How many numbers less than $1000$ with digit sum to $11$ and divisible by $11$

Question

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.

Answer 1

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 $8$.

Answer 2

$N=a+10b+100c$. $a-b+c=0$ or $a-b+c=11$. Also we have $a+b+c=11$. We get two cases:

• $a-b+c=0,\;a+b+c=11$ from which we get $2a+2b=11$; impossible.
• $a-b+c=11,\;a+b+c=11$ from which we get $a+c=11,\; b=0$. They are $209, 308, 407, 506, 605, 704, 803, 902$.