How do I find a solution to the congruence $103a\equiv 1\mod 101$?
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3$2a\equiv 1 \pmod{101}$ and $\gcd(2,101)=1$ so $a \equiv \frac{1}{2}\equiv \frac{102}{2} \equiv 51 \pmod{101}$ – Evariste Jun 12 '21 at 22:53
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See the linked dupes for many ways to solve linear congruences, e.g. the above comment uses the fifth method in my answer in the second dupe, that division by 2 is trivial for an odd modulus, since we can force the dividend to be even by adding (or subtracting) the modulus (and that's a special case of the next method, inverse reciprocity, which in turn is a special case of the extended Euclidean algorithm - the master method) – Bill Dubuque Jun 12 '21 at 23:08