Digit-Sum & Divisibility-by-11 Technique

Let the digits be $A,B,C$ (hundreds → units, with $A\ge1$).

We have two conditions:

From (1): $B = 18-(A+C)$.

$$ A - \bigl[18-(A+C)\bigr] + C ;\equiv; 0 \pmod{11} \Longrightarrow 2(A+C) - 18 \equiv 0 \pmod{11}. $$

Because $8\equiv7\pmod{11}$:

$$ 2(A+C) \equiv 7 \pmod{11}. $$

Multiply by the inverse of $2$ modulo 11 (which is $6$, since $2\cdot6\equiv1$):

$$ A+C ;\equiv; 7!\cdot!6 ;\equiv; 42 ;\equiv; 9 \pmod{11}. $$

But $A+C$ is a sum of two single digits, so its range is 0–18.

The only value in that range congruent to $9$ mod $11$ is $A+C = 9$.

With $A+C=9$ and $B=18-(A+C)=9$.

Thus $B$ is fixed at $9$, while $A$ and $C$ are non-negative digits summing to $9$:

$$ (A,C) = (1,8),,(2,7),,(3,6),,(4,5),,(5,4),,(6,3),,(7,2),,(8,1),,(9,0). $$

All nine pairs are valid because $A\ge1$ and $C$ can be $0$.

There are $\boxed{9}$ such integers.

Digit-Sum & Divisibility-by-11 Technique#