The value of `(2^(m).3^(2m-n).5^(m+n+3).6^(n+1))/(6^(m+1).10^(n+3).15^(m))`

Options

A. depend on m
B. depend on n
C. is zero
D. does not depend on m and n

Explanation

To solve the expression

\[ \frac{2^{m} \cdot 3^{2m-n} \cdot 5^{m+n+3} \cdot 6^{n+1}}{6^{m+1} \cdot 10^{n+3} \cdot 15^{m}}, \]

we will simplify it step by step.

### Step 1: Rewrite the terms in the numerator and the denominator

First, we can express \(6\), \(10\), and \(15\) in terms of their prime factors:

- \(6 = 2 \cdot 3\)
- \(10 = 2 \cdot 5\)
- \(15 = 3 \cdot 5\)

Now, substituting these into the expression:

**Numerator:**
\[ 6^{n+1} = (2 \cdot 3)^{n+1} = 2^{n+1} \cdot 3^{n+1} \]

So, the numerator becomes:
\[ 2^{m} \cdot 3^{2m-n} \cdot 5^{m+n+3} \cdot 2^{n+1} \cdot 3^{n+1} = 2^{m+n+1} \cdot 3^{2m-n+n+1} \cdot 5^{m+n+3} = 2^{m+n+1} \cdot 3^{2m+1} \cdot 5^{m+n+3} \]

**Denominator:**
\[ 6^{m+1} = (2 \cdot 3)^{m+1} = 2^{m+1} \cdot 3^{m+1} \]
\[ 10^{n+3} = (2 \cdot 5)^{n+3} = 2^{n+3} \cdot 5^{n+3} \]
\[ 15^{m} = (3 \cdot 5)^{m} = 3^{m} \cdot 5^{m} \]

So, the denominator becomes:
\[ 2^{m+1} \cdot 3^{m+1} \cdot 2^{n+3} \cdot 5^{n+3} \cdot 3^{m} \cdot 5^{m} = 2^{m+n+4} \cdot 3^{m+m+1} \cdot 5^{n+3+m} \]

### Step 2: Combine the numerator and denominator

Now we can rewrite the entire expression:

\[ \frac{2^{m+n+1} \cdot 3^{2m+1} \cdot 5^{m+n+3}}{2^{m+n+4} \cdot 3^{2m+1} \cdot 5^{m+n+3}} \]

### Step 3: Simplify the expression

Now, we can simplify this expression by subtracting the exponents of like bases:

- For base \(2\):
\[ 2^{(m+n+1) - (m+n+4)} = 2^{-3} \]

- For base \(3\):
\[ 3^{(2m+1) - (2m+1)} = 3^{0} = 1 \]

- For base \(5\):
\[ 5^{(m+n+3) - (m+n+3)} = 5^{0} = 1 \]

### Step 4: Final result

Thus, the entire expression simplifies to:
\[ \frac{2^{-3}}{1 \cdot 1} = 2^{-3} = \frac{1}{8} \]

### Conclusion

The final value of the given expression is:
\[ \frac{1}{8} \]

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