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Number TheoryHardFactors & Multiples

Factors & Multiples · Hard problem

Since gcd⁡(19,27)=1\gcd(19, 27) = 1, there are integers x,yx, y with 19x+27y=119x + 27y = 1. Find the solution with the smallest ∣x∣|x| (and, if two tie, the negative one), giving xx and yy.