Transcript for the Math Jam "MOEMS Teachers Math Jam" on Oct 11.

Math Jam hosted by marsbake (Marshalyn Baker ).

DPatrick (19:28:13)
Greetings and welcome to another MOEMS Teachers Math Jam.

DPatrick (19:28:21)
For those of you who are students and 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 (19:28:26)
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php

DPatrick (19:28:34)
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 (19:28:45)
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 (19:28:57)
I'll now turn things over to your instructor for today, Marshalyn Baker.

marsbake (19:29:07)
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.
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 (19:29:54)
We're going to build on our ideas starting with this problem.

marsbake (19:30:21)
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?

vishalarul (19:31:04)
6=1+5, 10=5+5, 28=5+23, 96=7+89

marsbake (19:32:17)
Is there more than one way to do any of these?

marsbake (19:32:40)
Do you agree with the answers given so far?

frostmourne (19:32:05)
one isnt a prime

Rocket95 (19:32:10)
6-2,3
10-3,7
28-11,17
96-7,89

JordanV (19:32:31)
28 also= 17 +11

saktamsase (19:32:59)
no, is one a prime number?

marsbake (19:33:55)
A few people have responded that one isn't prime. Many students think that it is. How do you respond?

Zenik (19:33:20)
6=3+3

Zenik (19:34:12)
it is

saktamsase (19:34:24)
1 is neither composite or prime

Rocket95 (19:34:08)
it isn't

marsbake (19:35:02)
Why isn't one prime?

Rocket95 (19:35:00)
sorry, my first one is wrong
2+3=5 not 6

frostmourne (19:35:30)
prime needs two factors

marsbake (19:35:51)
Can you explain more about the two factors that a prime needs?

vishalarul (19:36:00)
Distinct factors?

Rocket95 (19:36:06)
being a prime means that it is divisible by 1 and itself. 1 is only divisible by 1 or itself not and.

Rocket95 (19:36:40)
only two distinct factors

marsbake (19:37:50)
This is a definition I use: A prime number is a counting number that has exactly two distinct factors, 1 and the number itself. One is a unique number since it has exactly one factor. Thus, by definition, one is NOT a prime. Two is the only even prime number.

marsbake (19:38:40)
Let's explore another piece of this problem:What 3 prime numbers have a sum of 9' 21' 45'

JordanV (19:39:28)
9= 3+3+3 or 5+2+2

frostmourne (19:39:10)
3+3+3, 7+7+7, 3+11+31

marsbake (19:40:21)
Is there a way for students (or us) to be sure we have all of the different possibilities?

JordanV (19:40:24)
21 also= 17+2+2

Rocket95 (19:41:04)
list them out as you find them?

JordanV (19:41:19)
try everything possible combiniation

marsbake (19:41:37)
How can you be sure you get them all?

vishalarul (19:41:30)
Try an organized method.

marsbake (19:41:56)
Can you explain what that might look like?

Rocket95 (19:41:54)
a list?

vishalarul (19:39:04)
9=3+3+3, 21=7+7+7, 45=3+5+37

iamback (19:42:32)
prime factorization trees

JordanV (19:44:00)
try every possible combination of prime numbers less than the sum

RichKal-MOEMS (19:43:52)
21= 7+7+7 = 11+7+3 = 11+5+5 = 13+5+3 = 17+2+2. I'm afraid to list the possibilities for 45.

saktamsase (19:44:08)
23+11+11=45

marsbake (19:45:03)
You might also try the smallest prime with as many different combinations as you can, then go to the next, etc.

Rocket95 (19:45:19)
how about all the primes up to and including the squareroot of a number?

marsbake (19:45:59)
Great job with all the different combinations!

marsbake (19:46:35)
For those of you who need the answer to 45, here it is:45 = 2 + 2 + 41, 3 + 5 + 37, 3 + 11 + 31, 3 + 13 + 29, 3 + 19 + 23, 5 + 11 + 29, 5 + 17 + 23, 7 + 7 + 31, 7 + 19 + 19, 11 + 11 + 23, 11 + 17 + 17, 13 + 13 + 19

marsbake (19:47:04)
Our next problem focuses on factors.

marsbake (19:47:22)
Find the largest factor of 2520 that is not divisible by 6.

vishalarul (19:48:21)
280?

frostmourne (19:48:46)
315

marsbake (19:49:13)
In addition to giving an answer, can you tell us how you arrived at it?

iamback (19:49:35)
i agree with 315 by using a prime factor tree

vishalarul (19:49:48)
Take the prime factorization and just make sure that the result doesn't contain a multiple of 2 and 3.

