Question 3

We will check each statement to see if it is always true, sometimes true, or never true.

Step 1 — Sum of two even numbers

Let us consider two even numbers. An even number is any number that can be divided by 2.

We can write any even number as $2a$, where $a$ is a whole number. Let the two even numbers be $2a$ and $2b$. Their sum is $2a + 2b$.

$2a + 2b$

$= 2(a + b)$

The sum is always an even number because it has a factor of 2. For the sum to be a multiple of 3, the expression $2(a+b)$ must be divisible by 3. Since 2 is not divisible by 3, $(a+b)$ must be divisible by 3. This condition is not always true.

Let us look at some examples. Example 1: Let the even numbers be 6 and 12. Here, $a=3$ and $b=6$. So $a+b = 3+6=9$. Since 9 is a multiple of 3, the sum should be a multiple of 3. Sum $= 6 + 12 = 18$. 18 is a multiple of 3 ($18 = 3 \times 6$). This case is true.

Example 2: Let the even numbers be 4 and 10. Here, $a=2$ and $b=5$. So $a+b = 2+5=7$. Since 7 is not a multiple of 3, the sum should not be a multiple of 3. Sum $= 4 + 10 = 14$. 14 is not a multiple of 3. This case is false.

Since the statement is true in some cases and false in others, it is sometimes true.

$\boxed{\text{Sometimes True}}$

Step 2 — Divisibility by 18 and 9

The statement says: If a number is not divisible by 18, then it is also not divisible by 9. Let us think about numbers that are divisible by 9. These are $9, 18, 27, 36, 45, \dots$ Now let us think about numbers that are divisible by 18. These are $18, 36, 54, \dots$

We need to check if the statement holds for all numbers. Let us try to find a number that makes the statement false. A statement "If P, then Q" is false if P is true and Q is false. So, we need a number that is NOT divisible by 18 (P is true), BUT IS divisible by 9 (Q is false).

Consider the number 9. Is 9 not divisible by 18? Yes, 9 divided by 18 is not a whole number. So the first part (P) is true. Is 9 not divisible by 9? No, 9 is divisible by 9. So the second part (Q) is false. Since we found a number (9) where the first part is true and the second part is false, the entire statement is false for this number.

Let us use algebra. Let the number be $N$. If $N$ is divisible by 9, we can write $N = 9k$ for some whole number $k$. If $k$ is an even number, say $k=2m$, then $N = 9(2m) = 18m$. This means $N$ is divisible by 18. If $k$ is an odd number, say $k=2m+1$, then $N = 9(2m+1) = 18m + 9$. This number $N$ is divisible by 9. However, $N = 18m + 9$ is not divisible by 18, because it leaves a remainder of 9 when divided by 18. So, for any odd multiple of 9 (like 9, 27, 45), the number is divisible by 9 but not by 18. For these numbers, the statement "If a number is not divisible by 18, then it is also not divisible by 9" becomes: "If (True, because it's not divisible by 18), then (False, because it IS divisible by 9)". This makes the whole statement false. Since we can always find such numbers, the statement is never true.

$\boxed{\text{Never True}}$

Step 3 — Sum of two numbers not divisible by 6

The statement says: If two numbers are not divisible by 6, then their sum is not divisible by 6. Let us test this with examples.

Example 1: Let the numbers be 1 and 2. Is 1 not divisible by 6? Yes. Is 2 not divisible by 6? Yes. Their sum is $1+2=3$. Is 3 not divisible by 6? Yes. In this case, the statement is true.

Example 2: Let the numbers be 4 and 2. Is 4 not divisible by 6? Yes. Is 2 not divisible by 6? Yes. Their sum is $4+2=6$. Is 6 not divisible by 6? No, 6 IS divisible by 6. In this case, the statement is false.

Since the statement is true in some cases and false in others, it is sometimes true. We do not need complex algebra here, as simple examples show both true and false scenarios.

$\boxed{\text{Sometimes True}}$

Step 4 — Sum of a multiple of 6 and a multiple of 9

The statement says: The sum of a multiple of 6 and a multiple of 9 is a multiple of 3. Let us represent a multiple of 6 as $6a$, where $a$ is a whole number. Let us represent a multiple of 9 as $9b$, where $b$ is a whole number. Their sum is $6a + 9b$.

$6a + 9b$

We can factor out 3 from this sum.

$= 3(2a + 3b)$

Since $a$ and $b$ are whole numbers, $2a$ is a whole number, and $3b$ is a whole number. The sum $(2a + 3b)$ is also a whole number. So, the sum $6a + 9b$ can always be written as 3 multiplied by some whole number. This means the sum is always a multiple of 3.

Let us look at some examples. Example 1: Multiple of 6 is 12, multiple of 9 is 18. Sum $= 12 + 18 = 30$. 30 is a multiple of 3 ($30 = 3 \times 10$).

Example 2: Multiple of 6 is 6, multiple of 9 is 45. Sum $= 6 + 45 = 51$. 51 is a multiple of 3 ($51 = 3 \times 17$).

The statement is always true.

$\boxed{\text{Always True}}$

Step 5 — Sum of a multiple of 6 and a multiple of 3

The statement says: The sum of a multiple of 6 and a multiple of 3 is a multiple of 9. Let us represent a multiple of 6 as $6a$, where $a$ is a whole number. Let us represent a multiple of 3 as $3b$, where $b$ is a whole number. Their sum is $6a + 3b$.

$6a + 3b$

We can factor out 3 from this sum.

$= 3(2a + b)$

For this sum to be a multiple of 9, the expression $3(2a+b)$ must be divisible by 9. This means that $(2a+b)$ must be divisible by 3. This condition is not always true.

Let us look at some examples. Example 1: Let the multiple of 6 be 6, and the multiple of 3 be 3. Here, $a=1$ and $b=1$. So $2a+b = 2(1)+1=3$. Since 3 is a multiple of 3, the sum should be a multiple of 9. Sum $= 6 + 3 = 9$. 9 is a multiple of 9. This case is true.

Example 2: Let the multiple of 6 be 6, and the multiple of 3 be 6. Here, $a=1$ and $b=2$. So $2a+b = 2(1)+2=4$. Since 4 is not a multiple of 3, the sum should not be a multiple of 9. Sum $= 6 + 6 = 12$. 12 is not a multiple of 9. This case is false.

Since the statement is true in some cases and false in others, it is sometimes true.

$\boxed{\text{Sometimes True}}$

Answer

(i) Sometimes True (ii) Never True (iii) Sometimes True (iv) Always True (v) Sometimes True