The discussion focuses on identifying all three-digit numbers P that are divisible by 11, where the quotient of P divided by 11 equals the sum of the squares of its digits. The problem was successfully tackled by user MarkFL, showcasing a mathematical approach to solve the equation $$\frac{P}{11} = a^2 + b^2 + c^2$$, where a, b, and c are the digits of P. This exploration highlights the intersection of number theory and digit manipulation.
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MarkFL said: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.