Open Middle equations

Ramon Garcia brought a problem type called an Open Middle problem. He talked about how usually teachers give students problems that look like this:

1+2=?

There is one answer: 3. And you’re done. (It could be more challenging, but it would still have only one solution that makes the equation true.)

Sometimes teachers give problems like this:

? + ? = 5

Then we have more possible solutions:

2+3=5
6+(-1)=5
2.5+2.5=5
etc.

Ramon talked about the Open Middle problem format, which he saw when I did a workshop for students at the Borough of Manhattan Community College’s Adult Education Program recently, where Ramon teaches. He saw some of the solutions students came up with for this problem from a packet of student materials CUNY published recently:

Here are the solutions the class came up with:

Offhand, I wondered aloud how many solutions there are that make this equation true… and Ramon was off and running. Fast forward two months and Ramon shared this problem with us to begin the meeting:

We each came up with a few solutions to problem:

1+7=8
2+3=5
4+2=6
etc.

Martha wondered aloud how many solutions there are. And so we started trying to find out.

We each worked individually for a while, then Ramon shared a second sheet.

When we started to share solution strategies, there were different ways of thinking about the problem.

Organizing by sum

A few of us looked at all the solutions that add up to 9:

1+8=9
2+7=9
3+6=9
4+5=9
5+4=9 (We already have 4+5=9)

Then all the solutions that add up to 8:

1+7=8
2+6=8
3+5=8
4+4=8 (All three digits have to be different)

And all the solutions that add up to 7…

… Will there be 3 different solutions that add to 7?

We looked for a pattern in the number of solutions, but we were a little confused by what we found.

Then we tried starting with an addend, one at a time:

1+2=3
1+3=4
1+4=5
1+5=6
1+6=7
1+7=8
1+8=9
7 solutions

2+3=5
2+4=6
2+5=7
2+6=8
2+7=9
5 solutions

Hmm.

Then we tried the solution path Ramon suggested: Looking for solutions based on different number of possible digits

How many different solutions are there if we are allowed the digits 1-3?

1+2=3

That’s it.

How many different solutions are there if we are allowed the digits 1-4?

1+2=3
1+3=4

How many different solutions are there if we are allowed the digits 1-5?

1+2=3
1+3=4
1+4=5
2+3=5

Seems like there is a pattern here…

Todd shared his strategy of using an addition table

There are still some open questions that we are investigating:

• Is there an equation that we could use to predict the number of solutions for a given number of possible digits?
• How many solutions are there for _ _ + _ _ = _ _ ?
• How many solutions are there for _ _ _ + _ _ _= _ _ _ ?

This is a problem that is definitely accessible for all of our students. As Ramon said, they may need our help learning how to organize their explorations, but adding 1-digit or 2-digit numbers is something they can do. Students will come up with solutions to any of these open middle addition problems.

Whether they will find all the solutions is a question we want to find out! Bring it to your classroom and see what happens. Let us know how it goes.

We talked about the possibility of developing support and push questions for the problem which would provide assistance and extensions for students. For example, we could prepare cards with hints related to using a table, using smaller numbers, looking for patterns, using an addition table, using a hundred-chart, etc. and also create extension questions such as: What if addition was changed to subtraction? Multiplication? Etc.

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Post-script: Todd Orelli spent a few hours on the train back and forth to Long Island continuing his thinking on the problem and sent this (spoiler alert) fascinating exploration of the use of multiplication tables in this problem.

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Post-post-script: Todd’s work on the problem inspired me to continue with some ideas I had about the problem. More spoilers.