the main loop.
* The inner loop peels off one bit at a time from the divisor and the dividend.
    + The rest of the divisor, if any, always goes into `B1`.
    + If both the divisor and dividend were nonempty, one bit goes into `C1` and one into `C3`. The rest of the dividend goes into `B2`.
    + If only the divisor was nonempty, we've reached the end of the dividend, and it's time to print the remainder. The bit from the divisor goes into `C2`.
    + If only the dividend was nonempty, we've reached the end of the divisor; it's time to process the bits in `C3` or `C2` for output. The rest of the dividend goes into `C4`.
* If there are any bits in the `B` collectors, they cycle their contents back around to the `A` collectors and continue in the inner loop.
* Once the `A` and `B` collectors are all empty, the `C` collectors open and we proceed to the processing stage:
    + `C1` and `C4` dump their contents (the divisor and the remaining dividend, respectively) into `D1` and `D3`.
    + If `C2` is empty, we're still printing the quotient.
        - The contents of `C3` go up to the top right `=` switch. The first `1` bit passes straight through to `!` and is output.
        - When the `1` bit passes through, it activates the switch to point rightward, which sends all the subsequent bits off the board.
    + If `C2` is not empty, we're printing the remainder.
        - The first bit of `C2` is negated to a `0` and passed through the switch. The `0` goes on to `!` and is output.
        - When the `0` bit passes through, it activates the switch to point leftward. Now all the bits from `C3` go leftward from the switch and are redirected around into the `!`, outputting the entire remainder.
        - A copy of the first bit from `C2` is also sent into `D2`.
* Now the `D` collectors open.
    + If there is anything in `D2`, that means we just printed the remainder. The bit from `D2` hits the `@`, which terminates the program.
    + Otherwise, the contents of `D1` and `D3` loop back into `A1` and `A2` respectively, and the main loop starts over.
> 24 votes

# Answer
## Funciton, ~~224~~ 108 bytes

Byte count assumes UTF-16 encoding with BOM.

```
 ┌──┬───┐
┌┴╖╓┴╖ ┌┴╖
│%╟║f╟┐│÷╟┘
╘╤╝╙─╜│╘╤╝
 └────┴─┘

```

Try it online!

The above defines a function `f`, which takes two integers and returns both their division and their product (functions in Funciton can have multiple outputs as long as the sum of inputs and outputs doesn't exceed 4).

Using two input values for multiple purposes is actually quite trivial: you simply split off the connector with a T-junction at the value will be duplicated along both branches, which we can then feed separately to the built-ins for division and modulo.

It actually took me twice as long to figure out how to display the result to the user than just to implement the solution.

Also, Funciton has a built-in divmod, `÷%`, and amusingly the built-ins `÷` and `%` that my solution uses are implemented *in terms of* `÷%`. However, my function `f` above isn't *quite* identical to `÷%`: I had to swap the order of the inputs and although it seems like it should be easy to change that, so far I haven't been able to do so without increasing the byte count.
> 15 votes

# Answer
# brainfuck, ~~43~~ 41 bytes

```
,<,[>->+<[>]>>>>+<<<[<+>-]<<[<]>-]>>.>>>.

```

This uses a modified version of my destructive modulus algorithm on Esolangs.

The program reads two bytes – **d** and **n**, in that order – from STDIN and prints two bytes – **n%d** and **n/d**, in that order – to STDOUT. It requires a brainfuck interpreter with a doubly infinite or circular tape, such as the one on TIO.

Try it online!

### How it works

Before the program starts, all cells hold the value **0**. After reading **d** from STDIN (`,`), moving one step left (`<`) and reading **n** from STDIN (`,`), the tape looks as follows.

```
       v
A      B      C      D      E      F      G      H      J
0      n      d      0      0      0      0      0      0

```

Next, assuming that **n \> 0**, we enter the while loop

```
[>->+<[>]>>>>+<<<[<+>-]<<[<]>-]

```

which transforms the tape as follows.

First, `>->+<` advances to cell **C** and decrements it, then advances to cell **D** and increments it, and finally goes back to cell **C**. What happens next depends on whether the value of cell **C** is zero or not.

* If cell **C** hold a positive value, `[>]` (go right while the cell is non-zero) will advance to cell **E**.

`>>>>+<<<` advances to cell **J** to increment it, then goes back to cell **F**.

Since cell **F** will always hold **0**, the while loop `[<+>-]` is skipped entirely, and `<<` goes back to cell **D**.

Finally, since neither **D** nor **C** hold **0**, `[<]` (go left while the cell is non-zero) will retrocede to cell **A**.
* If cell **C** holds **0**, the loop `[>]` is skipped entirely; `>>>>+<<<` advances to cell **G** to increment it, then goes back to cell **D**.

At this point, **D** will hold **d** (in fact, the sum of the values in **C** and **D** will always be **d**), so `[<+>-]` (while **D** is positive, increment **C** and decrement **D**) will set **C** to **d** and **D** to **0**.

Finally, `<<` retrocedes to cell **B**, `[<]` (go left while the cell is non-zero) further left to cell **A**.

In both cases, `>-` advances to cell **B** and decrements it, and the loop starts over unless this zeroes it out.

After **k** iterations, the tape looks as follows.

```
       v
A      B      C      D      E      F      G      H      J
0      n-k    d-k%d  k%d    0      0      k/d    0      k-k/d

```

After **n** iterations **B** is zeroed out and we break out of the loop. The desired values (**n%d** and **n/d**) will be stored in cells **D** and **G**, so `>>.>>>.` prints them.
> 15 votes

# Answer
# APL (Dyalog), 5 bytes

-2 bytes thanks to @ngn

```
⌊÷,|⍨

```

This is an atop (2-train) of a fork (3-train), where the atop's right tine is a derived function (the result of an operator applied to a function):

```
       result
         ↑┌──────────┐
         ││    ┌────┐│┌──────┐ (derived function)
         │↓    │    ↓│↓      │╱
       ┌───┐ ┌───┐ ┌───┐ ╔═══╤═══╗
       │ ⌊ │ │ ÷ │ │ , │ ║ | │ ⍨ ║
       └───┘ └───┘ └───┘ ╚═══╧═══╝
              ↑ ↑         ↑     ↑ ╲
left argument ┴─────────────────┘  (operator)
                └─────────┴ right argument

```

floor of

`÷` division

`,` catenated to

`|` division remainder
with swapped arguments (APL modulus is "backwards")

Try it online!
> 11 votes

# Answer
# Regex (ECMAScript / Python), 57 bytes

```
(x(x*)),(x*?)(?=\1*$)(x?(x*))(?=\4*$)((?=\2+$)\2\5*$|$\4)

```

Try it online! \- ECMAScript
Try it online! \- Python

Takes its arguments in unary, as two strings of `x` characters whose lengths represent the numbers. The divisor comes first, followed by a `,` delimiter, followed by the dividend. The quotient and remainder are returned in the capture groups `\4` and `\3`, respectively.

I developed the basic form of this on 2014-04-03 while working on my abundant numbers regex. It had been a month earlier that teukon and I had independently come up with the multiplication algorithm described here, but up until this point we hadn't written or golfed any regex to find the unknown quotient of a known dividend and divisor. And it wasn't until dividing by \\$\sqrt 2\\$ that I adapted it to handle a dividend of zero correctly.

It is used, in its various forms (shown above or below), by these other regexes:

Commented and indented:

```
(x(x*)),          # \1 = divisor; \2 = \1-1; tail = dividend
(x*?)(?=\1*$)     # \3 = remainder of division; tail -= \3
(x?(x*))          # \4 = conjectured quotient - find the largest one that matches the
                  #      following assertions; \5 = \4-1, or 0 if \4==0; tail -= \4
(?=\4*$)          # assert tail is divisible by quotient
(
    (?=\2+$)      # assert tail is positive and divisible by divisor-1
    \2\5*$        # assert tail-(divisor-1) is divisible by quotient-1
|
    $\4           # if dividend == 0, assert that quotient == dividend
)

```

To show what's going on, here is a Python function that does division using the regex's algorithm:

```
def modulo_is_zero(n, modulus):
    return n==0 if modulus==0 else n>=0 and n % modulus == 0

def divide(dividend, divisor):
    if divisor==0:
        return
    remainder = dividend % divisor
    dividend -= remainder
    quotient = dividend
    if dividend != 0:
        while 1:
            if modulo_is_zero(dividend - quotient, quotient) and \
               modulo_is_zero(dividend - quotient, divisor-1) and \
               modulo_is_zero(dividend - quotient - (divisor-1), quotient-1):
                break
            quotient -= 1
            if quotient == 0:
                return
    return [quotient, remainder]

```

Try it online!

Notice that in regex, \\$0 \equiv 0 \pmod 0\\$. This is very convenient, as it allows a quotient of \\$1\\$ to be returned with no special case. The only special case we actually need to handle is when the dividend is zero.

First, to demonstrate that the correct answer satisfies the assertions used – assume the remainder has already been subtracted away, and we are dividing \\$B=C/A\\$:

\\$\begin{aligned} C &= AB \\ C-B &= (A-1)B \\ C-B-(A-1) &= (A-1)(B-1) \end{aligned}\\$

Thus,

\\$\begin{aligned} C &\equiv 0 \pmod A \\ C &\equiv 0 \pmod B \\ C-B &\equiv 0 \pmod {A-1} \\ C-B-(A-1) &\equiv 0 \pmod {B-1} \end{aligned}\\$

The last two can be simplified to:

\\$\begin{aligned} C &\equiv B \pmod {A-1} \\ C &\equiv A \pmod {B-1} \end{aligned}\\$

At this point it becomes apparent why neither of these two alone would be enough. If \\$A\<B\\$, it looks like the first one could yield false positives, and if \\$B\<A\\$, it looks like the second one could. It turns out that only one of them is needed if we're guaranteed the input meets certain constraints, but until recently I simply accepted this intuitively and from testing the algorithm up to large numbers.

