Pencils and Sharpeners

Let price of a pencil be $p$, then price of a sharpener is $p + 2$. Let Aron buy $n$ pencils and $s$ sharpeners. Aditya buys $2n$ pencils and $s - 10$ sharpeners. Total spent by Aron = $n p + s (p + 2)$. Total spent by Aditya = $2n p + (s - 10)(p + 2)$. Since total money spent is the same: $n p + s(p + 2) = 2n p + (s - 10)(p + 2)$ $10(p + 2) = n p \implies n p = 10p + 20 \implies n = 10 + \frac{20}{p}$ Since $n$ and $s$ must be positive integers, $p$ must be a factor of 20. Also, Aditya buys $s - 10$ sharpeners, so $s > 10$. Total pencils bought together $= n + 2n = 3n = 3 \left(10 + \frac{20}{p}\right) = 30 + \frac{60}{p}$. To minimize total pencils, we maximize $p$. Maximum possible value of $p$ dividing 20 is $20$. So minimum number of pencils $= 30 + \frac{60}{20} = 30 + 3 = 33$.