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Calculate the following equation $$85^{2014}\equiv x \bmod {82}$$

Answer is 73 in Wolphram Alpha, but I always lose myself doing the step by step

Michael Hardy
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Paulo Vandeveld
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  • One does not "calculate" equations. One may solve equations. Likewise one does not solve integrals, sums, or other expressions, although one might calculuate or evaluate them. – Michael Hardy Sep 09 '14 at 17:37
  • This might help: http://math.stackexchange.com/questions/81228/how-do-i-compute-ab-bmod-c-by-hand – Martin Sleziak Sep 09 '14 at 18:27

4 Answers4

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In mod $82$, we have $$85^{2014}\equiv 3^{2014}=3^{4\times 503+2}=(3^4)^{503}\cdot 3^2\equiv (-1)^{503}\cdot 9$$

mathlove
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Key Idea $\ $ If the modulus $\, = a^n \pm 1,\,$ then exponents on $\,a\,$ can be reduced mod $\,n,\,$ e.g. here

$\qquad\ \ {\rm mod}\,\ \color{#c00}{a^4\!+\!1}:\,\ (\color{#c00}{a^4\!+\!1}\!+\!a)^{j+4k} \equiv\, a^{j+4k}\equiv\, a^j (\color{#c00}{a^4})^k \equiv\, a^j (\color{#c00}{-1})^k $

Yours is $\ a = 3,\,\ j\!+\!4k = 2\! +\! 4\cdot 253 = 2014,\ $ so $\ 85^{2013}\equiv 3^2(-1)^{253}\equiv -9\pmod{82}$

Bill Dubuque
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$$85\equiv3\pmod{82}\implies85^{2014}\equiv3^{2014}\pmod{82}$$

$$2014\equiv14\pmod{\phi(82)}\implies3^{2014}\equiv3^{14}\pmod{82}$$

Now $3^4\equiv-1\pmod{82}\implies3^{14}=3^2(3^4)^3\equiv9(-1)^3\pmod{82}\equiv ?$

lab bhattacharjee
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In case $\phi$ is unknown to you ... (and I guess it is if you are asking the question). Try $85^x \pmod{82}$ for $x=0,1,2$ and so on, you should see a pattern to let you avoid calculation beyond some point.

GEdgar
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