Saudi Arabian Mathematical Competitions, 2015

The problem is equivalent to proving that $k$ is relatively prime with $p_{1} p_{2} \cdots p_{n}$ and such that for all $i = 2, \ldots, n$, none of the integers $k,\ k + p_{1}, \ldots, k + p_{1} \cdots p_{i-1}$ is divisible by $p_{i}$.

Let $i \in \{2, \ldots, n\}$. Notice that $p_{i} \geq i + 1$. So the numbers $-p_{1}, -p_{1} p_{2}, \ldots, -p_{1} p_{2} \cdots p_{i-1}$ will not cover all the non-zero residues modulo $p_{i}$. Let $k_{i}$ be one of these non-covered non-zero residues. Clearly, none of the numbers $k_{i}, k_{i} + p_{1}, \ldots, k_{i} + p_{1} \cdots p_{i-1}$ is congruent to $0$ modulo $p_{i}$.

This defines $k_{2}, k_{3}, \ldots, k_{n}$ and let $k_{1} = 1$. By the Chinese remainder theorem there exists an integer $k$ such that $k \equiv k_{i} \not\equiv 0 \pmod{p_{i}}$ for all $i = 1, 2, \ldots, n$. The integer $k$ is relatively prime with $p_{1} p_{2} \cdots p_{n}$. On the other hand, for all $1 \leq j < i \leq n$, we have $k + p_{1} p_{2} \cdots p_{j} \equiv k_{i} + p_{1} p_{2} \cdots p_{j} \not\equiv 0 \pmod{p_{i}}$.