Transcript for the Math Jam "MOEMS Teachers Math Jam" on Jan 23.

Math Jam hosted by marsbake (Marshalyn Baker ).

rrusczyk (19:31:30)
Greetings and welcome to a MOEMS Teachers Math Jam!

rrusczyk (19:31:36)
My name is Richard Rusczyk, and I work here at Art of Problem Solving.

rrusczyk (19:31:39)
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.

rrusczyk (19:31:45)
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:

rrusczyk (19:31:49)
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php

rrusczyk (19:31:54)
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.

rrusczyk (19:32:00)
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.

rrusczyk (19:32:04)
Now, on with the show!

rrusczyk (19:32:08)
I'll now turn things over to your instructor for today, Marshalyn Baker.

marsbake (19:32:20)
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.
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.
At tonight?s Math Jam, 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 (19:33:06)
Now let's get started with our first problem.

marsbake (19:33:20)
Roni starts with the number 5 and counts by 8?s. This results in the sequence 5, 13, 21, 29, 37, and so on. What is the twenty-fifth number in this sequence?

Cliff Matsuda (19:33:52)
25x8+5

marsbake (19:34:19)
Can you tell me a little more about how you arrived at this?

chaostheorydave (19:34:46)
the first differences are 8 and there's 25 items plus the starting of 5

marsbake (19:35:30)
How will this help you get the answer?

marsbake (19:35:50)
What patterns are you seeing? How can they help you solve the problem?

kevlarsen (19:35:35)
but 5 is the first term, meaning there are only 24 additional ones from the initial number

chaostheorydave (19:36:09)
each number increases by 8. There are 25 times it increases, plus you strated off with eight. So do 25*8 + 5

PI-Dimension (19:36:13)
arithmetic sequance

marsbake (19:36:43)
What is an arithmetic sequence?

Cliff Matsuda (19:36:58)
multipliacation sequence

marsbake (19:37:28)
Is an arithmetic sequence the same as a multiplication sequence?

PI-Dimension (19:37:23)
a sequence of +ing and -ing

chaostheorydave (19:37:14)
any sequence in which the terms are separated by an operation involving adittion only (or subtraction)

chaostheorydave (19:37:35)
no thats geometric

kevlarsen (19:37:46)
an arithmetic sequence is one where there is a constant difference between successive terms

chaostheorydave (19:38:00)
[i]geometric[/i] is when you multiply ie. 5,10,20

marsbake (19:38:44)
You have brought up some great vocabulary to clarify for yourselves and your students.

marsbake (19:39:28)
Some Vocabulary Associated with Sequences
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.
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.

marsbake (19:40:04)
So that being said, let's get back to determining the answer.

Cliff Matsuda (19:35:39)
actually if you count 5 as the 1st one then it would be (25-1)8+5

Suman Goel (19:35:07)
[b][i][/i][/b]205

KingRoy_2 (19:34:05)
Is it 197?

OmaAnnie (19:40:56)
24X8+5=197

RichKal-MOEMS (19:41:16)
Here's another way to see it. Every member of the sequence is 3 less than the corresponding multiple of 8. The 25th multiple is 200, so the 25th member of the sequence is 197.

marsbake (19:42:35)
As we continue to explore other problems, let's look for strategies that can help our students see the answer in a variety of ways.

marsbake (19:43:52)
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?

Whit (19:42:44)
In order to arrive at the answer, I made a t-chart to determine how the term and the answers are related.

Whit (19:44:48)
100 x 3 + 4

RichKal-MOEMS (19:45:11)
What's a t-chart?

chaostheorydave (19:45:30)
wait... a t-chart? isn't that reeaaally loooong for 100 elements?

kevlarsen (19:45:31)
Sometimes I figure out the ""zeroeth"" term, which in this case is 4. From there I would add the appropriate number of differences. In this example, that would be 4 + 100*3, or 304

Cliff Matsuda (19:45:06)
(100-1)3+7

marsbake (19:46:31)
Can you explain what some of your numbers represent? How would you explain this to elementary students?

slbwalker (19:45:05)
100 x 3 + 4 = 304

Whit (19:46:52)
I created two columns, the first contains the term # and the second contains the term. I used it to find patterns.

slbwalker (19:46:32)
Counting by 3s, but the first term is 4 more, so you add 4 to whatever nth term x 3 is

chaostheorydave (19:46:59)
100 elements after the zeroth adding 3 each term and then you started with seven

blackstallion (19:44:51)
307

KingRoy_2 (19:44:46)
So it is 304

marsbake (19:48:17)
We have these answers. Which would be correct and why would the other not be?

