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# 7^x mod 23=8

0
448
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7^x mod 23=8

Guest Nov 13, 2014

#5
+26637
+10

Any reason why x should be restricted to integers?

.

Alan  Nov 14, 2014
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#1
+19207
+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
+92217
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
+19207
+5

Hi Melody,

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

heureka  Nov 14, 2014
#4
+92217
+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
+26637
+10

Any reason why x should be restricted to integers?

.

Alan  Nov 14, 2014
#6
+92217
+5

None that I can see.  Thanks Alan :))

Melody  Nov 14, 2014

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