Re: s + t + st is an even integer, where s and t are positive integers
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08 Dec 2024, 15:28
We know $\(s+t+s t\)$ is an even integer, where $\(s\)$ and $\(t\)$ are positive integers; we need to compare the remainder when $\(s\)$ is divided by 2 with the remainder when $\(t\)$ is divided by 2 .
Since $\(s\)$ and $\(t\)$ are positive integers and $\(s+t+s t\)$ is even, the values of $\(s\)$ and $\(t\)$ must be both even (Note: - The sum of 3 even integers or 2 odd and one even integer is even. Here it is not possible to have two odd and one even value cause of the product of $\(s\)$ and $\(t\)$ )
So, the remainder when s or t is divided by 2 must zero in both the cases.
Hence column A quantity equals column B quantity, so the answer is (C).