, 2008

Since $a_1, a_2, a_3, a_4, a_5$ are terms of a geometric sequence, we can write $a_i = a_1 \cdot q^{i-1}$ for $i = 2, 3, 4, 5$ and for some real number $q$. But $q = \frac{a_2}{a_1}$ is a quotient of two positive integers, so $q$ is rational. Write $q = \frac{m}{n}$ as a reduced fraction. The terms of the sequence are
$$a_1, \frac{a_1 \cdot m}{n}, \frac{a_1 \cdot m^2}{n^2}, \frac{a_1 \cdot m^3}{n^3}, \frac{a_1 \cdot m^4}{n^4}.$$
Since all terms are positive integers and $m$ and $n$ are relatively prime, we see that $n^4$ is a divisor of $a_1$. So, we can write $a_1 = dn^4$, where $d$ is some positive integer. Therefore, the terms of the sequence are
$$dn^4, dmn^3, dm^2n^2, dm^3n, dm^4.$$
But since no prime number divides all five of these terms, we have $d = 1$, and the sequence becomes $n^4, mn^3, m^2n^2, m^3n, m^4$. We also know that the terms are all less than $2008$, so $m^4 < 2008$ and $n^4 < 2008$. This implies $m \le 6$ and $n \le 6$. But $a_1 = n^4$ is not divisible by $6$, so $n \le 5$. We also know that $a_2 = n^3m$ is divisible by $5$, $a_3 = n^2m^2$ is divisible by $4$ and $a_4 = m^3n$ is divisible by $3$, so the product $mn$ is divisible by $2$, $3$ and $5$, i.e. by $30$. At the same time $mn \le 6 \cdot 5 = 30$, so this product is equal to $30$. This implies $m = 6$ and $n = 5$. The terms of the sequence are $5^4 = 625$, $5^3 \cdot 6 = 750$, $5^2 \cdot 6^2 = 900$, $5 \cdot 6^3 = 1080$ and $6^4 = 1296$.