Selected Problems from the Final Round of National Olympiad

Consider the sum modulo $2$ and modulo $5$. As powers of odd numbers are odd and powers of even numbers are even, the number of odd summands equals the number of odd elements in set $\{1,\dots ,2011\}$. As there are an even number of odd elements in this set, the sum given in the problem is even.

Concerning modulo $5$, note that $0$ to every power is congruent to $0$ and $a^4$ is congruent to $1$, whenever $1\le a\le 4$. Thus for all $a$ such that $1\le a\le 20$, $(a+20{)}^{a+20}=a^{a+20}=a^a\cdot a^{20}=a^a\cdot (a^4{)}^5=a^a\cdot 1=a^a(\mathrm{mod}5)$. Hence the remainders modulo $5$ repeat periodically with the period $20$. As there are $100$ full periods and $100$ is divisible by $5$, the sum of the last $2000$ summands is congruent to $0(\mathrm{mod}5)$.

It remains to compute the remainders of the first $11$ summands: $1^1=1$, $2^2=4$, $3^3=27\equiv 2$, $4^4\equiv 4^0=1$, $5^5\equiv 0^1=0$, $6^6\equiv 1^2=1$, $7^7\equiv 2^3\equiv 3$, $8^8\equiv 3^0=1$, $9^9\equiv 4^1=4$, $10^{10}\equiv 0^2=0$, $11^{11}\equiv 1^3=1$.

Hence the overall sum is congruent to $3$ modulo $5$. Consequently, the last digit of this sum is $8$.