Odd-even

Number Theory Level 2

For positive integer $n$ :-

$f(n)=n \bmod 2$

$a,b,$ are positive integers ,such that $f(a)+f(b)=1.$

So , which of the choices is correct ?

f(a+b)=1

f(a\times b)=1

f(a^b)=1

f(2a+b)=1

1 solution

ابراهيم فقرا
Dec 24, 2018

$a$ or $b$ is odd and the other is even , so $a+b$ is odd , and $f(odd)=1$