If m and n are natural numbers such that n > 1, and mⁿ = 2²⁵ ×3⁴⁰, then m−n equals to.

The correct option is 209942.

We are given that $m$ and $n$ are natural numbers such that $n>1$, and we are given the equation:

$m^n=2^{25}\times 3^{40}$

To find the values of $m$ and $n$, we need to express the right-hand side in the form of ${\text{Base}}^{\text{Exponent}}$, where the exponent $n$ is a natural number greater than 1.
Let's find the greatest common divisor (GCD) of the exponents 25 and 40.
The factors of 25 are 1 and 5.
The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.
Thus, the greatest common divisor of 25 and 40 is:

$\text{GCD}(25,40)=5$

Since $n>1$ must be a common divisor of the exponents 25 and 40 for $m$ to be an integer (specifically, a natural number), $n$ must divide both 25 and 40. The common divisors of 25 and 40 are 1 and 5.
Since we are given that $n>1$, the only possible value for $n$ is:

$n=5$

Now, we rewrite the equation by expressing the exponents in terms of 5:

$m^5=2^{5\times 5}\times 3^{8\times 5}$

Using the laws of exponents, $a^{x·y}={\left(a^x\right)}^y$, we can rewrite the terms as follows:

$m^5={\left(2^5\right)}^5\times {\left(3^8\right)}^5$

Combining the terms under a single exponent, we get:

$m^5={(2^5\times 3^8)}^5$

Taking the 5th root on both sides, we find the value of $m$:

$m=2^5\times 3^8$

Let's calculate the values of $2^5$ and $3^8$:
$2^5=32$
$3^8={\left(3^4\right)}^2={81}^2=6561$

Now, calculate $m$ by multiplying these two values:

$m=32\times 6561$

Let's perform the multiplication:
$32\times 6561=32\times (6500+61)=208000+1952=209952$

So, we have:

$m=209952$

We are asked to find the value of $m-n$:
Substitute the values of $m$ and $n$:

$m-n=209952-5=209942$

Thus, $m-n$ equals 209942.