To determine how many 3-digit numbers have exactly 9 factors, we need to analyze how a number can have exactly 9 factors based on its prime factorization.
The number of factors of a number \( n \) which has the prime factorization \( p_1^{e_1} \times p_2^{e_2} \times \ldots \times p_k^{e_k} \) is given by the formula:
\[ f(n) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) \]
For \( f(n) \) to be exactly 9, we consider the possible decompositions of 9 into products of integers, because each \( e_i + 1 \) corresponds to one of the factors.
The factorizations of 9 are:
- \( 9 = 9 \): This corresponds to \( n = p_1^8 \), where \( p_1 \) is a prime number.
- \( 9 = 3 \times 3 \): This corresponds to \( n = p_1^2 \times p_2^2 \), where \( p_1 \) and \( p_2 \) are distinct prime numbers.
Next, we will evaluate both cases:
Case 1: \( n = p_1^8 \)
We need \( p_1^8 \) to be a 3-digit number.
We check the 3-digit range for possible values of \( p_1 \):
\[ 100 \leq p_1^8 < 1000 \]
Taking the eighth root:
\[ 100^{1/8} \approx 2.51 \quad \text{and} \quad 1000^{1/8} \approx 3.16 \]
Thus, \( p_1 \) can only be 3 (because \( p_1 \) must be a prime number).
Calculating \( 3^8 \): \[ 3^8 = 6561 \quad (\text{which is a 4-digit number}) \]
Since no valid \( p_1 \) satisfies this condition, there are no valid numbers from this case.
Case 2: \( n = p_1^2 \times p_2^2 \)
Now, we need to find distinct primes \( p_1 \) and \( p_2 \) such that:
\[ 100 \leq p_1^2 \times p_2^2 < 1000 \]
This can be rewritten as:
\[ 10 \leq p_1 \times p_2 < 31.62 \]
Thus, \( p_1 \times p_2 \) should be in the range \( [10,31] \).
Next, we can identify pairs of small prime numbers to calculate valid products.
The relevant primes are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31.
We need to find distinct pairs \( (p_1, p_2) \):
- \( (2, 3) \rightarrow 6 \)
- \( (2, 5) \rightarrow 10 \)
- \( (2, 7) \rightarrow 14 \)
- \( (2, 11) \rightarrow 22 \)
- \( (2, 13) \rightarrow 26 \)
- \( (2, 17) \rightarrow 34 \quad (\text{Invalid})\)
- \( (3, 5) \rightarrow 15 \)
- \( (3, 7) \rightarrow 21 \)
- \( (3, 11) \rightarrow 33 \quad (\text{Invalid})\)
- \( (5, 7) \rightarrow 35 \quad (\text{Invalid})\)
- \( (3, 13) \rightarrow 39 \quad (\text{Invalid})\)
- \( (5, 11) \rightarrow 55 \quad (\text{Invalid})\)
We can now calculate the valid distinct pairs:
- \( (2, 5) \rightarrow 10 \)
- \( (2, 7) \rightarrow 14 \)
- \( (2, 11) \rightarrow 22 \)
- \( (2, 13) \rightarrow 26 \)
- \( (3, 5) \rightarrow 15 \)
- \( (3, 7) \rightarrow 21 \)
The valid combinations of \( p_1 \) and \( p_2 \):
- \( (2, 5), (2, 7), (2, 11), (2, 13), (3, 5), (3, 7) \)
Now we need to check the square of their products to see if they make valid 3-digit numbers:
- For \( (2, 5) \): \( n = 2^2 \times 5^2 = 4 \times 25 = 100 \)
- For \( (2, 7) \): \( n = 2^2 \times 7^2 = 4 \times 49 = 196 \)
- For \( (2, 11) \): \( n = 2^2 \times 11^2 = 4 \times 121 = 484 \)
- For \( (2, 13) \): \( n = 2^2 \times 13^2 = 4 \times 169 = 676 \)
- For \( (3, 5) \): \( n = 3^2 \times 5^2 = 9 \times 25 = 225 \)
- For \( (3, 7) \): \( n = 3^2 \times 7^2 = 9 \times 49 = 441 \)
Thus the 3-digit numbers with exactly 9 factors are \( 100, 196, 225, 441, 484, \) and \( 676 \).
In conclusion, there are 6 3-digit numbers with exactly 9 factors.