Unlike the multiplication algorithm, this division algorithm can't simply be proved correct from the Chinese remainder theorem, because \\$C\\$ is now constant while \\$B\\$, which defines one of the moduli, is the unknown. But thanks to H.PWiz, we finally have a rigorous proof for why the generalized form of division always works, and rigorously defined thresholds for when the division works in its two shortened forms.

Now let's look at ways to shorten the regex. First, if we assume \\$C\ge A\\$, it becomes **50 bytes**:

```
(x(x*)),(x*?)(?=\1*$)(x(x*))(?=\4*$)(?=\2+$)\2\5*$

```

Try it online! \- ECMAScript
Try it online! \- Python
Try it online! \- Python equivalent

If we know that all divisors will be in the required range, the division regex can be shortened further.

With \\$C \equiv A \pmod {B-1}\\$ and \\$C \equiv 0 \pmod B\\$, we need to find the *largest* matching quotient.
With \\$C \equiv B \pmod {A-1}\\$, we need to find the *smallest* matching quotient.

---

Coming in at **42 bytes**, here is the case of using only \\$C \equiv A \pmod {B-1}\\$ and \\$C \equiv 0 \pmod B\\$:

```
(x(x*)),(x*?)(?=\1*$)(x(x*))(?=\4*$)\2\5*$

```

Try it online! \- ECMAScript

This can be shortened further, thanks to a trick found by Grimmy (and used in this post). I've used it to shorten factorial, Proth, \\$np\_n\\$, and consecutive-prime/constant-exponent regexes. As a standalone division regex, it comes in at **39 bytes**:

```
(x+),(x*?)(?=\1*$)((x*)(?=\1\4*$)x)\3*$

```

Try it online! \- ECMAScript
Try it online! \- Python
Try it online! \- Python non-regex equivalent

```
(x+),             # \1 = divisor
(x*?)(?=\1*$)     # \2 = remainder of division; tail -= \2
(                 # \3 = conjectured quotient - find the largest one that matches
    (x*)          # \4 = \3-1; tail -= \4
    (?=\1\4*$)    # assert tail-(quotient-1)-divisor is divisible by quotient-1
    x             # tail -= 1
)
\3*$              # assert both tail and dividend are divisible by quotient

```

The above algorithm is guaranteed to return the correct quotient if at least one of the following constraints is met (along with \\$C \equiv 0 \pmod A\\$):

* \\$A^2+2A \< 4C\\$ or equivalently \\$A+2 \< 4B\\$
* \\$A=C\\$ or equivalently \\$B=1\\$
    + This works because the regex subtracts \\$C-(B-1)-A\\$ in order to assert that it is \\$\equiv 0 \pmod {B-1}\\$, and that subtraction will always result in a non-match (as it can't have a negative result) if \\$A=C\\$, unless \\$B=1\\$.
* \\$A\\$ is a prime power
    + One of the cases in which this must be true, is when \\$C\\$ is semi-prime and neither of the two other conditions are met. It also must be true if \\$C\\$ is a prime power.

The proof of the first constraint follows. We start from the assertions made by the regex:

\\$\begin{aligned} \qquad C &\equiv 0 \pmod B \\ C &\equiv A \pmod {B-1} \end{aligned}\\$

Suppose the algorithm finds a \\$B'\\$ satisfying these moduli, with \\$C=A'B'\\$. Since the search for \\$B\\$ is done from largest to smallest, the only way for it to first find a \\$B'\ne B\\$ is with \\$B\<B'\\$. Then:

\\$\begin{aligned} \qquad C &\equiv 0 \pmod {B'} \\ C &\equiv A \pmod {B'-1} \\ A'B' &\equiv A \pmod {B'-1} \\ B' &\equiv 1 \pmod {B'-1} \end{aligned}\\$

So, by modular division of \\$A'B'/B'\\$,

\\$\qquad A' \equiv A \pmod {B'-1}\\$

Due to \\$AB=A'B'\\$ and \\$B\<B'\\$,

\\$\begin{aligned} \qquad A &\> A' \\ A &\ge A'+B'-1 \end{aligned}\\$

Due to the AM-GM inequality, we have:

\\$\begin{aligned} A'+B' &\ge 2\sqrt C \\ A'+B'-1 &\ge 2\sqrt C-1 \\ A &\ge 2\sqrt C-1 \\ {(A+1)}^2 &\ge {(2\sqrt C)}^2 \\ A^2+2A+1 &\ge 4C \end{aligned}\\$

So for a \\$B'\\$ to be found, \\$A^2+2A+1 \ge 4C\\$ must be true. Therefore, if \\$A^2+2A+1 \< 4C\\$, no such \\$B'\\$ will be found. But \\$A^2+2A+1 = 4AB\\$ has no solutions:

\\$\begin{aligned} A^2+2A+1 &\stackrel?= 4AB \\ A &\> 0 \\ A+2+{1\over A} &\ne 4B \end{aligned}\\$

So we can simplify the inequality that guarantees no \\$B'\\$ will be found:

\\$\begin{aligned} A^2+2A+1 &\le 4C \\ A^2+2A &\< 4C \end{aligned}\\$

And now the proof of the prime power constraint. Suppose that \\$A\\$ is a prime power and the algorithm has found a \\$B'\\$:

\\$\begin{aligned} \qquad A&=p^n \\ A'B'&=C\\ B'&\>B \\ A'&\<A \\ A' &= p^{n-k},\ k\>0 \end{aligned}\\$

So \\$B'\\$ must have gained the prime power factor that \\$A'\\$ lost:

\\$\begin{aligned} \qquad B' &= xp^k\\ B' &\equiv 0 \pmod p \\ A &\equiv 0 \pmod p \end{aligned}\\$

So \\$A\\$ and \\$B'-1\\$ are coprime, and we can do modular division by \\$A\\$, since it is not \\$\equiv 0 \pmod {B'-1}\\$:

\\$\begin{aligned} \qquad C &\equiv A \pmod {B'-1} \\ AB &= A\pmod {B'-1} \\ A &\equiv A \pmod {B'-1} \\ B &\equiv 1 \pmod {B'-1} \\ B &\< B' \\ B &\le B'-1 \\ B &= 1 \end{aligned}\\$

So for a \\$B'\\$ to be found when \\$A=p^n\\$, it must be the case that \\$B=1\\$, meaning \\$A=C\\$. But we already know that the algorithm works properly when \\$A=C\\$.

---

Coming in at **~~38~~ 30 bytes**, here is the case of using only \\$C \equiv B \pmod {A-1}\\$:

```
(x(x*)),(x*?)(?=\1*$)(x+?)\2+$

```

Try it online! \- ECMAScript
Try it online! \- Python
Try it online! \- Python non-regex equivalent

```
(x(x*)),          # \1 = divisor; \2 = \1-1; tail = dividend
(x*?)(?=\1*$)     # \3 = remainder of division; tail -= \3
(x+?)             # \4 = conjectured quotient - find the smallest one that matches
\2+$              # assert tail is positive and divisible by divisor-1

```

The above algorithm is guaranteed to return the correct quotient if at least one of the following constraints is met:

* \\$A^2 \> C \ge A\\$ or equivalently \\$A \> B \ge 1\\$
* \\$A=1\\$ or equivalently \\$B=C\\$

The proof of the first constraint follows. Starting from the assertion made by the regex:

\\$\qquad C \equiv B \pmod {A-1}\\$

Suppose the algorithm finds a \\$B'\\$ satisfying this modulus. Since the search for \\$B\\$ is done from smallest to largest, the only way for it to first find a \\$B'\ne B\\$ is with \\$B\>B'\>0\\$. Then:

\\$\begin{aligned} \qquad C &\equiv B' \pmod {A-1} \\ AB &\equiv B' \pmod {A-1} \\ A &\equiv 1 \pmod {A-1} \end{aligned}\\$

So, by modular division of \\$AB/A\\$,

\\$\begin{aligned} B &\equiv B' \pmod {A-1} \\ B &\> B' \\ B &\ge B' + A-1 \\ B-(A-1) &\ge B' \> 0 \\ B-(A-1) &\> 0 \\ B &\> A-1 \\ B &\ge A \end{aligned}\\$

So for a \\$B'\\$ to be found, \\$B \ge A\\$ must be true. Therefore, if \\$B \< A\\$, no such \\$B'\\$ will be found.

This regex was actually discovered just for the sake of this post, to complete the symmetry. Then the math above proved it could be shortened from 38 to just 30 bytes by dropping the assertion of \\$C \equiv 0 \pmod B\\$, which was not referenced anywhere in the proof; I hadn't even tried shortening it that much beforehand.

At the time of posting it had never yet been used in a larger regex. Since then I found the perfect use for it in Shift right by half a bit, shortening that regex by 141 bytes.

