Problem 1279

Given the conditions:
$$3a = b^3$$
$$5a = c^2$$
and the assumption that $a$ is divisible by $d^6$ where $d = 1$.

### Part 1: Prove that $a$ is divisible by $3$ and $5$.

1. From the equation $3a = b^3$, we can see that $a = \frac{b^3}{3}$. For $a$ to be an integer, $b^3$ must be divisible by $3$. Since $b$ is a positive integer, $b$ itself must be divisible by $3$. Let $b = 3k$ for some integer $k$. Then:
$$b^3 = (3k)^3 = 27k^3$$
Substituting back, we get:
$$a = \frac{27k^3}{3} = 9k^3$$
Thus, $a$ is divisible by $3$.

2. From the equation $5a = c^2$, we can see that $a = \frac{c^2}{5}$. For $a$ to be an integer, $c^2$ must be divisible by $5$. Since $c$ is a positive integer, $c$ itself must be divisible by $5$. Let $c = 5m$ for some integer $m$. Then:
$$c^2 = (5m)^2 = 25m^2$$
Substituting back, we get:
$$a = \frac{25m^2}{5} = 5m^2$$
Thus, $a$ is divisible by $5$.

Since $a$ is divisible by both $3$ and $5$, we conclude that $a$ is divisible by $15$.

### Part 2: Prove that the prime factors of $a$ are limited to $3$ and $5$.

1. From the previous part, we have:
$$a = 9k^3 \quad \text{and} \quad a = 5m^2$$
Since $a$ must satisfy both conditions simultaneously, we equate the two expressions:
$$9k^3 = 5m^2$$
This implies that $a$ must be a common multiple of $9$ and $5$. The least common multiple of $9$ and $5$ is $45$. Therefore, $a$ must be of the form:
$$a = 45n$$
for some integer $n$.

2. Since $a$ is of the form $45n$, the prime factors of $a$ are $3$ and $5$.

### Part 3: Find $a$.

1. From the equation $9k^3 = 5m^2$, we need to find integers $k$ and $m$ such that this equation holds. We rewrite it as:
$$9k^3 = 5m^2$$
This implies that $k$ and $m$ must be chosen such that both sides are equal. Since $9k^3$ and $5m^2$ must be equal, we need to find the smallest $k$ and $m$ that satisfy this condition.

2. By trial and error or by solving the equation, we find that the smallest values that satisfy this condition are $k = 5$ and $m = 3$. Substituting these values back, we get:
$$a = 9k^3 = 9 \cdot 5^3 = 9 \cdot 125 = 1125$$
$$a = 5m^2 = 5 \cdot 3^2 = 5 \cdot 9 = 45$$

Since both expressions must be equal, we find that the smallest value of $a$ that satisfies both conditions is $1125$.

The final answer is $\boxed{1125}$.