Question 1
What is the last two digits of the number
$$(11+12+13+...+2006)^2?$$
Solution
We have
\begin{align*}A&=11+12+13+...+2006\\&=2017\times 998\equiv 17\times 98 \text{(mod 100)}\\& \equiv 66 \text{(mod 100)}\end{align*}
Hence
$$A^2 \equiv 56 \text{(mod 100)}$$
Thus the last two digits of $A^2$ is $56$
Question 2
Find the last two digits of the sum
$$2005^{11}+2005^{12}+2005^{13}+...+2005^{2006}$$
Solution
Let:
$$B=2005^{11}+2005^{12}+2005^{13}+...+2005^{2006}$$
Forall $k \geq 2$, we have
$$2005^k \equiv 5^k \equiv 25 \text{(mod 100)}$$
Hence
$$B \equiv 25 \times 1996 \equiv 0 \text{(mod 100)}$$
The last two digits of $B$ is $00$
