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

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

### Step 2: Substitute the prime factors
Now, we substitute these into the expression:

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

\[ 6^{m+1} = (2 \cdot 3)^{m+1} = 2^{m+1} \cdot 3^{m+1} \]

\[ 10^{n+3} = (2 \cdot 5)^{n+3} = 2^{n+3} \cdot 5^{n+3} \]

\[ 15^{m} = (3 \cdot 5)^{m} = 3^{m} \cdot 5^{m} \]

### Step 3: Substitute back into the expression
Now substituting these into our original expression gives:

\[ \frac{2^{m} \cdot 3^{2m-n} \cdot 5^{m+n+3} \cdot (2^{n+1} \cdot 3^{n+1})}{(2^{m+1} \cdot 3^{m+1}) \cdot (2^{n+3} \cdot 5^{n+3}) \cdot (3^{m} \cdot 5^{m})} \]

### Step 4: Combine the terms
Now we can combine the terms in the numerator and denominator:

**Numerator:**
\[ 2^{m} \cdot 3^{2m-n} \cdot 5^{m+n+3} \cdot 2^{n+1} \cdot 3^{n+1} = 2^{m+n+1} \cdot 3^{2m-n+n+1} \cdot 5^{m+n+3} = 2^{m+n+1} \cdot 3^{2m+1} \cdot 5^{m+n+3} \]

**Denominator:**
\[ (2^{m+1} \cdot 3^{m+1}) \cdot (2^{n+3} \cdot 5^{n+3}) \cdot (3^{m} \cdot 5^{m}) = 2^{m+1+n+3} \cdot 3^{m+1+m} \cdot 5^{n+3+m} = 2^{m+n+4} \cdot 3^{2m+1} \cdot 5^{n+m+3} \]

### Step 5: Simplify the expression
Now we can simplify the expression:

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

This simplifies to:

\[ 2^{(m+n+1)-(m+n+4)} \cdot 3^{(2m+1)-(2m+1)} \cdot 5^{(m+n+3)-(m+n+3)} = 2^{-3} \cdot 3^{0} \cdot 5^{0} = \frac{1}{2^3} = \frac{1}{8} \]

### Final Answer
Thus, the value of the expression is

\[ \frac{1}{8} \]