



 
Hi Rick, Suppose you go on $R$ rides, play $G$ games and buy $T$ treats then \[R + G + T = 10.\] You have 48 tickets to spend so \[6R + 4G + 3T = 48.\] Solving the first equation for $R$ and substituting into the second equation gives \[6(10  G  T) + 4G + 3T = 48\] which simplifies to \[2G + 3T = 12.\] Thus you could have $G = 0$ and solve for $T$ and $R,$ or $T=0$ and solve for $G$ and $R.$ Also if you look at \[2G + 3T = 12\] you see that 2 divides the right side so 2 must also divide the left side. Since 2 divides $2G$ it must also divide $3T,$ and thus 2 divides $T$ and hence $T = 0, 2 \mbox{ or }4.$ The right side of \[2G + 3T = 12\] is also divisible by 3. Penny  


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