One of my all time favorites – The McNugget problem!!

This morning my younger son had some trouble with a problem on an old MOEMs test that is similar to the famous Chicken McNugget problem. Similar enough, actually, that I decided to put off our second day of divisibility rules until tomorrow in order to spend our time today talking about this problem.

Given that we just finished up a section on modular arithmetic, there is a bit of a connection to what we’ve been studying, but I mostly just wanted to talk McNuggets!

First up was talking through his approach to the MOEMs problem this morning. In this video we do get to the answer by checking every number until we get a bunch in a row. I wanted to let him talk through this approach just to show him that he really could solve this problem on his own.


Next up was an approach to the problem that was a little more systematic and also brought in some ideas about pattern recognition. When we write out our grid of numbers my younger son notices that above 13 you can just keep adding 3’s to one of the rows and get all of the remaining numbers.


The last part of talking about the MOEMs problem was bringing in modular arithmetic. The ideas here help us see why we could just keep adding 3’s in the last step:


Finally, the punchline – the Chicken McNugget problem. This one is a tiny bit more difficult because there are 3 sizes rather than just the 2 from the MOEMs problem, but we do manage to get to the end using basically the same ideas we talked about earlier in the project.


So, an accidentally fun morning with one of my all time favorite little math puzzles 🙂

Fun trapezoid coincidence on twitter tonight

Earlier tonight I saw this great question on twitter from Wendy Menard:

Dave Radcliffe posted a link to his really clever solution:

The discussion following Dave’s tweet was about why the triangles ACD and BCD in his picture have the same height (their areas are the same because the have the same base and the same height, but as Dave says in the discussion, the fact that the heights are the same isn’t obvious).

Well . . . funny enough before I saw Wendy Menard’s post tonight my son and I were talking through problem 20 from the 2008 AMC 10a:

Problem #20 from the 2008 AMC 10 A

Here’s the problem: Trapezoid ABCD has bases \overline{AB} and \overline{CD} and diagonals intersecting at K. Suppose that AB = 9, DC = 12, and the area of \triangle AKD is 24. What is the area of trapezoid ABCD?


Working through the AMC 10 problem you’ll see that triangles AKD and BKC have the same area. That fact gives a little insight into why the two shaded triangles in Dave’s picture have the same area (and why the heights of triangles ABD and BCD are the same) -> ABCD in Dave’s picture is a trapezoid because angle DAB is 72 degrees!

How fun that two problems brought to our attention almost randomly have really similar ideas that help you get to the solution!