The power of two is more fun than the power of one

The sum can be written as

$a = 2*(2^{0} - 2^{1} + 2^{2} - 2^{3} + ..... + 2^{2012}) =$

$2*\dfrac{(-2)^{2013} - 1}{(-2) - 1} = \dfrac{2}{3} (2^{2013} + 1)$ .

Now in order to find the last two digits of $a$ we will need to determine $a\mod 100$ . Now $2^{22} \equiv 2^{2}\mod 100$ , i.e., the remainders $\mod 100$ have a 'period' of $20$ , so

$2^{2002} \equiv 2^{2}\mod 100 \Longrightarrow 2^{2013} \equiv 2^{13}\mod 100 \equiv 92\mod 100$ .

Therefore $\dfrac{2}{3}(2^{2013} + 1) \equiv \dfrac{2}{3}(93) \equiv 62\mod 100$ .

Thus the last two digits of $a$ are $\boxed{62}$ .

I started by noticing $(2^3-2^2) + (2^5-2^4) = 20,$ and the next set of four powers of two would be $(2^7-2^6) + (2^9-2^8) = 2^4\cdot(2^3-2^2+2^5-2^4) = 16\cdot 20 = 320,$ and continuing this pattern we keep having as last digits ...20.

Since there are 2012 of such terms to account for (we skipped $2^1$ ), we would add $2012/4 = 503$ of these 20s. It is easy to see that this gives last digits ...60.

Finally, adding in $2^1 = 2$ , we get $\boxed{62}$ .

$2^1-2^2+2^3+...+2^{2013}=2^1+2^{2*1}+2^{2*2}+..+2^{2*1006}$ It is easy to realize that: ${2^{2*(10*i+1)}+2^{2*(10*i+2)}+2^{2*(10*i+3)}+...+2^{2*(10*i+10)}} \mod 100 = 0$ with i=0..99 So: $2^1-2^2+2^3+...+2^{2013} \mod 100$ $=2^1+2^{2002}+2^{2004}+...+2^{2012} \mod 100=\boxed{62}$

Lets try to sum in the usual way:

$S=(2^{1}+2^{3}+....2^{2013})-(2^{2}+2^{4}+.....2^{2012})$

$=2(2^{0}+2^{2}+....2^{2012})-(2^{0}+2^{2}+.....2^{2012})+1$

$=(2^{0}+2^{2}+2^{4}+.....2^{2012})+1$

The expression in () is an AP with a=1, r=2², n=1007

Which gives $S=\frac{4^{1007}-1}{3}+1$

Observe the following:

$\frac{4^{1}-1}{3}=01$

$\frac{4^{3}-1}{3}=21$

$\frac{4^{5}-1}{3}=341$

$\frac{4^{7}-1}{3}=...61$

$\frac{4^{9}-1}{3}=....81$

$\frac{4^{11}-1}{3}=.....01$

So, $\frac{4^{1007}-1}{3}=.......61$

Hence S will end in 61+1 = $\boxed{62}$

Since we only need the last 2 digits, we can do all of our computation in mod 100.

First, we find the residues (mod 100) of the powers of 2, just doubling each time and subtracting 100 if necessary:

$2^1\equiv 2\pmod{100}$

$2^2\equiv 4$

$2^3\equiv 8$

$2^4\equiv 16$

$2^5\equiv 32$

$2^6\equiv 64$

$2^7\equiv 28$

$2^8\equiv 56$

$2^9\equiv 12$

$2^{10}\equiv 24$

$2^{11}\equiv 48$

$2^{12}\equiv 96$

$2^{13}\equiv 92$

$2^{14}\equiv 84$

$2^{15}\equiv 68$

$2^{16}\equiv 36$

$2^{17}\equiv 72$

$2^{18}\equiv 44$

$2^{19}\equiv 88$

$2^{20}\equiv 76$

$2^{21}\equiv 52$

$2^{22}\equiv 2^2\equiv 4$

We find that every 20 powers of 2, the residue in mod 100 repeats.

