Solve $a\cdot 1111+b\cdot 111+c\cdot 11=9002$ 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 $a=a_{1}+1$
 A: 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 $(a,b,c)=(7,4,71)$.
A: 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).
A: $$a×1111+b×111+c×11=9002 \\
11 ( 101a +10b+c)= 9002-b$$
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.
