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Let n be a positive integer. If \(a\equiv (3^{2n}+4)^{-1}\pmod{9}\), what is the remainder when a is divided by 9?

 Jul 13, 2020
 #1
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I looked at some cases, and it always works out to 2.

 Jul 13, 2020
 #2
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No matter what n is, a = 7

7 mod 9 = 7

 Jul 13, 2020
 #3
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Let \(n\) be a positive integer. If \(a\equiv (3^{2n}+4)^{-1}\pmod{9}\),
what is the remainder when \(a\) is divided by \(9\)?

 

\(\begin{array}{|rcll|} \hline \mathbf{a} &\equiv& \mathbf{ (3^{2n}+4)^{-1}\pmod{9} } \\ && \boxed{ 3^{2n}+4 = \left(3^2\right)^n+4 = 9^n+4 } \qquad \gcd(9^n+4,9) = \gcd(4,9)=1!\\ a &\equiv& (9^n+4)^{-1}\pmod{9} \\ && \boxed{(9^n+4)^{-1}\pmod{9} = (9^n+4)^{\phi(9)-1} \quad | \quad \phi(9)=9*\left( 1-\dfrac{1}{3}\right)=6 \\ =(9^n+4)^{5} \\ \mathbf{(9^n+4)^{-1}\pmod{9} =(9^n+4)^{5}}\pmod{9} } \\ a &\equiv& (9^n+4)^{5} \pmod{9} \\\\ a &\equiv& \underbrace{\binom50 9^{5n}+\binom51 9^{4n}*4+\binom52 9^{3n}*4^2+\binom53 9^{2n}*4^3+\binom54 9^{n}*4^4}_{=0\pmod{9} }+ \binom55 4^5 \pmod{9} \\\\ a &\equiv& \binom55 4^5 \pmod{9} \\\\ a &\equiv& 4^5 \pmod{9} \\\\ a &\equiv& 1024 \pmod{9} \\\\ a &\equiv& 1024-113*9 \pmod{9} \\\\ \mathbf{a} &\equiv& \mathbf{7 \pmod{9}} \\ \hline \end{array}\)

 

laugh

 Jul 14, 2020

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