Positive integers and the number of their digits Let $a$, $b$, $c$ be positive integers and $s(a)$, $s(b)$, $s(c)$ denote the number of their digits (when the integers are written in decimal form) respectively. If, 
$s(a)+s(b)=a\qquad$
$a + b + s(c) = c\qquad$ 
and
$4 + s(a) + s(b) + s(c) = b \qquad$ 
then what would be the possible values of $a$, $b$, $c$?

If $a$ is large, then (1) says that $b\approx10^a$; then (3) says that $c\approx10^{10^a}$, and then (2) can't work out. Thus $a$ can't be large. Then the same arguments applied to (2) and (3) show that $b$ also can't be large, and then the same argument applied just to (2) shows that $c$ can't be large, either. Thus there are only finitely many solutions.
To get rough bounds on the magnitudes, note that $s(x)\le1+x/10$. Thus from (1)
$$s(b)=a-s(a)\ge a-\left(1+\frac a{10}\right)=\frac9{10}a-1\;.$$
and hence
$$
b\ge10(s(b)-1)\ge9a-20\;.
$$
Then substituting into (3) yields
$$
\begin{align}
s(c)
&=b-4-s(a)-s(b)
\\
&\ge b-4-\left(1+\frac a{10}\right)-\left(1+\frac b{10}\right)
\\
&
=\frac9{10}b-\frac1{10}a-6
\end{align}$$
and hence
$$
c\ge10(s(c)-1)\ge9b-a-70\;.
$$
But (2) yields
$$c=a+b+s(c)\le a+b+1+\frac c{10}$$
and thus
$$c\le\frac{10}9(a+b+1)\;.$$
Together, this is
$$\frac{10}9(a+b+1)\ge9b-a-70$$
or
$$
71b\le19a+640\;,
$$
so
$$
71(9a-20)\le19a+640\;,
$$
or
$$
a\le\frac{103}{31}\lt4\;.
$$
Since $a=1$ doesn't work in (1), that leaves $a=2$ or $a=3$. That implies $s(b)=1$ or $s(b)=2$, thus $b\le99$ and thus
$$c\le\frac{10}9(a+b+1)\lt\frac{10}9(4+99+1)=\frac{1040}9\lt1000\;.$$
Thus $s(c)\le3$, and then (3) yields $b\le10$, then (2) yields $c\le16$ and thus $s(c)\le2$, then (3) yields $b\le9$ and thus $s(b)=1$, which by (1) implies $a=2$. Now (3) becomes $b=6+s(c)$; that leaves only $b=7$ and $b=8$, and in both cases (2) yields $c\ge10$ and thus $s(c)=2$; then finally (3) yields $b=8$ and (2) yields $c=12$.
Thus the solution $a=2$, $b=8$, $c=12$ that Gerry found is the only solution.
