I've found this problem in my book and left clueless on how to make the right interpretation to come up with a solution. The situation involves a hockey match. Typically when you have a goals for and goals against table you already know the results of the match. But what is the recommended procedure when you don't?.

The problem is as follows:

A tv reporter covering the results of a hockey match group stages for the Summer Universiade in Taipei forgot to write down the results of each match. There were three teams involved, Netherlands, France and Norway. Before airing her segment on the news, she remembered to have a handwritten notes of the goals for and goals against between the teams mentioned. This table is shown in the diagram from below. The reporter tells the viewers that the three teams have played only once against each other and each team has played only two matches. If the only available footage shown on tv tells the match between France and Norway ended in a draw. What was the result of the match between Netherlands and Norway?.

Diagram of the problem

The alternatives given were as follows:

$\begin{array}{ll} 1.&3-1\\ 2.&2-1\\ 3.&3-2\\ 4.&1-3\\ 5.&4-3\\ \end{array}$

In this particular situation I don't know where to begin. As I mentioned in the opening statement of this question. My doubt arises from the fact that the results from each match is not known and only the one which is known is the one between France and Norway. Does it exist a way to uncover or unhide this from what is known?. I tried to make an equation, but then I figured out. What can I label $x$ ?.

Upon examination what I could come up with was to know that the only existing combinations between these three are:

Netherland - France

Netherland - Norway

France - Norway

From the goals for it is only known that each team did scored that amount of goals across the entire phase (I suppose that). So at this moment it is not known what exactly the match ended for each team.

I assumed:

For the goals scored in the match:

$\textrm{Netherlands - France = x}$

$\textrm{Netherlands - Norway = y}$

$x+y = 5$

For the goals received in the match:

$\textrm{Netherlands - France = a}$

$\textrm{Netherlands - Norway = b}$

$a+b = 5$

But at this point I got confused on how to use previous information. I suspect that the goals one received accounts for what other has made. But I don't know how to put this piece together with the fact of the draw between the teams as given in the clue from the problem.

Can somebody help me in the right direction for this problem?. I think an answer which would help me the most would be one which can be very detailed in step by step to put the pieces together. Not just solving the problem but more like teaching what should be found first and build that from there. Overall can somebody help me to find an answer for this problem the most easiest way possible?.


This is a very nice question.

Let's start with the known fact that France vs Norway ended in a draw. And let's say that draw is of a score of $a : a$.

Then we know the Netherlands vs France game's result must be: $6-a : 7-a$ since the total goal for France is 7 and goal against is 6.

Similarly, we get that Netherlands vs Norway ended in $5-a:4-a$.

Lastly, we know the goals for Netherlands is $5 = 5 - a + 6 - a$. Therefore, $a=3$ and you can solve for the score of Netherlands vs Norway $2:1$.

  • $\begingroup$ It's a delight the simplicity of your method. It appears to be that the key was to begin from the clue given at the beginning. Just as I mentioned I didn't know how to use it. I also believe that to use Netherlands as a pivot to compare the results helps a lot. This approach works well with three teams. How about four, can this method be extended? or will that require additional variables?. $\endgroup$ – Chris Steinbeck Bell Feb 15 '19 at 19:18
  • $\begingroup$ I would be interested in trying to solve a similar problem with more than 3 teams. Depending on the given information, this approach might or might not work. $\endgroup$ – Dubs Feb 15 '19 at 19:52

Here's a somewhat different approach, without algebraic equations.

We see that France and Norway together scored $7+411$ goals. All these goals were either in the France-Norway game, or they were scored against the Netherlands. Since we know $5$ were scored against the Netherlands, we see $6$ were scored in the France-Norway game, which must have ended in a $3-3$ tie.

France scored $7$ goals, and $3$ of them were scored against Norway; the other $4$ were scored against the Netherlands. Similarly, $6$ goals were scored against France, $3$ by Norway, so the Netherlands scored the other $3$. In the same way, we can $subtract$ $3$ from each of Norway's for and against totals to figure out the score in the Norway-Netherlands game.

France beat the Netherlands $4-3,$ and Norway lost to the Netherlands $1-2.$

The crux of this problem is that each team played only two games, so once you know the score in one of its games, you can figure out the score in the other.

