divisibilitymodular arithmetic
Prove that $53^{53}-33^3$ is divisible by $10$
I don't know modular arithmetic, so I tried things like that:
$53^3 \cdot 53^{50}-33^3=(33+20)^3 \cdot 53^{50}-33^3=(33+20)(33+20)(33+20)\cdot 53^{50}-33^3$
but I get stuck and don't know how to continue
$5\nmid (n^2+n+1)$ for all integer $n$. Because $$n^2+n+1 \equiv (n+3)^2 + 2 \pmod 5$$ So if $5\mid (n^2+n+1)$ for some $n$, then $n$ satisfies $(n+3)^2 \equiv 3 \pmod 5$. But $3$ is not quadratic residue modulo $5$. (You can check that $3$ is not quadratic residue easily.)
${\color{White}{\text{Proof without words.}}}$
Best Answer
One way to hide the omnipresence of modular arithmetic is to note that a number is divisible by $10$ if and only if the final digit is $0$; and that the final digit of any number formed by a $\{+,-,\times\}$-recipe depends only on the final digits of the ingredients.
Now, the final digits of the powers of $3$ are $1$, $3$, $9$, $7$, and then the final digits repeat, starting with $81=3^4$. So the $4k$-th power of any number ending in $3$ will end in $1$, and thus $53^{53}=53^{52}\cdot53$ ends in $3$. On the other hand, $33^{33}=33^{32}\cdot33$ also ends in $3$, their difference ends in zero. If, as @MR.BEAN stated, the problem is $53^{53}-33^3$, then since $33^3$ ends in $7$, it’s the sum that is divisible by $10$.