Problem 1091

Due to the mutual simplicity of $p$ and $q$, the specified $p+q-2$ numbers are pairwise distinct (in addition, they are different from numbers of the form $k / p+q$). Therefore, they divide the segment $[0,1]$ into $p+q-1$ segments. It is sufficient to prove that within each of these segments there is a point of the form $k / p+q$.

On a segment of the form $\left[{ }^{m} / p,{ }^{m+1} / p\right]$, such a point clearly exists, since its length is greater than $1 / p+q$. Consider a segment of the form $\left[{ }^{m} / p,{ }^{n} / q\right]$. It contains the point ${ }^{m+n} / p+q$, since the inequalities ${ }^{m / p}<{ }^{m+n} / p+q<n / q$ are equivalent to the inequality $m q<n p$, which is the inequality $m / p<n / q$. Similarly, segments of the form $\left[{ }^{a} / q,{ }^{b} / p\right]$ are considered. Send a comment

Prove that if $\left(a_{1}, a_{2}, \ldots, a_{n}\right)=1$, then the equation $a_{1} x_{1}+a_{2} x_{2}+\ldots+a_{n} x_{n}=1$ is solvable in integers.

## Hint

By induction, it is not difficult to prove a stronger statement: the equation $a_{1} x_{1}+a_{2} x_{2}+\ldots+a_{n} x_{n}=\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ is solvable in integers. For this, one must use problem $\underline{60521}$ b).