0

Find the largest positive integer $n$ such that $3^{1024} - 1$ is divisible by $2^n$.

I am trying all powers of $2$, beginning from $2^0$ onwards; I believe the answer will be $n=\infty$, since $3^{1024} - 1$ is an even number and it can be divided by $2$, so it is divisible by all powers of $2$?

Peter
  • 86,576
Math Tise
  • 717
  • 1
  • 5
  • 11
  • 1
    The exponent surely is finite because the number is finite. – Peter May 23 '18 at 09:29
  • The answer is $12$, but I only found that out with PARI/GP, not by hand. – Peter May 23 '18 at 09:32
  • I tried using logarithm, but I am still getting only one equation with two unknowns, which makes it difficult to solve. – Math Tise May 23 '18 at 09:33
  • @MathTise This does not work, you need modular arithmetic. Euler's theorem should be helpful. Another approach could be induction to show that $3^{(2^n)}-1$ is divisible by $2^{n+2}$ , but not by $2^{n+3}$ – Peter May 23 '18 at 09:34
  • I also got the answer n=12, but I got by software and not by hand. – Math Tise May 23 '18 at 09:35
  • As a suggestion, start with a smaller exponent for practice. $1024=2^{10}$ so, let's stick to powers of $2$. Work the problem for $3^2-1$ and $3^4-1$ just to get a sense of what is being asked. – lulu May 23 '18 at 09:36
  • Thank You lulu, that did help to some extent, to understand and work out the solution; however, unfortunately excel doesn't display numbers after 2^9, and neither does my calculator! :-( – Math Tise May 23 '18 at 09:41
  • Well, aside from the answers in the duplicate questions, you might try induction. Note that $3^{1024}-1=(3^{512}-1)\times (3^{512}+1)$ and so on. – lulu May 23 '18 at 10:11

0 Answers0