, 2017

Solution:
Let $x = b - a$ and $y = c - b$ so that $b = a + x$ and $c = a + x + y$. Then
$$2^{a} 3^{b} 5^{c} = 2^{a} 3^{a + x} 5^{a + x + y} = 30^{a} 15^{x} 5^{y}$$
and $a, x, y$ are any positive integers. Thus
$$\begin{aligned} \sum_{1 \leq a < b < c} \frac{1}{2^{a} 3^{b} 5^{c}} & = \sum_{1 \leq a, x, y} \frac{1}{30^{a} 15^{x} 5^{y}} \\ & = \sum_{1 \leq a} \frac{1}{30^{a}} \sum_{1 \leq x} \frac{1}{15^{x}} \sum_{1 \leq y} \frac{1}{5^{y}} \\ & = \frac{1}{29} \cdot \frac{1}{14} \cdot \frac{1}{4} \\ & = \frac{1}{1624} \end{aligned}$$