Transcript for the Math Jam "MOEMS Teachers Math Jam" on Jan 31.

Math Jam hosted by marsbake (Marshalyn Baker ).

DPatrick (19:28:19)
Greetings and welcome to another MOEMS Math Jam.

DPatrick (19:28:26)
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 (19:28:38)
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:

DPatrick (19:28:44)
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php

DPatrick (19:28:52)
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 (19:29:05)
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 (19:29:14)
I'll now turn things over to your instructor for today, Marshalyn Baker.

marsbake (19:29:33)
Welcome to tonight's Math Jam. I am Marshalyn Baker, a middle school math teacher at Messalonskee Middle School in Oakland, Maine. I have been a PICO for over 20 years. I started out using the Olympiad with my fifth graders when I taught at the elementary level, and happily, a middle school Olympiad 'followed me' when I moved to the 7th and 8th grade math position. I now have 6 teams from my school engaged in the Olympiads. To me, the very best part of participating in the Olympiads is the discourse I have with my students when the contest is over. I love seeing a variety of strategies that they use to solve the problems. I especially like it when one of them shares something that I never would have thought of as a way to solve the problem.

marsbake (19:31:19)
Tonight we are going to follow a theme that is a very important part of students' number sense: divisibility.

marsbake (19:32:06)
As teachers, we must create classroom environments where students explore, look at number forms, question, and engage in sense making using numbers.

marsbake (19:32:37)
If we foster this exploration, students create their own strategies that allow them to reason about numbers.

marsbake (19:33:11)
As we solve some of the problems tonight, let's think about how we can guide our students in their pursuit of greater number sense.

marsbake (19:33:21)
Here's the first problem for you to try:

marsbake (19:33:55)
When Jenna opens her favorite book, the product of the page numbers on the pages facing her is 420. What is the smaller of the two page numbers facing Jenna?

marsbake (19:34:55)
Before I post the answers, I would like you to share strategies you think students might try.

marsbake (19:36:38)
Besides the strategy of guess and check, what other strategies might you share with your students?

robinhe (19:36:03)
Since the page numbers are consecutive, we must find two consecutive numbers that have a product of 420.

patrick (19:37:20)
Possibly Factor the number 420

marsbake (19:37:56)
What ""rules"" might students use that could help them do that?

RichKal-MOEMS (19:37:57)
Does the number 420 offer a clue?

robinhe (19:38:21)
One of them must be a multiple of 10

marsbake (19:38:52)
How do you know that one must be a multiple of 10?

8759 (19:39:18)
Not necessarily; they could end in 4 and 5.

marsbake (19:41:02)
Zak has an interesting statement, but I don't want to post all of it. He says to look at 20 * 20 because it's close to 400.

patrick (19:40:48)
The only way to get the number to end in zero is if you multiply a factor of two and a factor of five [img id=em-11]

marsbake (19:42:42)
Good, Patrick. So how can we relate that to what Zac has said?

RichKal-MOEMS (19:42:29)
You mean multiple of two and multiple of five?

marsbake (19:43:55)
Oops...my students confuse the two math terms: factor and multiple all the time. Thanks for catching that, Richard!

RichKal-MOEMS (19:44:27)
They're not alone!

marsbake (19:45:41)
So we have some divisibility rules regarding multiples of 10.

marsbake (19:46:07)
Now I'll show all of Zak's earlier comment:

RichKal-MOEMS (19:45:47)
So how do we get 420?

marsbake (19:47:40)
Somehow I lost Zak's comment. He said that since 20* 20 = 400, he'd try 20 * 21.

marsbake (19:48:08)
Did anyone try to find the prime factorization of 420?

gmbpico (19:49:00)
factors of 420 - 7,5,3,2,2 - so diveide these factors in groups and mutiply with each other and see the two consecutive numbers like 7*3 =21 and 5*2*2= 20

Zak! (19:49:09)
RIght, or perhaps in the other line we could factor 420 = 42 * 10, but 42 is just over 4 times greater and we need the factors to differ by 1. So dividing 42 by 2 and multiplying 10 by 2 gives 420 = 20 * 21.

marsbake (19:50:27)
Great number sense...dividing by 2 with one side and multiplying by 2 on the other side. Keeping that balance of the prime factors.

marsbake (19:51:11)
Before I move on with another problem, is there any other strategy that could work besides guess and check and finding the factors of 420?

RichKal-MOEMS (19:50:34)
Interesting approach. 42*10 = 21* 20. Use the associative law!

