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