Question

Given that 2m-1 is an odd number and 3n-1 is an even number, which of the following are necessarily odd?
$1)n^2-4m+52)m^2-2n+23)6m^2-n-1$

• A. Only 2
• B. Both 1 and 2
• C. Only 3
• D. None of the above options

Answer 1

The correct option is D None of the above options

If 2m-1 is odd, m can be odd or even.

If 3y-1 is even, then n has to be odd.

$1)n^2-4m+5$ - n${}^2$ will be odd, 4m is even and 5 is odd. Therefore it is even.

$2)m^2-2n+2$ - m${}^2$ will be even/odd. Therefore it cannot be determined.

$3)6m^2-n-1$ - 6m${}^2$ is even, n is odd and 1 is also odd. Therefore it is even.