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 4 of 6: Case A: checking b=0,4,8 gives N=550 only

Previous step Next step

b=0:\ 2a^2=10a\Rightarrow a=5\Rightarrow N=550

Detailed analysis

For $b=0$ , the equation becomes $2a^2=10a$ , so $a=5$ (as $a\neq0$ ), giving digits $5,5,0$ , i.e. $N=550$ — and indeed $550/11=50=5^2+5^2+0^2$ . For $b=4$ , the equation becomes $a^2-a+14=0$ , which has negative discriminant and no real (let alone integer) root. For $b=8$ , the constraint $a+b\le9$ forces $a=1$ , but this does not satisfy the equation. So Case A gives only $N=550$ .