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Algebra Difficulty 5.0 AIME, harder Find the answer

The fraction 12015\frac{1}{2015} has a unique "(restricted) partial fraction decomposition" of the form 12015=a5+b13+c31\frac{1}{2015}=\frac{a}{5}+\frac{b}{13}+\frac{c}{31} where a,b,ca, b, c are integers with 0≤a<50 \leq a<5 and 0≤b<130 \leq b<13. Find a+ba+b.

A number or a short expression. Spacing and $ signs are ignored.

Solution

This is equivalent to 1=13⋅31a+5⋅31b+5⋅13c1=13 \cdot 31 a+5 \cdot 31 b+5 \cdot 13 c. Taking modulo 5 gives 1≡3⋅1a( mod 5)1 \equiv 3 \cdot 1 a (\bmod 5), so a≡2( mod 5)a \equiv 2(\bmod 5). Taking modulo 13 gives 1≡5⋅5b=25b≡−b( mod 13)1 \equiv 5 \cdot 5 b=25 b \equiv-b(\bmod 13), so b≡12( mod 13)b \equiv 12 (\bmod 13). The size constraints on a,ba, b give a=2,b=12a=2, b=12, so a+b=14a+b=14.

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