Problem 1581

To solve the problem, we need to find all positive integers $n$ such that the average of all positive divisors of $n$ is 42. We denote the sum of all positive divisors of $n$ by $\sigma(n)$ and the number of positive divisors of $n$ by $d(n)$. The average of all positive divisors of $n$ is given by:

$$A(n) = \frac{\sigma(n)}{d(n)}$$

Given $A(n) = 42$, we have:

$$\frac{\sigma(n)}{d(n)} = 42 \implies \sigma(n) = 42 \cdot d(n)$$

We need to find all $n$ such that $\sigma(n) = 42 \cdot d(n)$. We will consider different cases based on the number of divisors $d(n)$.

1. **Case $d(n) = 2$**:
- If $n$ has exactly 2 divisors, $n$ must be a prime number. Let $n = p$, where $p$ is a prime.
- Then, $\sigma(n) = 1 + p$ and $d(n) = 2$.
- We need $1 + p = 42 \cdot 2 = 84$, so $p = 83$.
- Thus, $n = 83$.

2. **Case $d(n) = 3$**:
- If $n$ has exactly 3 divisors, $n$ must be the square of a prime number. Let $n = p^2$, where $p$ is a prime.
- Then, $\sigma(n) = 1 + p + p^2$ and $d(n) = 3$.
- We need $1 + p + p^2 = 42 \cdot 3 = 126$.
- Solving $p^2 + p + 1 = 126$ gives no integer solutions for $p$.

3. **Case $d(n) = 4$**:
- If $n$ has exactly 4 divisors, $n$ can be either $p^3$ or $p_1 p_2$, where $p$ and $p_1, p_2$ are primes.
- For $n = p_1 p_2$:
- $\sigma(n) = (1 + p_1)(1 + p_2)$ and $d(n) = 4$.
- We need $(1 + p_1)(1 + p_2) = 42 \cdot 4 = 168$.
- Possible pairs $(1 + p_1, 1 + p_2)$ are $(3, 56), (4, 42), (6, 28), (7, 24), (8, 21), (12, 14)$.
- Valid pairs are $(3, 56)$ and $(12, 14)$:
- For $(3, 56)$, $p_1 = 2$ and $p_2 = 55$ (not prime).
- For $(12, 14)$, $p_1 = 11$ and $p_2 = 13$, so $n = 11 \cdot 13 = 143$.

4. **Case $d(n) = 6$**:
- If $n$ has exactly 6 divisors, $n$ can be $p^5$ or $p_1^2 p_2$, where $p$ and $p_1, p_2$ are primes.
- For $n = p_1^2 p_2$:
- $\sigma(n) = (1 + p_1 + p_1^2)(1 + p_2)$ and $d(n) = 6$.
- We need $(1 + p_1 + p_1^2)(1 + p_2) = 42 \cdot 6 = 252$.
- Possible pairs $(1 + p_1 + p_1^2, 1 + p_2)$ are $(3, 84), (4, 63), (6, 42), (7, 36), (9, 28), (12, 21)$.
- Valid pairs are $(3, 84)$ and $(12, 21)$:
- For $(3, 84)$, $p_1 = 2$ and $p_2 = 83$, so $n = 2^2 \cdot 83 = 332$.
- For $(12, 21)$, $p_1 = 3$ and $p_2 = 20$ (not prime).

5. **Case $d(n) = 8$**:
- If $n$ has exactly 8 divisors, $n$ can be $p^7$ or $p_1^3 p_2$ or $p_1 p_2 p_3$, where $p$ and $p_1, p_2, p_3$ are primes.
- For $n = p_1 p_2 p_3$:
- $\sigma(n) = (1 + p_1)(1 + p_2)(1 + p_3)$ and $d(n) = 8$.
- We need $(1 + p_1)(1 + p_2)(1 + p_3) = 42 \cdot 8 = 336$.
- Possible triplets $(1 + p_1, 1 + p_2, 1 + p_3)$ are $(2, 7, 24), (3, 4, 28), (3, 6, 14), (4, 6, 14)$.
- Valid triplets are $(2, 7, 24)$ and $(3, 6, 14)$:
- For $(2, 7, 24)$, $p_1 = 1$ (not prime).
- For $(3, 6, 14)$, $p_1 = 2$, $p_2 = 5$, $p_3 = 13$, so $n = 2 \cdot 5 \cdot 13 = 130$.

Summarizing all valid solutions:
- $n = 83, 143, 332, 130$.

The sum of all solutions is:
$$83 + 143 + 332 + 130 = 688$$