To find how many 4-digit numbers have exactly 9 divisors, we need to understand the structure of numbers with exactly 9 divisors and how it applies to our set.

Step 1: Understanding the Divisor Count

For a number to have exactly 9 divisors, the prime factorization pattern should allow it to have exactly 9 factors. The number of divisors of a number $n$ with prime factorization $n = p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_k^{e_k}$ is given by: $D(n) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$

To achieve exactly 9 divisors, the possible forms of $n$ are:

1. $n = p^8$, which gives $D(n) = 9$ (since $8 + 1 = 9$)
2. $n = p^2 \cdot q^2$, which also gives $D(n) = (2 + 1)(2 + 1) = 9$, where $p$ and $q$ are distinct primes.

Step 2: Applying the Forms to the 4-Digit Constraint

Now we need to check which numbers from our two possible forms result in 4-digit values:

1. For $n = p^8$:

   • We need $p^8$ to be a 4-digit number. Testing small prime values:
     • $2^8 = 256$ (not 4 digits)
     • $3^8 = 6561$ (4 digits)
     • $5^8 = 390625$ (too large)
   • Only $3^8 = 6561$ is a 4-digit number that satisfies this form.

2. For $n = p^2 \cdot q^2$:

   • We need $p^2 \cdot q^2$ to be a 4-digit number.
   • Testing combinations:
     • $2^2 \cdot 3^2 = 4 \cdot 9 = 36$ (too small)
     • $2^2 \cdot 5^2 = 4 \cdot 25 = 100$ (too small)
     • $2^2 \cdot 7^2 = 4 \cdot 49 = 196$ (too small)
     • $3^2 \cdot 5^2 = 9 \cdot 25 = 225$ (too small)
     • $3^2 \cdot 7^2 = 9 \cdot 49 = 441$ (too small)
     • $5^2 \cdot 7^2 = 25 \cdot 49 = 1225$ (4 digits)
     • $3^2 \cdot 11^2 = 9 \cdot 121 = 1089$ (4 digits)
     • $2^2 \cdot 17^2 = 4 \cdot 289 = 1156$ (4 digits)
     • $2^2 \cdot 13^2 = 4 \cdot 169 = 676$ (too small)
     • $5^2 \cdot 11^2 = 25 \cdot 121 = 3025$ (4 digits)
     • $7^2 \cdot 11^2 = 49 \cdot 121 = 5929$ (4 digits)
     • $3^2 \cdot 17^2 = 9 \cdot 289 = 2601$ (4 digits)
   • Valid numbers: $1225, 1089, 1156, 3025, 5929, 2601$

Step 3: Counting the Numbers

Adding both sets of numbers:

• From $p^8$: $6561$
• From $p^2 \cdot q^2$: $1225, 1089, 1156, 3025, 5929, 2601$

Thus, there are 7 four-digit numbers that have exactly 9 divisors: $6561, 1225, 1089, 1156, 3025, 5929,$ and $2601$.

Would you like more details on any specific part?

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1. How do we find the divisor count of a number?
2. Why do numbers of the form $p^8$ and $p^2 \cdot q^2$ have exactly 9 divisors?
3. What if we needed numbers with a different count of divisors?
4. Could these techniques apply to numbers with different digit constraints?
5. What other prime factorizations lead to exact divisor counts?

Tip: The number of divisors of a number can reveal a lot about its factorization structure, especially for determining divisor constraints like having exactly 9 divisors.