woh !!!!!!! it's high power

2012^2 last 2 digits = 44

2^2012 = 2^10^200 × 2^10 × 2^2

= 1024^200 × 1024 × 4

24^ even power ends in 76 and 24^odd power ends in 24

total of last 2 digits = 44 + (76 × 24 × 4) = 44 + 76 × 96 = 44 + 96 = 40 considering only last 2 digits,

$2012 \times 2012$ leaves $44$ such the last two digits. Well, for $2^{2012}$ , we should see the the numbers in the form $2^n$ : $2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5 = 32, 2^6 = 64, 2^7=128, 2^8 = 256, ..., 2^{12} = 4096...$ We can realize that all the numbers in the form $2^{4m}$ have $6$ as units digit. The tens digit grows in an arithmetic progression, with a common difference equal to $4$ , and the first term is $1$ . So:

$2^{2012} = (2^4)^{503}$ $a_n = 1 + 502\cdot 4$ $a_n = 1 + 2008 = 2009$

So, the tens digit is $9$ .

Finally, $44 + 96 = 140$ , and the last two digits is $40$ .

What is only sought is the last two digits of the sum, so let us only look for the last two digits of both $2012^2$ and $2^{2012}$ .

For $2012^2$ (we can just ignore the leftmost numbers),

2 0 1 2
2 0 1 2
-------
    2 4
    2
-------
    4 4

So, the last two digits of $2012^2$ is $44$ .

Next, we have to find the last two digits of $2^{2012}$ . We will start with $02$ , which is $2$ , just with a leading $0$ in the tens place. The zero will not matter anyway since $0 \times 2 = 0$ , that is, the tens digit remains a zero. If we are going to multiply $02$ by itself long enough, taking note of only the last two digits, you would see a pattern.

$\begin{aligned} 1^{st} term &= 02 \\ 2^{nd} term &= 04 \\ 3^{rd} term &= 08 \\ 4^{th} term &= 16 \\ 5^{th} term &= 32 \\ 6^{th} term &= 64 \\ 7^{th} term &= 28 \\ 8^{th} term &= 56 \\ 9^{th} term &= 12 \\ 10^{th} term &= 24 \\ 11^{th} term &= 48 \\ 12^{th} term &= 96 \\ 13^{th} term &= 92 \\ 14^{th} term &= 84 \\ 15^{th} term &= 68 \\ 16^{th} term &= 36 \\ 17^{th} term &= 72 \\ 18^{th} term &= 44 \\ 19^{th} term &= 88 \\ 20^{th} term &= 76 \\ 21^{st} term &= 52 \\ 22^{nd} term &= 04 \\ 23^{rd} term &= 08 \\ \vdots \end{aligned}$

Starting from the $2^{nd}$ term, we will have a repeating sequence of numbers. The sequence repeats every $20$ terms, that is, the number repeats in the $2^{nd}$ term, $22^{nd}$ term, $42^{nd}$ term, and so on. We can generalize this pattern by saying that, except for the first term, for all positive integers $k$ , all $(20n + k)^{th}$ terms are the same. Now, this implies that the $2012^{th}$ term is the same as the $12^{th}$ term, which is $96$ .

Thus, the last two digits of the sum in question is $44 + 96 \equiv \boxed{40}$ .

2^22 has 04 as last 2 digits

so 2^2002= 2^(91*22) will have 04 as unit digits

2^2012=(2^2002) (2^10)= 04 1024= 96 as last 2 digits

add 2012^2 to it 40 is last two digits.