Look at the following expressions and the visualisation. Write the corresponding explanation and examples.

Question

Homi · Accepted Answer

We will explore how adding two numbers, one a multiple of 4 and the other a multiple of 4 plus 2, always results in a number that is a multiple of 4 plus 2.

Step 1 — Explanation with Algebra and Visualisation

Let us consider two numbers. The first number is a multiple of 4. We can write it as $4p$ , where $p$ is any whole number. The second number is a multiple of 4 plus 2. We can write it as $4q + 2$ , where $q$ is any whole number.

We want to find the sum of these two numbers.

$4p + (4q + 2)$

First, we remove the brackets.

$= 4p + 4q + 2$

Next, we can factor out 4 from the terms $4p$ and $4q$ .

$= 4(p + q) + 2$

This final expression shows that the sum is a multiple of 4, plus 2. This means the sum will always leave a remainder of 2 when divided by 4.

The visualisation helps us understand this. The blue circles represent the first number, $4p$ . There are $p$ rows, and each row has 4 circles. So, there are $4 \times p$ blue circles. The yellow circles represent the second number, $4q + 2$ . There are $q$ full rows of 4 circles, and then 2 extra circles. So, there are $4 \times q + 2$ yellow circles.

When we add these two sets of circles, we combine them. We have $p$ rows of blue circles and $q$ rows of yellow circles. Together, these form $(p + q)$ full rows of 4 circles. The 2 extra yellow circles remain separate.

So, the total number of circles is 4 multiplied by the total number of full rows $(p + q)$ , plus the 2 remaining circles. This matches our algebraic result.

$\boxed{4(p + q) + 2}$

Diagram 1

Step 2 — Examples

Let us pick some values for $p$ and $q$ to see this in action.

First, let $p = 1$ and $q = 2$ . The first number, $4p$ , is $4 \times 1 = \mathbf{4}$ . The second number, $4q + 2$ , is $4 \times 2 + 2 = 8 + 2 = \mathbf{10}$ .

Now, let us add them:

$4 + 10$

$= 14$

Let us check this with our combined formula $4(p + q) + 2$ :

$4(1 + 2) + 2$

$= 4(3) + 2$

$= 12 + 2$

$= 14$

The results match!

Let us try another example. Let $p = 3$ and $q = 0$ . The first number, $4p$ , is $4 \times 3 = \mathbf{12}$ . The second number, $4q + 2$ , is $4 \times 0 + 2 = 0 + 2 = \mathbf{2}$ .

Now, let us add them:

$12 + 2$

$= 14$

Let us check this with our combined formula $4(p + q) + 2$ :

$4(3 + 0) + 2$

$= 4(3) + 2$

$= 12 + 2$

$= 14$

Again, the results match. In both examples, the sum is 14, which leaves a remainder of 2 when divided by 4 ( $14 = 4 \times 3 + 2$ ).

Answer

(i) Explanation with Algebra and Visualisation: 4p and (4q + 2) = 4p + (4q + 2) = 4p + 4q + 2 = 4(p + q) + 2.

(ii) Examples:

For $p=1, q=2$ : $4p = 4 \times 1 = 4$ $4q + 2 = 4 \times 2 + 2 = 10$ $4 + 10 = 14$ Using the combined form: $4(1 + 2) + 2 = 4(3) + 2 = 12 + 2 = 14$

For $p=3, q=0$ : $4p = 4 \times 3 = 12$ $4q + 2 = 4 \times 0 + 2 = 2$ $12 + 2 = 14$ Using the combined form: $4(3 + 0) + 2 = 4(3) + 2 = 12 + 2 = 14$

Question 16

Answer

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