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. Only 1
• E. None of the above options

Answer 1

The correct option is E 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

$n^2-4m+5$ (n${}^2$) will be odd. Hence 4m is even and 5 is odd$\stackrel{`}{x}$ therefore its even

$m^2-2n+2$ (m${}^2$) will be even/odd, therefore it cannot be determined

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