Find all three digit even numbers N such that 693 times N is a perfect square.
I have no idea how to go about this?
Please help urgently!
First of all observe that $\displaystyle 693 = 9 .11 . 7$. So if $\displaystyle 693N$ is a perfect square, then $\displaystyle 77|N$ since only one factor of 7 and 11. More clearly $\displaystyle N = 77k^2$.
Now choose even k such that $\displaystyle 77k^2$ is a three digit number
You will see that for k= 4, N is already a four digit number. Thus k=2 is the only answer.
Or $\displaystyle N = 77 (2)^2 = 308$
So the only three digit that satisfies is $\displaystyle N = 308$