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This question is just for fun: Tell me what the last digit of \(3^{1234567890}\) is. :)
Then tell me what the last two digits are.
And if you feel up to the challenge, then tell me what the last three digits are!
 2 years ago
 2 years ago
This question is just for fun: Tell me what the last digit of \(3^{1234567890}\) is. :) Then tell me what the last two digits are. And if you feel up to the challenge, then tell me what the last three digits are!
 2 years ago
 2 years ago

This Question is Closed

acrossBest ResponseYou've already chosen the best response.7
\[\text{Hint: }a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.\]
 2 years ago

muhammadturawaBest ResponseYou've already chosen the best response.0
quite tricky...:)
 2 years ago

myininayaBest ResponseYou've already chosen the best response.0
Is that fermat's or euler's thingy...I can't remember
 2 years ago

acrossBest ResponseYou've already chosen the best response.7
You are right. :) This is Euler's theorem, and ϕ is the totient function. Naturally, it is an augmentation of Fermat's little theorem.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
fermat's little theorem \[a^{p1}\equiv 1( \text{ mod } p)\] euler phi totient function \[a^{\phi(n)}\equiv 1 (\text{ mod } n)\] if \[(a,n)=1\]
 2 years ago

myininayaBest ResponseYou've already chosen the best response.0
I think @jamesj and @zarkon would love this question (maybelol) I will keep thinking on it though Great question @across
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
the last digit is 9, no?
 2 years ago

acrossBest ResponseYou've already chosen the best response.7
Yes it is. :) Did you use Euler's theorem?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
no I used a little logic no idea what that theorem is
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
\[3^0=1\]\[3^1=3\]\[3^2=9\]last digit is nine is the important point here\[3^3=27\]\[3^4=81\]and so the last digit repeats every four powers...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
1,3,9,7,1,3,... so 1234567890/4=308641927+2/4 and 3^2=9
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
I'm sure that's got a professional way of writing it with mod this and that, but that's beyond me lol
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
how you got the other digit is what I want to know
 2 years ago

acrossBest ResponseYou've already chosen the best response.7
That is genius. :) Do you know that you are intuitively deriving the above theorem? From it, it follows that to find out what the last two digits of that number are, you first need to observe that the sequence of two digits repeats every 40 powers. I will elaborate more on it a bit later. :)
 2 years ago

acrossBest ResponseYou've already chosen the best response.7
I am pretty sure you can figure out what those two digits are by having told you that up there. ;)
 2 years ago

acrossBest ResponseYou've already chosen the best response.7
@Zarkon, that is not it!
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
thank you for the compliment across, it means so much knowing who it's coming from :D
 2 years ago

acrossBest ResponseYou've already chosen the best response.7
^.^ And @Zarkon, you are right. Let us in in thy arcane ways!
 2 years ago

acrossBest ResponseYou've already chosen the best response.7
Anyway, :) to restate Euler's theorem:\[a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.\]And to define Euler's phifunction:\[\phi(n)=\text{number of positive integers relatively prime to }n.\]By the way, I will now say that what I am about to explain is a huge oversimplification of the number theory behind it all, and that I have boiled the entire process down to a series of mechanical steps stemming from said theory. Since we are trying to find out what the last few digits of \(a=3^{1234567890}\) are, all that we have to do is compute \(a\text{ }(\text{mod }10)\) for the last digit, \(a\text{ }(\text{mod }100)\) for the last two digits, and so on. Notice that \(\gcd(3,10)=\gcd(3,100)=\cdots=1\), so we can indeed make use of the above theorem. Because it really does not pertain to the problem at hand, I will just say that \(\phi(10)=4\), \(\phi(100)=40\), and so on. Therefore, we have that \(3^{\phi(10)}\equiv3^4\equiv1\text{ }(\text{mod }10)\) (check this to convince yourself that it is true), and it follows that, for the last digit, \[3^{1234567890}\equiv3^2\cdot(3^4)^{308641972}\equiv3^2\cdot(1)^{308641972}\equiv9\text{ }(\text{mod }10).\]Which is indeed our last digit. You can do the same thing for two digits:\[3^{1234567890}\equiv3^{10}\cdot(3^{40})^{30864197}\equiv3^{10}\cdot(1)^{30864197}\equiv49\text{ }(\text{mod }100).\]This can go on and on. For three digits, you will have to compute \(3^{290}\), which is doable by Windows' calculator. But there are better methods for bigger and bigger numbers. :)
 2 years ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Can I do the last 3 digits? Using Euler's theorem, we know that \(\phi(1000)=400\) so \(3^{400} \equiv 1 \mod 1000\). Now we calculate \(1234567890 \mod 400\). As it turns out, this is 290. Thus, we only have to calculate \(3^{290} \mod 1000\). Using successive squaring/fast powering, this is easy, and we get that the last three digits are 449.
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
It's good to mention Carmichael theorem in this context. http://en.wikipedia.org/wiki/Carmichael_function#Carmichael.27s_theorem
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
And if you are in a hurry, just a one liner in python: http://ideone.com/47UX7
 2 years ago

ParthKohliBest ResponseYou've already chosen the best response.0
Woot, solution number two!\[3^n \equiv 3^{n4} \times 3^4 \equiv 3^{n 4}\pmod{10} \]Suffice to say that \(3^{n} \equiv 3^{n  4k} \pmod {10}\) for all integer \(k\). Well, that's just the solution @TuringTest posted.
 one year ago
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