-1

Using the Euler's theorem: $x^{\varphi\left(m\right)}\equiv1 \pmod{m}$, we have: If $x\ =\ 12,\ m\ =\ 17,$ then ${12}^{\varphi\left(17\right)}\equiv1 \pmod{17}$.

Now I have to use that fact to determine the residue of ${12}^{482}$ (mod 17).

I'm thinking of this approach:

$482=256+128+64+32+2=2^8+2^7+2^6+2^5+2^1$

Thus ${12}^{482} = 12^{256} \cdot 12^{128} \cdot 12^{64} \cdot 12^{32} \cdot 12^{2} $

So the residue of ${12}^{482}$ (mod 17) can be obtained by multiplying together the residues of $12^{256}$, ... above and then taking the residue of that product.

But is this the simplest way? Have I made the best use of the Euler's theorem result above? Thank you so much!

L. F.
  • 1,940
hanamontana
  • 333
  • This'll give you the right answer, because except for the last factor, all the exponents are zero mod $16$. –  May 12 '21 at 11:05
  • Remember, since $\varphi(17)=16$, Euler's theorem tells us we can reduce the exponent $\pmod{16}$. –  May 12 '21 at 11:12
  • In other words $12^{482} \equiv 12^{480} \times 12^2 \equiv 12^2 \pmod{17},~$ because $12^{480} = 12^{16 \times 30} = \left[12^{16}\right]^{(30)} \equiv 1^{30} \equiv 1 \pmod{17}.$ – user2661923 May 12 '21 at 11:22
  • By the Theorem in the linked dupe $,12^{16}\equiv 1\Rightarrow 12^n \equiv 12^{n\bmod 16}.\ $ See the posts linked there for many worked examples. – Bill Dubuque May 12 '21 at 17:18

1 Answers1

0

Hi I think I got the right steps and want to share if they are exactly correct:

We have $\varphi\left(17\right)\ =\ 16$.

By Euler’s theorem, we have ${12}^{\varphi\left(17\right)}={12}^{16}\equiv1\left(mod\ 17\right).$

Hence, ${12}^{482}={({12}^{16})}^{30} \dot {12}^2\equiv{1^{30} \dot 12}^2 = 144 ≡ 8$ (mod 17)

Note: Not sure why my dot is up above my number?

hanamontana
  • 333