Divisibility and Multiples | IT

Question 11

Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers.

Solution

We will check each algebraic expression to see if it always results in an even number, no matter what integer values the letters represent.

Step 1 — Analyzing 2a + 2b

An even number is any integer that can be divided by 2 without a remainder. An odd number is any integer that is not even.

Let us consider the term $2a$ . Multiplying an even number by any integer always gives an even number. Since $2$ is an even number, $2a$ is always even.

Let us consider the term $2b$ . Similarly, $2b$ is always an even number.

When we add two even numbers, the result is always an even number. $2a + 2b = \text{Even} + \text{Even}$

$\boxed{\text{Always Even}}$

Step 2 — Analyzing 3g + 5h

Let us consider the term $3g$ . If $g$ is an odd number, $3g$ will be odd (e.g., $3 \times 1 = 3$ ). If $g$ is an even number, $3g$ will be even (e.g., $3 \times 2 = 6$ ). So, $3g$ can be odd or even.

Let us consider the term $5h$ . Similarly, $5h$ can be odd or even depending on $h$ .

If we choose $g=1$ (odd) and $h=2$ (even): $3g + 5h = 3(1) + 5(2)$ $= 3 + 10$ $= 13$ The number $13$ is an odd number.

$\boxed{\text{Not Always Even}}$

Step 3 — Analyzing 4m + 2n

Let us consider the term $4m$ . Since $4$ is an even number, $4m$ is always even.

Let us consider the term $2n$ . Since $2$ is an even number, $2n$ is always even.

When we add two even numbers, the result is always an even number. $4m + 2n = \text{Even} + \text{Even}$

$\boxed{\text{Always Even}}$

Step 4 — Analyzing 2u – 4v

Let us consider the term $2u$ . Since $2$ is an even number, $2u$ is always even.

Let us consider the term $4v$ . Since $4$ is an even number, $4v$ is always even.

When we subtract an even number from another even number, the result is always an even number. $2u - 4v = \text{Even} - \text{Even}$

$\boxed{\text{Always Even}}$

Step 5 — Analyzing 13k – 5k

First, we can simplify this expression by combining like terms. $13k - 5k = (13 - 5)k$ $= 8k$ Now, let us consider the term $8k$ . Since $8$ is an even number, multiplying it by any integer $k$ will always give an even number. So, $8k$ is always even.

$\boxed{\text{Always Even}}$

Step 6 — Analyzing 6m – 3n

Let us consider the term $6m$ . Since $6$ is an even number, $6m$ is always even.

Let us consider the term $3n$ . If $n$ is an odd number, $3n$ will be odd (e.g., $3 \times 1 = 3$ ). If $n$ is an even number, $3n$ will be even (e.g., $3 \times 2 = 6$ ). So, $3n$ can be odd or even.

If we choose $m=1$ and $n=1$ (odd): $6m - 3n = 6(1) - 3(1)$ $= 6 - 3$ $= 3$ The number $3$ is an odd number.

$\boxed{\text{Not Always Even}}$

Step 7 — Analyzing x² + 2

Let us consider the term $x^2$ . If $x$ is an odd number, $x^2$ will be odd (e.g., $1^2 = 1$ , $3^2 = 9$ ). If $x$ is an even number, $x^2$ will be even (e.g., $2^2 = 4$ , $4^2 = 16$ ). So, $x^2$ can be odd or even.

Let us consider the term $2$ . The number $2$ is always even.

If we choose $x=1$ (odd): $x^2 + 2 = 1^2 + 2$ $= 1 + 2$ $= 3$ The number $3$ is an odd number.

$\boxed{\text{Not Always Even}}$

Step 8 — Analyzing b² + 1

Let us consider the term $b^2$ . If $b$ is an odd number, $b^2$ will be odd. If $b$ is an even number, $b^2$ will be even. So, $b^2$ can be odd or even.

Let us consider the term $1$ . The number $1$ is always odd.

If we choose $b=2$ (even): $b^2 + 1 = 2^2 + 1$ $= 4 + 1$ $= 5$ The number $5$ is an odd number.

$\boxed{\text{Not Always Even}}$

Step 9 — Analyzing 4k × 3j

Let us consider the term $4k$ . Since $4$ is an even number, $4k$ is always even.

Let us consider the term $3j$ . This term can be odd or even, depending on $j$ .

When an even number is multiplied by any integer (whether odd or even), the product is always an even number. $4k \times 3j = \text{Even} \times (3j)$

$\boxed{\text{Always Even}}$

Answer

The algebraic expressions that give an even number for any integer values for the letter-numbers are:

(i) $2a + 2b$ (ii) $4m + 2n$ (iii) $2u - 4v$ (iv) $13k - 5k$ (v) $4k \times 3j$

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Question 11

Answer

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