Show $11^{11}+12^{12}+13^{13} =10k$ without direct calculation Prove that $11^{11}+12^{12}+13^{13}$ is divisible by $10$.
Obviously you could just put that in to a calculator and see the results, but I was wondering about some of the other approaches to this? I have not studied modulus', so if you could explain it without them, it would be better for me. Thanks!

You can easily prove it without any modular arithmetic. Just look at the last digits.
$$3^1 = \color{blue}{3} \quad 3^2 = \color{blue}{9} \quad 3^3 = 2\color{blue}{7} \quad 3^4 = 8\color{blue}{1} \implies 3, 9, 7, 1, 3, 9, 7, 1, ...$$
$$2^1 = \color{green}{2} \quad 2^2 = \color{green}{4} \quad 2^3 = \color{green}{8} \quad 2^4 = 1\color{green}{6} \implies 2, 4, 8, 6, 2, 4, 8, 6, ...$$
The pattern loops every $4^{th}$ exponent, as you can see. Notice that $13^{13}$ must end with whatever $3^{13}$ ends with, and $2^{12}$ must end with whatever $2^{12}$ ends with.
$$13 = \color{purple}{3}(4)+\color{blue}{1} \quad\text{Three loops done; first exponent}$$
$$12 = \color{purple}{2}(4)+\color{green}{4} \quad\text{Two loops done; fourth exponent}$$
Thus, $13^{13}$ ends with $\color{blue}{3}$ and $12^{12}$ ends with $\color{green}{6}$. $11^{11}$ obviously ends with $1$, so what does the unit digit become?
