A cool coincidence with the 2014 Putnam and an old blog post

A few weeks ago I wrote up a fun exercise that my younger son and I had worked through:

What About Fifty Ten?

That exercise explored a fairly straightforward situation: what two digit numbers are equal to 4 times the sum of their digits? There are a few: 12, 24, 36, and 48. My son saw the pattern that the first digit had to be twice the second digit, so I asked him: what about the number Fifty Ten?

My question is sort of silly, I know, but I wanted to get him thinking about place value. Fifty Ten would be the same as 60, but the sum of the digits of these two “numbers” is different.

This morning I got quite a surprise reading problem B1 from the 2014 Putnam Exam:

The 2014 Putnam exam at the Art of Problem Solving site

Essentially the question asks about this situation: if you allow 10 to be a digit in base 10, some positive integers will have a unique representation in the new number system and some won’t. For example, the number ten could be written in the usual way as 10, or in a new way as the single “digit” (10). The number 19, however, can only be written one way in this new number system. What are the positive integers will have a unique representation in this new number system?

What a cool coincidence – a question where fifty ten is actually a number!

How fun to have an old project with my younger son sort of overlap with a Putnam question 🙂

Updated to include talking through the problem with my son:

When I learn a little bit about talking math with my kids

Every day I select (essentially at random) an old AMC contest problem to go through with my older son. I do this mostly for a bit of variety in the math we do daily, but also for building up a little problem solving experience for him. Currently I’m picking problems from the old AMC 10s.

The problem we picked on Friday was #11 on the 2008 AMC 10a:

Follow this link to problem 11 from the 2008 AMC10a

Here is the problem without the link:

“While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing toward the shore at a constant rate of 4 miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?”

Upon first reading the problem my son thought that it was going to be pretty difficult. We talked through a little bit about what you needed to know in order to solve the problem, and he was surprised that the solution was actually not as difficult as he expected.

I told him that it reminded me of a problem that I had struggled with on an exam when I was in high school. That problem was involved escalators, but I didn’t remember the details except that the problem really gave me fits. This evening I saw Tracy Johnston Zager this post this crazy video on twitter and figured it must be some sort of sign!

 

So, after watching this video I had to find the problem that gave me trouble in high school! Eventually I did:

Follow this link to problem 10 from the 1987 AIME

And the problem without the link:

“Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al’s speed of walking (in steps per unit time) is three times Bob’s walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)”

It is interesting to reflect on the difference between these two problems
nearly 28 years just few short years after seeing the AIME problem for the first time.

Of course, you would expect a general problem #10 on the AIME to be more difficult than a general problem #11 on the AMC 10, but exactly what makes it more challenging? You go through pretty much exactly the same steps to solve both problems. The AIME problem is missing a bit of information about the speed of the walkers, but it turns out that knowing that specific information isn’t necessary to solve the problem. Missing that information does force you to work a bit more abstractly, though, so perhaps the obstacle that this missing information presents is the biggest difference.

Looking at the 2014 AMC 10a and AIME, I note that roughly 54% of the AMC 10a participants got #10 right while only about 7.5% of the AIME participants got #10 right on that exam. I don’t know how to go back to 2007 or 1987 to see the percentages for those exams, but I’d be surprised if the numbers were drastically different in those years. As I try to improve my own ability to talk about math with my kids, it is really instructive for me to see how changes that seem small from where I sit today make huge differences in difficulty of problems for kids learning math.