Last week’s AlgPoW, Don’t Be Square, was (I think) a simple problem hidden behind lots of complications. It is a perfect problem to think about different ways to look at a hard situation and think about how to use the Solve a Simpler Problem strategy to eventually crack the whole, hard problem.

The PoW went something like this:

√((3/2)*(4/3)*(5/4)*…*(a/b)) = 10√10.

If the whole problem were completely simplified, it might look something like this:

a/2 = 1000

b = a – 1

Find a and b.

Easy, right? But that’s not the problem we gave, on purpose. We gave a problem where the solver’s job is to look at a mess and say to themselves,

  • What’s making this problem hard?
  • What patterns can I find and write more compactly?
  • Are there other, simpler ways to write the same thing?
  • What can I learn from replacing hard numbers with easier ones?
  • Could I do the problem if…?
  • What can I learn from doing a few sample of this long pattern?
  • What happens if I try specific examples?

Those are all questions that are really valuable in all kinds of hard problem situations, whether the problem is hard because someone made it tricky for you, or because it’s one you stumbled across in daily life (like the problem I was stumped on for a while: how many times can 36 people in groups of 6 change groups before someone has to be in a group with someone they were with before?).

So, what happens when students ask and answer these questions? Here are some examples from the Don’t Be Square PoW.

Annemarie, from Rosemont School of the Holy Child:

When I first saw this problem, I was very confused. I thought that I was going to have to keep typing out the fractions in my calculator until I reached 1000, which would take a very long time.

See how she’s thinking, “so many fractions… that makes this problem hard!” Let’s see what she thinks to do to make the problem simpler.

So I started typing in the pattern, and stopped at 24/23. The answer was 12.

I recognized that 12 was 1/2 of 24, so I tried it again to see if the pattern worked.

Too many fractions, let’s try fewer, and (this is important) see what we notice. She noticed if she stopped at a = 24, the product was 12, which is half of 24.

This time, I did (3/2)(4/3). According to my prediction, it should equal 2, which it did.

Then I tried (3/2)(4/3)(5/4), which should equal 2.5. It did.

Annemarie didn’t stop at noticing, she formed a hypothesis and tested it with really simple, small values for a. Then, when it worked, she had an idea for how to apply her findings back to the main problem.

The first time I did the next part, I mixed up the numbers and got (500/499)

However when I checked my answer, I realized my mistake. The numerator is 2 times the result, not 1/2 of it.

At first she thought, I need to let a equal half of what I want to get under the radical. But then she checked her answer (so important when you are trying to generalize from a simpler version to the original hard problem!) and realized a needed to be twice what she wanted to have under the radical. I thought Annemarie’s story was a great narrative about using a “Solve a Simpler Problem” strategy.

Let’s look at another example from Rosemont. Jack set out to rewrite the problem (though he didn’t say how) and then to look for patterns. It sounds like he had the  ”What patterns can I find and write more compactly?” idea in mind:

I tried to find a pattern on the left side of the equation. I found that each time you add another fraction you get .5 more. 1000/0.5 is 2000 that means there will have to be 2000 fractions in this.

So Jack has turned the (3/2)*(4/3)*(5/4)*…*(a/b) into a more compact question, something like, “how many 0.5′s does it take to get to 1000?” Let’s see where he takes this:

Therefore a+b= 4005, because you start the fractions at 2.

Hmm… Now it seems like Jack isn’t able to apply his concept from the compact description of the pattern to the original problem. Has he lost track of how the pattern could be expanded? Maybe thinking about how many products it would take to get to 5 or 10 and checking that would be a good intermediate step for Jack to make sure his compact pattern can be applied to the enormous numbers in the problem?

Sarah, from East Hampton High School, did a nice job showing how she simplified the right side of the problem, and also had an interesting idea about the left side of the problem. Here’s her work:

10√10 = √1000
cause √10 x √100 = √1000
and the square root of 100 is 10,
so 10√10= √1000

(3/2)(4/3) —->(20/19)= 10
(21/20)(22/21)——>(40/39)= 10
(41/40)(42/41)——>(60/59)= 10

10x10x10= 1000 which is under the square root

so √(3/2)(4/3)(5/4)(6/5)…(a/b) = √1000 = 10√10

I think Sarah had a brilliant idea to look for groups of fractions that multiply to 10, and to look for 3 of those groups. I wonder if she tried the simpler problem of inspecting the first 10 products, then the first 20 products, etc. I agree with her that (3/2)(4/3)…(20/19) = 10 but I wonder if she overgeneralized when she thought of trying (21/20)(22/21)…(40/39). I wonder if she could find how far to extend the pattern of fractions starting at (21/20) to get it to multiply to 10? And if that would give her a pattern to extend to find the last 10? It’s a really neat pattern that she’s hit on, I think and I’m glad she thought to try simplifying the problem this way!

In general, I thought students had brilliant ways to look for patterns and simpler problems, whether it was noticing a + b has to be odd, a has to be b + 1, or that the stuff under the radicals has to equal 1000 on both sides. Sometimes looking at simpler problems, even when they weren’t perfectly executed, revealed useful hidden structures in the main problem! Nice work!