RichKal-MOEMS (19:47:49)
100 numbers implies 99 intervals of 3 each. 3x99+ 7 =304

slbwalker (19:48:08)
Some kids set up the table then ""test"" the ones that are in front of them... (first term....7 is 1 x 3 + 4 second term....10 is 2 x 3 + 4)

marsbake (19:48:50)
Can you tell us a bit more about this.

blackstallion (19:48:41)
100is for the # of digits, andx3 for the counting bythree, and there is only 99 spaces, so you minus 4

marsbake (19:49:36)
What do you think of this strategy?

marsbake (19:50:02)
Here is an interesting strategy:

kevlarsen (19:49:03)
With our elementary school, we have used the clock to represent modulo arithmetic. It's also handy for arithmetic sequences. The key is picking a starting point before we begin our ""laps"".

slbwalker (19:49:26)
They look at the relationship between the nth place and the numbers they already see...

marsbake (19:51:11)
Make a table and look for a pattern. The table should show the order of the numbers of the sequence as well as the numbers. Since successive numbers of the sequence increase by 3, show associated multiples of 3, M(3), in the table. Notice that the 100th multiple of 3 corresponds with the missing term of the sequence. Enter (300) on the bottom line of the table under the missing term. Also notice that each number of the sequence is 4 more than the corresponding multiple of 3. The required number is 304.

Whit (19:51:09)
I would like to hear more about the clock and laps.

kevlarsen (19:52:48)
Ok. Unlike an ordinary clock that has 12 spaces, our ""clock"" would have the same number of spaces as our differences...

blackstallion (19:53:48)
go on

kevlarsen (19:53:59)
in this case, it has 3 spaces...

chaostheorydave (19:54:16)
and then only care about where the hand is, not how many rotations it made

chaostheorydave (19:54:19)
right?

slbwalker (19:54:25)
How does the clock/lap strategy account for the extra 4?

kevlarsen (19:51:22)
So we pick the starting point that would allow us to get to the first point with one cycle, or difference.

kevlarsen (19:54:42)
Also, we need to pick our ""zeroeth"" point, or the starting number, which is 4.

KingRoy_2 (19:54:58)
Does it start at the biggening number?

kevlarsen (19:55:26)
so if you make laps, the numbers in the sequence are at the same place on the ""clock""

blackstallion (19:56:02)
now i understand

marsbake (19:56:10)
Perhaps this strategy about the clock and laps can be applied to our next problem.

marsbake (19:56:38)
If I start with 2 and count by 3?s until I reach 449, I will get: 2, 5, 8, 11, ?, 449 where 2 is the first number, 5 is the second number, 8 is the third number, and so forth. If 449 is the Nth number, what is the value of N?

KingRoy_2 (19:57:26)
First you have to do 449-2 and then divide the answer by 3

Cliff Matsuda (19:57:26)
449-2)div by 3

chaostheorydave (19:57:35)
one hundred forty nine. You subtract two from your 449 and divide that by three, working backwards tosolve the problem

Cliff Matsuda (19:57:49)
129

blackstallion (19:57:53)
104

kevlarsen (19:58:04)
With zeroeth number of -1, we would have 449 = -1 + N*3 . . . so 450 = 3N, or N = 150

marsbake (19:58:52)
We are getting quite a few different answers. What can we say about some of these? Do you see any misconceptions?

