Answer :
Given the conditions:
1. [tex]\( R + S \)[/tex] is an odd number.
2. [tex]\(-S\)[/tex] is also an odd number.
We need to determine which among the given expressions [tex]\( R+T \)[/tex], [tex]\( S+T \)[/tex], [tex]\( R \)[/tex], [tex]\( S \)[/tex], or [tex]\( T \)[/tex] must be an even number.
First, let's analyze the conditions and constraints provided:
#### Step 1: Understanding [tex]\( R + S \)[/tex]
- An integer [tex]\( R + S \)[/tex] being odd means one integer must be even and the other must be odd.
- This can be written as:
- [tex]\( R \)[/tex] is even and [tex]\( S \)[/tex] is odd, or
- [tex]\( R \)[/tex] is odd and [tex]\( S \)[/tex] is even.
#### Step 2: Understanding [tex]\(-S\)[/tex]
- [tex]\(-S\)[/tex] is odd. Importantly, the negative of an integer does not change its property of being even or odd.
- Therefore, if [tex]\(-S\)[/tex] is odd, [tex]\( S \)[/tex] must be odd.
We have just concluded:
- [tex]\( S \)[/tex] must be odd.
Since [tex]\( S \)[/tex] is odd and [tex]\( R + S \)[/tex] is odd, it follows that:
- [tex]\( R \)[/tex] must be even (because the sum of an even and an odd number is odd).
Now, let's evaluate the given expressions to identify which must be even.
#### Evaluating [tex]\(R + T\)[/tex]:
- Let's consider the parity of [tex]\( T \)[/tex] (even or odd).
- If [tex]\( T \)[/tex] is even, then [tex]\( R + T \)[/tex] is even (because even + even = even).
- If [tex]\( T \)[/tex] is odd, then [tex]\( R + T \)[/tex] is odd (because even + odd = odd).
Since [tex]\( T \)[/tex] can be either even or odd, [tex]\( R + T \)[/tex] can be either even or odd.
#### Evaluating [tex]\(S + T\)[/tex]:
- Since [tex]\( S \)[/tex] is established to be odd:
- If [tex]\( T \)[/tex] is even, [tex]\( S + T \)[/tex] is odd (odd + even = odd).
- If [tex]\( T \)[/tex] is odd, [tex]\( S + T \)[/tex] is even (odd + odd = even).
Since [tex]\( T \)[/tex] can be either even or odd, [tex]\( S + T \)[/tex] can be either odd or even.
#### Evaluating [tex]\( R \)[/tex]:
- [tex]\( R \)[/tex] is confirmed to be even based on the conditions provided.
#### Evaluating [tex]\( S \)[/tex]:
- [tex]\( S \)[/tex] is confirmed to be odd.
#### Evaluating [tex]\( T \)[/tex]:
- There is no restriction given explicitly for [tex]\( T \)[/tex], so [tex]\( T \)[/tex] can be either even or odd.
Considering all possible conditions, the only integer out of the provided options that must be even is:
[tex]\[ \boxed{R} \][/tex]
1. [tex]\( R + S \)[/tex] is an odd number.
2. [tex]\(-S\)[/tex] is also an odd number.
We need to determine which among the given expressions [tex]\( R+T \)[/tex], [tex]\( S+T \)[/tex], [tex]\( R \)[/tex], [tex]\( S \)[/tex], or [tex]\( T \)[/tex] must be an even number.
First, let's analyze the conditions and constraints provided:
#### Step 1: Understanding [tex]\( R + S \)[/tex]
- An integer [tex]\( R + S \)[/tex] being odd means one integer must be even and the other must be odd.
- This can be written as:
- [tex]\( R \)[/tex] is even and [tex]\( S \)[/tex] is odd, or
- [tex]\( R \)[/tex] is odd and [tex]\( S \)[/tex] is even.
#### Step 2: Understanding [tex]\(-S\)[/tex]
- [tex]\(-S\)[/tex] is odd. Importantly, the negative of an integer does not change its property of being even or odd.
- Therefore, if [tex]\(-S\)[/tex] is odd, [tex]\( S \)[/tex] must be odd.
We have just concluded:
- [tex]\( S \)[/tex] must be odd.
Since [tex]\( S \)[/tex] is odd and [tex]\( R + S \)[/tex] is odd, it follows that:
- [tex]\( R \)[/tex] must be even (because the sum of an even and an odd number is odd).
Now, let's evaluate the given expressions to identify which must be even.
#### Evaluating [tex]\(R + T\)[/tex]:
- Let's consider the parity of [tex]\( T \)[/tex] (even or odd).
- If [tex]\( T \)[/tex] is even, then [tex]\( R + T \)[/tex] is even (because even + even = even).
- If [tex]\( T \)[/tex] is odd, then [tex]\( R + T \)[/tex] is odd (because even + odd = odd).
Since [tex]\( T \)[/tex] can be either even or odd, [tex]\( R + T \)[/tex] can be either even or odd.
#### Evaluating [tex]\(S + T\)[/tex]:
- Since [tex]\( S \)[/tex] is established to be odd:
- If [tex]\( T \)[/tex] is even, [tex]\( S + T \)[/tex] is odd (odd + even = odd).
- If [tex]\( T \)[/tex] is odd, [tex]\( S + T \)[/tex] is even (odd + odd = even).
Since [tex]\( T \)[/tex] can be either even or odd, [tex]\( S + T \)[/tex] can be either odd or even.
#### Evaluating [tex]\( R \)[/tex]:
- [tex]\( R \)[/tex] is confirmed to be even based on the conditions provided.
#### Evaluating [tex]\( S \)[/tex]:
- [tex]\( S \)[/tex] is confirmed to be odd.
#### Evaluating [tex]\( T \)[/tex]:
- There is no restriction given explicitly for [tex]\( T \)[/tex], so [tex]\( T \)[/tex] can be either even or odd.
Considering all possible conditions, the only integer out of the provided options that must be even is:
[tex]\[ \boxed{R} \][/tex]
