, 2014

Each of the four integers has 1 as a divisor. Moreover, 2 is a common divisor of either $N$ and $N+2$ or $N+1$ and $N+3$. Hence, these four integers can have at most $6 \times 4 - 3 - 1 = 20$ different divisors altogether. Because there are exactly 20 different positive divisors, only one of the four integers can be divisible by 3, that is either $N+1$ or $N+2$ is divisible by 3.

As 27 is a divisor of one of the integers, and the integer has exactly six divisors, that number must be $3^5 = 243$. This can be seen, for example, by recognising that an integer $p_1^{a_1} \cdots p_n^{a_n}$, with primes $p_1 < \cdots < p_n$ and $a_i \ge 1$, has exactly $(a_1+1) \cdots (a_n+1)$ different positive divisors.

Hence $N+1 = 243$ or $N+2 = 243$. Because $243-2 = 241$ is a prime number (having only two positive divisors), we can only have $N = 242$.

It is easily checked that the four numbers $242 = 2 \cdot 11^2$, $243 = 3^5$, $244 = 2^2 \cdot 61$ and $245 = 5 \cdot 7^2$ have exactly 20 divisors, namely 1, 2, 3, 4, 5, 7, 9, 11, 22, 27, 35, 49, 61, 81, 121, 122, 242, 243, 244 and 245.