How many 4 digit numbers have exactly 9 divisors from the set (1, 2, 3, 4, 5, 6, 7,8, 9, 10}

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.

How many 4 digit numbers have exactly 9 divisors from the set (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)?

Learn how to find all 4-digit numbers that have exactly 9 divisors using concepts from number theory. This solution explores divisor counts using the divisor count formula and discusses the relevant prime factorization patterns. Suitable for students in Grades 9-12.

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