Transcript for the Math Jam "MOEMS Teachers Math Jam" on Jan 30.

Math Jam hosted by marsbake (Marshalyn Baker ).

rrusczyk (19:29:53)
Greetings and welcome to a MOEMS Teachers Math Jam!

rrusczyk (19:29:58)
My name is Richard Rusczyk, and I work here at Art of Problem Solving.

rrusczyk (19:30:05)
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 (19:30:08)
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 (19:30:11)
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php

rrusczyk (19:30:18)
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 (19:30:23)
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 (19:30:26)
Now, on with the show!

rrusczyk (19:30:31)
I'll now turn things over to your instructor for today, Marshalyn Baker.

marsbake (19:30:41)
Hello! My name is Marshalyn Baker. I?ll be your facilitator this evening for our second Math Jam. As a PICO (Person in Charge of the Olympiads) for over 20 years, I am still excited and enthused for each contest and the discourse it affords me in my middle school mathematics classroom.

marsbake (19:30:59)
Tonight our focus is Factors, Multiples, and Primes. As you share your strategies to solve problems, look for multiple representations and connections to other mathematics topics. Have fun and welcome to our virtual problem solving community.

marsbake (19:31:26)
Mathematician Christian Goldbach (1690-1764) made 2 conjectures (educated guesses) about whole numbers that no one has been able to prove. The first stated that every even number greater than 2 can be written as the sum of 2 prime numbers. The second conjecture is that every odd number greater than 5 can be written as the sum of 3 prime numbers.
What 2 prime numbers have a sum of 6? 10? 28? 96?

marsbake (19:32:05)
Be sure to share any strategies you use in addition to giving your answers.

marsbake (19:32:39)
Do your students have an misconceptions about this topic?

infinity4ever (19:32:00)
3 and 3, 5 and 5, 7 and 3, 43 and 43

marsbake (19:33:08)
Are there other ways to do these?

AlexaK (19:32:27)
1.)5,1

infinity4ever (19:32:29)
I just mentally found primes

minranli (19:33:19)
6=5+1, 10=3+7,28= 23+5, 96=93+3

daveshail (19:33:13)
3 and3, 5and5, 11 and 17,43 and 43

daveshail (19:33:20)
yes

daveshail (19:33:33)
11 and 17 for 28

marsbake (19:33:58)
Do you see any misconceptions yet?

infinity4ever (19:34:18)
what are misconceptions?

marsbake (19:34:41)
MIsconceptions are ideas students might have that are mathematically incorrect.

daveshail (19:34:21)
93 is not prime

marsbake (19:35:03)
How do you know that 93 isn't prime?

AlexaK (19:34:55)
yes, 93 is a composite number

marsbake (19:35:40)
As we also try to focus on the mathematical language, what would you expect your students to say to define prime and composite?

infinity4ever (19:35:09)
it's divisible by three

infinity4ever (19:35:58)
prime: a number with only 2 factors; one and itself

marsbake (19:36:42)
What do many students think a prime is? Based on the definition by infinity4ever, what could you say to those students?

infinity4ever (19:36:09)
composite: a number with more than 2 factors

minranli (19:36:47)
composite is a number that can be divided by itself and 2 other numbers

marsbake (19:37:07)
Just 2?

AlexaK (19:36:51)
prime is only divisible by one and itself and composite is divisible by one, itself and at least one other number

Carol_Denton (19:36:59)
A prime number is divisible by only one and itself. Composite numbers are ""complicated"" and have more factors than itself and one.

minranli (19:37:35)
itself and 1

marsbake (19:38:23)
I find many students come to me thinking 1 is a prime...one of the misconceptions I was speaking of. What do you say to that student?

