Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of 11.

Anyone know how to solve this one. We write N as abc then...

Since N is divisible by 11 we know that a - b + c is divisible by 11. So a-b+c = 0 or 11.

Then we also know that $\displaystyle a^2 + b^2 + c^2 = \frac{N}{11}$

I can't get any sort of 'nice' solution from this and just end up with having to check 18 cases. (a+c=b, sub into above than a=0,1,2,... etc)