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 October 9th, 2018, 06:22 PM #1 Newbie   Joined: Oct 2018 From: canada Posts: 3 Thanks: 0 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) Thanks
 October 9th, 2018, 06:43 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond 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.
 October 9th, 2018, 07:56 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra 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 $f$. Thanks from greg1313 Last edited by v8archie; October 9th, 2018 at 07:59 PM.

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