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Recipe

Find q and r from the division algorithm

Use when: you have integers $a$ and a positive $b$ and need the unique $q$ , $r$ with $a = bq + r$ and $0 \le r < b$

Goal: the exact quotient and remainder pair guaranteed by the division algorithm, even when $a$ is negative

Taught in Divisibility and the division algorithm, revisited

Quick recipe

Guess q by dividing a by b
Compute r = a - bq using that guess
Verify the remainder is in range before reporting the answer

1
Compute $a/b$ as a decimal and take the whole number part as a first guess for $q$ .

Example

For $a=100$ , $b=7$ : $100/7 \approx 14.28$ , so the whole-number part gives a first guess of $q=14$ .

Why: q has to be an integer, and the division algorithm's q is built directly from how many whole copies of b fit into a.
2
Subtract $bq$ from $a$ using the guessed $q$ , and check whether the result satisfies $0 \le r < b$ .

Example

For $a=-17$ , $b=5$ : $-17/5=-3.4$ , whole-number part gives $q=-3$ , and $r=-17-5(-3)=-17+15=-2$ , which is negative. Lower $q$ to $-4$ : $r=-17-5(-4)=-17+20=3$ , and now $0 \le 3 < 5$ .
- $a$ is zero or positive — the whole-number-part guess from Step 1 already gives a valid $r$ in range, no fix needed.
- $a$ is negative — the whole-number-part guess can leave $r$ negative, and $q$ must be lowered by 1 (made more negative) to fix it.
Why: the division algorithm requires $0 \le r < b$ exactly, and rounding toward zero on a negative $a$ produces a remainder outside that range.
3
Check the remainder actually falls between $0$ (inclusive) and $b$ (exclusive), $0 \le r < b$ , before treating it as final.

Example

For $a=100$ , $b=7$ : $r=100-7(14)=100-98=2$ , and $0 \le 2 < 7$ holds, so $q=14$ , $r=2$ is the finished answer.

Why: $q$ and $r$ only count as the division-algorithm pair once this range holds, otherwise it's some other, non-unique split of $a$ .

Common mistakes

Rounding $q$ to the nearest integer instead of taking the whole-number part (floor), which can overshoot on negative $a$ .
Skipping the re-check of $0 \le r < b$ after using a calculator's raw decimal quotient.
Reporting a negative remainder when $a$ is negative, instead of lowering $q$ by one.
Mixing up "b divides a" ( $r=0$ ) with "a divides b" ( $b \mid a$ means $r=0$ when a is the one being divided).