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across
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!
\[\text{Hint: }a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.\]
quite tricky...:)
Is that fermat's or euler's thingy...I can't remember
You are right. :) This is Euler's theorem, and ϕ is the totient function. Naturally, it is an augmentation of Fermat's little theorem.
fermat's little theorem \[a^{p-1}\equiv 1( \text{ mod } p)\] euler phi totient function \[a^{\phi(n)}\equiv 1 (\text{ mod } n)\] if \[(a,n)=1\]
I think @jamesj and @zarkon would love this question (maybe-lol) I will keep thinking on it though Great question @across
the last digit is 9, no?
Yes it is. :) Did you use Euler's theorem?
no I used a little logic no idea what that theorem is
\[3^0=1\]\[3^1=3\]\[3^2=9\]last digit is nine is the important point here\[3^3=27\]\[3^4=81\]and so the last digit repeats every four powers...
1,3,9,7,1,3,... so 1234567890/4=308641927+2/4 and 3^2=9
I'm sure that's got a professional way of writing it with mod this and that, but that's beyond me lol
how you got the other digit is what I want to know
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. :)
I am pretty sure you can figure out what those two digits are by having told you that up there. ;)
@Zarkon, that is not it!
thank you for the compliment across, it means so much knowing who it's coming from :D
^.^ And @Zarkon, you are right. Let us in in thy arcane ways!
Anyway, :) to re-state Euler's theorem:\[a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.\]And to define Euler's phi-function:\[\phi(n)=\text{number of positive integers relatively prime to }n.\]By the way, I will now say that what I am about to explain is a huge over-simplification 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. :)
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.
It's good to mention Carmichael theorem in this context. http://en.wikipedia.org/wiki/Carmichael_function#Carmichael.27s_theorem
And if you are in a hurry, just a one liner in python: http://ideone.com/47UX7
Woot, solution number two!\[3^n \equiv 3^{n-4} \times 3^4 \equiv 3^{n -4}\pmod{10} \]Suffice to say that \(3^{n} \equiv 3^{n - 4k} \pmod {10}\) for all integer \(k\). Well, that's just the solution @TuringTest posted.