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Weekly Problem 48 - 2009

Stage: 2 and 3 Short Challenge Level: Challenge Level:1

Suppose Dan gives Ann $x$ £$2$ coins and Ann gives Dan $y$ £$5$ coins. Then Dan has £$(18-2x+5y)$ and Ann has £$(40-5y+2x)$.

We need $18-2x+5y=40-5y+2x\Rightarrow 10y-4x=22\Rightarrow 5y-2x=11$. The solution to this equation which minimises $x+y$ is $x=2$, $y=3$, so the smallest number of coins that must change hands is $5$.

This problem is taken from the UKMT Mathematical Challenges.

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