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- Feb 14, 2012
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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.
Well done, MarkFL!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.