Question

27^54 - 13^54 is at least divisible by

Answer 1

$We \:know \:that , \\\pink {The \: expression \:a^{n} - b^{n} \:is \: always }\\\pink{divisible \:by \:(a-b) , if \:n\:is \:any} \\\pink{positive \:integer}$

$Given , \: expression \:27^{54} - 13^{54}$

$Here , a = 27, \:b = 13 , \: and \: n = 54$

Therefore.,

$27^{54} - 13^{54} \:is \: divisible \\by \:(27-13) = 14$

•••♪

Answer 2

Before we solve the problem, let us know some formulae:

1. $\mathsf{(a^{n}-b^{n})}$ is divisible by $\mathsf{(a-b)}$ where $\mathsf{n\in\mathbb{Z}^{+}}$.
2. $\mathsf{(a^{n}+b^{n})}$ is divisible by $\mathsf{(a+b)}$ where $\mathsf{n}$ is an odd integer.

We try to combine the above two rules:

$\quad$$\mathsf{(a^{2m}-b^{2m})}$ is divisible by both $\mathsf{(a-b)}$ and $\mathsf{(a+b)}$ where $\mathsf{m}$ is an odd integer

$\to \mathsf{(a^{2m}-b^{2m})}$ is divisible by $\mathsf{(a-b)(a+b)}$ where $\mathsf{m}$ is an odd integer

$\to \mathsf{(a^{2m}-b^{2m})}$ is divisible by $\mathsf{(a^{2}-b^{2})}$ where $\mathsf{m}$ is an odd integer.

Let us solve the given problem now:

$\therefore \mathsf{27^{54}-13^{54}}$

$\mathsf{=27^{2\times 27}-13^{2\times 27}}$

Comparing it with the above combined formula, we get:

$\mathsf{\quad a=27,\quad b=13,\quad m=27}$

$\quad$where $\mathsf{m=27}$ is an odd integer.

Thus $\mathsf{(27^{54}-13^{54})}$ is divisible $\mathsf{(a^{2}-b^{2})=(27^{2}-13^{2})}$

i.e., divisible by $\mathsf{(27+13)(27-13)}$

i.e., divisible by $\mathsf{(40\times 14)}$

i.e., divisible by $\bold{560}$

$\therefore \mathsf{(27^{54}-13^{54})}$ is at least divisible by $\bold{560}$

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