Problem 1

Determine all three-digit numbers

N

that are divisible by

11

, such that

N/11

equals the sum of the squares of the digits of

N

Step 6 of 6: Case B: solving b=3 (N=803) and ruling out b=7; final answer

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b=3:\ a^2-11a+28=0\Rightarrow a=7\Rightarrow N=803

Detailed analysis

For $b=3$ , the equation reduces to $a^2-11a+28=0$ , with roots $a=4$ or $a=7$ ; since Case B requires $a+b\ge10$ , i.e. $a\ge7$ , only $a=7$ is valid, giving digits $8,0,3$ , i.e. $N=803$ (check: $803/11=73=8^2+0^2+3^2$ ). For $b=7$ , the equation becomes $a^2-7a+26=0$ , whose discriminant is negative, so no solution. Having exhausted both cases, the complete answer is $N=550$ and $N=803$ .