Understanding Remainder With The Number Line

We saw how division can be seen as repeated subtraction, and how to perform a division operation using the number line.

Today we will see what are the ways we can make a given dividend divisible by a divisor by adding or subtracting another integer from the dividend.

We know that the number $38$ is not divisible by $6$. We will understand how much we need to

1. Add

2. Subtract

from $38$ to make it divisible by $6$. Again we will use the number line below.

As shown below, we will start taking jumps of $6$ starting at $0$. After $6$ hops we will reach $36$.

When we reach $36$ we will still have $2$ extra, therefore we cannot reach $38$ by taking hops of length $6$ starting at $0$.

If we remove (or subtract) the extra $2$ from $38$, then we will reach that number by taking hops of length $6$.

But if we take another hop of length $6$ at $36$ we will reach $4$ more from $38$, which will be $42$, and which will be divisible by $6$.

So, given $38$ as the dividend and $6$ as the divisor, we either need to:

- subtract the remainder, $2$, or

- add the difference of the divisor and the remainder, that is, $6 - 2 = 4$

to the the dividend to get a number that is divisible by $6$.

Let us try to answer the following questions:

- How much do we need to subtract from $101$ to make the result divisible by $12$

- How much do we need to add to $101$ to make the result divisible by $12$

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We know that when we divide $101$ by $12$ we get a remainder of $5$.

So either we need to subtract $5$ from $101$, or add $12 - 5 = 7$ to make the result divisible by $12$.

$101 - 5 = 96$, which is divisible by $12$, and,

$101 + 7 = 108$ which is again divisible by $12$

Let us try to solve one more problem.

A number, greater that $1000$, is divisible by $17$. $57$ is subtracted from this number. What will be the remainder when the result is divided by $17$?

We know that if any number, $N$ is divisible by $17$, then $N - 17$, $N - 17 \times 2$, $N - 17 \times 3$ will be divisible by $17$

We are subtracting $57$ from a number which is divisible by $17$. $57$ can be written as $51 + 6 = 17 \times 3 + 6$ from this number.

That means we are subtracting $6$ from a number which is divisible by $17$. If we divide the result by $17$ then we will get a remainder of $17 - 6 = 11$.

We will try to solve a little more complex problem using this approach:

What is the smallest number which when divided by $7$ leaves a remainder of $4$, when divided by $11$ leaves a remainder of $8$ and when divided by $15$ leaves a remainder of $12$?

The interesting observation above is that $7 - 4 = 3$, $11 - 8 = 3$ and $15 - 12 = 3$. We can see that the difference between the divisor and the remainder is same for all cases. So if we can find out the $LCM$ of $7$, $11$ and $15$ and subtract $3$ from that, the resulting number will

- leave a remainder of $7-3 = 4$ when divided by $7$,

- leave a remainder of $11 - 3 = 8$ when divided by $11$

- leave a remainder of $15 - 3 = 12$ when divided by $15$

That should fulfil the conditions given in the problem.

So, the $LCM$ of $7$, $11$ and $15$ is $1155$, subtracting $3$ from it we get $1152$

Hence, our answer is $1152$