1. **State the problem:** We have a positive integer $n$ with exactly 6 positive divisors. We list all positive divisors $d$ of $n$ and compute $\frac{n}{d}$ for each divisor $d$. Summing all these six values results in 84. We need to find $n$. 2. **Analyze the divisor count condition:** The number of divisors $d(n)$ of $n$ depends on the prime factorization: if $$n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k},$$ then $$d(n) = (a_1 + 1)(a_2 + 1) \cdots (a_k + 1).$$ Since $d(n) = 6$, the factorization possibilities are: - $n = p^5$ (so $a_1 = 5$), or - $n = p^1 q^2$ (so $(1+1)(2+1) = 2 \times 3 = 6$). 3. **Use the sum property:** Given divisors $d_1, d_2, ..., d_6$ of $n$, the sum is $$\sum_{i=1}^{6} \frac{n}{d_i}.$$ Since $\frac{n}{d_i}$ is also a divisor of $n$ (because divisors come in pairs $d$ and $\frac{n}{d}$), listing all divisors $d_i$ means the set $\{ \frac{n}{d_i} \}$ is the same as the set of divisors. Thus, $$\sum_{i=1}^{6} \frac{n}{d_i} = \sum_{i=1}^{6} d_i = 84.$$ This tells us the sum of divisors of $n$ is 84. 4. **Try the cases:** - Case 1: $n = p^5$. The divisors are $1, p, p^2, p^3, p^4, p^5$ Sum of divisors for $p^5$ is $$1 + p + p^2 + p^3 + p^4 + p^5 = \frac{p^6 - 1}{p - 1} = 84.$$ Try $p=2$: sum is $1 + 2 + 4 + 8 + 16 + 32 = 63 \neq 84$. Try $p=3$: sum is $1 + 3 + 9 + 27 + 81 + 243 = 364 \neq 84$. No prime $p$ will satisfy this easily for sum 84. - Case 2: $n = p^1 q^2$ with primes $p