---

So now let's prove that the generalized form of division works for all inputs. We have:

\\$\begin{aligned} \qquad C &\equiv 0 \pmod {B} \\ C &\equiv B \pmod {A-1} \\ C &\equiv A \pmod {B-1} \end{aligned}\\$

Suppose the algorithm finds a \\$B'\\$ satisfying these moduli, with \\$C=A'B'\\$. Since the search for \\$B\\$ is done from largest to smallest, the only way for it to first find a \\$B'\ne B\\$ is with \\$0\<B\<B'\\$. So now, doing the same steps we did for the two shortened forms of division:

\\$\begin{aligned} C &\equiv 0 \pmod {B'} \\ C &\equiv A \pmod {B'-1} \\ A'B' &\equiv A \pmod {B'-1} \\ B' &\equiv 1 \pmod {B'-1} \\ A' &\equiv A \pmod {B'-1} \\ A' &\< A \\ A'+B'-1 &\le A \end{aligned}\\$ \\$\begin{aligned} \ \\ C &\equiv B' \pmod {A-1} \\ AB &\equiv B' \pmod {A-1} \\ A &\equiv 1 \pmod {A-1} \\ B &\equiv B' \pmod {A-1} \\ B &\< B' \\ B+A-1 &\le B' \end{aligned}\\$

Putting it together,

\\$\begin{aligned} B+A-1 &\le B' \\ B+(A'+B'-1)-1 &\le B' \\ B+A'+B'-2 &\le B' \\ B+A' &\le B'-B'+2 \\ B+A' &\le 2 \\ B &\> 0 \\ A' &\> 0 \\ B = A' &= 1 \\ B' &= C \\ A &= C \end{aligned}\\$

So for a \\$B'\\$ to be found, \\$B'=C\\$ and \\$A=C\\$ must be true. But the regex tries to assert the modulus

\\$C-B'-(A-1) \equiv 0 \pmod {B'-1}\\$

But if \\$B'=C\\$ and \\$A=C\\$, this becomes

\\$\begin{aligned} C-C-(C-1) \equiv 0 \pmod {B'-1} \\ 0-(C-1) \equiv 0 \pmod {B'-1} \end{aligned}\\$

If \\$C\>1\\$, the regex will fail to subtract \\$C-1\\$ from \\$0\\$ before it can even try to assert the modulus, thus the search for \\$B\\$ will continue with other values, and the correct one will be found.

If \\$C=A=1\\$, the answer \\$B'=1\\$ is correct, as \\$B=1\\$ in this case.

Therefore the regex will always give the correct answer for \\$C\>0\\$, and since \\$C=0\\$ is handled as a special case, it will give the correct answer for all inputs.
> 11 votes

# Answer
# JavaScript (ES6), 17 bytes

Thanks to @Arnauld for golfing off one byte

```
x=>y=>[x/y|0,x%y]

```

Receives input in format (x)(y)

Gets floor of x/y by performing bitwise or
Gets remainder by x%y
Puts both values in an array so that they can both be returned

Try it online!
> 10 votes

# Answer
# Brachylog, 6 bytes

```
{÷|%}ᶠ

```

Try it online!

### Explanation

We abuse the metapredicate `ᶠ findall` to apply two different predicates to the Input list of two arguments:

```
{   }ᶠ         Findall for the Input [A,B] :
 ÷               Integer division
  |              Or…
   %             Modulo

```
> 9 votes

# Answer
# Java 8, 18 Bytes

`(a,b)->a/b+","+a%b`

This is a lambda expression of the type `BiFunction<Integer, Integer, String>`.

I'm surprised... this is actually a fairly concise solution for Java. Go lambda expressions!
> 7 votes

# Answer
# MATL, ~~12~~ 10 bytes

```
Qt:ie=&fhq

```

Input is `a`, then `b`. Output is remainder, then quotient.

Try it online!

### Explanation

This avoids both modulo and division. Instead it uses *array reshaping*:

1. Build an array of `a+1` nonzero elements.
2. Reshape as a 2D array of `b` rows. This automatically pads with zeros if needed.
3. The row and column indices of the last nonzero entry, minus `1`, are respectively the remainder and quotient.

Consider for example `a=7`, `b=3`.

```
Q    % Input a implicitly. Push a+1
     % STACK: 8
t:   % Duplicate. Range from 1 to that
     % STACK: 8, [1 2 3 4 5 6 7 8]
ie   % Input b. Reshape as a matrix with b rows (in column major order)
     % STACK: 8, [1 4 7;
                  2 5 8]
                  3 6 0]
=    % Compare for equality
     % STACK: [0 0 0;
               0 0 1;
               0 0 0]
&f   % Row and column indices (1-based) of nonzero element
     % STACK: 2, 3
hq   % Concatenate. Subtract 1. Implicitly display
     % STACK: [1 2]

```
> 7 votes

# Answer
# Regex (Perl / PCRE), ~~36~~ 35 bytes

```
x(x*),((x(?=((?(4)\4)\1)))*)\4?(x*)

```

Try it online!

Takes its arguments in unary, as two strings of `x` characters whose lengths represent the numbers. The divisor comes first, followed by a `,` delimiter, followed by the dividend. The quotient and remainder are returned in the capture groups `\2` and `\5`, respectively.

In contrast to the ECMAScript regex solution, this one doesn't have to do anything anywhere near as fancy or mathematically interesting. Just count the number of times \\$divisor\\$ fits into \\$dividend\\$ by splitting the divisor to keep two tandem running totals that are both subtracted from \\$dividend\\$, one that keeps subtracting \\$divisor-1\\$, and one that keeps subtracting \\$1\\$ and adding it to the total quotient. We must do a split like this, because regex refuses to repeat a zero-width group more than once (this, along with the limited space to work in, is exactly what prevents it from being Turing-complete).

I never wrote a division algorithm in any regex flavor besides ECMAScript before. So it's interesting to now know how they compare in golfed size.

```
x(x*),                    # \1 = divisor-1; tail = dividend
(                         # \2 = what will be the quotient
    (
        x                 # tail -= 1
        (?=
            (             # \4 = running total
                (?(4)\4)  # recall the previous contents of \4, if any
                \1        # \4 += divisor-1
            )
        )
    )*                    # Loop the above as many times as possible (zero or more); if
                          # it loops zero times, \4 will be unset (we'll treat that as 0)
)
\4?                       # tail -= \4, or leave tail unchanged if \4 is unset
(x*)                      # \5 = remainder

```

# Regex (Java), ~~41~~ ~~40~~ 38 bytes

```
x(x*),((x(?=(\4\1|(?!\3)\1)))*)\4?(x*)

```

Try it online!

This is a port of the Perl/PCRE regex to a flavor that has no conditionals. Emulating a conditional costs ~~5~~ 3 bytes here. The quotient and remainder are returned in the capture groups `\2` and `\5`, respectively.

```
x(x*),                  # \1 = divisor-1; tail = dividend
(                       # \2 = what will be the quotient
    (                   # \3 = the following (after the first iteration has finished),
                        #      which is always 1
        x               # tail -= 1
        (?=
            (           # \4 = running total
                \4      # recall the previous contents of \4 (only if it is set)
                \1      # \4 += divisor-1
            |
                (?!\3)  # Match this alternative only if this is the first iteration of
                        # the loop, meaning \4 is unset; \3 can never not match here if
                        # \3 is set (because \3==1) unless \1==0, in which case it
                        # doesn't matter if this alternative is taken instead of the
                        # above – in that case, the value of \4 isn't changed anyway.
                \1      # \4 = divisor-1
            )
        )
    )*                  # Loop the above as many times as possible (zero or more); if
                        # it loops zero times, \4 will be unset (we'll treat that as 0)
)
\4?                     # tail -= \4, or leave tail unchanged if \4 is unset
(x*)                    # \5 = remainder

```

# Regex (Python<sup>`regex`</sup> / Ruby), 43 bytes

```
x(x*),((x(?=((?(5)\5)\1))(?=(\4)))*)\4?(x*)

```

Try it online! \- Python <sub><sup>`import regex`</sup></sub>
Try it online! \- Ruby

This is a port of the Perl/PCRE regex to flavors that have no support for nested backreferences. Python's built-in `re` module does not even support forward backreferences, so for Python this requires `regex`.

Emulating a nested backreference by copying the group back and forth costs 8 bytes here. The quotient and remainder are returned in the capture groups `\2` and `\6`, respectively.

```
x(x*),                    # \1 = divisor-1; tail = dividend
(                         # \2 = what will be the quotient
    (
        x                 # tail -= 1
        (?=
            (             # \4 = running total
                (?(5)\5)  # recall the previous contents of \4 (as copied into \5) if any
                \1        # \4 += divisor-1
            )
        )
        (?=(\4))          # \5 = \4, to make up for Python's lack of nested backreferences
    )*                    # Loop the above as many times as possible (zero or more); if
                          # it loops zero times, \4 will be unset (we'll treat that as 0)
)
\4?                       # tail -= \4, or leave tail unchanged if \4 is unset
(x*)                      # \6 = remainder

```

# Regex (.NET), 29 bytes

```
(x+),(?=(\1)*(x*))((?<-2>x)*)

```

Try it online!

This uses .NET's Balanced Groups feature. It returns the quotient and remainder in the lengths of `\4` and `\3`, respectively.

```
(x+),                     # \1 = divisor; assert \1 > 0; tail = dividend
(?=
    (\1)*                 # push \2 onto the stack for each time \1 fits into dividend
    (x*)                  # \3 = remainder
)
((?<-2>x)*)               # \4 = quotient: pop all \2 from stack, doing \4 += 1 for each

```

# Regex  (.NET), 14 bytes

```
(x+),(\1)*(x*)

```

Try it online!

This uses .NET's Balanced Groups feature. It returns the quotient in the capture count of `\2`, and remainder in the length of `\3`.
> 7 votes

# Answer
## Mathematica, ~~20~~ 18 bytes

```
⌊#/#2⌋@Mod@##&

```

Minor abuse of the flexible output rules: the result is given as `div[mod]`, which will remain unevaluated. The individual numbers can be extracted with `result[[0]]` and `result[[1]]`.

And hey, it's only one byte longer than the ridiculously named built-in `QuotientRemainder`.

Mathematica, actually has a neat way to apply multiple functions to the same input, but it's three bytes longer:

```
Through@*{Quotient,Mod}

```
> 6 votes

# Answer
# Jelly, 3 bytes

```
:,%

```

Try it online!
> 6 votes

# Answer
# 05AB1E, 5 bytes

```
÷²¹%‚

```

Try it online!

