# Find All 3 digit Numbers

#### anemone

##### MHB POTW Director
Staff member
Determine all 3 digit numbers P which are divisible by 11 and where $$\displaystyle \frac{P}{11}$$ is equal to the sum of the squares of the digits of P.

#### MarkFL

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

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

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

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

and we require:

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

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

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

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

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

The only valid solution is:

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

and so the number is 550.

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

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

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

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

The only valid solution is:

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

and so the number is 803.

#### anemone

##### MHB POTW Director
Staff member
Let $A$, $B$, and $C$ be the 3 digits from left to right, so that:

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

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

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

and we require:

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

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

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

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

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

The only valid solution is:

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

and so the number is 550.

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

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

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

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

The only valid solution is:

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

and so the number is 803.
Well done, MarkFL!