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I wrote $13\times 12^{45}$ as $(12+1)\times 12^{45} = 12^{46}+12^{45}$ so I can use Fermat to say that $12^{46}$ equals $1$ in $\Bbb Z_{47} $.

Now I just have to find $12^{45}$ in $\Bbb Z_{47} $. I know that $12^2 = 144$ which equals $3$ in $\Bbb Z_{47} $, and by multiplying that again and again by $12$ and using some properties of congruences I got to $12^{45}$. I managed to see that the total remainder is $22$ (I'm not sure if that's correct).

Anyways, this method took me a lot of time and I'm sure that there is another way, faster way, to solve it. Any hint?

Mutantoe
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puradrogasincortar
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    $47$ is a prime, so $12^{46}\equiv1$. All you need is to find $12^{-1}$, which should not be too difficult. – Jyrki Lahtonen Sep 05 '17 at 11:18
  • Also, this umbrella thread explains many techniques for solving this type of problems. Many more are found by checking out the ones linked to that mother questions. I would close this as a duplicate of that one (or some other), but I have promised not to do that as a first voter. However, I will never upvote answers here, and reserve the right to downvote, if I judge that the answerer did not look for a dupe. – Jyrki Lahtonen Sep 05 '17 at 11:21
  • In a worst case, the exponent will be a prime number. The general method is convert the exponent to binary and apply recursively that $ x^{2n}=({x^2})^n $ and $ x^{n+1}=x*x^n $ doing calculations in the $ Z_{47}$ – pasaba por aqui Sep 05 '17 at 11:22
  • \cdot $\cdot$ and \times $\times$ are better for multiplication than * $*$ – gen-ℤ ready to perish Sep 05 '17 at 11:33

3 Answers3

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Since $12^{46} \equiv 1$, we have $12^{45} \equiv 12^{-1}$, which is most easily found using the extended Euclidean algorithm: You want an integer $x$ such that $12x\equiv 1$, which is the same as saying you want an $x$ such that there is an integer $n$ so that $$ 12x+47n = 1 $$ and the extended Euclidean algorithm is exactly what provides such an $x$ (and an $n$ as well, but you don't care about that).

Arthur
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Since $48\equiv1$ mod $47$, we have

$$13\cdot12^{45}\equiv48\cdot13\cdot12^{45}=52\cdot12^{46}\equiv5\cdot1=5\mod 47$$

Barry Cipra
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Note that since $47$ is prime, we have $12^{46} \equiv 1 \pmod{47}$ by Fermat's Little Theorem.

Thus, we can reduce the expression $13 \times 12^{45} = 12^{46} + 12^{45} \equiv 1 + 12^{-1} \pmod{47}$

To evaluate this, we note the fact that $12 \times 4 = 48 \equiv 1 \pmod{47}$. Thus, $4 \equiv 12^{-1} \pmod{47}$.

Substituting this in gives us $13 \times 12^{45} \equiv 1+4 = 5 \pmod{47}$.

Sharky Kesa
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