The least positive integer with which 661.25 should be multipled so that the product is a perfect square is __________

Options

A. 4
B. 5
C. both a and b
D. none of these

Explanation

To find the least positive integer with which 661.25 should be multiplied so that the product is a perfect square, we can follow these steps:

### Step 1: Convert the Decimal to a Fraction
First, we convert 661.25 into a fraction:
\[ 661.25 = \frac{66125}{100} \]

### Step 2: Factor the Numerator and Denominator
Next, we need to factor both the numerator (66125) and the denominator (100).

**Denominator:**
\[ 100 = 10 \times 10 = 10^2 = (2 \times 5)^2 = 2^2 \times 5^2 \]

**Numerator:**
To factor 66125, we can start dividing by prime numbers:
- 66125 is odd, so it is not divisible by 2.
- The sum of the digits (6 + 6 + 1 + 2 + 5 = 20) is divisible by 5, so we divide by 5:
\[ 66125 \div 5 = 13225 \]
- Now, we can divide 13225 by 5 again:
\[ 13225 \div 5 = 2645 \]
- Divide 2645 by 5 again:
\[ 2645 \div 5 = 529 \]
- Finally, we can factor 529, which is \(23 \times 23\) or \(23^2\).

So, we have:
\[ 66125 = 5^3 \times 23^2 \]

### Step 3: Combine the Factors
Now we can write the complete factorization of 661.25:
\[ 661.25 = \frac{5^3 \times 23^2}{2^2 \times 5^2} \]
This simplifies to:
\[ 661.25 = \frac{5^{3-2} \times 23^2}{2^2} = \frac{5^1 \times 23^2}{2^2} \]

### Step 4: Determine the Exponents
For a number to be a perfect square, all the exponents in its prime factorization must be even.

- The exponent of 5 is 1 (odd).
- The exponent of 23 is 2 (even).
- The exponent of 2 is 2 (even).

### Step 5: Find the Required Multiplication
To make the exponent of 5 even, we need to multiply by \(5^1\) (which is 5). This will make the exponent of 5 equal to 2 (even).

### Conclusion
Thus, the least positive integer with which 661.25 should be multiplied to make it a perfect square is:
\[ \boxed{5} \]