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  • 1 Post By v8archie
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October 9th, 2018, 06:22 PM   #1
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Question Golf Balls!

A worker is packaging 177 golf balls into sleeves. Each sleeve can hold 3 golf balls. There are two options for how many sleeves can be packaged into a box:

The first type of box can hold 4 sleeves
The second type of box can hold 5 sleeves

How many boxes and of which type are needed to package all of the golf balls?

(This problem is very simple to do using trial/error, but I'm looking for an algebraic solution)


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October 9th, 2018, 06:43 PM   #2
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Let f be the number of the first type of box and let s be the number of the second type of box.

12f + 15s = 177. By inspection, f = 1 and s = 11. I don't think there is enough information given to form a second equation relating boxes, balls and sleeves that isn't just a scalar multiple of the above equation.
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October 9th, 2018, 07:56 PM   #3
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Written as \begin{align}12(f+s)+3s &= 177 \\ 4(f+s) + s &= 59\end{align}
We see that we require $f+s < 15$ and that we pick $s$ to complete the sum for an appropriate value of $(f+s)$. This gives rise to solutions $(f,s) \in \{ (11,3), (6,7), (1,11) \}$, being the solutions for $(f+s)$ equal to 14, 13 and 12 respectively.

Lower values of $(f+s)$ don't have solutions as they require $s > f+s$ which implies negative values of $f$.
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Last edited by v8archie; October 9th, 2018 at 07:59 PM.
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