How many positive (integers) numbers less than $1000$ with digit sum to $11$ and divisible by $11$?
There are $\lfloor 1000/11 \rfloor = 90$ numbers less than $1000$ divisible by $11$.
$N = 100a + 10b + c$ where $a + b + c = 11$ and $0 \le a, b, c \le 9$
I got $\binom{13}{2} – 9 = 69$ solutions.
Best Answer
Digitsum is related to the modulo 9 operation. A weakening of the conditions given is that you are counting how many $0\leq n\leq 1000$ satisfy the coungruencies:
$\begin{array}{} n\equiv 2\pmod{9}\\ n\equiv 0\pmod{11}\end{array}$
By the chinese remainder theorem, we get that
$n\equiv 11\pmod{99}$
So, we can look at the possible solutions and trim the ones that don't meet the stronger requirement that the digit sum be $11$ (as opposed to $2$ or $20$ or $29$ or $37$)
We have the list then $\{11,110,209,308,407,506,605,704,803,902\}$
All but the first two have digitsum 11 (whereas the first two have only digit sum equaling 2).
The answer is then $8$.