Cono Sur Mathematical Olympiad

Let $p_1<p_2<p_3<\dots$ be the sequence of all prime numbers, and denote $s_n=a_1+a_2+\cdots +a_n$. Notice that $a_3=3$, and consequently $s_3=1+2+3=6=2\cdot 3=p_1{p}_2$. Suppose that for some $n$ we have $s_n=p_k{p}_{k+1}$. Then, the greatest prime divisor of $s_n$ is $p_{k+1}$, so $a_{n+1}=p_{k+1}$ and therefore $s_{n+1}=s_n+a_{n+1}=p_k{p}_{k+1}+p_{k+1}=(p_k+1)p_{k+1}$. Once again, the greatest prime divisor of this number is $p_{k+1}$, because $p_k+1\le p_{k+1}$ and so it cannot be divisible by any larger prime. Thus $s_{n+2}=(p_k+2)p_{k+1}$. This pattern holds until reaching step $j$ where, for the first time, $p_k+j$ is divisible by some prime number that is larger than $p_{k+1}$. Clearly this happens when $p_k+j=p_{k+2}$. In that moment we have $s_{n+j}=(p_k+j)p_{k+1}=p_{k+1}{p}_{k+2}$, and we are again in the same situation we were at the beginning. After this observation, since $s_3=2\cdot 3$, we can deduce that after 3 steps we get to $s_6=3\cdot 5$, after 4 more steps we get to $s_{10}=5\cdot 7$, and continuing in this way we get $$\begin{array}{rlrlrlrl}{s}_{16}& =7\cdot 11,& s_{22}& =11\cdot 13,& s_{28}& =13\cdot 17,& s_{34}& =17\cdot 19,\\ s_{40}& =19\cdot 23,& s_{50}& =23\cdot 29,& s_{58}& =29\cdot 31,& s_{66}& =31\cdot 37,\\ s_{76}& =37\cdot 41,& s_{82}& =41\cdot 43,& s_{88}& =43\cdot 47,& s_{98}& =47\cdot 53.\end{array}$$

Hence $a_{99}=53$, $s_{99}=48\cdot 53$ and finally $a_{100}=53$.