If $m$ and $n$ are natural numbers such that $n > 1$, and $m^{n}=2^{25} \times 3^{40}$, then $m-n$ equals

We have $m^n = 2^{25} \times 3^{40}$ where $m$ and $n$ are natural numbers and $n > 1$.

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To solve this question, we need to write

$m^n$ in the form $(2^a \times 3^b)^n$

When prime factorizations are equal, their exponents must match:

$an = 25$ (exponents of 2 must be equal)

$bn = 40$ (exponents of 3 must be equal)

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From $an = 25$ and $bn = 40$, we see that $n$ must divide both 25 and 40.

To find all values that divide both numbers, we need their Greatest Common Divisor (GCD).

Finding GCD(25, 40):

$25 = 5^2$

$40 = 2^3 \times 5$

Common prime factor: 5

Lowest power of 5: $5^1$

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The divisors of 5 are: 1 and 5

Since we're given $n > 1$, we have $n = 5$.

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Now with $n = 5$:

From $an = 25$: $a \times 5 = 25$, so $a = 5$

From $bn = 40$: $b \times 5 = 40$, so $b = 8$

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

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Let's calculate this:

$2^5 = 32$

$3^8 = 3^4 \times 3^4 = 81 \times 81 = 6561$

$m = 32 \times 6561 = 209952$

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$m - n = 209952 - 5 = 209947$