Transcript for the Math Jam "MOEMS Teachers Math Jam" on Oct 4.

Math Jam hosted by marsbake (Marshalyn Baker ).

DPatrick (19:28:36)
Greetings and welcome to our first MOEMS Teachers Math Jam of 2006-07!

DPatrick (19:28:45)
My name is Dave Patrick, and I work here at Art of Problem Solving.

DPatrick (19:28:55)
MOEMS is an outstanding way for students in grades 4-8 to get started with problem solving mathematics. Therefore, we have invited the people from MOEMS to host Math Jams to discuss how to use MOEMS to inspire students to tackle challenging problems.

DPatrick (19:29:07)
For those of you who may have showed up to see what AoPS classes are like, please view transcripts of other Math Jams from the past, by using the link below:

DPatrick (19:29:10)
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php

DPatrick (19:29:23)
Today's Math Jam will be considerably different than AoPS classes, as it is both for a program that focuses on an earlier age than most of our classes, and because it is aimed at teachers involved with MOEMS, instead of being aimed more at students.

DPatrick (19:29:38)
One note to our newcomers: This classroom is moderated, meaning that when you make a comment, it will come to the instructors. It will only appear in the classroom if the instructor decides to share it with the class. Therefore, some of your comments may appear after a considerable delay, or possibly not at all.

DPatrick (19:29:47)
Now, on with the show!

DPatrick (19:29:51)
I'll now turn things over to your instructor for today, Marshalyn Baker.

marsbake (19:30:01)
Welcome to tonight?s Math Jam. I am Marshalyn Baker, a middle school mathematics teacher at Messalonskee Middle School in Oakland, Maine. I have been a PICO for over 20 years at both the elementary level and now the middle level.

marsbake (19:30:28)
The best part about participating in the Math Olympiads is the discourse I have with my students after each contest is over. I am especially pleased when a student shares a strategy I would never have thought of.

marsbake (19:31:02)
At our first Math Jam of the school year, we are going to investigate number patterns. This is a very important step toward fostering algebraic thinking. Let?s build our own understanding of problem solving strategies as we explore number patterns together. When our understanding is strengthened, this transfers to richer experiences for our students.

marsbake (19:31:18)
Here's our first problem:

marsbake (19:31:54)
The following number sequence is formed by starting with 7 and then adding 3 to each term to get the next term: 7, 10, 13, 16, 19, ? The 1st term of the sequence is 7, the second term is 10, and so forth. What is the 100th term?

marsbake (19:33:30)
Before I share any answers, please tell me how you have solved the problem as well as your answer.

yuehao_xu (19:33:13)
Make a list and try to find a pattern in the number sequence?

renaekelly_2 (19:34:14)
307 I began by thinking each term was 3 more than the one before it, so I multiplied 100 by 3 and then added it to 7.

frozenorb3 (19:33:41)
7+[3(100-1)]=7+[3(99)]=7+297=304

marsbake (19:35:15)
We have a couple of suggested strategies, plus a couple of answers. What do you think?

2709 (19:34:35)
the first term is 3x2 +1, the second term is 3X3 +1, so each term is 3X (it's term number plus one) and then add one to that, so 304

rvaradan (19:34:28)
find a formula for nth term

marsbake (19:36:39)
Many of you are suggesting formulas. What if students haven't developed their algebraic thinking to that point yet?

2709 (19:36:07)
find the mutliple associated with the sequence

ceswarappa (19:36:40)
its an arithmetic sequence; the first term is 7 and the difference is 3 so 99(3)+7=304

marsbake (19:37:56)
Could you have students use a table? What would it look like?

2709 (19:37:56)
my students would make the list out to 100 terms and it's so easy to make a mistake doing that - I always suggest looking for patterns but it's hard for them

frozenorb3 (19:38:20)
students could use a table but it would take an awfully long time

marsbake (19:38:56)
Would they have to set up a table that used all of the numbers?

2709 (19:38:42)
a table would start with the term number and then the term in the second column - they could look for a relationship between the two columns

Hemamalini (19:39:16)
No they could do it in terms of 20

frostmourne (19:39:30)
so like go by 10s instead?

marsbake (19:40:18)
What would that look like?

Beverly Mairs_6 (19:41:20)
Using a T-table, have them notice that as x increases by 1, y increases by 3, when x increases by 2, y increases by 6,etc. Now find the ""skip"" from the previous number. From the 5th term, x increases by 95 so y increases by 3 x 95. 19 + 3 x 95 = 304

yuehao_xu (19:41:54)
1 2 3 4 5...
7 10 13 16 19...

rbowsend (19:41:54)
My students often think they can double the values in one column and double values in the other column. For example if the 4th term is 16, then the 8th term would be 32.

Hemamalini (19:41:15)
they could add 20*3 which is 60 to get each term\

Beverly Mairs_6 (19:42:22)
I prefer increasing by 10 or a multiple of 10 - fewer errors that way.

rplsparker (19:42:49)
Yes I agree totally with rbowsend

marsbake (19:44:40)
How would we help students who double the values to see that they have a misconception?

ceswarappa (19:43:53)
they could go in tens and use a formula like (term number-1)3

frozenorb3 (19:44:50)
Doubling the term doesn't get you the right answer though. For example, the second term is 10 and the 4th term is 16. Clearly, 16 is not twice as much as 10.

frozenorb3 (19:45:18)
Are we assuming that they do or do not know algebra?

marsbake (19:46:13)
This would be a problem that third or fourth graders could explore. We would assume initially that the students don't know algebra. We are looking for strategies that all students could use.

rplsparker (19:45:39)
They do not know algebra in 4,5,6

EatBeef (19:46:13)
I always ssume they do NOT know algebra.

frostmourne (19:46:43)
going by tens would help, i think

renaekelly_2 (19:47:04)
I think this age student only knows algebra intuitively and cannot necessarily use the language of mathematicians

marsbake (19:47:47)
Here is a table you might set up:

marsbake (19:49:36)
Order 1 2 3 4 5 ?100
Sequence 7 1013 16 19??
M(3) 3 6 9 12 15?(300)

marsbake (19:49:59)
Sorry the alignment is a bit messy...tables don't come through real well.

marsbake (19:51:24)
From looking at patterns in tables, students can begin to see relationships that eventually may help them articulate a rule. Later down the road, this will help them develop the formulas.

Rocket95 (19:50:22)
What does ""M"" stand for?

marsbake (19:51:49)
M is a multiple of 3.

RichKal-MOEMS (19:50:46)
Another way to see it is to ""downgrade"" the entire sequence by 4 so 7, 10, 13, 16, ? becomes 3, 6, 9, 12, ? ? the multiples of 3. The hundredth number would be 300. Reversing the downgrade, the sequence has to be 7, 10, 13, 16, ?, 304. I think of it as applying transformation geometry to numbers.

renaekelly_2 (19:51:59)
Ask students to find something interesting about the multiples and the order of the term along with the answer to the term

Rocket95 (19:51:59)
Oh

Scott Seyfried (19:51:49)
Why multiply term # by 3 (for M(3))

marsbake (19:53:44)
You are looking beyond the formulas to find strategies that all students can use to solve these problems! Awesome job!

marsbake (19:54:09)
Let's use a bit of scaffolding to our next problem.

marsbake (19:54:57)
Find the sum of the counting numbers from 1 to 25 inclusive. In other words, if S = 1 + 2 + 3 + ?+ 24, + 25, find the value of S.

marsbake (19:56:48)
Remember, let's look beyond the formula that an algebra student (or you ) might know. If you find that, while you wait, see if you can do the problem in another way.

LDodd (19:54:37)
Is there a way to see the original problem so that those of us that joined late can get caught up?

2709 (19:56:26)
that would be to look for a pattern - 25 +1, 24 +2, 23 + 3 and so on - all equal 26 so it just becomes a multiplication problem

rbowsend (19:56:55)
Rainbow math! 1 + 25, 2 + 24, 3 + 23, 12 pairs of 26 plus the middle number 13, 325

marsbake (19:57:25)
Can you explain what you mean by Rainbow math?

Rocket95 (19:54:27)
what is scaffolding?

marsbake (19:58:20)
Scaffolding is a way to build up to an idea. Start with an easy entry level and build to more complicated ideas.

yuehao_xu (19:57:43)
Gauss way of doing the addition number sequence.

rbowsend (19:58:30)
If you draw arcs from the first number to the last, the second to the second to last, etc makes a rainbow!

Beverly Mairs_6 (19:58:33)
(25/2) 26 = 325 Pair up the first & last, the second and next to last, etc. There are (25/2) pair of numbers, each with a sum of 26.

LDodd (19:58:00)
he says to group 25 +(1+24)+(2+23)+(3+22)+(4

frozenorb3 (19:58:57)
rbowsend is completely right

marsbake (19:59:31)
Can you explain the Gauss way of doing the addition number sequence?

frozenorb3 (19:59:29)
for younger students, rainbow math would be both an easy and a pretty way to look at solving the problem

mathgeek17 (19:59:50)
You take the largest number and add the smallest number then divide the sum by two. then you multiply those by the middle number and then add the middle number[img id=em-10]

renaekelly_2 (20:00:08)
I looked at how to put numbers in a sequence together to get easier numbers to add. So I paired:25,5; 24,6; 23,7 etc down to 16,14 and then was left with 15. Then I counted the number of pairs that made a sum of 30. I had 10 pairs of 30 and 15 left over, so my final sum is 315.

marsbake (20:00:59)
That's an interesting pairing...comments about the final sum?

yuehao_xu (20:00:32)
As same as the rainbow way, from the first to last...then times by how many pairs there is.

frostmourne (20:01:15)
you left out 1-4

2709 (20:01:38)
you need to make sure that students look at whether the sequence has an odd or even number of terms

marsbake (20:02:23)
How would you do that?

mathgeek17 (20:01:50)
the final sum is 325

frozenorb3 (20:02:21)
adding 1,2,3,4 and 15 would get you 25, therefore your final sum would be 325

Beverly Mairs_6 (20:01:48)
S = [N(F + l)]/2 Sum = Number of terms divided by 2 times the sum of the First and Last numbers. Divide by 2 to get the number of pairs of numbers. Gauss Method of Summation.

renaekelly_2 (20:03:05)
Whoops, I forgot to look at that. I would need to add my 4,1 to get 5 and my 2,3 to get another 5 so my revised answer would be 10 more:325

Rocket95 (19:57:44)
It would be equal to ""sigma 25"" which would be... 325

marsbake (20:03:52)
Could you explain sigma 25 to those who don't understand that?

rbowsend (20:03:20)
When there are odd terms it might make more sense to ""save"" the 1 until last and pair 2 + 25, 3 and 24. 12 pairs of 27 plus the beginning 1

frozenorb3 (20:03:49)
isn't sigma a little complicated?

RichKal-MOEMS (20:03:56)
So: applying rainbow math to the first problem, the sum of all the terms of 7, 10, 13, 16, ?, 304 would be (7+304) + (10+301) + (13+298 etc. That is 50 pairs because there are an even number of terms: 50*311 = 15,550.

2709 (20:03:33)
i like the idea of using the ""rainbow"" idea and letting them discover what happens when their are odd or even numbers of pairs - I would give them several to look and and explore

2709 (20:04:32)
yes! sigma is to hard!

marsbake (20:05:34)
Is there a way to explain that in elementary terms, so those students who could be challenged, could understand the idea?

frozenorb3 (20:05:30)
well, can you explain sigma anyways?

2709 (20:05:48)
I also like the idea someone suggested of making easy sums

rbowsend (20:06:30)
Is there a geometric model that corresponds to the outcomes? Look, sum of 1 is 1x1, the sum of 1+2 is 1x2, the sum of 1 + 2 + 3 is 2x3...

Rocket95 (20:06:59)
sigma is when you add all the integers under and including that number to get the answer of sigma n.

frozenorb3 (20:07:38)
what I would do is ask them to find the sum of an easy sequence such as 1+2+3+4+5+6, then try teaching them to solve it with rainbow math

frozenorb3 (20:08:16)
this way, students will understand that both ways work that when you get to complicated sequences, one way is faster than the other

marsbake (20:08:42)
As we get ready to go to another problem, here is a bit of history for those of you who didn't know this:

marsbake (20:09:16)
Do you know the story about Karl Friedrich Gauss (1777-1855)? He was one of the world?s greatest mathematicians and a child prodigy. The story is told that at age nine, Gauss?s teacher gave his class the tedious task of adding the natural numbers from 1 to 100. Gauss found the sum in a matter of seconds. How did he do it?

marsbake (20:09:41)
Perhaps he noticed a pattern. To add the first 100 numbers, pair 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on. There are 50 * 101 = 5050, the sum of the first 100 numbers.

marsbake (20:10:31)
How can you apply Gauss?s idea to find the sum of the first 10 natural (counting) numbers? The first 20? The first 30? What pattern do you notice in the 3 sums? Why is that so?

frozenorb3 (20:10:26)
wow, Gauss is a genius

marsbake (20:13:14)
What if you change it and scaffold a bit easier...the first 10, then next 10, the next 10?

frozenorb3 (20:11:58)
the first 10: 1+10=11, 10/2=5, 11x5=55
the first 20: 1+20=21, 20/2=10, 21x10=210
the first 30: 1+30=31, 30/2=15, 31x15=465

Beverly Mairs_6 (20:12:59)
First 10 = 5 x 11. First 20 = 10 x 21. First 30 = 15 x 31. So first 40 would be 20 x 41.

lauren1*_2 (20:12:06)
(1 + 10)*5, (1 + 20)*10, (1 + 30)*15?

2709 (20:14:36)
divide the number of terms by 2 and multiply it by the last term + 1

Rocket95 (20:14:45)
and the first 50 would be 25 x 51.

RichKal-MOEMS (20:13:14)
The other way that mathematicians think Gauss might have done it is: to 1+2+3+...+100 add 100+99+98+...+1. You get 100*101, but that's the series twice, so divide by 2. 100 * 101 / 2 = 5050.

renaekelly_2 (20:15:26)
So do you take half of the term number and multiply that by the sum of the first and last two numbers? Half of 10 is 5, and 1 plus 10 is 11 so 5 times 11 is 55, 20 numbers would be 10 times 20 plus 1 or 210, etc

rbowsend (20:15:29)
I really like to encourage my students to vocalize the pattern in words like 2709 did.

2709 (20:15:47)
the sequence is always 55 + 100 (10 X 10) since in the second set of terms each one was increased by 10

marsbake (20:16:32)
Let's now try a different kind of number pattern:

marsbake (20:16:56)
Write the next three terms of the following geometric sequence
2, 6, 18, 54, 162, ?
What is the 10th term of the sequence above?

frozenorb3 (20:15:18)
Thank you everyone for participating in this Math Jam! I have learned new approaches to easy problems which I never would've considered before.

mathgeek17 (20:18:35)
the tenth term is 118098. 6 is 2*3 squared, 18 is 2*3 cubed and so on.[img id=em-4]

rbowsend (20:18:50)
39366 each new term is the previous multiplied by 3.

frozenorb3 (20:19:20)
163x3=489
489x3=1467
1467x3=4401
tenth sequence is 2[3^(n-1)]
2[3^(10-1)]
2[3^(9)]
2x(19683)
39366[b]39366[/b]

marsbake (20:20:15)
If and when you wonderful problem solvers exit, please click on the leave button. But don't go yet!

iamback (20:19:46)
i get 39,366 when multiplying each number by 3

frostmourne (20:19:45)
did u guys actually calculate it out???

mathgeek17 (20:20:23)
sorry, 6 is 2*3 to the first, and 18 is 2*3 squared

marsbake (20:21:50)
V

marsbake (20:21:59)
My district has been focusing on literacy in the last few years. Students need to be proficient in communicating mathematics. A major piece is developing strong vocabulary. You might highlight the following as you explore number patterns with your students.

marsbake (20:22:20)
In mathematics, a sequence is a set of numbers that is ordered according to some rule. Each number in the sequence is a term. An arithmetic sequence has the same constant difference between any two consecutive terms. A geometric sequence is patterned by multiplication by a constant number, which results in a constant quotient between consecutive terms.

