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 Question from Rick, a parent: Have 48 tickets to spend, have to spend them all in exactly 10 events. Rides cost 6 tickets games cost 4 tickets & treats cost 3 tickets

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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