Waste less time on Facebook — follow Brilliant.
×

Modulo Problems

Again I am back with my lame questions....they are so lame that you would doubt if I am a legit level 5 In Number theory.

Find 3^2012's last 2 digits and also try to find its last 3 digits is the question

I know its mod 100 but i am not able to simplify it.

I would love if someone could tell me NOT ONLY HOW TO SOLVE THIS, BUT ALSO HOW TO SOLVE ANY SUCH PROBLEM USING MOD 100, 1000....AND ALSO HOW TO DO MODULAR EXPONENTIATION FAST. Thanks

Note by Krishna Ar
2 years, 12 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

I have completely different approach to the question to make it look very simple, \({ 3 }^{ 2012 }\quad could\quad be\quad written\quad as:\\ { { (3 }^{ 4 }) }^{ 503 }\\ it\quad is\quad quite\quad interesting\quad to\quad note\quad that\quad 81\quad multiplied\quad by\quad 81\quad henceforth\quad changes\\ its\quad ten's\quad digit\quad after\quad 5\quad cyclic\quad repetitions\quad i.e.\quad 8,6,4,2,0,8........\\ unit\quad digit\quad will\quad always\quad be\quad 1\\ \Rrightarrow divide\quad power\quad by\quad 5\quad i.e.\frac { 503 }{ 5 } R=3\\ find\quad the\quad third's\quad ten's\quad digits\quad i.e.\quad 4\\ answer\quad will\quad be \boxed{41}\) Utkarsh Rajput · 2 years, 11 months ago

Log in to reply

If you're not already aware, @Finn Hulse created this Modular Arithmetic set for you.

For further resources, you can look at #Modular Arithmetic, or do a search and filter by tags. @Akshaj Kadaveru 's note on Modular Arithmetic is a detailed introduction. Calvin Lin Staff · 2 years, 12 months ago

Log in to reply

@Calvin Lin Yes, sir. I am aware of this. But thanks to Nathan Ramesh I came to know of the Euler Theorem. Since then, i learnt a lot more about this topic and am yet learning. And I haven't seen a note about Euler's theorem anywhere, thus didnt know till recently. Thanks :) Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar Check out my set Olympiad Number Theory for similar notes.

You could also do a search for Euler's Theorem, and filter by notes, which will display my note on Euler's Theorem. Calvin Lin Staff · 2 years, 12 months ago

Log in to reply

@Calvin Lin @Calvin Lin - I extremely apologize for the inadvertent error of mine while setting up the answer of the question- "Integral Divisors". Your answer was right. Kindly change it as required. Thanks Krishna Ar · 2 years, 12 months ago

Log in to reply

@Calvin Lin Akshaj is the bomb diggity. :D Finn Hulse · 2 years, 12 months ago

Log in to reply

@Finn Hulse I somehow don't really understand that comment. Yuxuan Seah · 2 years, 12 months ago

Log in to reply

@Yuxuan Seah Oh I'm just saying he's awesome. :D Finn Hulse · 2 years, 12 months ago

Log in to reply

@Finn Hulse Oh haha :D Yuxuan Seah · 2 years, 11 months ago

Log in to reply

@Finn Hulse @Daniel Liu @Nathan Ramesh please help me out..is it some kind of a cycle or is it modular exponentiaiton...please tell me! :) Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar Use something called eulers totient theorem, which states \(3^{2012}\equiv 3^{2012\pmod{\phi(100)}}\equiv 3^{2012\pmod{40}}\equiv 3^{12}\equiv (3^3)^4\equiv 27^4\equiv 729^2\equiv 29^2\equiv 41 \pmod{100}\)

Where \(\phi(n)\) denotes the number of numbers less than \(n\) that are relatively prime to \(n\). Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Thanks :) ,....Thanks a lot!!!!!!!...Could you tell me how to learn more about these? For eg:- I knew this function but not this theorem :(....So any form of resource would be appreciated. @Nathan Ramesh Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar There isn't anything in particular that I think you have to read. I learned it all from browsing brilliant problems and googling things I don't know Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Hey, Is this the only way to do it? Doesn't modular exponentiation work here? I am just asking becuase I want to know a different way...:) Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar What I did is essentially the same thing. Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Ok. Yes. SO this works only if 3 is coprime to 100. So, for other cases you have to use the lengthy modular exponentiation right? And was this too easy a problem? :/ Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar No, give me another example and I will show you Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh @Nathan Ramesh @Finn Hulse pL answer Krishna Ar · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Hi @Nathan Ramesh - Take a look at this one- calculating \( 7^{1728} mod 1000\) . Using the euler theorem, you get that phi (1000)= 400. so this is equal to \( 7^{128} mod1000\) . How do I simplify it further.? @Finn Hulse you too have a look at this please. And i would like some further genralization on how to simplify such large mod powers. :) Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar No te that it equals \(49^{64}=(50-1)^{64}\) you can expand it with binomial theorem but note that \(50^3\equiv 0\pmod{1000}\) so all the first 62 terms cancel. The \(50^2\) term also is 0 mod 1000 then you can just do the last two terms easily :) Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh OK. Why do the first 62 terms cancel...I didnt get it could you explain again @Nathan Ramesh Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar When you expand it with binomial theorem the first 62 terms are all 0 mod 100 because they contain a 50^3 Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Okay, the last two don't and all I have to do is find their sum and get it mod 1000 ...rite? @Nathan Ramesh Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar Yes, correct. It is simple. Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Wow! Thank you. So they're like 64c63 into 50 and 1 into 1...so it becomes 3201 mod 1000 equal to 201. but answer is 801...am i wrong somewer? @Nathan Ramesh Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar Well since it's minus like (50-1) not (50+1) it's \(-50^1*\dbinom{64}{63}*11^{63}+1^{64}=-3199\equiv 801\pmod{1000}\) Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Oh! My bad, :P.....So can binomial theorem used to simplify modular exponentiation...an genral rule? Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar 7 is a good number Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Cyclic its decimal representations are you mean :P ? Krishna Ar · 2 years, 12 months ago

Log in to reply

@Krishna Ar No 49 is one less than 50 Nathan Ramesh · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Oh...Ok,...you meant that :) Krishna Ar · 2 years, 12 months ago

Log in to reply

@Nathan Ramesh Oh! Well :) Krishna Ar · 2 years, 12 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...