AlexaK (19:37:56)
no it can be more than 2

infinity4ever (19:38:43)
1 has only one factor

minranli (19:38:47)
1 is not prime

Cliff Matsuda (19:39:15)
1 is not prime bcuz it is not divisible by any # other than 1

AlexaK (19:39:01)
noe is is pnly divisible by 1 while prime numbers need 2 factors

AlexaK (19:39:25)
i meant one

marsbake (19:40:02)
What strategies would students use to solve this problem?

infinity4ever (19:40:15)
mental math! =D

Carol_Denton (19:40:15)
0 and 1 are special numbers. Think of the prime factor tree. The smallest number, or ""root"" of the tree is a 2. Two is the smallest prime number. Think of our definition of a prime number.

marsbake (19:41:23)
Let's get back to just 96. What are all the ways we can represent it as the sum of two primes? And tell me your strategy. We have mental math. What else?

math maniak (19:42:05)
I think many students would start by listing primes to have a pool of numbers to manipulate

marsbake (19:42:53)
That's a great strategy. If you have a list of the primes, it does become easier to manipulate the numbers!

infinity4ever (19:42:37)
another strategy could be to start from the smallest prime: 2, and see if the other addend is prime

minranli (19:42:54)
i guess the long way would be to find all the prime numbers and see which ones would sum 96

marsbake (19:43:43)
So I have seen: mental math, list...if it's organized in some way like starting with smaller primes, that would help.

Carol_Denton (19:43:52)
Yes, listing numbers in descending order on one side of a line and then corresponding numbers on the other side would work if they cross off non-primes.

RichKal-MOEMS (19:43:57)
Interesting that 6 and 28 are perfect numbers.

AlexaK (19:44:05)
you could make a math tree

marsbake (19:44:18)
Great job. Let's try another problem with a different focus.

marsbake (19:44:59)
Before I get the next problem, what else could you say about the perfect numbers?

Carol_Denton (19:44:24)
Is the sieve of Erasthenes helpful here?

AlexaK (19:45:18)
what are perfect numbers

minranli (19:45:44)
what's the sieve of Erathenes?

marsbake (19:46:51)
Perfect numbers have factors that add up to the number (don't include the number).

RichKal-MOEMS (19:46:35)
A perfect number is one which is the sum of ALL its factors, excepting itself. ^ is perfect because 1+2+3 = 6

marsbake (19:48:03)
The sieve of Erathostenes (I don't think I spelled it right) is a way to sift out numbers. Do a 100 chart, 10 by 10 and start crossing out multiples of two in one color, multiples of 3 in another color, and keep doing that with primes. Circle any primes left.

Cinderella (19:47:44)
The Sieve of Erathenes helps you to identify all the prime numbers. You eliminate the number 1 and the other composite numbers.

RichKal-MOEMS (19:47:13)
I mean 6, not ^

AlexaK (19:48:07)
cool

marsbake (19:48:32)
Here we go for the next problem...really this time!

marsbake (19:48:51)
Find the largest factor of 2520 that is not divisible by 6.

Cliff Matsuda (19:48:24)
how is 28 perfect?

marsbake (19:49:43)
28 = 1 + 2 + 4 + 7 + 14

infinity4ever (19:49:27)
21

marsbake (19:50:19)
How did you get 21? What strategy (ies) did you use? How are you sure you have the right answer?

marsbake (19:50:35)
How would your students do this problem?

AlexaK (19:50:25)
63

minranli (19:50:33)
what is an easy way to solve this problem?

AlexaK (19:50:46)
sorry, i meant 40

infinity4ever (19:50:56)
i made a factor tree, and i got 20, 6 and 21 as factors. since 20X21 is divisible by 6, i just thought of 21

infinity4ever (19:51:26)
i guess that doesn't work too well

marsbake (19:51:45)
What do you mean by that?

AlexaK (19:52:02)
yes i did mean 63

infinity4ever (19:52:03)
i have a better way: prime factorize it

marsbake (19:53:18)
If you are doing a factor tree, look back at the original question. We are looking for the largest factor that is not divisible by 6. How could we do the factor tree to find this answer?

infinity4ever (19:53:02)
2520= 2^3 X 3^2 X 5 X 7

marsbake (19:54:17)
Okay... it looks like infinity got 3 factors of 2, two factors of 3, a factor of 5 and one of 7. Now how can we use this to help us find what we are looking for?

infinity4ever (19:54:13)
you take the largest combination of those factors and make sure you don't have a 2 AND a 3, so 5 X 7 X 9 = 315

math maniak (19:54:45)
By manipulating the prime factor to get the largest number not divisible by 6- but I only got up to 35 are you using calculators?!

