Find a $x$ such that $2^{2015}xequiv 1 pmod{13}$ [on hold]
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Since 13 is prime number using little Fermat's theorem $2^{12}equiv 1 pmod {13}$ then $2^{2015}equiv 2^{12cdot167+11}equiv 2^{11} pmod{13}$ then $2^{2015} x equiv 2^{11} x equiv 1 pmod{13}$ so then $xequiv 2$ , is this ok?
discrete-mathematics modular-arithmetic divisibility
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Bill Dubuque
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Marko Škorić
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