Whit (19:58:18)
The clock will get you the answer, but it may not be the most effiient strategy for this problem.

blackstallion (19:58:49)
no, 114

blackstallion (19:59:04)
i added wrong

chaostheorydave (19:59:43)
can blackstallion or cliff matsuda explain theirs? i don't understand how they got theirs, and could maybe help them

KingRoy_2 (19:59:44)
you must do 149+1 because of the 2 you subtracted.

kevlarsen (20:00:10)
Or, as Rich explained earlier, each term is 1 less than the corresponding multiple of three. If we had a starting point of 0, the 150th term would be 450, and one less equals 449.

Cliff Matsuda (20:00:27)
N-1)3+2

kevlarsen (20:01:00)
there's a connection between Rich's offset and my zeroeth point, and his multiplier and my number of spaces on the clock.

marsbake (20:01:25)
Can you explain that further?

Cliff Matsuda (20:01:56)
N=150

slbwalker (20:02:47)
449 + 1 then divided by 3 because if you look at the chart given....2 + 1 divided by 3 is 1 (first place) then, 5 + 1 divided by 3 is 2 (second place), etc.

kevlarsen (20:02:59)
I would say theo most intuitive way to see the clock would have been if there were no offset, or starting at zero. My ""zeroeth"" point is the amount that each multiple differs from the ordinal multiple...

kevlarsen (20:03:14)
150 represents how many ""laps"" we made.

marsbake (20:04:35)
Great discussion!!! Here's what I had prepared (and yes, for you that want the answer)...Arrange the terms of the given sequence in order. Since the consecutive terms of the sequence increase by 3?s, also display the corresponding multiples of 3 in the table. Notice that each multiple of 3 is 1 more than the corresponding term of the sequence. Then the multiple of 3 that corresponds to 449 is 1 more than 449 or 450. Since 450 is the 150th multiple of 3, N must be 150.

kevlarsen (20:04:15)
I find another connection where kids often stumble on counting the number of pages in a section...

kevlarsen (20:04:40)
If I ask how many pages are there from page 1-10, they usually correctly say, ten...

marsbake (20:05:05)
Let's move on and tier our skills further.

marsbake (20:06:24)
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.

KingRoy_2 (20:07:51)
(25+1)+(24+2)........(12+14)+13=S

minranli (20:08:05)
S=325

kevlarsen (20:08:20)
We can pair up 1 and 24, 2 and 23, 3 and 22, and so on, up to 12 and 13. That's 12 pairs that total 25, plus the final value of 25, or 13*25 = 325.

KingRoy_2 (20:08:26)
So the answer is 26*12+13

marsbake (20:08:48)
Can you explain this further?

smerd (20:08:48)
(26 * 12) + 13

slbwalker (20:09:28)
there are 13 sets of 25 (making 25 with 1 + 24, 2 + 23, etc. so... 325

marsbake (20:10:08)
How would we explain (26*12) + 13?

smerd (20:10:01)
12 pairs that total 26 (1 + 25, 2+ 23....to 12+14) and then 13

Whit (20:10:19)
where does the 26 come from in 26 *12

chaostheorydave (20:10:37)
there is the first&last term, etc.., plus the middle term that won't pair with anything

Cliff Matsuda (20:10:39)
12 pairs of 26 with the odd 13 in the middle

chaostheorydave (20:07:57)
Gauss's formula(i think it's called that) states that th sum of the first n natural numbers is equal to (n+1)*(n/2)

marsbake (20:11:58)
Are you all familiar with Gauss?

KingRoy_2 (20:12:22)
no

kevlarsen (20:12:25)
Yes. And he figured this out when he was very young!

chaostheorydave (20:12:23)
i am; it's my [i]favorite[/i] formula :)

marsbake (20:13:11)
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?
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.
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?

Whit (20:11:48)
Gotcha, so you the difference in the number models come from how you pair the terms.

kevlarsen (20:10:51)
If we extend to the numbers from 1 to n, we have two cases. If n is odd, then we have (n-1)/2 pairs that total n, plus one more n, or (n-1)/2 + n = n(n+1)/2.

kevlarsen (20:11:23)
If N is even, we have n/2 pairs that total n+1, or n(n+1)/2. Both yield the same result.

marsbake (20:14:46)
Let's try another type of sequence.

marsbake (20:15:19)
Write the next three terms of the following geometric sequence
2, 6, 18, 54, 162, ?
What is the 10th term of the sequence above?

marsbake (20:17:00)
What do you notice is happening between the terms?

