To find the value of $m$, we need to determine the total power of $10$ present in the prime factorization of the right-hand side (RHS) of the equation.

Step 1: Prime factorize the terms on the RHS

• $75^{25} = (3 \times 25)^{25} = (3 \times 5^2)^{25} = 3^{25} \times 5^{50}$
• $25^{32} = (5^2)^{32} = 5^{64}$
• $32^{75} = (2^5)^{75} = 2^{375}$

Step 2: Combine the prime factors Multiplying these together, we get: $RHS = 3^{25} \times 5^{50} \times 5^{64} \times 2^{375}$ $RHS = 2^{375} \times 3^{25} \times 5^{114}$

Step 3: Determine the maximum power of $10$ in the RHS A factor of $10$ is formed by multiplying a $2$ and a $5$. The number of $10$s we can form is limited by the smaller of their exponents.

• The power of $2$ is $375$.
• The power of $5$ is $114$.

The minimum of these is $114$. Therefore, we can factor out $10^{114}$: $RHS = (2^{114} \times 5^{114}) \times 2^{375 - 114} \times 3^{25}$ $RHS = 10^{114} \times 2^{261} \times 3^{25}$

Step 4: Compare with the left-hand side (LHS) The LHS of the equation is given as: $LHS = 10^m \times 1000 \times n$ $LHS = 10^m \times 10^3 \times n = 10^{m+3} \times n$

Equating LHS and RHS: $10^{m+3} \times n = 10^{114} \times (2^{261} \times 3^{25})$

The problem states that $n$ is not divisible by $10$, which means $n$ cannot contain both $2$ and $5$ as prime factors. The remaining term $(2^{261} \times 3^{25})$ has no factors of $5$, so it is indeed not divisible by $10$. This means all the powers of $10$ have been completely factored out into the $10^{114}$ term.

Thus, we can equate the powers of $10$: $m + 3 = 114$ $m = 114 - 3$ $m = 111$

Conclusion: The value of $m$ is $111$.