Can You Find All 3 Digit Numbers Divisible by 11 with a Special Property?

[anemone]

Determine all 3 digit numbers P which are divisible by 11 and where $$\frac{P}{11}$$ is equal to the sum of the squares of the digits of P.

 

[MarkFL]

Spoiler

Let $A$, $B$, and $C$ be the 3 digits from left to right, so that:

$$P=100A+10B+C$$

If P is divisible by 11, then we must have:

$$A+C-B=11k$$ where $$0\le k\in\{0,1\}$$

and we require:

$$\frac{P}{11}=A^2+B^2+C^2$$

Case 1: $$k=0\,\therefore\,B=A+C$$

$$100A+10(A+C)+C=11\left(A^2+(A+C)^2+C^2 \right)$$

$$11(10A+C)=11\left(2A^2+2AC+2C^2 \right)$$

$$10A+C=2A^2+2AC+2C^2$$

The only valid solution is:

$$A=5,\,C=0\implies B=5$$

and so the number is 550.

Case 2: $$k=1\,\therefore\,B=A+C-11$$

$$100A+10(A+C-11)+C=11\left(A^2+(A+C-11)^2+C^2 \right)$$

$$11(10A+C-10)=11\left(2A^2+2AC-22A+2C^2-22C+121 \right)$$

$$2A^2+2AC-32A+2C^2-23C+131=0$$

The only valid solution is:

$$A=8,\,C=3\implies B=0$$

and so the number is 803.

 

[anemone]

Spoiler

Let $A$, $B$, and $C$ be the 3 digits from left to right, so that:

$$P=100A+10B+C$$

If P is divisible by 11, then we must have:

$$A+C-B=11k$$ where $$0\le k\in\{0,1\}$$

and we require:

$$\frac{P}{11}=A^2+B^2+C^2$$

Case 1: $$k=0\,\therefore\,B=A+C$$

$$100A+10(A+C)+C=11\left(A^2+(A+C)^2+C^2 \right)$$

$$11(10A+C)=11\left(2A^2+2AC+2C^2 \right)$$

$$10A+C=2A^2+2AC+2C^2$$

The only valid solution is:

$$A=5,\,C=0\implies B=5$$

and so the number is 550.

Case 2: $$k=1\,\therefore\,B=A+C-11$$

$$100A+10(A+C-11)+C=11\left(A^2+(A+C-11)^2+C^2 \right)$$

$$11(10A+C-10)=11\left(2A^2+2AC-22A+2C^2-22C+121 \right)$$

$$2A^2+2AC-32A+2C^2-23C+131=0$$

The only valid solution is:

$$A=8,\,C=3\implies B=0$$

and so the number is 803.

Well done, MarkFL!(Clapping)