Next, we rearrange the original equation to $2^{2013}-2^1-2^3-2^5...-2^{2011}$ . We can do this because $2^{n+1}$ is twice $2^n$ , so $2^n-2^{n+1}=-2^n$ .

In mod 100, this is equivalent to $2^{13}-2^1-100(2^3)-100(2^5)-100(2^7)-100(2^9)-100(2^{11})-100(2^{13})-$

$101(2^{15})-101(2^{17})-101(2^{19})-100(2^{21})\equiv 2^{13}-2^1-2^{15}-2^{17}-2^{19}\equiv$

$92-2-68-72-88\equiv \fbox{62} \pmod{100}$ .

First, rewrite the sum as

$\begin{aligned} & 2^{1}-2^{2}+2^{3}-2^{4}+2^{5}-\ldots +2^{2013}\\ =~& 2^{1}+(-2^{2}+2^{3})+(-2^{4}+2^{5})+\ldots +(-2^{2012}+2^{2013})\\ =~& 2+2^{2}+2^{4}+2^{6}+\ldots +2^{2012}\\ =~& 2+4^{1}+4^{2}+4^{3}+\ldots +4^{1006}. \end{aligned}$

Now, note that $4^{1}+4^{2}\equiv 20 \pmod{100}$ . Multiplying by $4^{2}$ gives us $4^{3}+4^{4}\equiv 320\equiv 20 \pmod{100}$ . Continuing this process yields $4^{2n-1}+4^{2n}\equiv 20 \pmod{100}$ . Therefore

$\begin{aligned} &~ 2+(4^{1}+4^{2})+(4^{3}+4^{4})+\ldots +(4^{1005}+4^{1006})\\ \equiv &~ 2 + \underbrace{20+\ldots +20}_{503} \pmod{100}\\ =&~ 10062\\ \equiv &~ \boxed{62} \pmod{100}. \end{aligned}$

xt=0; a= [4 8 16 32 64 28 56 12 24 48 96 92 84 68 36 72 44 88 76 52]; for y= 2:2013; if rem (y,20) == 0 c= 20 ; else c = rem (y,20); end x=a(c); xt = xt+x;

end disp (rem(xt+2,100))

Once again, not the best way to find a solution, as it certainly doesn't prove the existence of the solution:

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format long g, clear, clc
b = 0;

for (i = 1:100)
    a = ((-1)^(i-1))*(2^i);
    b = b + a
end

% 2260
% 00012487486362498625

The comments at the bottom, in order, are (respectively) the pattern of the one's place digit, and the pattern of the ten's place digit. This works until until scientific notation kicks in, so $i$ to 100 is actually quite overkill.

Since the first pattern has 4 numbers, 2013 (2012 divisible by 4) ends on the first number, which is 2.

The second pattern has 20, and after 2000 it ends on the 13th number, which is 6.

So the sum of the sequence will end with '62'

(2^n+2)/3, where n%4=2..... the second last digit when (2^n+2)/3 is ((n-2/4)%5)*(2)...In our case n=2014.... So ((2014-2/4)%5)(2)=(503%5)(2) =(3)(2)=6...Here the last digit of 2^n is 4..by n%4...So (2^n+2) has 6 as its last digit..:. (2^n+2)/3 has 2 as its last digit..So last 2 digits are 62..

Let's calculate just remainders mod 4 and 25. Mod 4 is easy: the remainder is 2. Mod 25: subtract $1=2^0$ from the sum and notice that it can be rewritten as $\sum_{k=0}^{1006}4^k$ . When we notice that $4^5=1024\equiv -1$ , we know that any 10 consecutive summands give a remainder of 0, so the whole sum's remainder is $\sum_{k=0}^6{4^k}\equiv 1+4+16+14+6-1-4=11$ ; the original sum's remainder then must have been 12.

Chinese remainder theorem guarantees that there's exactly one number $\ge 0, < 100$ that satisfies the two congruences; that number is easily guessed to be 62 (not 12, definitely not 37 or 87).