<s>[INST] 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!
 [/INST] 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?
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