Q.Write pseudocode that will perform the following:

The Division Algorithm: From Sharing to a Precise Rule

Think about what division really means. When you divide 17 apples among 5 friends, you can't give everyone a whole number of apples and have nothing left — unless you start cutting fruit. Each friend gets 3 apples, and 2 apples remain. That's the heart of the division algorithm: you can always split a number into equal groups, with a remainder smaller than the group size.

The word "algorithm" sounds intimidating, but it's just a step-by-step process. Here, the process is: given any two positive integers (the dividend and the divisor), you can find a unique quotient and a unique remainder. The remainder must be smaller than the divisor — otherwise you could give out one more whole group.

The Intuition

Imagine a number line. You start at 0 and take jumps of size $b$ (the divisor). You want to land as close as possible to $a$ (the dividend) without overshooting. The number of jumps you take is the quotient $q$, and the distance you still need to cover to reach $a$ is the remainder $r$.

For $a = 17$ and $b = 5$:

• Jump 1: 0 → 5
• Jump 2: 5 → 10
• Jump 3: 10 → 15
• Jump 4 would take you to 20, which overshoots 17.

So $q = 3$ jumps, and you've covered $3 \times 5 = 15$. The remaining distance is $17 - 15 = 2$, so $r = 2$. Notice $2 < 5$ — you can't make another full jump.

The Precise Statement

Two things make this powerful: existence (you can always find such $q$ and $r$) and uniqueness (there is only one correct pair). The condition $0 \le r < b$ is what forces uniqueness — if you allowed $r$ to be as large as $b$ or more, you could "steal" one more group and get a different $q$.

Why "Algorithm"?

The name comes from the process you use to find $q$ and $r$: keep subtracting $b$ from $a$ until what's left is less than $b$. The number of subtractions is $q$, and what's left is $r$. For large numbers, you'd use long division instead, but the idea is identical.

A Simple Example

Divide 43 by 6.

• How many full groups of 6 fit into 43? $6 \times 7 = 42$, and $6 \times 8 = 48$ is too big. So $q = 7$.
• Remainder: $43 - 42 = 1$.
• Check: $1 < 6$, so it's valid.
• Result: $43 = 6 \times 7 + 1$.

What About Negative Numbers? …