Discrete Math - Day 40

Today was our first day back from a week long break. As you can imagine, the first class back can be an adjustment for students, who may not be fully awake and geared up for school.

I briefly revisited where we had left off before break, reminding students that we needed to understand modular arithmetic and equations in order to be able to decipher coded text. With that introduction, I launched into our first problem.

It was the start of the baseball season this past weekend, so the first problem had a nice tie-in. Fred's Baseball Cards looks at the issue of determining the number of baseball cards that Fred could have, knowing that as he divided his cards into piles of various sizes that he would either have a card left over or not. Specifically, there is one card left over when dividing the piles by 2 cards, 3 cards, and 4 cards. Piles of seven cards divide out evenly, leaving no remainder.

This problem falls under a class of problems that embody the results of the Chinese Remainder Theorem. This problem is a first look and we'll be looking at the Chinese Remainder Theorem and its applications in more depth. The purpose of this first problem is to get students thinking about the patterns and relationships that they see.

Typically, it does not take too long for students to find that 49 is a value that satisfies the given conditions of the problem. I then ask students if there are any other values that satisfy the conditions as 49 is a paltry number of cards for a baseball card collection.

Students started exploring and it was interesting to see how much they struggled with finding additional values. They would come up with a value and I would ask if the had checked that all the conditions specified were met. Typically one or more of the conditions failed. Other students were confused as to how to determine if the remainder was one when using their calculator. There appeared to be a lack of understanding between the remainder, the fraction, and the decimal representations of division. While I have investigations and problems that I use in college-level math education courses, these are capabilities that students coming out of an Algebra II course should possess.

Some students started to look at multiplying 49 by 7 and claimed that this worked. I wrote three of these values down: 49, 343, 2401. I asked students if this worked and if so why. No response was forthcoming. I asked if students found any other values that worked. At this point these were the only candidates. Many other values that students found still were not satisfying all of the given conditions.

I told the class there was at least one value between 49 and 343 that worked. Students started, reluctantly, to look for a working value. Finally, a student found that 133 satisfied all of the conditions. I asked them to find more. The group immediately thought to look at the difference between 133 and 49. Using 84, they then tried the next few values to determine that 217 and 301 also worked.

I wrote these values on the board and asked the group to explain what they had found. We then had a linear equation B(x) = 84x + 49 that characterized the values that satisfied the problem's conditions. I turned to the multiples of 7 list that was still on the board and asked if 343 worked. Students found that this value did not work since dividing by 4 left a remainder of 3. They then checked 2401 and saw that this did work. The result is that values that are powers of 49 satisfy the given conditions.

The next question that was posed concerned the value of 84. Why are the differences between successive values 84? After some thought one student suggested that 2, 3, 4, and 7 are all factors of 84. I then asked if any other numbers fit this condition. Students tried some values but didn't find any.

I asked students to consider the prime factorization of 84 = 2² x 3 x 7. If any of these prime values are removed, we no longer have a number divisible by all of the values 2, 3, 4, and 7. It turns out that 84 is the least common multiple or LCM of 2, 3, 4, and 7. One student remarked that the LCM was like the reverse of the GCD. In fact, we will be exploring the relationship between these two values down the road.

I asked students to record their thoughts about working through this problem in their notebooks. We'll explore some more problems along these lines in the next class.

Visit the class summary for a student's perspective and to view the lesson slide.