Set of OSK 2016

It seems intuitive to work backwards, being careful to note the strict inequalities. The largest $e$ can be is $99.$ Then, the largest $d$ can be is $5\cdot 99 -1 = 494.$ Then, the largest $c$ can be is $4 \cdot 494 - 1 = 1975.$ Finally, $b$ can be $3\cdot 1975 - 1 = 5924,$ so $a$ can be at most $2\cdot 5924 - 1 = \boxed{11847}.$

Of course, there is a nice generalizable answer (given the process we used above) in terms of $n$ and $N$ given $a_{k-1} < ka_{k}$ for $k=2,3,\ldots,n$ and $a_n \le N.$ Specifically, the largest possible value of $a_1$ will be $\boxed{n! \cdot N - \sum_{k=1}^{n-1}k!}.$ Here, $n=5$ and $N=99,$ so we have $5! \cdot 99 - (1! + 2! + 3! + 4!) = 11847.$