Let's break down the problem step by step.

Given:

• A positive integer $n$ has 8 factors.
• $3n^2$ has 28 factors.
• $7n^3$ has 80 factors.
• We need to find how many factors $21n^4$ has.

Step 1: Prime Factorization of $n$

Let the prime factorization of $n$ be: $n = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}$ where $p_1, p_2, \dots, p_k$ are distinct primes and $e_1, e_2, \dots, e_k$ are non-negative integers.

The number of factors of a number is given by the formula: $\text{Number of factors of } n = (e_1 + 1)(e_2 + 1) \dots (e_k + 1)$

Step 2: Apply the first condition

We are told that $n$ has 8 factors. Therefore, we have: $(e_1 + 1)(e_2 + 1) \dots (e_k + 1) = 8$ The factorizations of 8 are:

• $8 = 8 \times 1$, which corresponds to $n = p_1^7$,
• $8 = 4 \times 2$, which corresponds to $n = p_1^3 p_2^1$,
• $8 = 2 \times 2 \times 2$, which corresponds to $n = p_1^1 p_2^1 p_3^1$.

Now, let's check the other conditions to narrow it down.

Step 3: Apply the second condition

We are told that $3n^2$ has 28 factors. The prime factorization of $3n^2$ is: $3n^2 = 3 \times (p_1^{2e_1} p_2^{2e_2} \dots p_k^{2e_k})$ The number of factors of $3n^2$ is: $(1 + 1)(2e_1 + 1)(2e_2 + 1) \dots (2e_k + 1)$ We are given that this equals 28: $(1 + 1)(2e_1 + 1)(2e_2 + 1) \dots (2e_k + 1) = 28$ Simplifying: $2(2e_1 + 1)(2e_2 + 1) \dots (2e_k + 1) = 28$ $(2e_1 + 1)(2e_2 + 1) \dots (2e_k + 1) = 14$ The factorizations of 14 are:

• $14 = 14 \times 1$, which corresponds to $2e_1 + 1 = 14$ and $2e_2 + 1 = 1$, i.e., $e_1 = 6, e_2 = 0$,
• $14 = 7 \times 2$, which corresponds to $2e_1 + 1 = 7$ and $2e_2 + 1 = 2$, i.e., $e_1 = 3, e_2 = 0$,
• $14 = 2 \times 7$, which corresponds to $2e_1 + 1 = 2$ and $2e_2 + 1 = 7$, i.e., $e_1 = 0, e_2 = 3$.

Step 4: Apply the third condition

We are told that $7n^3$ has 80 factors. The prime factorization of $7n^3$ is: $7n^3 = 7 \times (p_1^{3e_1} p_2^{3e_2} \dots p_k^{3e_k})$ The number of factors of $7n^3$ is: $(1 + 1)(3e_1 + 1)(3e_2 + 1) \dots (3e_k + 1)$ We are given that this equals 80: $(1 + 1)(3e_1 + 1)(3e_2 + 1) \dots (3e_k + 1) = 80$ Simplifying: $2(3e_1 + 1)(3e_2 + 1) \dots (3e_k + 1) = 80$ $(3e_1 + 1)(3e_2 + 1) \dots (3e_k + 1) = 40$ The factorizations of 40 are:

• $40 = 40 \times 1$, which corresponds to $3e_1 + 1 = 40$,
• $40 = 10 \times 4$.

Step 5: