## Sunday, December 27, 2009

## Monday, November 9, 2009

### mastermind

## Tuesday, November 3, 2009

### Tootsie Roll Game (“Iterated Sharing” from Solve This)

The group now follows these instructions: Give half of your candy to the person on your left and half to the person on your right (and hence receive a supply of candy from the people on your left and right). Do this simultaneously. Now recount your candy supply. If you now have an odd number of pieces, take an extra piece of candy from the reserve supply. This boosts your pile back up to an even number of pieces and everyone is ready to perform the maneuver again.

What happens to the distribution of candy among the group if this maneuver is performed over and over again? Will people be forever taking extra pieces from the center, so everyone’s amount of candy will grow without bound? Or will the distribution stabilize or equalize in some sense? Will one person end up with all the candy? Might “clumps” of candy move around the circle with each iteration or some strange oscillatory pattern emerge? Is it possible to predict what the result will be?

Taking it further: What happens if instead of adding pieces, you eat any odd piece of candy to bring your pile back down to an even number?

Hint: Based on the investigations my class performed, and the patterns they found, I encourage you to examine what the range of distributions is. For example, if you started with 0 – 18 pieces (a range of 18), what happens? And is it different if you started with 0-20 pieces (a range of 20)?

My classes enjoyed this activity, though my advanced class looked into in much further than my lower level class. They tried several rounds to test and prove conjectures. It was a difficult game to play with a class larger than 12 students. I tried putting them into two groups, but this was difficult since I was facilitating both groups. It was also difficult to play with desks in a circle. Playing on the floor or at a large table would have made things easier.

## Monday, October 26, 2009

### Math Circle Documents

-Kendall

## Monday, October 12, 2009

456

987

1443

2430

3874

Then add the sum to the 2nd addend to get a 4th number. Now add the 3rd and 4th to get a 5th, keep going until you have 10 numbers. How can you quickly get the sum of all 10? Trick: 11 times 7th number = sum of all 10. Kendall’s problem is a challenge for his 13-year-old son Benjamin. Kendall also asks for help to decide favorites from Jackson to take to ND where he’ll present about TCs, and for suggests about other patterns that are like this, or ways to extend this. Maybe there is a similar pattern with 3 numbers to start.

Ariane asks for more information about Kendall’s pattern, and Lynne suggests understanding the pattern using letters instead of specific numbers.

## Tuesday, September 29, 2009

### 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.

## Monday, September 28, 2009

### Bead Activity

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!

## Wednesday, September 23, 2009

### Math Bracelets

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.

## Monday, September 21, 2009

### Divide and Conquer

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.

## Wednesday, September 16, 2009

### 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! 7^{20 }ends in 001 or 7^{20 }≡ 1^{ }mod 1000, then the pattern begins to repeat. That means that 7^{20 }≡ (7^{20 })^{n} ≡ 1^{ }mod 1000. So, think of 7^{9999} as 7^{20 }x7^{20 }x7^{20}…7^{19} . so 7^{9999 } 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 7^{9999 }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?