Cliff Matsuda (20:16:53)
2x 3 to the 9th power

KingRoy_2 (20:16:53)
2*3^(N-1)

Whit (20:17:26)
They arei beng multiplied by 3.

jax (20:17:20)
x3

marsbake (20:18:24)
What would KingRoy's formula yield for an answer and how could we explain his algebraic rule?

KingRoy_2 (20:18:04)
So 2*3^9=2*19683=39366

kevlarsen (20:18:07)
Here, my zeroeth term is 2/3, so the 10th term will be 2/3 * 3^10= 39366

marsbake (20:19:48)
While we are finishing up this question, how many of you are PICO's (persons in charge of Math Olympiads)?

minranli (20:19:20)
the 10th term is 39376

Whit (20:19:59)
I am .

mryilmaz (20:20:10)
me

kevlarsen (20:20:04)
I am

smerd (20:20:05)
I am

Cliff Matsuda (20:20:12)
I am(volunteer parent)

marsbake (20:21:10)
How would you encourage students who aren't quite ready for the algebraic formula to solve this problem?

marsbake (20:22:12)
Would a table be of any help? How about using a simpler problem? Other strategies?

Whit (20:21:49)
I am sitting here with my T-chart, but am still having trouble.

marsbake (20:22:43)
Perhaps that's where we use a simpler problem (to show the value of the t chart.)

Cliff Matsuda (20:22:34)
see that the pattern is a multiple of 3 times the previous number

kevlarsen (20:22:27)
We make a table or list.

marsbake (20:23:05)
Great strategies.

marsbake (20:23:18)
Let's go for our final challenges of the evening.

marsbake (20:23:42)
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?

marsbake (20:25:36)
Do any of the strategies we've already used this evening help us with this one?

Whit (20:25:41)
I am using the vcounting number strategy from before...

smerd (20:25:44)
Gauss?

kevlarsen (20:26:22)
yes, this is sum of consecutive numbers

minranli (20:26:28)
we can calculate 1+2+3+..........+30 and then multiply that by 12 for each month. then we add 31to each of the months with 31 days.........

marsbake (20:27:18)
Are you sure that the grandmother starts over each month?

KingRoy_2 (20:26:42)
(1+2+3........+31)7+(1+2+3+4+...............+30)4+(1+2+3+4+...........+29)=V

RichKal-MOEMS (20:27:21)
and subtract 59 for February

marsbake (20:27:39)
Why?-

Cliff Matsuda (20:25:24)
364x182+365

kevlarsen (20:25:51)
365*366/2 = 66795 cents, or $667.95

Cliff Matsuda (20:26:10)
365-1)365/2

kevlarsen (20:27:51)
ah yes, good problem!

marsbake (20:28:31)
I'm going to let you wrestle a bit more on your own with this one since our time is about over.

marsbake (20:28:46)
I have one more that is awesome to leave you with.

marsbake (20:30:36)
As we near the end of our session, I would like to share a number pattern comic that was recently (9-10-06) in the newspaper. Did you see that Bill Amend is helping math teachers by posing problems to his characters (and the public)? Here's the problem:
A math teacher offers to assign one second of homework the first week of school, two seconds the second week, four seconds the third, and so on.?
If the amount of homework doubles every week, is this something you would agree to for the duration of the 36-week school year?

How would you answer? What would your students say? How would you present this to your class?
Have fun with number patterns. Yes, math (and number patterns) are all around us!

marsbake (20:31:05)
Thanks for being a great discussion participant. I hope you learned a bit more to help your students.

marsbake (20:31:23)
Thank you so much for contributing to tonight?s Math Jam. Please join us again next week as we explore factors, multiples, and primes.