Problem 1601

To determine all three-digit numbers $N$ such that $N$ is divisible by 11 and $\frac{N}{11}$ is equal to the sum of the squares of the digits of $N$, we can proceed as follows:

1. **Express $N$ in terms of its digits:**
Let $N = 100a + 10b + c$, where $a, b, c$ are the digits of $N$ and $a \neq 0$ since $N$ is a three-digit number.

2. Condition for divisibility by 11:
A number is divisible by 11 if the alternating sum of its digits is divisible by 11. Therefore, we need:
$$a - b + c \equiv 0 \pmod{11}$$

3. Condition for the sum of squares:
We are given that:
$$\frac{N}{11} = a^2 + b^2 + c^2$$
Thus:
$$N = 11(a^2 + b^2 + c^2)$$

4. Combine the conditions:
Substitute $N = 100a + 10b + c$ into the equation $N = 11(a^2 + b^2 + c^2)$:
$$100a + 10b + c = 11(a^2 + b^2 + c^2)$$

5. **Check possible values of $a$:**
Since $N$ is a three-digit number, $a$ ranges from 1 to 9. We will check each case to see if there are valid solutions.

### Case 1: $a = 1$
$$100(1) + 10b + c = 11(1^2 + b^2 + c^2)$$
$$100 + 10b + c = 11(1 + b^2 + c^2)$$
$$100 + 10b + c = 11 + 11b^2 + 11c^2$$
$$89 + 10b + c = 11b^2 + 11c^2$$
Testing values of $b$ and $c$ within the range 0 to 9, we find no valid solutions.

### Case 2: $a = 2$
$$100(2) + 10b + c = 11(2^2 + b^2 + c^2)$$
$$200 + 10b + c = 11(4 + b^2 + c^2)$$
$$200 + 10b + c = 44 + 11b^2 + 11c^2$$
$$156 + 10b + c = 11b^2 + 11c^2$$
Testing values of $b$ and $c$ within the range 0 to 9, we find no valid solutions.

### Case 3: $a = 3$
$$100(3) + 10b + c = 11(3^2 + b^2 + c^2)$$
$$300 + 10b + c = 11(9 + b^2 + c^2)$$
$$300 + 10b + c = 99 + 11b^2 + 11c^2$$
$$201 + 10b + c = 11b^2 + 11c^2$$
Testing values of $b$ and $c$ within the range 0 to 9, we find no valid solutions.

### Case 4: $a = 4$
$$100(4) + 10b + c = 11(4^2 + b^2 + c^2)$$
$$400 + 10b + c = 11(16 + b^2 + c^2)$$
$$400 + 10b + c = 176 + 11b^2 + 11c^2$$
$$224 + 10b + c = 11b^2 + 11c^2$$
Testing values of $b$ and $c$ within the range 0 to 9, we find no valid solutions.

### Case 5: $a = 5$
$$100(5) + 10b + c = 11(5^2 + b^2 + c^2)$$
$$500 + 10b + c = 11(25 + b^2 + c^2)$$
$$500 + 10b + c = 275 + 11b^2 + 11c^2$$
$$225 + 10b + c = 11b^2 + 11c^2$$
Testing values of $b$ and $c$ within the range 0 to 9, we find that $b = 5$ and $c = 0$ works:
$$225 + 10(5) + 0 = 11(5^2 + 0^2)$$
$$275 = 275$$
Thus, $N = 550$ is a solution.

### Case 6: $a = 6$
$$100(6) + 10b + c = 11(6^2 + b^2 + c^2)$$
$$600 + 10b + c = 11(36 + b^2 + c^2)$$
$$600 + 10b + c = 396 + 11b^2 + 11c^2$$
$$204 + 10b + c = 11b^2 + 11c^2$$
Testing values of $b$ and $c$ within the range 0 to 9, we find no valid solutions.

### Case 7: $a = 7$
$$100(7) + 10b + c = 11(7^2 + b^2 + c^2)$$
$$700 + 10b + c = 11(49 + b^2 + c^2)$$
$$700 + 10b + c = 539 + 11b^2 + 11c^2$$
$$161 + 10b + c = 11b^2 + 11c^2$$
Testing values of $b$ and $c$ within the range 0 to 9, we find no valid solutions.

### Case 8: $a = 8$
$$100(8) + 10b + c = 11(8^2 + b^2 + c^2)$$
$$800 + 10b + c = 11(64 + b^2 + c^2)$$
$$800 + 10b + c = 704 + 11b^2 + 11c^2$$
$$96 + 10b + c = 11b^2 + 11c^2$$
Testing values of $b$ and $c$ within the range 0 to 9, we find that $b = 0$ and $c = 3$ works:
$$96 + 10(0) + 3 = 11(0^2 + 3^2)$$
$$99 = 99$$
Thus, $N = 803$ is a solution.

### Case 9: $a = 9$
$$100(9) + 10b + c = 11(9^2 + b^2 + c^2)$$
$$900 + 10b + c = 11(81 + b^2 + c^2)$$
$$900 + 10b + c = 891 + 11b^2 + 11c^2$$
$$9 + 10b + c = 11b^2 + 11c^2$$
Testing values of $b$ and $c$ within the range 0 to 9, we find no valid solutions.

Thus, our only solutions are $N = 550$ and $N = 803$.

The final answer is $\boxed{550}$ and $\boxed{803}$.