stonecold wrote:
Here we the easiest way out is use the examples
Now by using 3/11 ; 3/9 and 3/12 => all options can be eliminated except the last one => E
I have got one more;
let X=nY+R (n= some quotient, R=remainder)
or X=nY+2k+1 (2k+1=odd remainder)
Now if Y is even;
X=nY+2k+1=even+even+1=odd
and if Y is odd;
X=(odd or even)+(even)+1
=(odd or even)
So,
at least 1 of X and Y will always be Odd.
Kudos plz if you find this solution useful...
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