Consider the divisors of $30$: $1,2,3,5,6,10,15,30$. 
It can be seen that for every divisor $d$ of $30$, $d + 30 / d$ is prime.


Find the sum of all positive integers $n$ not exceeding $100\,000\,000$ such that for every divisor $d$ of $n$, $d + n / d$ is prime.

To solve the problem, we need to find the sum of all positive integers $ n $ not exceeding $ 100,000,000 $ such that for every divisor $ d $ of $ n $, the expression $ d + \frac{n}{d} $ is prime.

### Analyzing the Condition $ d + \frac{n}{d} $ is Prime

For any $ n $, the divisors $ d $ and $ \frac{n}{d} $ pair up, and $ d + \frac{n}{d} $ should be a prime number. This implies the structure of $ n $ must be such that the sum $ d + \frac{n}{d} $ is prime for all divisors $ d $.

Let us first investigate a number with exactly two distinct prime factors, i.e., $ n = p_1^{a_1} \times p_2^{a_2} $.

Given the condition, $ d + \frac{n}{d} $ must be prime for all divisors $ d $. Consider the following cases:

1. **Case $ n = p^k $ for some prime $ p $:**
    - Here, the divisors of $ n $ are $ 1, p, p^2, \ldots, p^k $.
    - The sums are $ 1 + p^k $, $ p + p^{k-1} $, $ p^2 + p^{k-2} $, ..., all of which must be prime.
    - This situation generally requires $ n $ to be of a special form.

2. **Case $ n = p_1 \times p_2 $ where $ p_1 $ and $ p_2 $ are distinct primes:**
    - Here, divisors are $ 1, p_1, p_2, p_1p_2 $.
    - The sums are $ 1 + p_1p_2 $, $ p_1 + p_2 $, $ p_2 + p_1 $, and $ p_1p_2 + 1 $, which are unlikely to all be primes.

The analysis shows that the number must be of the form $ p^k $ where $ p $ is a prime number. We need to check which powers $ p^k $ satisfy the given condition.

### Determining Valid Values of $ p^k $
For the simplest case:
- $ n = p $ where $ p $ is prime, we have divisors $ 1 $ and $ p $, and $ 1 + p $ must be prime. This condition trivially holds if $ p = 2 $ (since $ 1 + 2 = 3 $ is prime).

To explore higher powers:
- If $ n = p^2 $, the divisors are $ 1, p, p^2 $.
- $ d + \frac{p^2}{d} $ gives $ 1 + p^2 $, $ p + p $, and $ p^2 + 1 $. $ 2p $ must be prime, which is impossible unless $ p = 2 $.

Thus, $ n = 2^k $ is a suitable candidate.

### Check the Condition for $ n = 2^k $
For $ n = 2^k $, divisors are $ 1, 2, 2^2, \ldots, 2^k $:
- $ 1 + 2^k $, $ 2 + 2^{k-1} $, ..., all need to be primes.

Specifically:
- For $ k = 1 $, $ 2 $, $ 4 $, etc., the conditions work well.

### Sum All Valid $ 2^k $
To satisfy the condition, $ n = 2^k $, where $ 2^k \leq 100,000,000 $.

Find the maximum $ k $:
\[
2^k \leq 100,000,000 \implies k = \left\lfloor \log_2(100,000,000) \right\rfloor = 26
\]
So $ k $ ranges from 0 to 26.

Sum:
\[
\text{Sum} = 2^0 + 2^1 + \cdots + 2^{26} = 2^{27} - 1 = 134217727
\]

Therefore, the sum of all such numbers $ n $ is $ \boxed{134217727} $.

Consider the divisors of 30: 1, 2, 3, 5, 6, 10, 15, 30. It can be seen that for every divisor d of 30, d + 30/d is prime. Find the sum of all positive integers n not exceeding 100,000,000 such that for every divisor d of n, d + n/d is prime.

Explore the problem of finding the sum of all integers n under 100 million, where for each divisor d of n, the expression d + n/d is prime. The solution involves number theory concepts such as prime numbers and divisors, making it suitable for advanced math students.

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