Prove that $10152^8-10887^8+27195^8$ is divisible by $26460$.

The very first thing that we need to do is to factorize $26460$ as [MATH]\color{yellow}\bbox[5px,purple]{26460=2^2\cdot 3^3\cdot 5\cdot 7^2}[/MATH]. The next thing we have to accomplish is to show that $10152^8-10887^8+27195^8$ is divisible by $2^2\cdot 3^3\cdot 5\cdot 7^2$, either

1. all at once or
2. separately.

But, I am sure you will also work out the prime factors for the other three numbers

$10152=2^3\cdot 3^3\cdot 47$; $10887=3\cdot 19\cdot 191$; $27195=3\cdot 5\cdot 7^2\cdot 37$

$3^a+3^b+3^c = 7299$

Find the total number of positive integers ordered pairs of the equation $3^a+3^b+3^c = 7299$.

WLOG, let $c>b>a$.

Rewrite the RHS of the given equation as the product of two factors, we have:

$7299 = 3^a + 3^b + 3^c$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3^a\left(1 + \dfrac{3^b}{3^a} +\dfrac{3^c}{3^a}\right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3^a(1 + 3^{b-a} + 3^{c-a})$