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In a sport, assume a team can either score 3 points or 7 points. Find the smallest example of a team winning the game by 5 points. Yes, you can cheat and guess that the smallest example is the winning team scoring 12 points off of 4 three-pointers and the losing team scoring 7 points off of 1 seven-pointer, but how can you mathematically get to that answer without any guessing?

user3247160
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  • how is the size of an example determined? – Asinomás Feb 02 '14 at 06:03
  • If they both have a three-pointer, then there is another solution without the three-pointer. The same for a seven-pointer. So one side has only three-pointers, and the other side has only seven-pointers. So there are two options: Start with 5, and add 3s until you reach a multiple of 7; (5+3+3+3=7+7) or start with 5, and add 7s until you reach a multiple of 3.(5+7=3+3+3+3) – Empy2 Feb 02 '14 at 06:04
  • @user4140 by the sum of the scores of both teams – user3247160 Feb 02 '14 at 06:04

4 Answers4

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Theorem: in the smallest combination both players will score in only one way:

Proof:Suppose both teams scored 7 pointers. Subtract 7 from both scores to get a smaller working scoreboard. Do the same with 3.

So now we know one scores only with 7 and the other only with 3. Prove (0,5) is impossible. prove (2,7) is impossible. Show (7,12) is possible.

Asinomás
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Let $x_1,x_2$ be the number of points won by winning team and $y_1,y_2$ be the number of points won by losing team. $$(3x_1+7x_2)-(3y_1+7y_2) = 5$$ $$3(x_1-x_2) + 7(y_1-y_2) = 5$$

Going by last digits, you can arrive at the smallest win by saying that $3(x_1-x_2)$ should end in $2$ and $7(y_1-y_2)$ should end in $7$.This makes four $3$ points and one $7$ point(total $5$).Or $3(x_1-x_2)$ should end in $9$ and $7(y_1-y_2)$ should end in $4$.This makes three $3$ points and two $7$ points(total $5$). Hence : $$(x_1-x_2) = 4$$ $$x_1=4, x_2 = 0$$ $$(y_1-y_2) = 1$$ $$y_1=1,y_2=0$$ Or $$(x_1-x_2) = 3$$ $$x_1=3, x_2 = 0$$ $$(y_1-y_2) = 2$$ $$y_1=2,y_2=0$$

lsp
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Let winning team score $x_1$ 3-pointers and $y_1$ 7-pointers and similarly with the losing team. Then the condition imposed is $(3x_1 + 7y_1) - (3x_2 + 7y_2) = 5$ and we are supposed to minimize $(3x_1 + 7y_1) + (3x_2 + 7y_2)$. This is equivalent to $z_1 - z_2 = 5$ and minimizing $z_1 + z_2$ whose solution is to simply minimize $z_2$. In our case, the second term's minimum is $0$ but then $z_1 = 5$ which cannot be written as a sum of $3's$ and $7's$. Next is $3$ which fails and finally we reach $z_2 = 7$ for which $z_1 = 12$ can be written as $4*3$.

Sandeep Thilakan
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(This is an alternative way of describing Sandeep's answer.) You want to find the smallest number $p$ so that both $p$ and $p+5$ can be expressed as a sum of $3s$ and $7s$. This would mean you want to write $5$ as some $3s$ minus some $7s$ or vice versa with the smallest possible value subtracted, because that number is the smallest possible $p$. A graph of $a3+b7=5$ shows this to be $5=3+3+3+3-7$, so the loser's score will be $7$ and the winner's $12$.

Steve Kass
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