If `(241)/(4000)=(241)/(2^mxx5^n)` then find the value of m + n, where m and n are non-negative integers.

Options

A. 10
B. 8
C. 6
D. 7

Answer

To solve the equation \(\frac{241}{4000} = \frac{241}{2^m \times 5^n}\) and find the value of \(m + n\), we need to express \(4000\) in terms of its prime factors.

### Step 1: Factorize \(4000\)
First, we can break down \(4000\) into its prime factors:
\[ 4000 = 4 \times 1000 \]
Next, we can further factor \(4\) and \(1000\):
\[ 4 = 2^2 \quad \text{and} \quad 1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3 \]

### Step 2: Combine the factors
Now, we can combine these factors:
\[ 4000 = 4 \times 1000 = 2^2 \times (2^3 \times 5^3) = 2^2 \times 2^3 \times 5^3 \]
Combining the powers of \(2\):
\[ 4000 = 2^{2+3} \times 5^3 = 2^5 \times 5^3 \]

### Step 3: Set up the equation
Now we can rewrite the original equation:
\[ \frac{241}{4000} = \frac{241}{2^5 \times 5^3} \]
This means that \(m = 5\) and \(n = 3\).

### Step 4: Calculate \(m + n\)
Now, we can find \(m + n\):
\[ m + n = 5 + 3 = 8 \]

### Final Answer
Thus, the value of \(m + n\) is:
\[ \boxed{8} \]