The number `N=6^(log_(10)40). 5^(log_(10)36)` is a natural number ,Then sum of digits of N is :

Answer

To solve the problem \( N = 6^{\log_{10} 40} \cdot 5^{\log_{10} 36} \), we will use properties of logarithms.

### Step-by-step Solution:

1. **Rewrite the expression using logarithmic properties**:
\[ N = 6^{\log_{10} 40} \cdot 5^{\log_{10} 36} \]
We can use the property \( a^{\log_b c} = c^{\log_b a} \) to rewrite \( 5^{\log_{10} 36} \):
\[ 5^{\log_{10} 36} = 36^{\log_{10} 5} \]
Thus, we can rewrite \( N \) as:
\[ N = 6^{\log_{10} 40} \cdot 36^{\log_{10} 5} \]

2. **Express 36 in terms of 6**:
Since \( 36 = 6^2 \), we can substitute:
\[ N = 6^{\log_{10} 40} \cdot (6^2)^{\log_{10} 5} \]
This simplifies to:
\[ N = 6^{\log_{10} 40} \cdot 6^{2 \log_{10} 5} \]

3. **Combine the exponents**:
Since the bases are the same, we can add the exponents:
\[ N = 6^{\log_{10} 40 + 2 \log_{10} 5} \]

4. **Use the property of logarithms**:
We can use the property \( a \log_b c = \log_b(c^a) \) to combine the logarithms:
\[ N = 6^{\log_{10}(40 \cdot 5^2)} \]
Since \( 5^2 = 25 \), we have:
\[ N = 6^{\log_{10}(40 \cdot 25)} \]

5. **Calculate \( 40 \cdot 25 \)**:
\[ 40 \cdot 25 = 1000 \]
Therefore:
\[ N = 6^{\log_{10} 1000} \]

6. **Evaluate \( \log_{10} 1000 \)**:
Since \( 1000 = 10^3 \), we have:
\[ \log_{10} 1000 = 3 \]
Thus:
\[ N = 6^3 \]

7. **Calculate \( 6^3 \)**:
\[ 6^3 = 216 \]

8. **Find the sum of the digits of \( N \)**:
The digits of \( 216 \) are \( 2, 1, \) and \( 6 \). Therefore, the sum of the digits is:
\[ 2 + 1 + 6 = 9 \]

### Final Answer:
The sum of the digits of \( N \) is \( 9 \).