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across
 4 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!
across
 4 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!

This Question is Closed

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

muhammadturawa
 4 years ago
Best ResponseYou've already chosen the best response.0quite tricky...:)

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0fermat's little theorem \[a^{p1}\equiv 1( \text{ mod } p)\] euler phi totient function \[a^{\phi(n)}\equiv 1 (\text{ mod } n)\] if \[(a,n)=1\]

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1the last digit is 9, no?

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

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

TuringTest
 4 years ago
Best 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...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.11,3,9,7,1,3,... so 1234567890/4=308641927+2/4 and 3^2=9

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

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

across
 4 years ago
Best ResponseYou've already chosen the best response.7That 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. :)

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

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

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

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

across
 4 years ago
Best ResponseYou've already chosen the best response.7Anyway, :) 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. :)

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0Can 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.

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

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Woot, 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.
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