Bead Bracelets with fifth graders.

Taking an idea from Ariane, I modified the activity to multiplying beads rather than adding beads. I did this simply to relate to what we are doing in class. So most of the class was working with multiplication. However, I did have a group that did the addition as written up. The questions were posted and I added a few questions:

What patterns do you notice?

If you used only odd (or only even numbers) could you make a prediction about the ones digit in the next answer?

If you used odd and even numbers could you make a prediction about the ones digit in the next answer?

All of the students enjoyed the activity. The higher students, who tend to be bored during regular math instruction and need to be challenged, enjoyed looking for patterns and relationships. The students who tend to struggle enjoyed that they had an entry point.

The answers to the questions were not as thoughtful and deep as one would hope. That was to be expected due to the age of the students and the lack of exposure of problem solving and justifying or proving their reasoning.

I also did a quick poll as to whether the students would like to do this activity again, never again, or if I made them. The class was pretty much divided.

Bead Activity

My students used the bead activity as an introduction to patterns/algorithms;i.e. order of operations. The students so enjoyed the activity. I had them write down their two numbers on a sheet of paper and do the algorithm to find when the two numbers repeat. After completing the pencil/paper task the students actually made their necklace/bracelet from beads and pipe cleaners. The task became quite competitive and students worked diligently!

It gave me chill bumps to see the interaction and chat regarding algorithms.

This also led into the "Math in Action" photo for the new concept. The students were to take a picture of an algorithm and use pencil/paper to describe the algorithm. . . rec'd some real unique ideas!

Math Bracelets

Wow! My students were hooked on the math bracelet activity! They wanted to try multiplying the numbers after they were successful with adding the numbers. Two students now have a contest going to see who can make the longest bracelet. The numbers were great to set the pattern and practice some basic computation. My enrichment class expanded the activity by creating a color code for the beads. One student suggested using to beads when we get to double digit answers. I was able to get some digital photos of the colored bracelets. Hopefully, I will be able to share them next time we meet.
One of my goals was to share this activity with our special education teacher. I was also able to share it with my team. They thought it was great.

Divide and Conquer

Divide and Conquer is a game played by subtracting proper divisors from the initial number. One is always a proper divisor and no number can be divided by itself. The game is won when a player presents his/her opponent with one.
As we worked with 100, which has several proper divisors, we tried strategies that included making a list and drawing a diagram on graph paper. In the list we found a proper divisor of 100, and then subtracted the divisor from 100. We repeated the process with each new difference until one player could present his opponent with one. Similarly, we drew a 10 x 10 grid on graph paper and took turns crossing off proper divisors until one square remained.
We discussed that many middle school students will learn how to play this game with the initial number of 30, They will memorized the specific steps for 30 and attempt to generalize those exact steps to other numbers. The conclusion was that in order for students to truly demonstrate their understanding of factors in this game, they need to play it with other numbers.
I shared this game with my summer school students. They were thrilled to be able to beat the teacher.

Finding the last digits of 7 to the 9999th power

Now here is a question. Suppose you took 7 to the 9999^th power. That would be a number with 8450 digits. In 12 point type, that is a number about 70 feet long. What are the last three digits of that number? Why would we ask such a question? No reason, that’s just what we do.

One problem solving strategy is to create an organized list and look for patterns. Using Excel, here are the first 10 powers of 7 (formulas are listed in column C).

	A	B
1	7	7
2	49	=A1*7
3	343	=A2*7
4	2401	=A3*7
5	16807	=A4*7
6	117649	=A5*7
7	823543	=A6*7
8	5764801	=A7*7
9	40353607	=A8*7
10	282475249	=A9*7

No obvious patterns seem to jump out and the numbers are already getting pretty big. Is there a better way? Asking for the last three digits is the same as asking for the remainder when you divide by 1000. Excel has a modulo function we can use to find the last three digits of the powers of 7. We only need to multiply these values by 7 to find the last the last three digits of the next power of 7 and the mod function returns only the last three digits.

As a shot in the dark, what is the value of the first 25 values for the powers of 7 mod 1000? It would be very cool is some relatively small power of 7 ended in 001 (or written in mod notation: 7^x mod 1000 ≡ 1) since powers of that number would also end in 001.

A	B	C
1	7
2	49	=MOD(B1*7,1000)
3	343	=MOD(B2*7,1000)
4	401	=MOD(B3*7,1000)
5	807	=MOD(B4*7,1000)
6	649	=MOD(B5*7,1000)
7	543	=MOD(B6*7,1000)
8	801	=MOD(B7*7,1000)
9	607	=MOD(B8*7,1000)
10	249	=MOD(B9*7,1000)
11	743	=MOD(B10*7,1000)
12	201	=MOD(B11*7,1000)
13	407	=MOD(B12*7,1000)
14	849	=MOD(B13*7,1000)
15	943	=MOD(B14*7,1000)
16	601	=MOD(B15*7,1000)
17	207	=MOD(B16*7,1000)
18	449	=MOD(B17*7,1000)
19	143	=MOD(B18*7,1000)
20	1	=MOD(B19*7,1000)
21	7	=MOD(B20*7,1000)
22	49	=MOD(B21*7,1000)
23	343	=MOD(B22*7,1000)
24	401	=MOD(B23*7,1000)
25	807	=MOD(B24*7,1000)

Holy Math Smack! 7²⁰ ends in 001 or 7²⁰ ≡ 1 mod 1000, then the pattern begins to repeat. That means that 7²⁰ ≡ (7²⁰ )ⁿ ≡ 1 mod 1000. So, think of 7⁹⁹⁹⁹ as 7²⁰ x7²⁰ x7²⁰…7¹⁹ . so 7⁹⁹⁹⁹ ends on the 19^th step of a repeating 20 step cycle. That value, according to the Excel sheet, is 143. QED. (that is Latin abbreviation you put at the end of a proof. I think it means “Thank God its over” but you should google it to be sure).

So even though 7⁹⁹⁹⁹ is impossibly large on the scale of 8^th grade math, we can make some statements about it, such as the last three digits have to be 143. Now, can you find the first three digits?

Clock Arithmetic

The problem above makes use of a repeating pattern in the last three digits of the powers of 7.

Repeating patterns such at this come up often enough to be interesting. Here is a simple case that would be accessible to 4^th or 4^th graders.

My friend, Mary O’Dell works in a candy shop. Part of her job is to bag the candy when a fresh batch comes out of the kitchen. As one of the benefits of working at the shop, she gets to keep any candies left over after bagging – that is, if there are some candies left but not enough to fill a whole bag, she gets to keep them. Today she is bagging candies 9 to a bag and the cooks made 665 in this batch. How many candies does Mary get to keep? What are some numbers that would leave Mary several candies? What are some numbers that would leave her with no leftovers at all?

Clock arithmetic is a common example. If it is 8:00 am now, what time will it be 133 hours later?

A mathematician would ask, “what are the patterns in these kinds of problems?” The first thing we’d need is a notation – a way of writing these ideas down. In this case us a mod notation. If we want to look at Mary O’Dell bagging 9 candies per bag then we can start with the number of candies which was 665 in the example above, then designate the number is a bag (or the number in any group) by using the notation “mod 9”. The notation would be 665 mod 9. We use a congruence notation rather than an equal sign because we are trying to say these are numbers with the same remainder when divided by 9 and these numbers are not equal, their remainder is when they are divided by 9. We use the notation ≡ and we read it as “is congruent to” so we would write 665 mod 9 ≡ 8 and we read it as “666 mod 9 is congruent to 8”