HMMT February, 2015

Solution:

Answer: 14 This is equivalent to $1=13 \cdot 31 a+5 \cdot 31 b+5 \cdot 13 c$. Taking modulo $5$ gives $1 \equiv 13 \cdot 31 a \pmod{5}$, and since $13 \cdot 31 \equiv 3 \pmod{5}$, we have $1 \equiv 3a \pmod{5}$, so $a \equiv 2 \pmod{5}$. Taking modulo $13$ gives $1 \equiv 5 \cdot 31 b \pmod{13}$, and $5 \cdot 31 = 155 \equiv -1 \pmod{13}$, so $1 \equiv -b \pmod{13}$, thus $b \equiv 12 \pmod{13}$. The size constraints on $a, b$ give $a=2, b=12$, so $a+b=14$.