Question

If $53^{53}-33^{3}$ is divided by $10,$ then the remainder obtained is

Answer 1

We know that,
$x^{n}-y^{n}=\left(x - y\right)\left(x^{n - 1} + x^{n - 2} y + . . . . + x y^{n - 2} + y^{n - 1}\right)$
i.e. $x^{n}-y^{n}$ is divisible by $x-y$
$53^{53}-33^{3}=53^{53}-3^{53}+3^{53}-33^{3}+3^{3}-3^{3}$
$=\left(5 3^{53} - 3^{53}\right)+\left(3^{53} - 3^{3}\right)-\left(3 3^{3} - 3^{3}\right)$
Clearly, $1^{s t}$ bracket is divisible by $50$ and the last one is divisible by $30.$ So, both are divisible by $10.$
$3^{53}-3^{3}=3^{3}\cdot 3^{50}-3^{3}=3^{3}\left(3^{50} - 1\right)$
$=3^{3}\left(\left(10 - 1\right)^{25} - 1\right)=27\left(10 - 1\right)^{25}-27=27\left(10 k - 1\right)-27$
$=270k-54=270k-60+6=10\mu +6.$
Hence, the remainder obtained is $6.$