I think this belongs in pre olympiad. One of the primes must be $2$ due to parity. Then to maximize the year, we need the other two primes to be as close as possible, which follows from a smoothing argument. Then the last tico year is $2 \cdot 31 \cdot 41$.

So the next one is $2\cdot 29\cdot 43 = 2494$ and the next and last one is $2\cdot 31\cdot 41 = 2542$. A challenging problem for an interstate competition ...

Indeed, every year the first problem of this contest tends to be very easy; this is somewhat of a convention. You might find the rest of the problems more interesting (day 2 will be uploaded tomorrow).