if the sum of the digits of $a+b$, $b+c$ and $a+c$ is $<5$, then is the sum of the digits of $a+b+c$ less or equal than 50? This is a problem from a math olympiad: 
Let $S(x)$ denote the sum of the digits of a natural number $x$ in its decimal representation. Do there exist three natural numbers $a$, $b$ and $c$ with $S(a+b)<5$, $S(a+c)<5$, $S(b+c)<5$ and $S(a+b+c)>50$?
My attempt: 
NOTATION: Given a real number $x$, $[x]$ is the integer part of $x$.
REMARK 1: Let $x$ be a natural number with $2$ digits. Then the sum of its two digits is 
$$ \left[\frac{x}{10}\right]+x-\left[\frac{x}{10}\right]\cdot 10=x-9\cdot \left[\frac{x}{10}\right]. $$
REMARK 2: If $x$ is a real number with $\left[ x\right]\geq1$, then $\left[ 2x\right]\leq 3\left[ x\right]$.
Let $a$, $b$ and $c$ be any natural numbers. Write
$$ a=\sum_{i=0}^{n-1}a_i 10^i,\quad b=\sum_{i=0}^{n-1}b_i 10^i, \quad c=\sum_{i=0}^{n-1}c_i 10^i, $$
where $a_i,b_i,c_i\in\{0,\ldots,9\}$ and $a_n\neq0$ or $b_n\neq0$ or $c_n\neq0$. By Remark 1,
$$ S(a+b+c)=\sum_{i=0}^{n-1}\left(a_i+b_i+c_i-9\cdot \left[ \frac{a_i+b_i+c_i}{10}\right]\right), $$
$$ S(a+b)=\sum_{i=0}^{n-1}\left(a_i+b_i-9\cdot \left[ \frac{a_i+b_i}{10}\right]\right), $$
$$ S(b+c)=\sum_{i=0}^{n-1}\left(b_i+c_i-9\cdot \left[ \frac{b_i+c_i}{10}\right]\right), $$
$$ S(a+c)=\sum_{i=0}^{n-1}\left(a_i+c_i-9\cdot \left[ \frac{a_i+c_i}{10}\right]\right). $$
By Remark 2, if $\left[ (a_i+b_i+c_i)/10\right]\geq1$, then $$3\left[(a_i+b_i+c_i)/10\right]\geq \left[ 2(a_i+b_i+c_i)/10\right]\geq \left[ (a_i+b_i)/10\right]+\left[ (b_i+c_i)/10\right]+\left[ (a_i+c_i)/10\right].$$ Thus,
$$ \begin{align*}
S(a+b+c)= {} & \sum_{i=0}^{n-1}\left(a_i+b_i+c_i-9\cdot \left[ \frac{a_i+b_i+c_i}{10}\right]\right) \\\\
= {} & \frac12\sum_{i=0}^{n-1}(a_i+b_i)+\frac12\sum_{i=0}^{n-1}(b_i+c_i)+\frac12\sum_{i=0}^{n-1}(a_i+c_i)\\
&\qquad-9\sum_{\substack{{0\leq i\leq n-1}\\{(a_i+b_i+c_i)/10\geq1}}}\left[ \frac{a_i+b_i+c_i}{10}\right] \\\\
\leq {} & \frac12\sum_{i=0}^{n-1}(a_i+b_i)+\frac12\sum_{i=0}^{n-1}(b_i+c_i)+\frac12\sum_{i=0}^{n-1}(a_i+c_i)-3\sum_{i=0}^{n-1}\left[ \frac{a_i+b_i}{10}\right]\\
&\qquad-3\sum_{i=0}^{n-1}\left[ \frac{b_i+c_i}{10}\right]-3\sum_{i=0}^{n-1}\left[ \frac{a_i+c_i}{10}\right] \\\\
= & \frac12\sum_{i=0}^{n-1}\left(a_i+b_i-6\left[ \frac{a_i+b_i}{10}\right]\right)+\frac12\sum_{i=0}^{n-1}\left(b_i+c_i-6\left[ \frac{b_i+c_i}{10}\right]\right)\\
&\qquad+\frac12\sum_{i=0}^{n-1}\left(a_i+c_i-6\left[ \frac{a_i+c_i}{10}\right]\right) \\\\
= {} & \frac{S(a+b)+S(b+c)+S(a+c)}{2}+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{a_i+b_i}{10}\right]\\
&\qquad+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{b_i+c_i}{10}\right]+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{a_i+c_i}{10}\right] \\\\
\leq {} & \frac{S(a+b)+S(b+c)+S(a+c)}{2}+\frac{3}{2}S(a+b)+\frac{3}{2}S(b+c)+\frac{3}{2}S(a+c) \\\\
= {} & 2(S(a+b)+S(b+c)+S(a+c))\;.
\end{align*}
$$ Then, if $S(a+b)<5$, $S(a+c)<5$, $S(b+c)<5$, we have $S(a+b+c)<2(5+5+5)=30$. 

Counterexample (the sum is 51):
a = 555 444 555 5
b = 555 555 444 5
c = 444 555 555 5
a+b = 1111 000 000 0
a+c = 1000 000 111 0
b+c = 1000 111 000 0
a+b+c = 1555 555 555 5 
