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I found a math problem for 2nd graders in Norway. The article only presented drawing a table of everything and simply looking at it, instead of a mathematical approach, like a formula. I'm not sure what kind of solution is needed, so I cannot google the type of math needed. It goes like this:

-There are eight benches in a park.
-On each of the benches, there are either 1, 2 or 3 people. So no bench is empty.
-All together there are 19 people on the benches

How many benches have only two people?

The answer is 5: http://tinyimg.io/i/1SJcOzb.png

Is there a way to figure this out with a formula?

  • Why do you say the answer is $5$? If $(a,b,c)$ means there are $a$ benches with one person, $b$ with two, $c$ with three then $(1,3,4),(2,1,5)$ and $(0,5,3)$ all work. – lulu Nov 19 '18 at 20:58
  • This sort of problem is really meant to be played with. Algebraic machinery is going to come down to checking cases (not difficult, but probably not what second graders are going to do). – lulu Nov 19 '18 at 21:00
  • @lulu lol yeah, I just posted the answer they said it was. I of course see now that I can take away one person from any bench with two people and add to another with only two people. – Streching my competence Nov 19 '18 at 21:04
  • Exactly. $\quad$ – lulu Nov 19 '18 at 21:14

2 Answers2

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If there are $x$ benches with one person, $y$ with two and $z$ with three then you have the simultaneous equations $$x+y+z=8$$ $$x+2y+3z=19$$

You can eliminate one term to give (depending on which is eliminated)$$y +2x = 5$$ $$y+2z=11$$ $$z-x=3$$

so if $x,y,z$ are non negative integers, then from the first of these $x$ can only be any of $0,1,2$, requiring $z=3,4,5$ and $y=5,3,1$ respectively

Henry
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  • This is more than an American 2nd grader would be able to handle. I can't speak to Norway, but I bet they're waaay smarter. ;-) – Russ Nov 19 '18 at 21:15
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    Amazing! Thanks. @Russ: Haha, I doubt it. I believe they don't start with theoretical math until university, to my understanding from the article mentioning that text based problems in such low grades are a problem. I know for a fact they don't in Sweden at least, as one of my math teachers in 12th grade math explained that it's an issue in Sweden and that this harms kids ability to learn in other subjects too, and that it's pretty unique compared to the rest of the world. Sweden and Norway are very similar. – Streching my competence Nov 19 '18 at 21:23
  • @Strechingmycompetence it varies in the U.S. In the DC area, many schools participate in Math Olympiads, and problems similar to this might be presented to 4th or 5th graders: "how many ways are there to arrange 19 people into 8 benches such that ..." They would not check the work on this, only the answer, but they would review the problem with the class afterward. However, Math Olypiad is optional for most, and usually is given to the advanced students that go to a special class instead of the usual math class. – Russ Nov 20 '18 at 13:09
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But it's not true as the problem is stated. The stated solution is listed as

\begin{array}{r|cccccccc|} \text{Benches} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ \hline & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc \\ \text{people} & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc \\ & \bigcirc & \bigcirc & \bigcirc \\ \hline \end{array}

And there are five benches with only two people.

But what is wrong with this answer?

\begin{array}{r|cccccccc|} \text{Benches} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ \hline & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc \\ \text{people} & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc\\ & \bigcirc & \bigcirc & \bigcirc & \bigcirc \\ \hline \end{array}

Or this answer?

\begin{array}{r|cccccccc|} \text{Benches} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ \hline & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc \\ \text{people} & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc \\ & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc\\ \hline \end{array}