Cinderella (19:55:12)
That's what I got. I used the factor tree.

infinity4ever (19:55:17)
no!

RichKal-MOEMS (19:55:18)
To avoid divisibility by 6, keep the powers of 2 in one factor and the powers of 3 in the other. Then arrange the others to meet the condition of largest.

Cinderella (19:55:38)
I got 315 by dividing by 2 three times.

marsbake (19:56:11)
That's an interesting strategy...get all of the 2's out or all of the 3's out and look to see what is left.

marsbake (19:57:16)
If I take all 2's out, I am left with the 3's, 5, and 7 or if I take all the 3's out, I'm left with 2's, 5 and 7. Which is greater?

Cinderella (19:56:28)
That's what I did!

infinity4ever (19:57:26)
the first choice

AlexaK (19:56:06)
I am confused.

marsbake (19:58:18)
Can anyone explain the last strategy that Cinderella used in another way?

Cliff Matsuda (19:57:47)
3x3>2x2x2 so 9x5x7=315

marsbake (19:58:49)
Great illustration of this!

Cinderella (19:58:41)
I divided by 2 until I got 315. Then divided that into 2520 and got 8. I realized that 315 was not evenly divisible by 6.

minranli (19:58:50)
I am confused too

marsbake (19:59:38)
Cliff is saying that 9 is bigger than 8, so use 3 * 3 with 5 and 7.

marsbake (20:00:12)
Remember, you can't have a factor of 6, so no combination of any 3 and 2 together can be used since 3 * 2 = 6.

Carol_Denton (19:59:54)
It is the age-old 6th grade stumper: is 2 to the 3rd power greater than or less than3 to the 2nd power? 2x2x2=8 and 3x3=9, so 9x35(the product of 5x7) is larger than 6x35.

math maniak (19:59:59)
I did the prime factorization, only I did not notice after taking 2 2's out I was left with 315. Nice illustration.

Cliff Matsuda (20:00:14)
since you cannot use 2 and 3 together

marsbake (20:00:43)
Great job...we're now ready for a more challenging problem.

marsbake (20:01:02)
There are two sizes of tables in a banquet hall. One size seats exactly 5 people; the other size seats exactly 8 people. At tonight?s banquet, exactly 79 people will be seated at less than one dozen tables and there will be no empty places. How many tables of each size will there be?

marsbake (20:01:52)
One more strategy for the previous problem too good to leave behind.

RichKal-MOEMS (20:00:33)
2520 ?3?3 = 280, which < 315

marsbake (20:02:37)
Remember to share your strategies for solving the problem.

Cinderella (20:03:26)
I would take 8x8=64 subtract from 79 get 15. So 3 tables of 5 and 8 tables of 8.l

marsbake (20:03:51)
Why did you do it like that?

Cliff Matsuda (20:03:59)
5 can only end with 5 or 0 so the 8 table must end in 4 or 9; 8 cannot end in 9 so it must be 4 and 3x8 =24

minranli (20:04:24)
8 is larger, so i would try and use it the most

infinity4ever (20:04:27)
no multiples of 8 end in 9, so there must be an odd number of tables that seat 5.

Cinderella (20:04:40)
Figured out the multiples of 8 until I found a number that would subtract to give me a multiple of 5.

Carol_Denton (20:04:27)
11 5 seater tables and 3 8 seater tables. I took 79 and divided by 5 until I had a remainder of a product of 8.

AlexaK (20:05:15)
that is the same thing i did.

marsbake (20:05:53)
So let's look at the different ways that people solved this problem.

marsbake (20:06:36)
I see people using their divisibility knowledge. There are people that looked at how numbers ended. They tried larger numbers. What else?

math maniak (20:05:49)
Did anyone solve it algebraically?

marsbake (20:07:26)
Even though someone might want an algebraic solution, let's hold that for near the end of the strategies. We want to find a variety of ways that any learner can do this.