RichKal-MOEMS (19:52:20)
Zak! said to look at 20*20 first. That's a third way.

marsbake (19:53:19)
Somehow, I deleted another solution by Zak. We'll get it back.

marsbake (19:55:09)
Zak suggested trying an algebraic solution x times x+1. What do you think about that? Would middle schoolers be able to do this? How could we help them?

marsbake (19:56:03)
Can we go back to our vocabulary in the question: consecutive numbers?

Zak! (19:56:02)
I had deleted it but you could try n * (n+1) = 420 or n**2 + n - 420 = 0 and then factor that equation to get (n+21)*(n-20)=0 so that n = 20 or n = -21, the later of which isn't right. So n = 20 and n+1 = 21 is a solution. But this involves algebra which not all of the students have had.

marsbake (19:57:12)
Is there a way to begin to lead them to an algebraic solution, even if they haven't had algebra?

marsbake (19:58:10)
Perhaps try a simpler problem with consecutive numbers?

marsbake (19:58:26)
Great discussion and strategies. Let's try another problem.

RichKal-MOEMS (19:58:41)
n(n+1) = 420 just means the product of two CONSECUTIVE numbers is 420. Find them.

gmbpico (19:58:20)
This is first year as PICO. I have 22 kids in my team (after a few drop outs) ,50% are 4th graders other 5th. When I post a probelm like this to them, only a few ( maybe 4 out of 22) will be interested in solving the problem with me on the white board. The others don't seem to care. Any advise on how to keep everyones focus on the white board?

marsbake (19:59:36)
Before I send the next problem, would anyone like to respond to gmbpico?

marsbake (20:00:38)
You might try to have them in groups. Put those 4 that are usually engaged as a team leader. Have them help the others with the problems or make a challenge out of it.

marsbake (20:01:02)
My kids love competitions amongst tables.

marsbake (20:01:14)
New problem:The number 33,822 is divisible by each of 2, 3, 6, and 9. What is the next larger whole number also divisible by 2, 3, 6, and 9?

marsbake (20:02:05)
While you are working on this problem, Zak has a response to gmbpico:

Zak! (20:01:13)
I usually give the kids a problem to work on and then ask for a volunteer with the correct answer (or one close) to either come to the board or say in words how they did it.

marsbake (20:04:32)
What strategies might students try to solve this?

robinhe (20:02:28)
The LCM of those numbers is 18, so add 18 to 33822

Zak! (20:04:37)
First, we must add a number divisible by 2, 3, 6, and 9. The smallest such number is 18. So the next number is 33,840.

marsbake (20:06:45)
If students don't remember the LCM strategy, what direction could you steer them in? Could we break Zak's answer apart and start looking at every 2 numbers from 33822, then every 3rd number, etc?

Cynthia (20:05:37)
Then students could test that number using divisibility rules.

gmbpico (20:05:46)
thanks RichKal. I will call you. I have tried working in group method, calling them to the board method- by the end of the class I question myself if I have achived anything.

RichKal-MOEMS (20:07:41)
Never thought of that!

marsbake (20:09:14)
Here's a follow up: What is the smallest number larger than 1 that leaves a remainder of 1 when divided by 10 or 12 or 18?

marsbake (20:10:19)
Can you take some of the earlier dialogue and see possible strategies to use?

marsbake (20:11:32)
Can the LCM be used to answer this? How about divisibility rules?

RichKal-MOEMS (20:11:20)
If kids try to test multiples of 10 or 12 or 18, it's best to look at multiples of 18 and check 10 and 12 against it.

marsbake (20:12:28)
Students often mention that method when they share their strategies. It's an efficient one!

RichKal-MOEMS (20:12:59)
The shortest way: LCM (18,12) is 36. Then LCM (36,10) is 180 -- that's 36 x 5.

RichKal-MOEMS (20:13:33)
Oh. And then add one.

Cynthia (20:13:48)
I tried used the idea of prime factorization to come up with the LCM of 180 and then added 1

marsbake (20:14:38)
Could you explain that further, Cynthia?

marsbake (20:16:18)
Have any of you used a Venn diagram to do this?

marsbake (20:17:52)
You can put the factors of one number in one circle, the factors of the other number in the other circle, and in the intersected circle, put the factors they share. That is your GCF in the intersected circle and all the numbers in all circles in the Venn are your LCM.

marsbake (20:18:03)
Here's problem number 3:

marsbake (20:18:07)
The five-digit number 83,9A2 is divisible by 12. What digit does A represent?

marsbake (20:19:49)
Cynthia sent me this response:

Cynthia (20:18:02)
I was looking for the prime factors that would be needed for the LCM for 10, 12 and 18. That would be 2*2*3*3*5. A

marsbake (20:20:12)
Thanks, Cynthia.

marsbake (20:20:35)
Any strategies for problem number 3?