~~05AB1E has a bug, so implicit input doesn't work :(~~ Emigna noted that inputs are often pushed in reverse.
> 6 votes

# Answer
## Jellyfish, 14 bytes

```
p
m
,|S
% i
Ei

```

Try it online!

### Explanation

Jellyfish is a beautiful language when it comes to applying multiple functions to the same input. The language is 2D and all binary functions look south for one input and east for another. So by approaching one value from the west and from the north, we can feed it to two functions without having to duplicate it in the code.

The two `i`s in the program are replaced with the two input values when the program starts. Now `%` is division. It takes one input directly from the east, and when going south it hits the `E` which redirects that search east as well. So both inputs get fed to `%` as arguments.

`|` is the built-in for modulo, which basically does the same thing, but ends up looking south for both in puts.

We concatenate both results into a pair with `,`. Then `m` is the floor function (which we need because `%` is floating-point division) and finally we print the result with `p`.
> 6 votes

# Answer
# Cubix, 12 ~~13~~ bytes

```
;W@o,I|\S%;O

```

Which maps onto the following cube

```
    ; W
    @ o
, I | \ S % ; O
. . . . . . . .
    . .
    . .

```

Try it here

Explanation with steps as executed
`,I|I,` \- starts with an superflous integer divide, gets the first integer from input, reflects back and gets the next integer from input, then divides again
`O;` \- Output the result of the integer division and pop it
`%` \- do the mod. This could be done later, but ended up here
`S\o` \- Add space character to stack, redirect up and output space
`W;` \- Shift left and pop the space from the stack
`O|@` \- Output the mod previously calculated, pass through the horizontal reflector and halt.
> 6 votes

# Answer
# sed, 36 bytes

35 bytes of code, +1 for the `-r` flag.

```
:a;s/^(1+)( 1*)\1/\1\2x/;ta;s/.* //

```

Takes input in unary, space-separated, with the smaller number first. Outputs as unary, with the quotient first in `1`s and the remainder second in `x`s. (If this isn't acceptable, let me know and I'll change it to space-separated `1`s like the input.)

### Explanation

```
:a;                                  Define label a
   s/            /     /;            Perform this substitution:
     ^(1+)                           Match the first unary number...
          ( 1*)                      ... followed by a space and 0 or more 1s...
               \1                    ... followed by the the first group again
                  \1\2x              Keep the first two parts unchanged; replace the third
                                     with an x
                         ta;         If the substitution succeeded, goto a
                            s/.* //  After the loop is over, remove the first number

```
> 6 votes

# Answer
# Brain-Flak, ~~56~~ 54 bytes

```
({}<>)<>([()]{()<(({})){({}[()])<>}{}>}<><([{}()]{})>)

```

Try it online!

*-2 bytes thanks to Wheat Wizard*

## Explanation

The current best known integer division and modulo in Brain-Flak are very similar (in fact the currently used integer division is just a modification I made on feersum's modulo).

**Comparison of modulo and integer division:**
```
Modulo:   ({}(<>))<>     {   (({})){({}[()])<>}{} }{}<> ([{}()]{})
Division: ({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><  {}   {} >)

```

Conveniently, the integer division program uses only the third stack for storing data while the modulo program uses only the normal two stacks for storing data. Thus by simply running them both at the same time they do not collide at each other.

**Combination of modulo and integer division:**
```
Modulo:   ({}(<>))<>     {   (({})){({}[()])<>}{} }{}<> ([{}()]{})
Division: ({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><  {}   {} >)

Combined: ({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><([{}()]{})>)

```

Finally, both the integer division and modulo programs used in this combination were designed to be stack clean (not leave garbage on the stacks/not depend on the (non)existence of values on the stacks other than their input) but that is not necessary for this problem. Thus we can save two bytes by not bothering to pop the zero at the end of the main loop and another two bytes by not pushing zero at the start, instead relying on the zero padding on the bottom of the stacks.

**This gives us the final program:**
```
({}<>)<>([()]{()<(({})){({}[()])<>}{}>}<><([{}()]{})>)

```

For the explanation for the integer division program see feersum's answer

### Integer Division Explanation Coming Soon...
> 6 votes

# Answer
# Python 2, 20 bytes

```
lambda a,b:(a/b,a%b)

```

Try it online!

### Built-in, 6 bytes

```
divmod

```

Try it online!
> 5 votes

# Answer
# OIL, ~~134~~ ~~106~~ ~~103~~ 102 bytes

Takes the input from stdin, the two numbers seperated by a newline. Outputs the result of the integer division, then a newline, and then the remainder.

This is one of the most complicated OIL programs I've ever written, as OIL lacks builtins for division, remainder, addition, substraction, and so on. It works with the primitive way of doing division: repeated nested decrementation.

I present the code in an annotated format, with comments in the style of scripting languages. Before executing, the comments have to be removed.

```
5  # read input into lines 0 and 2

5
2
0  # the zero to compare to (nops)
1  # make a backup of the second input at line 3
2
3
10 # check if the second input is 0. %
4
2
24 # if so, jump to 24 (marked with §)
13 # else, go on
10 # check if the first input is zero &
4

31 # if so, jump to 31 (marked with $)
18 # else, go on
9  # decrement both numbers

9
2
6  # jump to line 8 (marked with %)
8
8  # increment the value in line 1 (initially a zero) §
1
1  # "restore the backup"
3
2
6  # jump to line 13 (marked with &)
13
10 # is the second number zero? $
4
2
42 # if so, jump to 42 (marked with +)
36 # else go on
9  # decrement both the second number and the backup
2
9
3
6  # jump to 31 (marked with $)
31
4  # print the division +
1
11 # a newline
4
3  # and the remainder (from the backup)

```

edit: Shaved off 3 more bytes by moving a "constant" to a one-digit location (less bytes to reference), and then implicit-ing 2 zero-locations (By using an empty line instead. One of them I could have done before).

edit: And another byte by making the initial zero implicit. We really only need a single literal zero.
> 5 votes

# Answer
# Excel 2013, ~~31 30~~ 26 bytes

```
=INT(A1/B1)&","&MOD(A1;B1)

```

# Explanation

Input is in cell `A1` and `B1`. This simply returns the return values of the `FLOOR` and `MOD` function, which are for flooring the division and for the remainder. These values are separated by a comma.
> 5 votes

# Answer
# Brain-Flak, ~~168 148~~ 110 bytes

I guess I should have checked the Wiki first

```
(({})(<({}(<(({})<>)>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>)>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<>)

```

## Format:

```
Input:    Output:
A (18)    remainder (2)
B (4)     division  (4)

```

Try it online!

```
(({})(<           # Copy A
({}(<             # Pick up A
(({})<>)          # Copy B to the other stack
>))               # Put A on top of a 0 on the second stack
                  # At this point the stacks look like this:   A
                                                               0
                                                             B B
                                                               ^

<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>) # Positive division from the wiki
>))                                          # Put down A on top of a 0
                                             # The stack now: A
                                                              0
                                                            Div B
                                                              ^

<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<>)     # Modulo from the wiki

```
> 5 votes

# Answer
# Retina, 14 bytes

Let's abuse the input/output formats!

```
(.*)¶(\1)*
$#2

```

Takes input as `b\na`, in unary, using for unary digit any single non-digit, non-newline character. Outputs the quotient in decimal, immediately followed by the remainder in unary, using the same character as the input.

Try it online!

`(.*) ¶(\1)*` matches the first number , then a newline (¶ is Retina's shorthand for \n), then the first number again as many times as possible. The number of matches of the second group will be the result of the division, and the part not matched will be the remainder.

With `$#2`, we replace everything that was matched in the previous line with the number of captures of the second group, and get then our result.
> 5 votes

# Answer
## Labyrinth, 11 bytes

```
?:?:}/!\{%!

```

Try it online!

### Explanation

```
?:   Read a and duplicate.
?:   Read b and duplicate.
}    Move a copy of b over to the auxiliary stage.
/    Compute a/b.
!    Print it.
\    Print a linefeed.
{    Get back the other copy of b.
%    Compute a%b.
!    Print it.

```

The IP then hits a dead end, turns around and the program terminates due to the attempted division by zero when `%` is executed again.
> 4 votes

# Answer
# C, 32 bytes

```
f(a,b){printf("%d %d",a/b,a%b);}

```

Try it online!
> 4 votes

# Answer
# ArnoldC, ~~286~~ 283 bytes

```
HEY CHRISTMAS TREE c
YOU SET US UP 0
HEY CHRISTMAS TREE d
YOU SET US UP 0
GET TO THE CHOPPER c
HERE IS MY INVITATION a
HE HAD TO SPLIT b
ENOUGH TALK
GET TO THE CHOPPER d
HERE IS MY INVITATION a
I LET HIM GO b
ENOUGH TALK
TALK TO THE HAND c
TALK TO THE HAND d
YOU HAVE BEEN TERMINATED

```

Try it online!

### How It Works

```
HEY CHRISTMAS TREE c      //DECLARE VARIABLE c = 0
YOU SET US UP 0
HEY CHRISTMAS TREE d      //DECLARE VARIABLE d = 0
YOU SET US UP 0

GET TO THE CHOPPER c      /*
HERE IS MY INVITATION a      SET c = a/b
HE HAD TO SPLIT b
ENOUGH TALK                */

GET TO THE CHOPPER d      /*
HERE IS MY INVITATION a      SET d = a mod b
I LET HIM GO b
ENOUGH TALK                */

TALK TO THE HAND c        // PRINT c
TALK TO THE HAND d        // PRINT d
YOU HAVE BEEN TERMINATED  //END

```

### Output Format

```
a/b
a mod b

```
> 4 votes

# Answer
# Quipu, 54 bytes

```
\/\/[]\n[]
    1&/\1&
    []  []
    //  %%
    /\  /\

```

Try it online!

