To solve this problem, we need to determine how many 4-digit numbers have exactly 9 divisors and that all divisors are from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Step-by-Step Solution:

1. Understanding Divisor Count for Integers:

   • For a number $N$ to have exactly 9 divisors, it must be in a form that aligns with divisor rules.
   • If $N = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$ (where $p_i$ are primes and $e_i$ are their respective powers), the number of divisors of $N$ is given by $(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$.
   • To get exactly 9 divisors, we need:
     • $9 = 9 \times 1$ (which implies $N = p^8$) or
     • $9 = 3 \times 3$ (which implies $N = p_1^2 \times p_2^2$, where $p_1$ and $p_2$ are distinct primes).

2. Focus on 4-Digit Numbers:

   • We need $N$ to be a 4-digit number (between 1000 and 9999).

3. Case Analysis:

   • Case 1: $N = p^8$:
     • The only 4-digit number of this form would be $2^8 = 256$, which is not a 4-digit number.
   • Case 2: $N = p_1^2 \times p_2^2$:
     • We need two distinct primes $p_1$ and $p_2$ such that $N = p_1^2 \times p_2^2$ results in a 4-digit number.
     • We must also ensure that all divisors of $N$ come from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Calculation for Case 2:

Let’s examine potential pairs $(p_1, p_2)$:

• Possible primes are limited to those within the divisor set, namely: $\{2, 3, 5, 7\}$.
• We test pairs:
  • $N = 2^2 \times 3^2 = 4 \times 9 = 36$ (not 4-digit)
  • $N = 2^2 \times 5^2 = 4 \times 25 = 100$ (not 4-digit)
  • $N = 2^2 \times 7^2 = 4 \times 49 = 196$ (not 4-digit)
  • $N = 3^2 \times 5^2 = 9 \times 25 = 225$ (not 4-digit)
  • $N = 3^2 \times 7^2 = 9 \times 49 = 441$ (not 4-digit)
  • $N = 5^2 \times 7^2 = 25 \times 49 = 1225$ (4-digit and valid)

After testing, we find that:

• The only 4-digit number satisfying these conditions is 1225.

Conclusion:

Only one 4-digit number, 1225, has exactly 9 divisors and meets the given criteria.

Would you like more details on this solution, or do you have any questions?

Here are five related questions to explore this topic further:

1. What are the rules for determining the number of divisors of any integer?
2. Can a number with 9 divisors ever be a 5-digit number?
3. How do we find all divisors of a given number systematically?
4. What happens if we change the condition to finding numbers with exactly 8 divisors?
5. How would the solution change if we only considered prime numbers up to 20?

Tip: The divisor count formula is useful in many contexts, especially when dealing with factors or prime powers in number theory problems.