If $$a$$ and $$b$$ are positive integers, is $$a-b$$ odd?
>(1) $$\frac{a}{b}$$ is an odd integer.
>(2) $$\frac{b}{a}$$ is an odd integer.

Incorrect.
[[snippet]]
Stat. (1) indirectly tells you that since $$\frac{a}{b}$$ is an integer, $$b$$ is a factor of $$a$$. The variable $$a$$ can be represented as follows:
>$$a = b \cdot \text{(an odd number)}$$.
Thus, there are two cases to consider:
>If $$a$$ is odd, then $$b$$ must also be odd because only odd $$\cdot$$ odd $$=$$ odd. In this case, $$a-b$$ is odd $$-$$ odd $$=$$ even.
>If $$a$$ is even, then $$b$$ must also be even because even $$\cdot$$ odd $$=$$ even. In this case also, $$a-b$$ is even $$-$$ even $$=$$ even.
Therefore, $$a-b$$ must be even. That's a clear "No," so **Stat.(1) → Always No → S → AD**.

Correct.
[[snippet]]
Stat. (1) indirectly tells you that since $$\frac{a}{b}$$ is an integer, $$b$$ is a factor of $$a$$. The variable $$a$$ can be represented as follows:
>$$a = b \cdot \text{(an odd number)}$$.
Thus, there are two cases to consider:
>If $$a$$ is odd, then $$b$$ must also be odd because only odd $$\cdot$$ odd $$=$$ odd. In this case, $$a-b$$ is odd $$-$$ odd $$=$$ even.
>If $$a$$ is even, then $$b$$ must also be even because even $$\cdot$$ odd $$=$$ even. In this case also, $$a-b$$ is even $$-$$ even $$=$$ even.
Therefore, $$a-b$$ must be even. That's a clear "No," so **Stat.(1) → Always No → S → AD**.
The same goes for Stat. (2):
If $$\frac{b}{a}$$ is odd, then $$a$$ is a factor of $$b$$, and they are either both even or both odd.
In either case, $$a-b$$ is even. Hence, **Stat.(2) → S → D**.

Incorrect.
[[snippet]]
Stat. (1) indirectly tells you that since $$\frac{a}{b}$$ is an integer, $$b$$ is a factor of $$a$$. The variable $$a$$ can be represented as follows:
>$$a = b \cdot \text{(an odd number)}$$.
Thus, there are two cases to consider:
>If $$a$$ is odd, then $$b$$ must also be odd because only odd $$\cdot$$ odd $$=$$ odd. In this case, $$a-b$$ is odd $$-$$ odd $$=$$ even.
>If $$a$$ is even, then $$b$$ must also be even because even $$\cdot$$ odd $$=$$ even. In this case also, $$a-b$$ is even $$-$$ even $$=$$ even.
Therefore, $$a-b$$ must be even. That's a clear "No," so **Stat.(1) → Always No → S → AD**.

Incorrect.
[[snippet]]
Stat. (2) indirectly tells you that since $$\frac{b}{a}$$ is an integer, $$a$$ is a factor of $$b$$. The variable $$b$$ can be represented as follows:
>$$a = b \cdot \text{(an odd number)}$$.
Thus, there are two cases to consider:
>If $$b$$ is odd, then $$a$$ must also be odd because only odd $$\cdot$$ odd $$=$$ odd. In this case, $$a-b$$ is odd $$-$$ odd $$=$$ even.
>If $$b$$ is even, then $$a$$ must also be even because even $$\cdot $$ odd $$=$$ even. In this case also, $$a-b$$ is even $$-$$ even $$=$$ even.
Therefore, $$a-b$$ must be even. That's a clear "No," so **Stat.(2) → Always No → S**.

Incorrect.
[[snippet]]
Stat. (1) indirectly tells you that since $$\frac{a}{b}$$ is an integer, $$b$$ is a factor of $$a$$. The variable $$a$$ can be represented as follows:
>$$a = b \cdot \text{(an odd number)}$$.
Thus, there are two cases to consider:
>If $$a$$ is odd, then $$b$$ must also be odd because only odd $$\cdot$$ odd $$=$$ odd. In this case, $$a-b$$ is odd $$-$$ odd $$=$$ even.
>If $$a$$ is even, then $$b$$ must also be even because even $$\cdot$$ odd $$=$$ even. In this case also, $$a-b$$ is even $$-$$ even $$=$$ even.
Therefore, $$a-b$$ must be even. That's a clear "No," so **Stat.(1) → Always No → S → AD**.

Statement (1) ALONE is sufficient, but Statement (2) alone is not sufficient to answer the question asked.

Statement (2) ALONE is sufficient, but Statement (1) alone is not sufficient to answer the question asked.

BOTH Statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.

EACH statement ALONE is sufficient to answer the question asked.

Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.