RichKal-MOEMS (20:21:29)
2*3*3*3*3 is 162, not 163.

renaekelly_2 (20:21:32)
I first looked for what was done to the term before the number in question to get that number. What was done to 2 to get 6? What was done to 6 to get 18 etc. When I found that the numbers were 3 tiimes bigger than the previous number, I then was able to find the next three numbers in the sequence.162 times 3= 486 times 3=1458 times 3=4374

rbowsend (20:22:48)
Saw that first was 2 x 1, then 2x3, 2x9, 2x27 The base of three was the key. The pattern became 2 x 3^0, 2 x 3^1, 2 x 3^2...or two times 3 raised to it's place in the sequence minus one.

rbowsend (20:23:17)
what a nice beginning for those minds that will encounter exponential growth someday!

marsbake (20:24:09)
To continue our scaffolding, here's one more level

marsbake (20:24:50)
Grandmother?s Problem
Your grandmother has given you a unique gift. Starting in the new year, she will give you 1 cent on the first day of each month, 2 cents on the second day of each month, 3 cents on the third day of each month, and so forth. Based on the 2007 calendar, what can you expect to receive from your grandmother at the end of the year?

lauren1* (20:25:16)
WHOA, that's a lot of money!

frozenorb3 (20:25:17)
wow...that requires some thinking

RichKal-MOEMS (20:17:30)
You all should know that Marsbake -- Marshalyn Baker -- is running for a seat on the Board of Directors of the National Council of Teachers of Mathematics (NCTM). Those of us who have worked with her think she'd be a great addition to the Board.

frozenorb3 (20:26:08)
hint: you can break each month into an arithmetic sequence

frozenorb3 (20:23:19)
whoops, sorry, my mistake

Rocket95 (20:23:06)
1 2 3 4 5 6 7 8 9 10
2 6 18 54 162 486 1458 4374 13122 39366
It would be 39366.

rbowsend (20:26:24)
That adds a whole new twist because the months don't all have the same number of days!

marsbake (20:27:22)
The problem is a bit different that increasing each day through the whole 365 days.

RichKal-MOEMS (20:27:21)
After getting 30 or 31? the last day of a month, the first of the new month would be disappointing!

rbowsend (20:27:54)
You could fall back on 12(1+2+3+...+30) We already solved that one, and then adjust for the months with 31 and 28.

marsbake (20:29:10)
I have been sooooo impressed with your problem solving strategies. Many of you have extended your own thinking to find other strategies.

Beverly Mairs_6 (20:28:18)
Similar to a recent Fox Trot cartoon about a math teacher who assigns 1 second of homework on the first day and doubles it each day. How much HW would be assigned on the 30th day of class? My students loved that one !!

2709 (20:29:42)
it's a rainbow math problem - * 7 months of 31, 4 months of 30 and 1 of 28

marsbake (20:30:20)
The cartoon was in on Sept. 10th in the Sunday paper.

marsbake (20:30:57)
This will give you something to ponder as you leave us. Remember to click on leave.Thank you so much for contributing to tonight?s Math Jam. Please join us again next week as we explore factors, multiples, and primes.

renaekelly_2 (20:31:48)
Figuring out how much money you would make each month using the idea of ""thirty days has Sept. April June and Nov. you know you have four months with 30 days, 7 months with 31, and one with 28. Using that strategy you could use what we learned about patterns to find the answer.

marsbake (20:32:23)
Have a great rest of the night.

frozenorb3 (20:31:35)
Thank you for having us!

Rocket95 (20:32:32)
the answer to Beverly Mairs_6 would be 536870912 seconds.