### Explanation

Each pair of columns constitutes a "thread." Threads are executed one at a time, top to bottom, left to right. Each thread also stores a value (initially 0).

Each of **threads 0 and 1** uses the `\/` knot to read a number from stdin and stores that number as its value.

**Thread 2** does the following:

* `[]`: Get the value of thread N, where N is the previous value of this thread. Here, there is no previous value, so we get the value of thread 0.
* `1&`: Push a 1.
* `[]`: Get the value of thread N, where N is the 1 we just pushed.
* `//`: Integer-divide the last two values.
* `/\`: Output the result.

**Thread 3** sets its value to a string containing a newline (`\n`) and then outputs it (`/\`).

**Thread 4** does the same thing as thread 2, but with modulo (`%%`) instead of division.
> 4 votes

# Answer
# \>\<\>, ~~27 26~~ 16 + 1 = 17 bytes

```
:r:{%:n','o-$,n;

```

# Note

* Input using the `-v` flag, see TIO for an example.
* This outputs the remainder first, then a comma and lastly the integer division.

Try it online!

# Explanation

Note that the stack starts as `A, B`, where `A` and `B` represent the first and second input, because of the `-v` flag used.

```
:r:{%:n','o-$,n; # Explanation
:r:{             # Do some stack modifications to prepare it for
                 #    next part
                 #    (Stack: B, A, A, B)
    %            # Take the modulo of the top two items
                 #    (Stack: B, A, A%B)
     :           # Duplicate it
                 #    (Stack: B, A, A%B, A%B)
      n          # Pop once and output as number
                 #    (Stack: B, A, A%B)
       ','o      # Print separator
                 #    (Stack: B, A, A%B)
           -     # Subtract the modulo from the first input
                 #    (Stack: B, A-A%B)
            $,   # Swap the two inputs so they are back in order
                 #     and divide, so we get an integer
                 #    (Stack: floor(A/B))
              n; # Output that as a number and finish.

```
> 3 votes

# Answer
# Haskell, 21 bytes

```
a#b=(div a b,mod a b)

```

Try it online! Example usage: `13#2` returns `(6,1)`. Yes, this is pretty boring, however slightly more interesting than the `divMod` build-in which works the same.

While we are at it, there is also `quot`, `rem` and `quotRem` which behave the same on natural numbers as `div`, `mod` and `divMod`. However, for negative inputs the result of `mod` has the same sign as the divisor, while the result of `rem` has the same sign as the dividend. Or, as it is put in the Prelude documentation, `quot` is integer division truncated toward zero and `div` is integer division truncated toward negative infinity.

---

How about no `div` or `mod` build-ins?

### No build-ins, ~~36 32~~ 31 bytes

```
a#b|a<b=(a,0)|m<-a-b=(+1)<$>m#b

```

Try it online! Example usage: `13#2` returns `(1,6)`, that is the `mod` result is first and the `div` result second. If `a` is smaller `b`, then `a mod b` is `a` and `a div b` is `0`, so `(a,0)` is returned. Otherwise recursively compute `mod` and `div` of `a-b` and `b`, add `1` to the division result and keep the remainder.

Adding 1 to the division result is achieved by using `<$>`, which is commonly used as `map` to map functions over lists, but works on tuples too, however the function is applied to the second tuple element only.

Edit: Saved one byte thanks to xnor!
> 3 votes

# Answer
# C, 21 bytes

```
#define f(a,b)a/b,a%b

```

A macro that replaces f(a,b) with the 2 terms comma separated. Though you'd better be passing it to a function or else there's no way to pick the 2 apart.

Try It Online
> 3 votes

# Answer
# ReRegex, ~~75~~ 49 Bytes.

```
#import math
s(\d+),(\d+)e/$1\/$2,$1%$2/s#input
e

```

## Explanation

This simply performs the following Regex Substitution

* `s(\d+),(\d+)e` / `$1/$2,$1%$2`

And the input line is simply

`s(?#input)e`.

The `s...e` takes the input in the form of `123,123` and wraps it, so we know to execute a function on it. In particular, line 2, which changes `123,456` to `123/456,123%456`. Then, all of it is taken care of by the math library.

Try it online!
> 3 votes

# Answer
# SWI Prolog, 109 bytes

```
p(A):-print(A).
d(F,S,0,F):-S>F.
d(F,S,D,R):-G is F-S,d(G,S,E,R),D is E+1.
d(F,S):-d(F,S,D,R),p(D),p(-),p(R).

```

Output:

```
?- d(255,25).
10-5
true .
?- d(5,7).
0-5
true .

```

Description:

Simple recursive algorithm without builtin division or modulo. It simply counts "how many times fits the Second number into the First one?" and reports the result (unified to D) with the remainder (R).

//edit: removed unnecessary spaces
> 3 votes

# Answer
# x86-16 machine code, 5 bytes (non-competing?)

```
88 1E 0105  MOV  BYTE PTR[AAM1+1], BL   ; modify second byte of AAM opcode for divisor
        AAM1:
D4 ?        AAM                         ; ASCII Adjust AX After Multiply

```

Input numbers in `AL` and `BL`. Output division in `AH`, remainder in `AL`.

This "abuses" the x86 "ASCII adjust after multiply" instruction intended to facilitate multiplication of binary coded decimal values (BCD), to convert results to base-10 values. These instructions were deprecated in x64, however as an under-documented feature this behavior can be altered by using self-modifying code to perform arbitrary base conversion of variable input data.

**Caveat:**

The challenge does state `Builtins that return both the result of the division and the remainder are forbidden.` These operations are always done at the same time in x86 machine code so not clear if this would disqualify the platform. This answer, however, does not use the x86's `DIV` instruction. Question posed to OP. :)
> 3 votes

# Answer
# Whitespace, 68 bytes

```
[S S S N
_Push_0][S N
S _Dupe_0][T    N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve_input1][S N
S _Dupe_input1][S N
S _Dupe_input1][T   N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve_input2][T S T S _Integer_divide][T    N
S T _Print_as_integer][S S S T  S T S N
_Push_10_newline][T N
S S _Print_as_character][S S S N
_Push_0][T  T   T   _Retrieve_input1][S N
S _Dupe_input1][T   T   T   _Retrieve_input2][T S T T   _Modulo][T  N
S T _Print_as_integer]

```

Letters `S` (space), `T` (tab), and `N` (new-line) added as highlighting only.
`[..._some_action]` added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Prints with a newline delimiter (could alternatively be a tab for the same byte-count).

**Explanation in pseudo-code:**

```
Integer a = STDIN as integer
Integer b = STDIN as integer
Print a integer-divided by b
Print "\n"
Print a modulo-b

```

**Example program flow (inputs `255` and `25`):**

```
Command   Explanation              Stack          Heap                STDIN  STDOUT  STDERR

SSSN      Push 0                   [0]
SNS       Duplicate 0              [0,0]
TNTT      Read STDIN as integer    [0]            [{0:255}]           255
TTT       Retrieve from heap #0    [255]          [{0:255}]
SNS       Duplicate 255            [255,255]      [{0:255}]
SNS       Duplicate 255            [255,255,255]  [{0:255}]
TNTT      Read STDIN as integer    [255,255]      [{0:255},{255:25}]  25
TTT       Retrieve from heap #255  [255,25]       [{0:255},{255:25}]
TSTS      Integer divide (255/25)  [10]           [{0:255},{255:25}]
TNST      Print as integer         []             [{0:255},{255:25}]         10
SSSTSTSN  Push 10                  [10]           [{0:255},{255:25}]
TNSS      Print as character       []             [{0:255},{255:25}]         \n
SSSN      Push 0                   [0]            [{0:255},{255:25}]
TTT       Retrieve from heap #0    [255]          [{0:255},{255:25}]
SNS       Duplicate 255            [255,255]      [{0:255},{255:25}]
TTT       Retrieve from heap #255  [255,25]       [{0:255},{255:25}]
TSTT      Modulo (255%25)          [5]            [{0:255},{255:25}]
TNST      Print as integer         []             [{0:255},{255:25}]         5
                                                                                error

```

Program stops with an error because no exit is defined.

Since the inputs are guaranteed to be non-negative, I'm using the first input as heap-address to store the second input to save bytes.
> 3 votes

# Answer
# Grok, 19 bytes

```
:Yp:Y%zp/Yp1%-I,Wzq

```

This outputs `remainder,division`. Additionally, inputs must be passed through STDIN; it can't be piped from another command.

Explanation:

```
:Yp                    # Takes the first integer from STDIN and duplicates it.
   :Y                  # Takes the second integer from STDIN and duplicates it to the register.
     %z                # Modulos the two numbers and outputs the remainder.
       p               # Moves the second number from the register to the Document (stack).
        /              # Performs float division on the two numbers.
         Yp1%-         # Floor function shamelessly stolen from https://stackoverflow.com/a/5123037/10363591
              I,W      # Pushes , to the stack and outputs it.
                 zq    # Outputs quotient and terminates.

```
> 3 votes

# Answer
# Piet \+ ascii-piet, 18 bytes (14×2=28 codels)

```
TABlIiIAQJICSBqKkk

```

Try Piet online!

Separates by \x01.

-22 bytes thanks to Aiden Chow
> 3 votes

# Answer
# TypeScript's Type System, 78 bytes

```
//@ts-ignore
type M<A,B,T=[]>=A extends[...B,...infer I]?M<I,B,[...T,1]>:[T,A]

```

Try it at the TypeScript Playground!

This submission is a generic type taking in two numbers in unary and outputting a tuple of two unary numbers, the first being the division and the second the remainder.

