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I need help calculating the following: $$146^{11} \pmod{12}$$ I think the answer is $8$, but I'm not sure. Can anyone help?

Thank you very much.

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  • $\begingroup$ Have you tried? If you are having trouble, show us your process. We cannot just answer it for you. $\endgroup$ – Julian Rachman Jan 20 '15 at 16:57
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    $\begingroup$ $146\equiv 2\pmod{12}$. Look at the powers of 2 mod 12 and see if you can spot a pattern. $\endgroup$ – Rick Decker Jan 20 '15 at 17:01
  • $\begingroup$ Hint: Try using binomial theorem. $\endgroup$ – AvZ Jan 20 '15 at 17:25
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146 $\equiv2$ mod12, then 146^{11}$\equiv$ $2^{11}$ mod 12 and $2^{11}$$\equiv$-4 mod 12 which is same as 8

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$$\begin{align}146^{11} & \pmod{12}\\ 146& = 12 \times 12 + 2\\ 146^1& = 2 \pmod{12} \\ 146^2& = 2^2 \pmod{12} \\ 146^4& = 2^4 \pmod{12}\\ 2^4& = 4 \pmod{12}\\ 2^5& = 8 \pmod{12}\\ 2^6& = 4 \pmod{12}\\ 2^7& = 8 \pmod{12}\\ \end{align}$$

From here I can conclude that 4 and 8 are repeating and 4 is when the number is on even and 8 is when number is odd, from here I get $146^{11} = 8 \pmod{12}$

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  • $\begingroup$ I updated your answer into MathJax formulas. I can follow your reasoning and you seem to have the main ideas correctly, but only because I know the answer anyway. You could probably benefit from adding a little more explanation into future answers, both to clarify the process and as a self-check that you really understand how you are reaching your answer. $\endgroup$ – Joffan Jan 20 '15 at 18:13
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We can write this as
$146^{11}=(2+144)^{11}=(2+12^{2})^{11}={11\choose0}2^{11}+{11\choose1}2^{10}12^{2}+{11\choose2}2^{9}12^{4}+\ldots+{11\choose11}2^{0}12^{22}$
Furthermore, we can write this as
$2^{11}+12k$ where $k$ is an positive integer.
Since we know that $2^{11}=2048$, we can now simply divide this by $12$ and get the remainder $8$ without using any modular arithmetic.

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$$146\equiv2\pmod{12}\implies146^{11}\equiv2^{11}$$

Now as $(2^{11},12)=4=2^2,$ let us find $2^{11-2}\pmod{\dfrac{12}4}$ i.e., $2^9\pmod3$

Now, $2\equiv-1\pmod3\implies2^9\equiv(-1)^9\equiv-1$

As $a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod{m\cdot c},$

$2^2\cdot2^9\equiv2^2(-1)\pmod{3\cdot2^2}$

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