Problem 1168

To find the value of the sum
$$\sum_{1 \le a < b < c} \frac{1}{2^a 3^b 5^c},$$
we need to consider all possible triples $(a, b, c)$ where $a, b, c$ are positive integers such that $a < b < c$.

1. **Sum over $c$:**
For a fixed $a$ and $b$, $c$ can take any value greater than $b$. Therefore, the sum over $c$ is:
$$\sum_{c=b+1}^{\infty} \frac{1}{5^c} = \frac{1}{5^{b+1}} \sum_{k=0}^{\infty} \frac{1}{5^k} = \frac{1}{5^{b+1}} \cdot \frac{1}{1 - \frac{1}{5}} = \frac{1}{5^{b+1}} \cdot \frac{5}{4} = \frac{1}{4 \cdot 5^b}.$$

2. **Sum over $b$:**
For a fixed $a$, $b$ can take any value greater than $a$. Therefore, the sum over $b$ is:
$$\sum_{b=a+1}^{\infty} \frac{1}{3^b} \cdot \frac{1}{4 \cdot 5^b} = \frac{1}{4} \sum_{b=a+1}^{\infty} \frac{1}{(3 \cdot 5)^b} = \frac{1}{4} \sum_{b=a+1}^{\infty} \frac{1}{15^b} = \frac{1}{4 \cdot 15^{a+1}} \sum_{k=0}^{\infty} \frac{1}{15^k} = \frac{1}{4 \cdot 15^{a+1}} \cdot \frac{1}{1 - \frac{1}{15}} = \frac{1}{4 \cdot 15^{a+1}} \cdot \frac{15}{14} = \frac{1}{14 \cdot 15^a}.$$

3. **Sum over $a$:**
Finally, $a$ can take any positive integer value. Therefore, the sum over $a$ is:
$$\sum_{a=1}^{\infty} \frac{1}{2^a} \cdot \frac{1}{14 \cdot 15^a} = \frac{1}{14} \sum_{a=1}^{\infty} \frac{1}{(2 \cdot 15)^a} = \frac{1}{14} \sum_{a=1}^{\infty} \frac{1}{30^a} = \frac{1}{14 \cdot 30} \sum_{k=0}^{\infty} \frac{1}{30^k} = \frac{1}{14 \cdot 30} \cdot \frac{1}{1 - \frac{1}{30}} = \frac{1}{14 \cdot 30} \cdot \frac{30}{29} = \frac{1}{14 \cdot 29} = \frac{1}{406}.$$

The final answer is $\boxed{\frac{1}{406}}$.