Solve $a\cdot 1111+b\cdot 111+c\cdot 11=9002$

Question

I would like to find all $(a,b,c)\in\mathbb{N}^3$ such that $a\times 1111+b\times 111+c\times 11=9002$.

It is obvious that $a+b+c \equiv 2 \pmod{10}$. I made a little computer program, that basically looped over all possiblities and I found no answer, so I just have to prove it...
I am thinking of an inductive solution, something like:
If no solution is found for $a=a_{1}$, then there is no solution for $a=a_{1}+1$

Answer 1

Rewrite the equation as

$$1000a+100(a+b)+11(a+b+c)=9002$$

and now look for solutions of $1000A+100B+11C=9002$ with $0\le A\le B\le C$ (or $0\lt A\lt B\lt C$ if you want positive integer solutions). From $11C\equiv2$ mod $100$, we see $-C\equiv99C\equiv18$ mod $100$, so $C\equiv82$ mod $100$. Thus, since $882\cdot11=9702\gt9002$, we have

$$C\in\{82,182,282,382,482,582,682,782\}$$

From these we get

$$10A+B= \begin{cases} 81\\70\\59\\48\\37\\26\\15\\4 \end{cases}$$

and appropriate values for $0\le A\le B$ can be easily worked out, and then values of $(a,b,c)$ obtained. For example $(A,B,C)=(7,11,82)$ corresponds to $(a,b,c)=(7,4,71)$.

Answer 2

Working $\bmod 11$, you have:
$9002\equiv 4$
$11\equiv 0$
$111\equiv 1$
$1111\equiv 0$

So we need $b\equiv 4\bmod 11$ (so assuming $a,b,c$ are non-negative, we have $b\in \{4,15,\ldots,81\}$). Any of these $b$ values produces one or more solutions; you can assume $a=0$, solve directly for $c$ and then find solutions with larger $a$ (if the initial $c$ is big enough).

Answer 3

$$a×1111+b×111+c×11=9002 \\ 11 ( 101a +10b+c)= 9002-b$$

It follows that $b=11k+4$ for some $k \in \mathbb N$. The equation then becomes $$101 a+10b +c=818-k \\ 101 a+111k +c=778\\ $$

The problem boils down to solving $$101a+111k \leq 778$$ as you can set $c=778-101a+111k$.

I claim that $$101a+111k \leq 778 \Leftrightarrow a+k \leq 7$$

$\Rightarrow$ $$101a+111k \leq 778 \Rightarrow 101a+101k \leq 778 < 808 \Rightarrow a+k <8$$

$\Leftarrow$

$$101a+111k \leq 111(a+k) \leq 777 \leq 778$$

So you need to count how many solutions does the equation $a+k \leq 7$ have, which is very easy to count.