Let's analyze: Suppose this three-digit number divided by 11 has a quotient of $10a + b$, where $a$ is the tens digit of the quotient, and $b$ is the units digit of the quotient. If $a + b \geqslant 10$, then the original number is $100(a + 1) + 10(a + b - 10) + b$; if $a + b < 10$, then we have $(a + 1)^2 + (a + b - 10)^2 + b^2 \geqslant (a + 1)^2 + 1 + (11 - a)^2 - 2\left(a - \frac{15}{2}\right)^2 + 10a + \frac{21}{2} > 10a + b$. Therefore, if (1) holds, it can only be that $a + b = 10$. Substituting $b = 10 - a$ into (1) yields the unique positive integer solution $a = 7, b = 3$.
Now consider the second scenario. In this case, by the problem's conditions, we have $a^2 + (a + b)^2 + b^2 = 10a + b$.
If $a + b > 5$, then $a^2 + (a + b)^2 + b^2 = 2(a + b) \cdot a + 2b^2 > 10a + b$, so (2) can only hold if $a + b \leqslant 5$. Note that in (2), the left side and $10a$ are both even, so $b$ must also be even. If $a + b < 5$, then $b$ can only be 2. Substituting $b = 2$ into (2) does not yield an integer solution. Therefore, $a + b = 5$.
Substituting $b = 5 - a$ into (2) yields the unique positive integer solution $a = 5, b = 0$.
In summary, the only three-digit numbers that meet the requirements are 550 and 803.
16.2 .41 ** Find all such three-digit numbers: it is divisible by 11, and the quotient equals the square of the sum of the digits of this three-digit number.



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