[Number Theory]
Problem: Compute the third least positive integer $n$ such that each of $n, n+1$, and $n+2$ is a product of exactly two (not necessarily distinct) primes.

Solution: Define a positive integer $n$ to be a semiprime if it is a product of exactly two (not necessarily distinct) primes. Define a lucky trio to be a sequence of three consecutive integers, $n, n+1, n+2$, each of which is a semiprime. Note that a lucky trio must contain exactly one multiple of 3. Also note that the middle number in a lucky trio must be even. To see this, note that if the first and last numbers in a lucky trio were both even, then exactly one of these numbers would be a multiple of 4 . But neither $2,3,4$ nor 4,5,6 is a lucky trio, and if a list of three consecutive integers contains a multiple of 4 that is greater than 4 , this number cannot be a semiprime. Using this conclusion and because $3,4,5$ is not a lucky trio, it follows that the middle number of a lucky trio cannot be a multiple of 4 . Hence it is necessary that a lucky trio has the form $4 k+1,4 k+2,4 k+3$, for some positive integer $k$, with $2 k+1$ being a prime. Note that $k \neq 1(\bmod 3)$ because when $k=1$, the sequence $5,6,7$ is not a lucky trio, and when $k>1,4 k+2$ would be a multiple of 6 greater than 6 , hence it cannot be a semiprime. Trying $k=2,3,5,6,8,9, \ldots$ allows one to eliminate sequences of three consecutive integers that are not lucky trios, and if lucky trios are ordered by their least elements, one finds that the first three lucky trios are 33,34,35; 85,86,87; and 93,94,95. Hence the answer is 93.

Final answer: 93