Solutions to $100a+10b+c=11a^2+11b^2+11c^2$ and $100a+10b+c=11a^2+11b^2+11c^2+ k$ The problem states that

Find all three digit natural numbers such that when the number is divided by $11$ gives the quotient equal to sum of squares of their digits

Since there is no information about whether remainder is $0$ or not , I firstly assumed that the question is talking about the numbers perfectly divisible by $11$
Now I have $80$ numbers left , I can check them separately , but it will be lengthy
I made an equation $$100a+10b+c=11a^2+11b^2+11c^2$$
Rearranged and got $$a(11a-100)+b(11b-10)+c(11c-1)=0$$
I have $10$ values for $a,b,c=\{1,2,3,4,5,6,7,8,9,0\}$
I made a table corresponding to $a(11a-100),b(11b-10),c(11c-1)$ and found that only for $a=b=5,c=0$ and $a=8,b=0,c=3$ are giving their sum $0$ , hence $550$ and $803$ are the only numbers satisfying given property and divisible by 11.
Now I have two questions:

$1.)$Is my way and my answer correct? If no, then where have I misunderstood?
$2.)$What about the numbers which are not divisible by $11$?
As mentioned by an answer er of this post , there are six such numbers which are not divisible by $11$ , but still give the quotient the sum of square of their digits. But answer er found it using a computer program, which is not suitable for pen paper mathematics. So how can I find those six numbers ?


This might not be the best way but here is how I would do it.
Assume the remainder to be $\lambda$. Remember that $\lambda$ can take any value from 0 to 10.
Note that $100a + 10b +c = 11a^2 + 11b^2 + 11c^2 + \lambda$ can be rewritten as
$$(11a - 50)^2 + (11b - 5)^2 + (11c - \frac{1}{2})^2 = 2525 + \frac{1}{4} - \lambda$$
For the remainder of the answer, I'll assume $\lambda = 0.$ The other cases for $\lambda$ can be handled similarly.
Since each of the terms on the left hand side are positive, we get that $c \leq 4$
We can take 5 cases based upon each of the five possible values for $c$ and solve for $a$ and $b$.
I won't go through all of the cases but in each one of them we should be able to form upper bounds on the values of $a$ and $b$. Furthermore, the search space can be pruned even further by making use of the parity of $a$ and $b$.
This method is not very different from your "table generating" method. The only difference is that the search space is pruned quite a bit.
Also, if the number is itself divisible by 11, we can use the property mentioned in the comments to prune the search space even more.
The process can be repeated for other values of $\lambda$. The upper bounds should not differ a lot but the individual solutions will.
