# Coin Piles: Sufficient Conditions Let \( a \) and \( b \) be the number of coins in the two piles where \( a \le b \), \( 3 \mid (a + b) \), and \( a \ge b/2 \). We will now show that there exist nonnegative integers \( x \) and \( y \) such that both coin piles become empty after performing: * \( x \) moves where we remove \( 1 \) coin from the first pile and \( 2 \) coins from the second pile. * \( y \) moves where we remove \( 2 \) coins from the first pile and \( 1 \) coin from the second pile. That is, we will show that there exist nonnegative integers \( x \) and \( y \) such that they satisfy the equations: \begin{align*} a &= x + 2y, \\ b &= 2x + y. \end{align*} We will show this by constructing \( x \) and \( y \) as follows: \[ x = \frac{2b - a}{3}, \qquad y = \frac{2a - b}{3}. \] ## Nonnegativity of \( x \) and \( y \) Since \( a \le b \), we get \( 2b \ge a \). Thus \( x \ge 0 \). Since \( a \ge b/2 \), we get \( 2a \ge b \). Thus \( y \ge 0 \). ## Integral Values of \( x \) and \( y \) \begin{align*} a + b &\equiv 0 \pmod{3} \implies -a \equiv b \pmod{3} \implies 2b - a \equiv 3b \equiv 0 \pmod{3}. \\ a + b &\equiv 0 \pmod{3} \implies -b \equiv a \pmod{3} \implies 2a - b \equiv 3a \equiv 0 \pmod{3}. \end{align*} ## Satisfying The Equations \begin{align*} x + 2y &= \frac{2b - a}{3} + \frac{4a - 2b}{3} = a, \\ 2x + y &= \frac{4b - 2a}{3} + \frac{2a - b}{3} = b. \end{align*}