The probability that a randomly chosen 3-digit number has exactly 3 factors is k/900 then k is______.

Answer

To solve the problem, we need to find the probability that a randomly chosen 3-digit number has exactly 3 factors. Here’s a step-by-step solution:

### Step 1: Determine the total number of 3-digit numbers
The range of 3-digit numbers is from 100 to 999.

\[ \text{Total 3-digit numbers} = 999 - 100 + 1 = 900 \]

### Step 2: Understand the condition for having exactly 3 factors
A number has exactly 3 factors if and only if it is the square of a prime number. This is because the factors of \( p^2 \) (where \( p \) is a prime) are \( 1, p, p^2 \), which gives us exactly 3 factors.

### Step 3: Identify the prime numbers whose squares are 3-digit numbers
We need to find prime numbers \( p \) such that \( p^2 \) lies between 100 and 999.

- The smallest prime number whose square is a 3-digit number is \( 11 \) (since \( 11^2 = 121 \)).
- The largest prime number whose square is still a 3-digit number is \( 31 \) (since \( 31^2 = 961 \)).

### Step 4: List the prime numbers between 11 and 31
The prime numbers in this range are:
- 11
- 13
- 17
- 19
- 23
- 29
- 31

### Step 5: Count the prime numbers
We have the following primes: \( 11, 13, 17, 19, 23, 29, 31 \).

Counting these, we find there are 7 prime numbers.

### Step 6: Calculate the probability
The probability \( P \) that a randomly chosen 3-digit number has exactly 3 factors is given by the ratio of the number of favorable outcomes (which are the squares of the 7 prime numbers) to the total number of outcomes (which are the 900 three-digit numbers).

\[ P = \frac{7}{900} \]

### Step 7: Identify \( k \)
In the problem, it is stated that the probability can be expressed as \( \frac{k}{900} \). Here, \( k = 7 \).

Thus, the final answer is:

\[ \boxed{7} \]