Cynthia (20:21:01)
My math teachers in junior high taught so many concepts using prime factorization. It is a strategy that really stayed with me :)

marsbake (20:21:56)
Golden apples for them. Do you have any special memories that you'd like to share?

Zak! (20:20:19)
Divisible by 12 means divisible by 3 and divisible by 4. The divisibility by 3 means the sum of the digits must be divisible by 3 or 22 + A must be divisible by 3. So A = 2, or 5, or 8. Divisibility by 4 means the last 2 digits must be divisibly by 4 so the only one is 52, so that A = 5.

marsbake (20:23:59)
When I go over divisibility ?tests,? I ask students if they see connections. Why is the test for 6 connected to tests for 2 and 3? Why are the tests for 4 and 8 connected to 2? What do 2, 4, and 8 have in common? Would this pattern continue with a divisibility rule for 16?

RichKal-MOEMS (20:23:58)
Zak!, why did you choose 3 and 4 instead of 6 and 2, or ... ?

Cynthia (20:24:24)
Sorry, I think I accidentally hit LEAVE instead of SEND

Zak! (20:25:19)
Because I know the divisibility rule for 3 and 4, but I don't know it for 6.

Zak! (20:25:42)
Actually I do, it is just 3 AND 2.

marsbake (20:27:04)
Have you ever done the Sieve of Eratosthenes? This allows students to look for patterns with multiples of numbers. They can use this to see that all even numbers end in 0, 2, 4, 6, or 8. They can see that all multiples of 4 end in 0, 4, 8, etc. If you carry it out to over 100, they see that it?s 104, 108, 112, etc. thus a test being developed for any power of two.

RichKal-MOEMS (20:26:53)
So why 3 and 4?

Zak! (20:28:30)
Oh, because 3 * 4 = 12 and if a number is divisibly by 12, then it must also be divisible by 3 and 4.

RichKal-MOEMS (20:29:08)
I always saw the test for 4 as checking the emainder after division by 20.

marsbake (20:29:51)
Can you explain that further, Rich?

Zak! (20:29:52)
Yes that is probably better, Thanks.

RichKal-MOEMS (20:30:03)
So 235472 is the same as just 72 which goes to 12. So it's divisible by 4.

marsbake (20:30:53)
As we are getting close to the end of the session, I have one last challenging problem for you:

marsbake (20:31:10)
A4273B is a six digit number in which A and B are digits. If the number is divisible by 72 without a remainder, what values do A and B each have?

marsbake (20:31:57)
What divisibility tests would students use to help them solve this problem?

RichKal-MOEMS (20:34:24)
72 = 2x2x2x3x3 = 8x9, so use tests for 8 and 9.

Cynthia (20:34:26)
divisibility by 8 is really important to get the last digit

Cynthia (20:35:16)
and then I would use the div. rule for 9 to get A

marsbake (20:37:26)
Anyone have a solution?

RichKal-MOEMS (20:37:31)
8 divides 1000, so it divides all multiples of 1000. Then by the distributive law, it also divides the last 3 digits. So... 73B is a multiple of 8

Zak! (20:37:42)
As with the 20 test, we could check the remainder when divided by 200, or 13B. Using the ""20"" test we need check 1B. so B = 6.

Cynthia (20:37:59)
A = 5 ; B = 6

marsbake (20:38:50)
I would like to thank all of you for your participation. I have a great ending comment to share from Cynthia:

Cynthia (20:40:06)
marsbake - this is my first session but it was such fun and very informative! Thank you. I know there is only one more class. Do you know if this series will be offered next year?

marsbake (20:41:17)
Cynthia appreciated that there are several ways to solve Olympiad problems and there is something for all kids to try.

RichKal-MOEMS (20:40:27)
A very nice job, Marshalyn! Congratulations and Thanks.

Cynthia (20:41:10)
did you mean about how some kids have such great solutions?

marsbake (20:42:23)
That's so true. I learn as much from my kids as they do from me. It is wonderful to have them share a strategy I never thought of.

marsbake (20:43:03)
I wish you all the best in solving those great Olympiad problems. Happy problem solving and thanks for allowing me to moderate this session. You were wonderful participants.

Cynthia (20:42:09)
I am always amazed at how certain kids solve these really difficult problems

RichKal-MOEMS (20:41:56)
Very likely, starting in September for more sessions.

RichKal-MOEMS (20:43:08)
I've seen kids come up with as many as 17 ways to solve a problem. Mindblowing!

Cynthia (20:43:19)
Thank you!!