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 1 of 6: Express N through the two digits of the quotient N/11

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In plain words

The condition links $N$ 's digits to $N/11$ , so it helps to name the quotient's own digits $a,b$ and rebuild $N$ from them — that turns a statement about $N$ 's digits into an equation purely in $a$ and $b$ .

Q=\dfrac{N}{11}=10a+b,\qquad N=11Q=100a+10(a+b)+b

Detailed analysis

Since $11\mid N$ , write the quotient as a two-digit number $Q=N/11=10a+b$ , with $a\in\{1,\dots,9\}$ and $b\in\{0,\dots,9\}$ (a three-digit $N$ gives $Q$ between $10$ and $90$ , so it always has exactly two digits). Then $N=11Q=100a+10(a+b)+b$ : if $a+b\le 9$ this already displays the ordinary decimal digits of $N$ as $a$ , $a+b$ , $b$ ; if $a+b\ge 10$ there is a carry into the hundreds digit, handled in a later step.