RichKal-MOEMS (20:06:15)
The maximum is 11 tables. Suppose each has 8 people. Thus: 88 people. But 79 is 9 less than 88. Replace some 8s with 5s, and each of those tables has 3 people fewer. To subtract 9 from 88, change 3 tables. Thus: 3 tables seat 5 and the remaining 8 tables seat 8.

infinity4ever (20:07:12)
i tried, but since it's LESS than a dozen, not equal to a dozen, i got confused and messed up

Cliff Matsuda (20:07:43)
started doing it by algebra but had 2 variables so quit

marsbake (20:08:51)
How could you use a table to solve this problem?

KingRoy (20:08:33)
I did what Rich did.

Carol_Denton (20:09:33)
You can always use repeated subtraction or repeated addition which is very tedious but preferred by some students. Once comfortable with multiplication and division, shorthand for these, it is much quicker.

marsbake (20:10:07)
That's a great strategy...it lets any level of learner into the problem and able to solve it.

infinity4ever (20:09:39)
make a table with 2 columns, and try 1 table for the 8-people column. if that doesn't work keep trying, with 2 tables, then 3 tables, and so forth

Cinderella (20:10:23)
You could list the multiples of 5 and the multiples of 8 and then figure out when they add up to 79

Carol_Denton (20:10:28)
5xA + 8xB = 79

marsbake (20:11:20)
Even though this has two variables, students could substitute different numbers into the equation and see what works.

minranli (20:10:44)
on one side, we could put multiples of 5 and multiples of 8 on the other. then we put one 5 and ten 8 and so on until you add up to 79

Carol_Denton (20:11:35)
Take the above chart heading and sub in #s and say if it works or not.

Cliff Matsuda (20:10:44)
I forgot that needed to be <12 tables; so if 8 needs to end in 4 than next multiple would be 8x8=64 and remainder would be 15 in 3 tables of 5

marsbake (20:12:15)
Here's another algebraic form to play with iffff your students are at that level.

infinity4ever (20:08:37)
x+y<12 and 5x+8y=79

marsbake (20:12:38)
Great job. What about any misconceptions with this problem?

AlexaK (20:13:09)
people forgot it had to be less than 12 tables

math maniak (20:13:41)
You cant solve simultaneously with and inequality.--can you?

minranli (20:13:47)
how do we know 5 goes with x and 8 goes with y?

Carol_Denton (20:13:52)
Less than is a key. It is a multi-step problem and kids often stop before they finish or else they don't sub their answer back into the problem to see if it checks out.

marsbake (20:14:38)
That's so true. Many students don't try their answers back in the problem to see if they check out. Do any of your students confuse factors and multiples? How do you address this?

infinity4ever (20:14:06)
x+y<12, so 5x+5y<60 5x+8y=79, so 3y must be less than 29, the maximum y could be could be 8

infinity4ever (20:14:58)
yay! [img id=em-10]

marsbake (20:15:29)
We all can identify with that, can't we?

marsbake (20:15:51)
Let's move to a problem that talks about multiplication strings.

marsbake (20:16:10)
3 * 3, 3 * 3 * 3, and 3 * 3 * 3 * 3 are ?multiplication strings of two 3?s, three 3?s, and four 3?s respectively. When each string multiplication is done, 3 * 3 ends in a 9, 3 * 3 * 3 ends in 7, and 3 * 3 * 3 * 3 ends in 1. In what digit will a multiplication string of thirty-five 3?s end?

