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7^x mod 23=8

Guest Nov 13, 2014

Best Answer 

 #5
avatar+27035 
+10

Any reason why x should be restricted to integers?

Modulo 23

.

Alan  Nov 14, 2014
 #1
avatar+20024 
+10

7^x mod 23=8

$$\\ 7^{20+0*22} \mod 23 = 8 \\
7^{20+1*22} \mod 23 = 8 \\
7^{20+2*22} \mod 23 = 8 \\
... \\
7^{20+n*22} \mod 23 = 8 \\
\boxed { 7^{20+n*22} \mod 23 = 8 \quad n\ge 0 \quad x = 20+n*\phi(23) \quad \phi{(23)} = 22
}$$

$$\phi{()} = Eulers\ phi-function$$

heureka  Nov 13, 2014
 #2
avatar+93644 
0

there's that phi function again - it is sneaking in every where.

 

Thanks Heureka.  

How did you get that original 7^20 ?

Melody  Nov 14, 2014
 #3
avatar+20024 
+5

Hi Melody,

$$\small{\text{
$
\begin{array}{|l|c|r|}
\hline
7^0...7^{21}&& 1...22 \\
\hline
7^0 & mod\ 23 & 1 \\
\hline
7^1 & mod\ 23 & 7 \\
\hline
7^2 & mod\ 23 & 3 \\
\hline
7^3 & mod\ 23 & 21 \\
\hline
7^4 & mod\ 23 & 9 \\
\hline
7^5 & mod\ 23 & 17 \\
\hline
7^6 & mod\ 23 & 4 \\
\hline
7^7 & mod\ 23 & 5 \\
\hline
7^8& mod\ 23 & 12 \\
\hline
7^9& mod\ 23 & 15 \\
\hline
7^{10}& mod\ 23 & 13 \\
\hline
7^{11}& mod\ 23 & 22 \\
\hline
7^{12}& mod\ 23 & 16 \\
\hline
7^{13}& mod\ 23 & 20 \\
\hline
7^{14}& mod\ 23 & 2 \\
\hline
7^{15}& mod\ 23 & 14 \\
\hline
7^{16}& mod\ 23 & 6 \\
\hline
7^{17}& mod\ 23 & 19 \\
\hline
7^{18}& mod\ 23 & 18 \\
\hline
7^{19}& mod\ 23 & 11 \\
\hline
7^{\textcolor[rgb]{1,0,0}{20}}& mod\ 23 & \textcolor[rgb]{1,0,0}{8} \\
\hline
7^{21}& mod\ 23 & 10 \\
\hline
\end{array}
$
}}$$

heureka  Nov 14, 2014
 #4
avatar+93644 
+5

okay Heureka so you did it the long way.  I thought there might have been a short cut.

Thank you :)

Melody  Nov 14, 2014
 #5
avatar+27035 
+10
Best Answer

Any reason why x should be restricted to integers?

Modulo 23

.

Alan  Nov 14, 2014
 #6
avatar+93644 
+5

None that I can see.  Thanks Alan :))

Melody  Nov 14, 2014

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