Open middle problems are problems that have a “closed beginning” and a “closed end.” This means that all students are given the same problem at the start, and they all finish with the same answer or similar answers. The “open middle” refers to how there are many different ways to tackle and work through the problem. Students may solve open middle problems using a variety of approaches, and they may find they prefer one strategy over others for some problems.

Open middle problems are probably most associated with Robert Kaplinsky, a US-based math educator. Many of the problems that Kaplinsky offers, including the ones on OpenMiddle.com, involve arranging digits from 0 to 9 or from a specific range to accomplish a task. Some problems stipulate that each digit may be used only once.

Here’s an example of such a problem:

Sample answers: I put the 5 and the 4 in each hundreds place since they’re the two biggest numbers. Then I put the next two biggest numbers, 3 and 2, in each tens place and the two smallest numbers, 1 and 0, in each ones place. That made 531 + 420, which equals 951.

OR I used 520 + 431 to make 951. OR I used 530 + 421 to make 951. OR I used 521 + 430 to make 951.

Another problem might look like this:

Choose and arrange four of the digits from 1 to 9 to make the answer as close to 250 as possible. You may use each of your chosen digits only once.

Sample answer: I know that 400 – 100 = 300 and 400 – 199 = 201, so I might want a number close to 400 (but not 400 because I can’t use zeros) and another number close to 150 to subtract. I thought I would try 402 – 153, so all six digits are different. The result is 249, which is really close to 250.

I also tried 403 – 152, which is 251, which is also really close.

I don’t think it could be exactly 250 since the two ones digits would be the same and that is not allowed.

Open middle problems can support several aspects of the curriculum, including geometry, statistics and probability, ratios, algebra, and measurement. The problem below, for example, looks at operations with fractions.

Choose and arrange four of the digits from 1 to 9 to get the smallest answer possible. You may use each of your chosen digits only once.

Sample answer: I think both fractions have to be small to make the smallest sum. I know a fraction is smaller when the numerator is much smaller than the denominator, so the smallest fraction I can make with the digits is (frac{1}{9}). The second smallest fraction I can make with the leftover digits is (frac{2}{8}). (frac{1}{9}) + (frac{2}{8}) = (frac{13}{36}), which is about 0.361.

I also tried (frac{1}{8}) + (frac{2}{9}) which makes (frac{25}{72}), or about 0.347, which is smaller than 0.361. My answer is (frac{1}{8}) and (frac{2}{9}).

The questions usually require some thinking or reasoning and not just computational skills. Having various methods of solving a problem helps students to see multiple perspectives and encourages flexibility in their thinking. Being able to recognize different solution paths and how each of these strategies may be connected can lead to stronger, deeper understanding. Open middle problems are a valuable and versatile addition to any math classroom’s tool kit.