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Find All 3 digit Numbers

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,690
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

Administrator
Staff member
Feb 24, 2012
13,775
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.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,690
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!(Clapping)