  • $\begingroup$ Your approach is faster than the other methods, but I'm still struggling on how you get there. In other words, in a problem like this, what I want to do is find the number of goals between the teams lets say second and third so I could compare it with the first. In this case the remaining goals which don't account for Netherlands must involve the other two. $\endgroup$ – Chris Steinbeck Bell Feb 15 '19 at 20:08
  • $\begingroup$ Since the problem mentions that it was a tie, so it should be understood as half of that value hence $3$. Since France has $GF=7$ this meant for the game involving Netherlands (the only left choice) $x+3=7$ therefore $x=4$ and $GA$ is $y+3=6$ so $y=3$. $\endgroup$ – Chris Steinbeck Bell Feb 15 '19 at 20:08
  • $\begingroup$ Conversely for Norway in the match against Netherlands, from the result in $GF$ is $x+3=4$ so $x=1$ and $GA$ is $y+3=5$ so $y=2$ Finally $1$ accounts for the goal made in that match and the other $2$ the received from Netherlands, so the result was $2-1$. I've just hope to be understanding it right. Am I?. I had to use the $x$ and $y$ variables as an aid to avoid confusion. $\endgroup$ – Chris Steinbeck Bell Feb 15 '19 at 20:09
  • $\begingroup$ Yes, it seems to me that you are right. Since you had a little trouble following my answer, I'll add a bit more explanation. $\endgroup$ – saulspatz Feb 15 '19 at 20:36

Let $Ab$ represent the number of goals made by $A$ against team $b$ and use the abbreviations $L,F,W$ to refer to Netherlands, France, and Norway respectively. So $Lf$ is the number of goals scored by Netherlands against France for example.

We are told from the table the following information:

$$\begin{cases} Lf+Lw = 5\\Fl+Wl = 5\\ Fl+Fw = 7\\Lf+Wf = 6\\Wf+Wl = 4\\Fw + Lw = 5\end{cases}$$

We are also told one additional piece of information. The match between France and Norway ended in a draw, giving us the equation:

$$Fw -Wf= 0$$

Including this last equation, this is a linear system of seven equations with six unknowns. We may then apply Gaussian Elimination to find the solution(s) if any exist.

Letting the columns be in the order $Lf, Lw, Fl, Fw, Wf, Wl$ we have the augmented matrix:


After row reducing this leaves us with the augmented matrix:


From this, we can read out the values for each of $Lf,Lw,\cdots$ and we learn that the result of the match between Netherlands and Norway was $2$-$1$

  • $\begingroup$ Aren't there 7 equations with the drawn game included? Do you need that one? In this particular case or in general for problem like this? $\endgroup$ – Ned Feb 15 '19 at 19:19
  • $\begingroup$ @JMoravitz I'm a bit confused with the upper and lowercase variables you used. Why should be represented $Ab$ the number of goals?. Why the product?. Does it mean a convertion factor?. Let's say $L\times \textrm{goals made against france} + L \times \textrm{goals made against norway}=5$ $\endgroup$ – Chris Steinbeck Bell Feb 15 '19 at 19:26
  • $\begingroup$ @JMoravitz But what then is the uppercase (also goals?), but this wouldn't produce a squared?. This part is where I'm stuck at. Hope you can help me to clear out this doubt. $\endgroup$ – Chris Steinbeck Bell Feb 15 '19 at 19:28
  • $\begingroup$ @ChrisSteinbeckBell I did not intend it to be interpreted as a product, but rather as a variable represented by two letters. Just like how you might use equations like $\text{Speed}\times \text{Time}=\text{Distance}$ where the variable is written like $\text{Speed}$ rather than shortening to a single character like $s\times t = d$. There is no multiplication happening in my work above, just two-letter variable names. $\endgroup$ – JMoravitz Feb 15 '19 at 19:33
  • $\begingroup$ @JMoravitz Upon reading again your answer it seems that this was the intended meaning. Sorry I'm slow at catching up these things. But why using two letters?. Because the two digits would represent together the result of the match?. $\endgroup$ – Chris Steinbeck Bell Feb 15 '19 at 19:36

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