Solving linear congruences

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I am trying to solve $25x\equiv15\pmod{29}$

I multiply both sides by $7$ which makes the L.h.S congruent to $1x \pmod{29}$

From this I have that $7\times25x\equiv7\times15\pmod{29}$

I am really confused with where to go from here. Some help would be much appreciated :)

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As $(25,15)=5$ and $(5,29)=1,$ we can immediately divide either sides by $5$ to get $\displaystyle 5x\equiv3\pmod{29}$

Multiply either sides by $6$ as $\displaystyle 5\cdot6\equiv1\pmod{29}$