1-2-3-4 Nostalgia

$3^{1234}\\=9^{617}\\=(10-1)^{617}$

Using binomial expansion,

$(10-1)^{617}=\binom{617}{0}10^{617}-\binom{617}{1}10^{616}+...-\binom{617}{615}10^{2}+\binom{617}{616}10-\binom{617}{617}$ $=100k+6170-1=100\alpha+69$

Therefore, $3^{1234}\equiv 69\mod 100$

So the last two digits are $\boxed{69}$

3^0 = 1

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729

3^7 = 2187

3^8 = 6561

3^9 = 19683

3^10 = 59049

3^11 = 177147

3^12 = 531441

3^13 = 1594323

3^14 = 4782969

3^15 = 14348907

3^16 = 43046721

3^17 = 129140163

3^18 = 387420489

3^19 = 1162261467

3^20 = 3486784401

3^21 = 10460353203

3^22 = 31381059609

3^23 = 94143178827

3^24 = 282429536481 .

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Notice how the pattern of the last two digits repeats from the power 20. If you take 1234/20 you get 14 as a remainder, so: 3^1234 will have the same last 2 digits as 3^14

Hence the answer is 69

Divide and Conquer:
$3^4 \equiv 81\; (mod\; 100)$ $3^8\equiv 81 \times 81 \equiv 61 \; (mod\; 100)$ $3^{10}\equiv 9 \times 61 \equiv 49 \; (mod\; 100)$ $3^{20}\equiv 49^2 \equiv 1 \; (mod\; 100)$ $3^{1234}\equiv (3^{20})^{61}(3^{14})\equiv (3^4)(3^{10}) \equiv 81 \times 49 \equiv 69 \; (mod \; 100)$

$\boxed{69}$ . $(Ans.)$

Euler's Theorem (A little advanced approach for an elementary problem):

We will use Euler's totient function to count how many integers are relatively prime to $100$ .

$\phi (100) =\phi (2^25^2)= 100(1-\frac12)(1-\frac15)=40$ .

Now, Euler's theorem implies that, $3^{40}\equiv 1\; (mod\; 100)$ .

So, $3^{1234}\equiv (3^{40})^{30}(3^{34})\equiv (3^{40-6})\; (mod\; 100)$

Now, our duty is curbed down to find the inverse of $3^6$ , where $3^6\equiv (27)^2 \equiv 29 \;(mod \; 100)$ .

The inverse, my friends, is easily found to be $\boxed{69}$ . The last step is left as a homework.