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