Carol_Denton (20:17:04)
Factor Tree--Factory--Break it down! is a chant I repeat until they do it automatically. It only makes since because I tell them a story about Henry Ford making an assembly line for individual car parts rather than the entire car before finishing. I chant Multiple--Mulitiply--get Bigger! to remind them multiples are the big #

marsbake (20:17:53)
Neat analogy! I like that!

infinity4ever (20:17:18)
3 ends in 3, 9 ends in 9, 27 ends in 7, 81 ends in 1 so 35/4 = 8 r. 3, so the answer is the 3rd in the pattern, which is 7

daveshail (20:18:56)
the last diguits has a pattern

Cliff Matsuda (20:19:09)
3Y<19 I think?

marsbake (20:19:38)
What do you mean by that?

daveshail (20:19:43)
9,7,1,3 9,7,1,3 and so on

marsbake (20:20:07)
How can you now use this pattern?

minranli (20:20:05)
if the number of 3s is divisible by 2, then it ends in 9, if divisible by 3, it ends in 7, divisible by 4 is one and the cycle starts all over again

AlexaK (20:20:28)
you count until you get 35

daveshail (20:20:28)
35th will have end digit as 9

marsbake (20:21:37)
The cycle idea is a good one...then look at minranli's additional pattern.

KingRoy (20:21:49)
35th answer will have an ending digit of 7

minranli (20:22:19)
so it is 9, 7, 1, 3, and since there's 4 numbers, 4 goes into 37 nine times with 1 left, so the cycle starts all over again. the next one would end in 9

math maniak (20:22:15)
Following infinity4ever's logic the 3rd number, for the remainder, in the pattern is a 1 not a 7. (9,7,1,3)

AlexaK (20:23:19)
why 37?

daveshail (20:23:10)
sorry

infinity4ever (20:23:27)
don't you mean 3,9,7,1 starting from a string of 1 three?

KingRoy (20:23:39)
It starts at 3 which is 3 so the beginning number is 3 not 9

daveshail (20:23:55)
yes foolowing the pattern the 35th wiil have 7 at the end

Cliff Matsuda (20:23:58)
pattern of 4 so 35 divide by 4 =8R3

marsbake (20:24:38)
Great discussion to observe the cycles and patterns and arrive at the answer of 7!

marsbake (20:24:56)
Time is really flying by, and we have time for one more problem.

marsbake (20:25:15)
At a Fourth of July parade, the local scout troop found that they could arrange themselves in rows of exactly 6, exactly 7, or exactly 8 and no one would be left over. What is the least number of scouts in the troop?

infinity4ever (20:25:50)
168 people... the LCM of 6,7, and 8 is 168

marsbake (20:26:17)
Why do you use the least common multiple?

minranli (20:26:47)
well 6 multipled by 7 is not divisible by 8 so 6*7*8 would probaly work

infinity4ever (20:26:51)
the question asks for the least number of scouts, so we use the least common multiple that is divisble by each 6,7, and 8

marsbake (20:27:14)
Is there another way to do this problem?

marsbake (20:27:40)
Could a student not familiar with LCM approach the problem?

marsbake (20:28:12)
Before we leave for the evening, could you let us know if you are a PICO (Person in charge of the Math Olympiads)?

Carol_Denton (20:27:49)
48 because the LCM of 6 7 and 8 is 48. I found the LCM of 6 and 7 and then started at the next closest multiple of 8

KingRoy (20:28:14)
6*7=42 42*4=168

infinity4ever (20:28:18)
they could make a list of multiples of 6, of 7, and of 8 and find the first number in all three lists

KingRoy (20:28:25)
iam

Carol_Denton (20:29:02)
I am also a PICO

sbmoemsp (20:28:57)
I am PICO

RichKal-MOEMS (20:29:04)
I'd start lists with the 8, not the 6, to minimize the arithmetic.

math maniak (20:29:32)
I'm a PICO

Cliff Matsuda (20:29:52)
I am PICO

marsbake (20:30:14)
What wonderful participation! Thank you for your contributions to our virtual classroom discourse. It is so exciting to come together as a community of learners to share our problem solving strategies. I hope you have some new ideas to take back to your students. Be sure to join Eric O?Brien and Jim Matthews in the upcoming weeks as they continue this mathematical journey with you. Have a great night.
Mathly,
Marshalyn