Equation: $[3*(k \mod 4)] \mod 4 = 3$

It's relatively easy to check the equation for the possible values of $k \mod 4$. Is there a more elegant way to calculate the solution, for example by using modulo properties?

  • $\begingroup$ Welcome to Mathematics Stack Exchange. If $3k\equiv3 \pmod 4$ then $3\times3k\equiv3\times3\pmod 4$ so $9k\equiv 9\pmod 4$ so $k\equiv 1\pmod 4$ [I chose to multiply by $3$ because $3^{-1}\equiv3\pmod 4$] $\endgroup$ – J. W. Tanner Oct 16 at 13:58

To solve $3k\equiv3\pmod4$, multiply both sides by the inverse of $3\pmod 4$ (which is $3$):

$3k\equiv3\pmod4\implies 3^{-1}3k\equiv3^{-1}3\pmod4\implies k\equiv1\pmod 4$

  • $\begingroup$ And what about a situation in which $[3*(k \mod 4)] \mod 4 = 1$? Or any other number which wouldn't have 3 as a factor? $\endgroup$ – DaddyMike Oct 16 at 18:50
  • $\begingroup$ The solution to $3k\equiv 1\pmod4$ is $k\equiv 3^{-1}1\equiv 3\pmod 4$; $3k\equiv0\pmod4$ is $k\equiv3^{-1}0\equiv0\pmod4$; $3k\equiv2\pmod4$ is $k\equiv3^{-1}2\equiv 3\times2\equiv6\equiv2 \pmod4$ $\endgroup$ – J. W. Tanner Oct 16 at 19:18
  • $\begingroup$ You could also note that $3k\equiv a\pmod 4$ means $-k\equiv a \pmod 4$ so $k\equiv -a \pmod 4$ $\endgroup$ – J. W. Tanner Oct 16 at 19:30

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