Explanation:

```
//@ts-ignore           // ignore any compiler errors in this type
type M<
  A,                   // A is the numerator
  B,                   // B is the denominator
  T = []               // T is the # of times recursed (division result)
> =
  A extends [...B,     // if B fits in A (A >= B):
    ...infer I         //   set I to the difference (A - B)
  ] ? M<               //   recurse with:
        I,             //     I as the numerator
        B,             //     B as the denominator
        [...T, 1]      //     T with 1 appended (i.e. increment)
      >
    : [T, A]           // otherwise return [T, A]

```

This uses tail recursion to increase the maximum recursion depth from 50 to 999.
> 3 votes

# Answer
# Thunno 2, 3 bytes

```
ç¦Œ

```

Try it online!

* `ç`: Apply two commands parallelly:
    + `¦`: swapped integer division
    + `Œ`: swapped modulo
> 3 votes

# Answer
# Cardinal, ~~34~~ ~~32~~ 31 bytes

```
%:~:v
,0.M#
-
-
>8\ < <
^+/'Jx^

```

Try it online!

Input is in the format b first, a second
Output is in the format a%b, a/b

## Explanation:

```
%:~:v
    #

```

Sets second input as active value, first input as inactive value then sends this pointer in two directions

```
,0.M

```

Gets the active value mod inactive value, outputs then outputs a space

```
  .
-
-
>8\ < <
^+/'Jx^

```

Counts the number of times the inactive value can be subtracted from the active value before it falls to \<=0 before outputting the count.
> 2 votes

# Answer
# MATL, 5 bytes

```
/k&G\

```

Try it out at **MATL Online!**

**Explanation**

```
        % Implicitly grab the two inputs as numbers
/       % Divide them
k       % Round down the result
&G      % Grab the two inputs again
\       % Compute the remainder

```
> 2 votes

# Answer
# PowerShell, ~~38~~ 31 bytes

```
$a,$b=$args;($a-$a%$b)/$b;$a%$b

```

Try it online!

Sheesh, this is icky. So, PowerShell (helpfully) returns floating point values when doing division if it doesn't divide evenly. Sometimes, this is a Good Thing, but other times (like here) it's very not. So, you'd figure *"Oh, let's just toss an `[int]` cast and call it good, right?"* Nope. Casting from a `[double]` to an `[int]` in PowerShell does banker's rounding, so for input `5, 7` we would get `1` back, not `0`. As a result, we need to subtract the remainder (from the modulo), then calculate the division, and then calculate the modulo again. Yay!

Both results are left on the pipeline, and output is implicit.

*Saved 7 bytes thanks to PeterTaylor being smarter than me.*
> 2 votes

# Answer
# Gol\>\<\>, 11 bytes

```
I:I:@$S,N%h

```

Online interpreter
> 2 votes

# Answer
# Pyke, 3 bytes

```
'f%

```

Try it online!
> 2 votes

# Answer
# Ouroboros, 15 bytes

```
r.r.@/Inao\%n1(

```

Takes the numbers in reverse order (e.g. `10 42`). Try it here.

### Explanation

```
r.r.             Read a number, duplicate, read a number, duplicate
    @            Rotate a copy of the first number to the top of the stack
     /I          Divide and truncate to integer
       n         Output as number
        ao       Push 10 and output as character (newline)
          \%     Swap the remaining two values and take the mod
            n    Output as number
             1(  Push 1 and swallow that many characters from the end of the program,
                 halting execution

```
> 2 votes

# Answer
**Ruby, 27 bytes**

```
a,b=$<.map &:to_i
p a/b,a%b

```

Reads from standard input, stops when it reaches EOF, and only considers the two integers on the first two lines of its input.
> 2 votes

# Answer
# dc, 12 bytes

```
?sadla/rla%f

```

Explanation:

```
?            # Read input as space-separated integers, push to stack
 sa          # Store second to register a
   d         # Duplicate first
    la       # Load second
      /      # Push first / second
       r     # Swap (stack now first/second, first)
        la   # Load second
          %  # Push first % second
           f # Print stack (first/second, first%second, but in reverse)

```

Try it online!

With builtins, it's just `?~f`. I also tried reading each integer on its own line, but it's the same length (`?d?dsa/rla%f`).
> 2 votes

# Answer
# , 5 bytes

```
/.m,%

```

`/` divide the arguments

`.m` apply **m**inimum

`,%` append modulus

Try it online!
> 2 votes

# Answer
# NO!, 118 bytes

It is too insane to do this challenge with NO! which is why it's non-competing (not really, the language is too young to compete in the bloodthirsty arena known as PPCG)

```
NOOOOOOOOOOOO?NOOOOOOOOOOO
NOOOOOOOOOOOO?NOOOOOOOOOOO
NOOOOO?yes!yess
NOOOOOOOOOOOOOOOOO?yes!yess
NOOOOOOOO?nooo!noooo

```

Please say you don't want an explanation (one added anyway)

```
NOOOOOOOOOOOO?                   Create an integer from
              NOOOOOOOOOOO                              STDIN

NOOOOOOOOOOOO?                   Create an integer from
              NOOOOOOOOOOO                              STDIN

NOOOOO?                          Divide               by
       yes!                             line 1 result
           yess                                          line 2 result

NOOOOOOOOOOOOOOOOO?              Modulo               by
                   yes!                 line 1 result
                       yess                              line 2 result

NOOOOOOOO?                       Output               and
          nooo!                         line 3 result
               noooo                                      line 4 result

```

A `yes` is used to denote a line number (number determined by number of `s`s) but the `output` command takes either numbers or line numbers as arguments and `noo!nooo` is shorter than `yess!yesss`
> 2 votes

# Answer
## Alice, ~~12~~ 10 bytes

```
 \%.\
O:io

```

Try it online!

### Explanation

Alice does have a divmod built-in, which would allow a solution in at most 7 bytes:

```
</
iMOP

```

(The `P` could be any character whatsoever, so I thought I'd put the image of Apple-branded cleaning products in your head. You're welcome.)

But without that built-in, things certainly get a bit more interesting.

Since Alice also has separate built-ins for division and modulo, the program is entirely linear, but the instruction pointer takes a fairly tricky path through the code. Here is the breakdown of the program flow:

```
\    Reflect to NE. Switch to Ordinal
     Reflect off boundary --> SE.
i    Read all input as a string.
     Reflect off boundary --> NE.
.    Duplicate the input string.
     Reflect off boundary --> SE.
     Reflect off corner --> NW.
.    Duplicate the input string. (Irrelevant)
     Reflect off boundary --> SW.
i    Try to read more input, but that just pushes "".
     Reflect off boundary --> NW.
\    Reflect to S. Switch to Cardinal.
:    Implicitly discard the empty string and convert a copy of the input
     to the two integers it contains and compute the result of their integer division.
     IP wraps back to the first line.
\    Reflect to NW. Switch to Ordinal.
     Reflect off boundary --> SW.
O    Implicitly convert the division result to a string and print it with a
     trailing linefeed.
     Reflect off corner --> NE.
\    Reflect to E. Switch to Cardinal.
%    Implicitly convert a copy of the input to the two integers it contains
     and compute the result of their modulo.
.    Duplicate the modulo result. (Irrelevant)
\    Reflect to NE. Switch to Ordinal.
     Reflect off corner --> SW.
o    Implicitly convert the modulo reuslt to a string and print it without a
     trailing linefeed.

```

At this point we're basically done. The IP does a few more rounds through the code and pushes and prints a few more empty strings, but none of that affects the output. Eventually, the code will attempt a division by zero which terminates the program. The instructions executed by Alice after the last meaningful `o` are:

```
%::%o\\i..i\:

```

The exact movement of the IP through the grid and what exactly these commands do to the stack are left as an exercise to the reader...
> 2 votes

# Answer
# MathGolf, 4 bytes

```
‼/%]

```

Inputs in the order `b a`, output as a list containing two integers.

Try it online.

**Explanation:**

```
‼   # Apply the following two commands to the stack separately:
 /  #  Division (which will be integer division if the given arguments are integers)
 %  #  Modulo
    #  (both builtins will use the implicit input-integers for their two required arguments)
  ] # Then wrap all values on the stack into a list
    # (after which the entire stack joined together is output implicitly as result)

```

For an actual character-separated string, we could use one of these instead for 5 bytes: `‼/% \` (space separator); `‼/%n\` or `‼/%]n` (newline separator); `‼/%⌂\` (`*` separator).
> 2 votes

# Answer
# Haskell, 20 bytes

```
(<*>).(<*>)[div,mod]

```

Arguments are passed as singleton lists, for example: `((<*>).(<*>)[div,mod]) [10] [3]` returns `[3,1]`.

The `<*>` operator takes a list of function and a list of values, and applies each function to each value. This function puts `div` and `mod` in a list, and then passes two arguments to them using `<*>`.
Haskell also has the builtin `divMod` which does the same thing (it takes two integers and returns a tuple).
> 2 votes

# Answer
# J-uby, 10 bytes

```
-[+:/,+:%]

```

Attempt This Online!

## Explanation

The built-in way would be `:divmod` (7 bytes) but since that’s forbidden we have:

```
-[ +:/, +:% ]

-[    ,     ]  # Apply each function to the input and return an array of the results
   +:/         # Divide (`+` splats an array into separate arguments)
        +:%    # Modulo

```
> 2 votes

# Answer
# Scratch, 60 bytes

Scratchblocks Syntax:

```
define(x)(y
say(join(join([floor v]of((x)/(y)))[,])((x)mod(y

```

*+12 bytes due to an error reported by xigoi*
> 2 votes

# Answer
# Japt, 7 bytes

```
£rYguiz

```

Try it

```
£rYguiz     :Implicit input of array U
£           :Map each element at index Y
 r          :  Reduce U by
  Yg        :    Index Y into
    uiz     :      prepend "z" (floor division) to "u" (modulo)

```
> 2 votes

# Answer
# Thunno `DK`, 4 bytes

(Actually \\$4\log\_{256}(96)\approx\\$ 3.29 bytes but the leaderboard doesn't take decimals)

```
|X$x

```

Attempt This Online!

