To determine the smallest number of primes that could divide an integer with 20 positive divisors, let's analyze the relationship between the divisors and the prime factors.

Step 1: Divisor Count Formula

For an integer $n$ with prime factorization: $n = p_1^{e_1} \cdot p_2^{e_2} \cdot \dots \cdot p_k^{e_k}$ where $p_1, p_2, \dots, p_k$ are distinct primes and $e_1, e_2, \dots, e_k$ are their respective powers, the total number of divisors $d(n)$ of $n$ is given by: $d(n) = (e_1 + 1)(e_2 + 1) \dots (e_k + 1)$ In this case, we are given that $d(n) = 20$.

Step 2: Finding Factor Combinations of 20

To satisfy $d(n) = 20$, we need to find combinations of $(e_1 + 1), (e_2 + 1), \dots, (e_k + 1)$ that multiply to 20. Let's examine possible factorizations of 20:

1. $20 = 20 \times 1$ — This would mean $n$ is of the form $p^{19}$, with only one prime. However, $p^{19}$ is too large.
2. $20 = 10 \times 2$ — This suggests $n = p_1^9 \cdot p_2^1$, with two primes.
3. $20 = 5 \times 4$ — This suggests $n = p_1^4 \cdot p_2^3$, also with two primes.
4. $20 = 4 \times 5$ — This is identical to the previous option.
5. $20 = 2 \times 2 \times 5$ — This suggests $n = p_1^1 \cdot p_2^1 \cdot p_3^4$, with three primes.

Step 3: Minimizing the Number of Primes

From the above combinations, we can see that the smallest number of primes needed is 2, achieved by either:

• $n = p_1^9 \cdot p_2^1$, or
• $n = p_1^4 \cdot p_2^3$.

Thus, the smallest number of primes that could divide an integer with exactly 20 positive divisors is:

$\boxed{2}$

Would you like further explanation or examples? Here are related questions you may explore:

1. How would the answer change if the integer had 12 divisors instead?
2. How can the divisor formula be applied to find the total divisors of any composite number?
3. What if the integer needed exactly 30 divisors instead of 20?
4. Are there integers with a specific number of divisors for a single prime factor?
5. What is the minimum value of such an integer with exactly 20 divisors?

Tip: Knowing the divisor formula helps in solving problems involving prime factor counts efficiently.