HMMT February

Algebra Difficulty 5.0 AIME, harder Find the answer

The fraction $\frac{1}{2015}$ has a unique "(restricted) partial fraction decomposition" of the form $\frac{1}{2015}=\frac{a}{5}+\frac{b}{13}+\frac{c}{31}$ where $a, b, c$ are integers with $0 \leq a<5$ and $0 \leq b<13$ . Find $a+b$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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

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