#### Explanation

```
|X   # Integer division, store in X
  $x # Modulus, push X
     # Implicit output of the stack

```
> 2 votes

# Answer
# ARBLE, 14 bytes

```
a//b..","..a%b

```

6 bytes longer than I'd like, but I couldn't find a shorter way to deliminate the two as `a//b,a%b` simply returns the two functions.

Try it online!
> 2 votes

# Answer
# Uiua, 4 bytes

```
⌊⊃÷◿

```

Try it!

```
⌊⊃÷◿
 ⊃    # fork (in this case apply two functions to the same two arguments)
   ◿  # mod
  ÷   # divide
⌊     # floor

```
> 2 votes

# Answer
# GolfScript, 6 bytes

```
.~/p~%

```

Try it online!

*I'm literally on an answer spree!*

-3 thanks to Peter Taylor.
> 1 votes

# Answer
## Batch, 37 bytes

```
@set/ad=%1/%2,m=%1%%%2
@echo %d% %m%

```
> 1 votes

# Answer
# Microscript, 11 bytes

```
ivissCl/pl%

```
> 1 votes

# Answer
# brainfuck, 47 bytes

```
>,>,-[+<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>>>>]<<.>.

```

Input/output in byte-values.

Try it online! \- Computes `109,10->10,9`, which in characters corresponds to `m,\n->\n,\t`

This is simply the esolang brainfuck divmod algorithm, with a wrapper for doing input/output and dealing with the case where divisor=1. I'm sure that someone could modify the algorithm to make it work with any input without the need for my boilerplate, but at least now there's a brainfuck answer^^
> 1 votes

# Answer
# Mathematica, ~~32~~ 31 bytes

*Thanks to Martin Ender for calming the code down to the tune of 1 byte!*

```
0@#2//.a_@b_/;b>=#:>(a+1)[b-#]&

```

Just to mess with the language. Pure function taking the two positive integer arguments in the opposite (counterintuitive) order, and returning the quotient `q` and the remainder `r` in the same style, `q[r]`, as in Martin Ender's Mathematica answer. While that answer is shorter, this one is ... more contrary? It implements repeated subtraction on expressions of the form `a[b]`.
> 1 votes

# Answer
## QBIC, 12 bytes

```
::?a'\`b,a%b

```

Explanation:

```
::     Gets two integers from the command line, a and b
?      PRINT
 a'\`b   Integer division in QBasic is '\', however, '\' is the command for ELSE in QBIC
         To tell QBIC not to swap in ELSE at the '\', we need a code literal.
         Everything from the ' to the `is passed on to QBasic without being parsed.
 ,a%b    Also print a MOD b

```
> 1 votes

# Answer
# Pip, 8 bytes

```
Pa//ba%b

```

Try it online!

Explanation:

```
           The input is read automatically into the vars a and b
     a%b   calculate a%b. Since this is the last thing we do, this is printed implicitly.
 a//b      We also want to show a // b (double slashes is integer div).
P          So we need an explicit PRINT command.

```
> 1 votes

# Answer
# QBasic 4.5, 23 bytes

```
INPUT a,b
?a\b,a MOD b

```

This speaks for itself: get two numbers (input needs to be separated by comma) and print their integer division and modulo. QBasic doesn't have a shorthand for `MOD`, the `%`symbol is reserved for defining integers.
> 1 votes

# Answer
# , 9 bytes

```
(<.@%,|~)

```

Usage:

```
   5 (<.@%,|~) 7
0 5
   5 (<.@%,|~) 1
5 0
   18 (<.@%,|~) 4
4 2
   255 (<.@%,|~) 25
10 5

```

Explanation:

In J a verb is like a function. Verbs are monadic `(f x)` or dyadic `(x f y)` and builtin verbs share the same symbol for different monadic and dyadic cases (e.g. `x % y` is division, `% x` is reciprocal).

`(<.@% , |~)` is a train of 3 verbs (fork in J terminology). A fork is: `x (f g h) y` = `(x f y) g (x h y)`. Our verbs are:

* `|~` \- `|` is modulo in J, but arguments are for some reason reverse to what you'd expect. We need the adverb `~` to reverse the arguments: `x f~ y` = `y f x`
* `<. @ %` \- `x % y` is division. `@` is a conjunction denoting function composition: `x (f @ g) y` = `f (x g y)` (J applies right-to-left). `<. x` is floor.
* `,` \- Append.

The parenthesis are needed to form the 3-verb train. No parenthesis would parse as `x <.@% (, (|~ y))`.
> 1 votes

# Answer
## REXX, 20 bytes

```
arg a b
say a%b a//b

```
> 1 votes

# Answer
# Fortran 95, 45 bytes

```
function f(i,j)
write(*,*)i/j,modulo(i,j)
end

```
> 1 votes

# Answer
# jq, 25 characters

```
def f(a;b):a/b|floor,a%b;

```

Sample run:

```
bash-4.3$ jq -n 'def f(a;b):a/b|floor,a%b;f(255;25)'
10
5

```

On-line test
> 1 votes

# Answer
# Gema, 25 characters

```
* *=@div{*;*} @mod{$1;$2}

```

Sample run:

```
bash-4.3$ gema '* *=@div{*;*} @mod{$1;$2}' <<< '255 25'
10 5

```
> 1 votes

# Answer
# AHK, 35 bytes

```
a=%1%
b=%2%
Send,% a//b ","Mod(a,b)

```

AutoHotkey assigns numbers 1-n as variable names for the incoming parameters. It causes some problems when you try to use those in functions because it thinks you mean the literal number 1 instead of the variable *named* 1. The best workaround I can find is to assign them to different variables.
> 1 votes

# Answer
# Forth, 4 bytes

```
/MOD

```

Top of stack will be the quotient, then the remainder.
> 1 votes

# Answer
# TI-Basic (TI-84 Plus CE): 18 bytes

```
Prompt A
Prompt B
{int(A/B),remainder(A,B

```

`Prompt`, `A`, `B`, newline, `{`, `int(`, `/`, `,`, and `)` are all one-byte tokens, but `remainder(` is a two-byte token.

`Prompt` prompts you for the numbers.

`int(A/B)` computes the floored division of A and B

`remainder(A,B` computes remainder when A is divided by B

`{` causes the values of `int(A/B)` and `remainder(A,B)` to be stored to a list. This list is implicitly returned, as it is the last evaluated value in the program.

The returned list is printed as `{divison remainder}`, with a space between the numbers.

Note: TI-Basic does not require closing parentheses or brackets in most cases; one exception being here with `int(A/B)`: the `)` is needed so as not to pass the value after the comma as a second argument to `int(` (which would raise an Error: Syntax).
> 1 votes

# Answer
# Taxi, 1189 bytes

```
"," is waiting at Writer's Depot.Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Cyclone.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l 1 l 2 r.Pickup a passenger going to What's The Difference.Pickup a passenger going to Divide and Conquer.Pickup a passenger going to Divide and Conquer.Go to Divide and Conquer:n 2 r 2 r 1 r.Pickup a passenger going to Trunkers.Go to Trunkers:e 1 r 3 r 1 l.Pickup a passenger going to Cyclone.Go to Cyclone:w 2 r.Pickup a passenger going to Multiplication Station.Pickup a passenger going to Multiplication Station.Go to Multiplication Station:s 1 l 2 r 4 l.Pickup a passenger going to What's The Difference.Go to Cyclone:s 1 r 2 l 2 r.Pickup a passenger going to The Babelfishery.Go to What's The Difference:n 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:e 3 r.Pickup a passenger going to Post Office.Go to Writer's Depot:n 1 l 1 l 2 l.Pickup a passenger going to Post Office.Go to The Babelfishery:n 1 r 2 r 1 r.Pickup a passenger going to Post Office.Go to Post Office:n 1 l 1 r.

```

Try it online!

---

With line breaks:

```
"," is waiting at Writer's Depot.
Go to Post Office:w 1 l 1 r 1 l.
Pickup a passenger going to The Babelfishery.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 l 1 r.
Pickup a passenger going to Cyclone.
Pickup a passenger going to Cyclone.
Go to Cyclone:n 1 l 1 l 2 r.
Pickup a passenger going to What's The Difference.
Pickup a passenger going to Divide and Conquer.
Pickup a passenger going to Divide and Conquer.
Go to Divide and Conquer:n 2 r 2 r 1 r.
Pickup a passenger going to Trunkers.
Go to Trunkers:e 1 r 3 r 1 l.
Pickup a passenger going to Cyclone.
Go to Cyclone:w 2 r.
Pickup a passenger going to Multiplication Station.
Pickup a passenger going to Multiplication Station.
Go to Multiplication Station:s 1 l 2 r 4 l.
Pickup a passenger going to What's The Difference.
Go to Cyclone:s 1 r 2 l 2 r.
Pickup a passenger going to The Babelfishery.
Go to What's The Difference:n 1 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:e 3 r.
Pickup a passenger going to Post Office.
Go to Writer's Depot:n 1 l 1 l 2 l.
Pickup a passenger going to Post Office.
Go to The Babelfishery:n 1 r 2 r 1 r.
Pickup a passenger going to Post Office.
Go to Post Office:n 1 l 1 r.

```
> 1 votes

# Answer
# Triangular, ~~21~~ 19 bÿtes

```
$\:A@$U.vS/p%_<%mp<

```

Prints the result of division, a newline, then the remainder. Formats into this triangle:

```
     $
    \ :
   A @ $
  U . v S
 / p % _ <
% m p < ÿ ÿ

```

ÿ is the no-op that is automaticallÿ inserted when there is no code to fill the smallest possible triangle.

Trÿ it online!

Explanation:

~~I don't know whÿ `\` is the first character...~~ Removed it and managed to save two bÿtes. Still don't know whÿ it was there in the first place. ಠ\_ಠ

Directionals:

```
     .
    \ .
   . . .
  . . v .
 / . . . <
. . . <

```

This is how the interpreter sees the code, without directionals:

```
$:$S_%pUA@pm%

```
* `$:` reads integer input to the stack and duplicates it. Stack contains `input1 input1`.
* `$S` reads another input to the stack, then stashes it in memorÿ. Stack contains `input1 input1 input2`, and memorÿ contains `input2`.
* `_` divides the top two stack values. Now the stack contains `input1 input1/input2`.
* `%` prints the top of stack as an integer.
* `p` pops the top of stack. Now the stack contains `input1`.
* `U` pulls memorÿ onto the stack. Now the stack contains `input1 input2`.
* `A` pushes 10 to the stack. Stack contains `input1 input2 10`.
* `@` prints it as ASCII (newline).
* `p` pops it - stack contains `input1 input2`.
* `m` performs modulus on the top two stack values. Stack now contains `input1%input2`.
* `%` prints the top of stack as an integer.
* the IP then runs off the plaÿing field, and the program terminates.
> 1 votes

# Answer
# Implicit, 9 bytes

```
$$|/%;_@9

```

Try it online!

```
$$|/%;_@9
$$         read two integer inputs
  |        duplicate stack
   /       division
    %      print
     ;     pop
      _    modulus
       @9  print HTAB (delimiter)

```
> 1 votes

# Answer
# Whispers v2, 77 bytes

```
> Input
> Input
>> 1÷2
>> ⌊3⌋
>> 1%2
>> Output 4
>> Output 5
>> Then 6 7

```

Try it online!
> 1 votes

# Answer
## Keg, ~~12~~ 10 bytes

Other than the creator, I think only I am actively using it. Therefore it is rarely seen on this site; in addition it is esoteric.

Takes 2 inputs separated with newlines.

```
¿¿:^:"$/"%

```

Explanation

```
¿¿#          Takes 2 integer inputs: e.g.     stack [1, 2]
  :#         Duplicate the top item.          stack [1, 2, 2]
   ^#        Reverse the stack.               stack [2, 2, 1]
    :#       Duplicate the stack.             stack [2, 2, 1, 1]
     "#      Right shifts (put top to bottom) stack [1, 2, 2, 1]
      $#     Swap top two items               stack [1, 2, 1, 2]
       /#    Divide top two                   stack [1, 2, 0.5]
         "#  Right shifts                     stack [0.5, 1, 2]
          %# Modulos top two items            stack [0.5, 1]

# Stack is implicitly outputted

```

TIO
> 1 votes

# Answer
# 33, 11 bytes

```
OcOcsdoilro

```

Try it online!

## Explanation

```
OcOc        | Harvest input from user
    s       | Store "a" for later use
     doi    | Print the integer division followed by a newline
        l   | Get "a" back
         ro | Print the remainder of the division

```
> 1 votes

# Answer
# Vyxal, 3 bytes

```
₍ḭ%

```

Try it Online!

Try a test suite!

The header and footer are for formatting purposes. Otherwise takes arguments in reversed order.

## Explained

```
₍ḭ%
₍   # parallel apply the next two commands and collect into a list:
 ḭ% # integer divide and modulo

```
> 1 votes

# Answer
# Factor, 20 bytes

Stack shuffling:

```
[ 2dup /i -rot mod ]

```

Try it online!

## Explanation

```
        ! 24 9
2dup    ! 24 9 24 9
/i      ! 24 9 2
-rot    ! 2 24 9
mod     ! 2 6

```

# Factor, 22 bytes

Data flow combinators:

```
[ [ /i ] [ mod ] 2bi ]

```

Try it online!

# Factor, 25 bytes

Local variables:

```
[| a b | a b /i a b mod ]

```

Try it online!

# Factor, 4 bytes

Built-in:

```
/mod

```

Try it online!
> 1 votes

# Answer
# Google Sheets, 41 bytes

```
=concatenate(MOD(A1,A2)," ",QUOTIENT(A1,A2

```

`A1 = divident` and `A2 = diviser`

Last parentheses are auto completed.
> 1 votes

# Answer
# Desmos, 28 bytes

```
f(a,b)=[floor(a/b),mod(a,b)]

```

Try It On Desmos!
> 1 votes

# Answer
# Go, 38 bytes

```
func(a,b int)(int,int){return a/b,a%b}
```

Attempt This Online!

Anonymous function that takes 2 ints and returns 2 ints.
> 1 votes

# Answer
# APL, 6 bytes

```
{0⍵⊤⍺}

```

Definitely not the shortest possible solution, but the classic one, should be compatible with any APL implementation.
> 1 votes

# Answer
# Arturo, 15 bytes

```
$[x y][x/y x%y]

```

Try it!
> 1 votes

# Answer
# Trilangle 1.3, 9 bytes

```
??!z,!:%@

```

Try it on the online interpreter!

TODO: Add colorized explanation

```
??        Read two numbers in
   z      Duplicate them both
      :   Compute the quotient
  !       Print the quotient followed by a newline
    ,     Pop the quotient off the stack
       %  Compute the remainder
     !    Print the remainder (followed by a newline)
        @ Terminate the program

```

The `z` instruction (DP2) makes this fairly straightforward. Without that, the answer is a bit more involved:

## Trilangle 1.0, 13 bytes

```
??j2S,':%@2!!

??            Read two numbers in
   2          Duplicate the one on top of the stack
      '   2   Push the number 2...
  j           ...and use it to index into the stack, copying the bottom to the top
    S         Swap the top two values, so the stack is [a b a b] instead of [a b b a]
     , :   !  Compute, print, and pop the quotient
        %   ! Compute and print the remainder
         @    Terminate the program

```
> 1 votes

# Answer
# CJam, 7 bytes

```
q_~/n~%

```

Try it online!

`n` = print with trailing newline

-2 thanks to Martin Ender.
-3 thanks to Peter Taylor and Basic Sunset.<sup>I was thinking eval first then the rest.</sup>
> 0 votes

# Answer
# Wonder, 13 bytes

```
@@[/#1#0%#1#0

```

Usage:

```
((@@[/#1#0%#1#0)10)5

```

Simply a curried lambda that returns a list with both the division and modulus. I'm looking for a golfier way to do this, something that doesn't involve lambdas.

Be sure to do `tK1000` beforehand to view list items in the output.
> 0 votes

# Answer
# PHP, 35 Bytes

```
[,$a,$b]=$argv;echo$a/$b^0,_,$a%$b;

```
> 0 votes

# Answer
# Forth, 25 bytes

```
: f 2dup / -rot mod ;

```

**Try it online**

Input like `a b`, output like `remainder quotient`.
> 0 votes

# Answer
# Elixir, 24 bytes

```
&{div(&1,&2),rem(&1,&2)}

```

Anonymous function which uses the capture operator and returns a tuple containing the results.

Full program with test cases (and yes, the `.` in the function call is required!):

```
f =
&{div(&1,&2),rem(&1,&2)}

# test cases:
IO.inspect f.(5,7)      # 0,5
IO.inspect f.(5,1)      # 5,0
IO.inspect f.(18,4)     # 4,2
IO.inspect f.(255,25)   # 10,5

```

Try it online on Elixir Playground.
> 0 votes

# Answer
# Lua, 40 Bytes

```
a=arg;io.write(a[1]//a[2],",",a[1]%a[2])

```
> 0 votes

# Answer
# Math++, 18 bytes

```
?>a
?>b
_(a/b)
a%b

```
> 0 votes

# Answer
# ForceLang, 106 bytes

```
def N io.readnum()
def S set
S W gui.show
S a N
S b N
W S c math.floor a.mult b.pow -1
S c c.mult b
W a+-c

```
> 0 votes

# Answer
# Sinclair ZX80 16-bit Integer BASIC (4K ROM), ~152 BASIC bytes used (listing)

```
 1 LET R=0
 2 PRINT "ENTER A NUMBER THEN DIVISOR"
 3 INPUT Z
 4 INPUT D
 5 PRINT Z;"/";D;"=";
 6 GO SUB 9
 7 PRINT R;" REMAINDER ";Z
 8 STOP
 9 FOR I=1 TO 0
10 LET Z=Z-D
11 LET R=R+1
12 LET I=(Z>D)+(Z=D)
13 NEXT I
14 RETURN

```

Some notes:

Because the ZX80 (with the old ROM) can only handle 16-bit signed integer numbers, your range is -32768 to +32767 for your inputs. There is no native modulo function, nor does there seem to be a `>=` (if there is then I can't find it) - hence why this had to be split in line 12.

I haven't handled the division by zero in the program, but I guess that it entering `0` as your divisor will keep the sub-routine from line 9 in an infinite loop (as it should), except for the 16-bit integer range, so `Z/0` should be `Z r ∞`.

The byte count is only an approximation for now.
> 0 votes

# Answer
# Pyth, 7 bytes

```
/QKE%QK

```

Online interpreter
> 0 votes

# Answer
## `r`, 3 bytes

```
/@m

```

## Explanation

<pre><code>abab    % Implicit ops
    /   % Divide a by b.                Stack: a b (a/b)
     @  % Roll down to show 2 operands. Stack: (a/b) a b
      m % Modulo the operands.          Stack: (a/b) (a%b)

% Output the whole stack
```
</code></pre>
> 0 votes

---
Tags: code-golf, arithmetic, division
---