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!
$13^4$ ends with $1$ so does $13^{12}$ and thus $13^{13}$ ends with $3$

$12^4$ ends with $6$ so $12^{12}$ also ends with $6$

$11^{11}$ ends with $1$ so your statement is valid.
Assuming you have knowledge of modular arithmetic.

$11\equiv 1\mod 10\implies 11^{11}\equiv 1\mod 10$.

$12\equiv 2\mod 10\implies 12^6\equiv 2^6 \mod10\equiv 4\mod 10$ $\implies 12^{12}\equiv 4^2\mod 10\equiv 6\mod 10$. 

$13\equiv 3 \mod 10\implies 13^4\equiv 3^4 \mod 10\equiv 1 \mod 10\implies 13^{13}\equiv 3\mod 10.$

So $11^{11}+12^{12}+13^{13}\equiv 1+6+3 \mod 10\equiv 0 \mod 10.$

Hence $11^{11}+12^{12}+13^{13}$ is divisible by 10.