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

#NumberTheory #ModularArithmetic #HelpMe! #brillianthelp #Pleasehelp

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 7 years ago

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 - 7 years ago

Akshaj is the bomb diggity. :D

Finn Hulse - 7 years ago

I somehow don't really understand that comment.

Yuxuan Seah - 7 years ago

@Yuxuan Seah – Oh I'm just saying he's awesome. :D

Finn Hulse - 7 years ago

@Finn Hulse – Oh haha :D

Yuxuan Seah - 7 years ago

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 - 7 years ago

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 - 7 years ago

@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 - 7 years ago

@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 - 7 years ago

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 - 7 years ago

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 - 7 years ago

@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 - 7 years ago

@Nathan Ramesh – Oh! Well :)

Krishna Ar - 7 years ago

@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 - 7 years ago

@Krishna Ar – What I did is essentially the same thing.

Nathan Ramesh - 7 years ago

@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 - 7 years ago

@Krishna Ar – No, give me another example and I will show you

Nathan Ramesh - 7 years ago

@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 - 7 years ago

@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 - 7 years ago

@Nathan Ramesh – OK. Why do the first 62 terms cancel...I didnt get it could you explain again @Nathan Ramesh

Krishna Ar - 7 years ago

@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 - 7 years ago

@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 - 7 years ago

@Krishna Ar – Yes, correct. It is simple.

Nathan Ramesh - 7 years ago

@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 - 7 years ago

@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 - 7 years ago

@Nathan Ramesh – Oh! My bad, :P.....So can binomial theorem used to simplify modular exponentiation...an genral rule?

Krishna Ar - 7 years ago

@Krishna Ar – 7 is a good number

Nathan Ramesh - 7 years ago

@Nathan Ramesh – Cyclic its decimal representations are you mean :P ?

Krishna Ar - 7 years ago

@Krishna Ar – No 49 is one less than 50

Nathan Ramesh - 7 years ago

@Nathan Ramesh – Oh...Ok,...you meant that :)

Krishna Ar - 7 years ago

@Nathan Ramesh – @Nathan Ramesh @Finn Hulse pL answer

Krishna Ar - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$