Even-Odd

to understand B clearly, assure two different value sets for n and then arrive on 2n

let me try;
Assume n as Odd - let's say 5, divisible by 2 different integers (factors); 1 & 5
so 2n is 10 - having 4 factors; 1,2,5,10

Now Assure even - let's say 6 - 4 factors; 1,2,3,6
so 2n is 12 - 6 factors; 1,2,3,4,6,12

Tried with couple of other odd and even nos and found the same result.

IMO B

What is OA?

i think this is B because to have twice the number of factors if multiplied by 2, n has to have factors which are distinct and prime (not including 2)


meaning that it will be odd since all prime numbers other than 2 are odd.

robust wrote:i think this is B because to have twice the number of factors if multiplied by 2, n has to have factors which are distinct and prime (not including 2)


meaning that it will be odd since all prime numbers other than 2 are odd.

OA is B but can you explain.
I am not clear how you proved that n has to have factors which are distinct and prime

I tried using a table for the second seems to help gather my ideas quite clearly..

n 2n nf 2nf
3 6 2 4
4 8 3 4
5 10 2 4
9 18 3 6

here nf = factors of n
2nf - factors of 2n

you see a pattern emerges. only for odd numbers it seems true.[/list]

vish150783 wrote:I tried using a table for the second seems to help gather my ideas quite clearly..

n 2n nf 2nf
3 6 2 4
4 8 3 4
5 10 2 4
9 18 3 6

here nf = factors of n
2nf - factors of 2n

you see a pattern emerges. only for odd numbers it seems true.[/list]

I am not clear how you have calculated nf and 2nf and what pattern emerges