Divisibility and Multiples | IT

Question 12

Context: The algebraic expressions from the previous page are:

Q. Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression.

Solution

Understand the Question

An integer is even if it can be written in the form $2k$ , where $k$ is an integer (i.e., it is a multiple of $2$ ).
Key properties of even and odd numbers:
- Any integer multiplied by an even number is always even.
- The sum or difference of two even numbers is always even ( $ext{Even} imes ext{Integer} = ext{Even}$ , $ext{Even} \pm ext{Even} = ext{Even}$ ).
- Any expression with an overall common factor of $2$ (or a multiple of $2$ ) always results in an even number.

(i) Determine and explain whether $6p + 4q$ always gives an even number.

Step 1 · Explain and verify $6p + 4q$

Since $6$ and $4$ are even numbers, $6p$ and $4q$ are both even for any integers $p$ and $q$ . The sum of two even numbers is always even: $6p + 4q = 2(3p + 2q)$

Examples:

For $p = 1$ and $q = 2$ :

\begin{aligned} 6(1) + 4(2) &= 6 + 8 \\ &= 14 \end{aligned}

For $p = -3$ and $q = 5$ :

\begin{aligned} 6(-3) + 4(5) &= -18 + 20 \\ &= 2 \end{aligned}

Answer

(i) Always even, since $6p + 4q = 2(3p + 2q)$ is a multiple of $2$ .

(ii) Determine and explain whether $10x - 2y$ always gives an even number.

Step 1 · Explain and verify $10x - 2y$

Since $10$ and $2$ are even numbers, $10x$ and $2y$ are both even for any integers $x$ and $y$ . The difference between two even numbers is always even: $10x - 2y = 2(5x - y)$

Examples:

For $x = 3$ and $y = 4$ :

\begin{aligned} 10(3) - 2(4) &= 30 - 8 \\ &= 22 \end{aligned}

For $x = -2$ and $y = 7$ :

\begin{aligned} 10(-2) - 2(7) &= -20 - 14 \\ &= -34 \end{aligned}

Answer

(ii) Always even, since $10x - 2y = 2(5x - y)$ is a multiple of $2$ .

(iii) Determine and explain whether $8a \times b$ always gives an even number.

Step 1 · Explain and verify $8a \times b$

Since $8$ is even, $8a$ is even for any integer $a$ . Multiplying an even number by any integer $b$ always results in an even number: $8a \times b = 2(4ab)$

Examples:

For $a = 2$ and $b = 5$ :

\begin{aligned} 8(2) \times 5 &= 16 \times 5 \\ &= 80 \end{aligned}

For $a = -1$ and $b = -3$ :

\begin{aligned} 8(-1) \times (-3) &= -8 \times (-3) \\ &= 24 \end{aligned}

Answer

(iii) Always even, since $8a \times b = 2(4ab)$ is a multiple of $2$ .

(iv) Determine and explain whether $2(m^2 + n^2)$ always gives an even number.

Step 1 · Explain and verify $2(m^2 + n^2)$

For any integers $m$ and $n$ , the term $(m^2 + n^2)$ is an integer. Multiplying any integer by $2$ always produces an even number.

Examples:

For $m = 3$ and $n = 4$ :

\begin{aligned} 2(3^2 + 4^2) &= 2(9 + 16) \\ &= 2(25) \\ &= 50 \end{aligned}

For $m = -2$ and $n = -1$ :

\begin{aligned} 2((-2)^2 + (-1)^2) &= 2(4 + 1) \\ &= 2(5) \\ &= 10 \end{aligned}

Answer

(iv) Always even, as it is directly a multiple of $2$ .

(v) Determine and explain whether $4(p + q + r)$ always gives an even number.

Step 1 · Explain and verify $4(p + q + r)$

For any integers $p, q, r$ , their sum $(p + q + r)$ is an integer. Since $4 = 2 \times 2$ , any multiple of $4$ is also a multiple of $2$ and therefore always even: $4(p + q + r) = 2[2(p + q + r)]$

Examples:

For $p = 1, q = 1, r = 1$ :

\begin{aligned} 4(1 + 1 + 1) &= 4(3) \\ &= 12 \end{aligned}

For $p = -4, q = 2, r = 0$ :

\begin{aligned} 4(-4 + 2 + 0) &= 4(-2) \\ &= -8 \end{aligned}

Answer

(v) Always even, since $4(p + q + r) = 2[2(p + q + r)]$ is a multiple of $2$ .

(vi) Determine and explain whether $12k - 8m$ always gives an even number.

Step 1 · Explain and verify $12k - 8m$

Since $12$ and $8$ are both even, $12k$ and $8m$ are both even for any integers $k$ and $m$ . The difference of two even numbers is always even: $12k - 8m = 2(6k - 4m)$

Examples:

For $k = 1$ and $m = 1$ :

\begin{aligned} 12(1) - 8(1) &= 12 - 8 \\ &= 4 \end{aligned}

For $k = -3$ and $m = 2$ :

\begin{aligned} 12(-3) - 8(2) &= -36 - 16 \\ &= -52 \end{aligned}

Answer

(vi) Always even, since $12k - 8m = 2(6k - 4m)$ is a multiple of $2$ .

(vii) Determine and explain whether $2(a + b + c + d)$ always gives an even number.

Step 1 · Explain and verify $2(a + b + c + d)$

For any integers $a, b, c, d$ , the sum $(a + b + c + d)$ is an integer. Any integer multiplied by $2$ is always an even number.

Examples:

For $a = 2, b = 5, c = -3, d = 1$ :

\begin{aligned} 2(2 + 5 + (-3) + 1) &= 2(7 - 3 + 1) \\ &= 2(5) \\ &= 10 \end{aligned}

For $a = 0, b = 0, c = 0, d = 7$ :

\begin{aligned} 2(0 + 0 + 0 + 7) &= 2(7) \\ &= 14 \end{aligned}

Answer

(vii) Always even, as it is a multiple of $2$ .

(viii) Determine and explain whether $(x - y)^2 - (x + y)^2$ always gives an even number.

Step 1 · Expand, simplify, and verify $(x - y)^2 - (x + y)^2$

Expand using $(A - B)^2 = A^2 - 2AB + B^2$ and $(A + B)^2 = A^2 + 2AB + B^2$ :

\begin{aligned} (x - y)^2 - (x + y)^2 &= (x^2 - 2xy + y^2) - (x^2 + 2xy + y^2) \\ &= x^2 - 2xy + y^2 - x^2 - 2xy - y^2 \\ &= -4xy \end{aligned}

Since $-4xy = 2(-2xy)$ is a multiple of $2$ , it will always be an even number for all integers $x$ and $y$ .

Examples:

For $x = 3$ and $y = 1$ :

\begin{aligned} -4(3)(1) &= -12 \end{aligned}

For $x = -2$ and $y = 5$ :

\begin{aligned} -4(-2)(5) &= 40 \end{aligned}

Answer

(viii) Always even, since $(x - y)^2 - (x + y)^2 = -4xy = 2(-2xy)$ .

Common Mistakes

Sign Errors in Algebraic Expansion: Forgetting to distribute the negative sign across all terms inside the second bracket when expanding $-(x^2 + 2xy + y^2)$ .
Integer Domain Assumption: Assuming parity properties hold for non-integers; even and odd definitions strictly require the variables to be integers.
Algebraic Proof vs Examples: Relying solely on numerical examples instead of factoring out $2$ to prove algebraically that the expression is always even.

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What’s my number? What’s my name?

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Q40

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Q41

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Question 12

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