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I understand through trial and error that d=91, but trial and error is extremely inefficient and slow. I understand that you can use the extended Euclidean Algorithm to find d faster, but I'm encountering some unexpected problem. For some reason, I'm unable to move past the first step, could someone point out what I'm missing.

19d $\equiv$ 1 (mod 96)

96 = 5*19 + 1

$1 = 96 - 5\times19 = 3\times2^5 - 5\times19$

Rosan
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    $5 \times 19 \equiv (-1) \pmod{96} \implies (-5) \times 19 \equiv 1 \pmod{96}$. Then $(91) \equiv (-5) \pmod{96}.$ – user2661923 Apr 25 '21 at 07:52
  • How did you get -1? – Rosan Apr 25 '21 at 15:24
  • By your own analysis, $(5 \times 19) + 1 = 96 \implies (5 \times 19) = (96 - 1).$ This implies that $(5 \times 19) \equiv (96 - 1) \equiv (-1) \pmod{96}.$ – user2661923 Apr 25 '21 at 20:54
  • Also, $(5 \times 19) = (96 - 1) \implies ([-5] \times 19) = (1 - 96).$ This implies that $([-5] \times 19) \equiv (1 - 96) \equiv (1) \pmod{96}.$ – user2661923 Apr 25 '21 at 20:57

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