$3^{1234}$ ?

What are the last two digits ofThe answer is 69.

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Nice. Can we use binomial expansion for $7^{1234}$ ? We have already considered this solution (before we posted ours, when we were discussing various approaches), and decided to not post it, because, we could not stand this procedure for other cases; in short, we could not generalize. Can you help?

Sheikh Asif Imran Shouborno
- 6 years, 5 months ago

$7^{1234}=49^{617}\\=(50-1)^{617}\\\binom{617}{0}50^{617}-\binom{617}{1}50^{616}+...+\binom{617}{616}50-\binom{617}{617}$

$=617×50-1\\=3084$

So last two digits would be $\boxed{84}$

I can't generalize it, but this can be used for almost every number I can think of right now.

Can you give me an example of such number?

Pranjal Jain
- 6 years, 5 months ago

Thanks. There are infinite examples. Are you asking for a typical exponential representation such as $123^{456}$ ?

Sheikh Asif Imran Shouborno
- 6 years, 5 months ago

@Sheikh Asif Imran Shouborno – Square up 123 and it would become $(10k-1)^{2}$ . Follow the same process!

I mean an example where such manipulation won't work

Pranjal Jain
- 6 years, 5 months ago

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

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This solution is right but is very impractical. Since you're only worrying about the last two digits, you can just keep the last two digits of $3^x$ , this simplifies the arithmetic calculation by a lot. And it's easy to make an error in the calculation which will lead to a wrong value of cycle.

i solved it by the same way

Sando Black
- 6 years, 5 months ago

**
Divide and Conquer:
**

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

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

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**
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.

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You could simplify your work by apply $3^{34} = 9^{17} = (10 - 1)^{17} \equiv -1 + 170 \equiv 69$

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

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 $\boxed{69}$