Problem 2. Solve in the set of positive rational numbers the equation:

$$
\begin{array}{r}
\frac{x+1}{2}+\frac{x+5}{3}+\frac{x+11}{4}+\frac{x+19}{5}+\frac{x+29}{6}+\frac{x+41}{7}+\frac{x+55}{8}+\frac{x+71}{9}+\frac{x+89}{10}=45 . \\
\text { Mathematical Gazette nr. 10/2013 }
\end{array}
$$
## Problem 2.

We have: $1+2+3+\cdots+9=45$ (1p)

The equation can be written as:

$$
\begin{aligned}
& \left(\frac{x+1}{2}-1\right)+\left(\frac{x+5}{3}-2\right)+\left(\frac{x+11}{4}-3\right)+\left(\frac{x+19}{5}-4\right)+\left(\frac{x+29}{6}-5\right)+\left(\frac{x+41}{7}-6\right)+\left(\frac{x+55}{8}-7\right)+ \\
& \left(\frac{x+71}{9}-8\right)+\left(\frac{x+89}{10}-9\right)=0(3 p) \\
& \Leftrightarrow \frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}+\cdots+\frac{x-1}{10}=0 \quad(1 p) \quad \Rightarrow(x-1) \underbrace{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots \frac{1}{10}\right)}_{\neq 0}=0(1 \mathrm{p}) \\
& \Leftrightarrow x-1=0 \Rightarrow x=1 \text { (1p) }
\end{aligned}
$$