1

So the question is: $$547^{384^{273}} \equiv x\pmod {11}$$The teacher sent us the solution and started with $384^{273}=4^{273}=a\pmod{10}.$ And I couldn't understand why he wrote $\pmod{10}$ there. Can someone explain it to me please? I appreciate all the help.-

Jyrki Lahtonen
  • 133,153
  • 3
    You use Euler's theorem (https://en.wikipedia.org/wiki/Euler%27s_theorem) and observe that $\varphi(11)=10$. – Andrea Mori Apr 29 '21 at 07:37
  • thank you i got it :) –  Apr 29 '21 at 07:42
  • 3
    See this thread for examples of several techniques that help you here. I would consider this as a duplicate of that one, but won't vote to close as such without hearing other opinions. For example, there are only a few answers explaining what to do with iterated powers there. – Jyrki Lahtonen Apr 29 '21 at 08:27

1 Answers1

2

Euler's theorem says that "for given positive integer $n,m$, if $\gcd(n,m)=1$, then $m^{\phi(n)}=1(\mod n)$ where $\phi(n)$ is the number of positive integers smaller than $n$ and gcd is 1". Since 11 is prime number, you can check the fact that gcd(547,11)=1 and $547^{10}=1(\mod 11)$. Let $384^{273}=10k+r$ where k,r are nonnegative integers and r is smaller than 10. Then $547^{384^{273}}=547^{10k+r}=\left(547^{10}\right)^{k}×547^{r}=547^{r}(\mod 11)$. Moreover, you can simplify it as $(-3)^{r}$ since 547=550-3=50×11-3.

Now, let's calculate it explicitly. You know that $384^{273}=4^{273}$. So if you want to simplify it, you have to know $\phi(10)$. Since 1,3,7,9 do not have any divisor of 10 and there is no more positive integer $k$ smaller than 10 such that $\gcd\left(k,10\right)=1$, you can conclude that $\phi(10)=4$. Since 273=4×68+1, what you want extremely simplified as $(-3)^{4^{1}}=(-3)^{4}=81(\mod 11)$. Division algorithm 81=7×11+4 gives the answer 4.

Mvaldi
  • 278