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 $a$, $b$, $c$?

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What do the dots mean in $a$...? – Gerry Myerson Oct 30 '12 at 12:01
If I ignore the dots, one solution is $a,b,c=2,8,12$. – Gerry Myerson Oct 30 '12 at 12:03
Note that combining the first and the second gives $s(a)+s(b)+s(c)+b=c$, so we have $4+c=2b$. That makes one variable disappear, and it implies that $s(b)-s(c)\in\{-1,0,1\}$. This gives 3 cases. – barto Oct 30 '12 at 12:46

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.

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 Nice solution. I think it's impossible to get there without estimating the values using powers of $10$. – barto Oct 30 '12 at 12:59 Wonderful solution , lately I have also found the same , I was having difficulty at first because I was using 10n ≥ 10^s(n) , though this is stronger than 10 + x ≥ 10 s(x) it did not help. – Souvik Dey Oct 31 '12 at 2:48 @Souvik: It's not strictly stronger -- your right-hand side is stronger, but my left-hand side is stronger, so the inequalities are incomparable. – joriki Oct 31 '12 at 3:48 @joriki: Not comparable? The inequalities are n ≥ 10( s(n) - 1) and n ≥ 10 ^( s(n) -1) , and since 10^k ≥ 10k for any non-negative integer k , the former follows readily from the later. – Souvik Dey Nov 1 '12 at 3:10