How many positive integers less than 1000 are multiples of 11 whose square roots are whole numbers.

Options

A. 2
B. 4
C. 8
D. 11

Answer

To solve the problem of finding how many positive integers less than 1000 are multiples of 11 whose square roots are whole numbers, we can follow these steps:

### Step 1: Identify the condition for square roots
We need to find numbers that are perfect squares and also multiples of 11. A perfect square can be expressed as \( n^2 \), where \( n \) is a whole number.

### Step 2: Express the perfect square as a multiple of 11
Since we are looking for perfect squares that are multiples of 11, we can express this as:
\[ n^2 = 11k \]
where \( k \) is a positive integer.

### Step 3: Determine the form of \( n \)
For \( n^2 \) to be a multiple of 11, \( n \) itself must also be a multiple of 11. Therefore, we can express \( n \) as:
\[ n = 11m \]
where \( m \) is a whole number.

### Step 4: Substitute \( n \) back into the equation
Substituting \( n \) into the equation for \( n^2 \):
\[ n^2 = (11m)^2 = 121m^2 \]

### Step 5: Set the condition for \( n^2 < 1000 \)
Now we need to find values of \( m \) such that:
\[ 121m^2 < 1000 \]

### Step 6: Solve for \( m^2 \)
Dividing both sides by 121:
\[ m^2 < \frac{1000}{121} \approx 8.26 \]

### Step 7: Find the maximum integer value for \( m \)
Taking the square root of both sides:
\[ m < \sqrt{8.26} \approx 2.87 \]
Thus, the possible integer values for \( m \) are 1 and 2.

### Step 8: Calculate the corresponding \( n^2 \) values
Now we can calculate \( n^2 \) for \( m = 1 \) and \( m = 2 \):
- For \( m = 1 \):
\[ n^2 = 121 \times 1^2 = 121 \]
- For \( m = 2 \):
\[ n^2 = 121 \times 2^2 = 484 \]

### Step 9: List the valid perfect squares
The valid perfect squares that are multiples of 11 and less than 1000 are:
- \( 121 \)
- \( 484 \)

### Step 10: Count the valid numbers
Thus, there are **2 positive integers less than 1000 that are multiples of 11 and whose square roots are whole numbers**.

### Final Answer
The answer is **2**.

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