marsbake (19:50:16)
That's an interesting strategy...can you tell us why that would work?

frostmourne (19:50:52)
everything that is divisible by 2 and 3 are divisible by 6

vishalarul (19:51:16)
If there are no multiples of 2 and 3, you can be sure that the result would not contain any multiples of 6. For example, the correct answer is 3^2*5*7.

marsbake (19:52:26)
Another strategy would involve dividing out as many 2's as possible from 2520 or as many 3's as possible. Then, 2520 divided by 2 = 1260; 1260 divided by 2 = 630; 630 divided by 2 = 315; and 2520 divided by 3 = 840 and 840 divided by 3 = 280. Thus, the largest factor not divisible by 6 is 315.

iamback (19:52:43)
is 3^2*5*7 a factor of 2520? how so?

vishalarul (19:53:21)
Since the prime factorization is 2^3*3^2*5*7.

LDodd (19:53:42)
(1,2520), (2,1260), (3,840), (4, 630), (5,540), (6,420), (7,360), (8,315)

LDodd (19:54:13)
I believe it is 315 the method I used was plug and chug.

RichKal-MOEMS (19:54:23)
Eric, I would -- and did.

Rocket95 (19:54:18)
why do you choose to get rid of 2^3?

vishalarul (19:54:57)
Choosing to get rid of the 3^2 would get a smaller answer. So get rid of the 2^3.

Rocket95 (19:55:28)
what is plug and chug?

marsbake (19:56:24)
Great questions and sharing. Here's a multiples question: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?

LDodd (19:56:38)
Plug and chug: I started listing the pairs and then dividing the larger one by six. I stopped when I got to one that was not divisible. I plugged a number in and chugged along.

marsbake (19:57:26)
I like your strategy name. Students would remember that one!

marsbake (19:58:05)
Again, be sure to explain how you've arrived at your answers. And can you find the answer in another way?

marsbake (19:59:57)
I have had a couple of algebra responses. Before sharing those, try to find a way that elementary students could solve this.

frostmourne (19:57:29)
8 8-people tables and 3 5-people tables

marsbake (20:00:22)
How did you get that?

marsbake (20:01:39)
Can anyone verify that frostmourne's answer is correct? And how would you get that?

frostmourne (20:01:36)
plug and chug

LDodd (20:01:42)
79= 45+24 so 9 5 people tables and 3 8 people tables. I started by looking at combinations that equaled 79 that were already multiples of 5. I threw out the ones ending in 0 because there weren't any values that would add with 8

marsbake (20:02:32)
Can you explain more about throwing out the ones ending in 0 by giving an example?

Rocket95 (20:02:02)
8*8=64
5*3=15
64+15=79

marsbake (20:03:35)
How did you keep track of the combinations?

LDodd (20:03:59)
79=20 + 59 No multiple of 8 ends in an odd number so I can't use values that end in an odd number.

vishalarul (20:04:08)
Make an organized list and keep subtracting multiples of 8.

Rocket95 (20:03:51)
are you talking to me?

eobrien19 (20:04:04)
Looking at the problem in terms of modular arithmetic is a very interesting method for solving this problem.

marsbake (20:05:26)
Can you explain more about modular arithmetic for those who aren't familiar with that? What level of student would be able to do that?

JordanV (20:05:32)
Consider multiples of 8 that whose last digit is 4 or 9 only. All others won't work since multiples of 5 end in 0 or 5.

marsbake (20:06:44)
There was a graphing solution...would that person send back your strategy. I accidently lost it.

eobrien19 (20:05:15)
By subtracting groups of 5 from the 79, we get answers ending in 4's and 9's. We work until we get a number ending in 4 that is a multiple of 8.

vishalarul (20:07:18)
I said that graph 5x+8y=79 and x+y<=12. Find the ordered pair that works.

marsbake (20:07:56)
That's an interesting strategy. What level of students might be able to do that?

marsbake (20:08:41)
You could also do multiples of 5 in one column, multiples of 8 in another column and look for the two to add up to 79. See what number of tables that yields

marsbake (20:09:36)
Let's look at another problem.

marsbake (20:10:00)
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?

marsbake (20:11:07)
One more thought on the previous problem:

eobrien19 (20:10:30)
A more simplified strategy for graphing . . . the x-axis refers to Tables of 8. 0 tables of 8 means 11 tables of 5 or 55 diners. 11 tables of 8 means 0 tables of 5 or 88 diners. Look for where the line crosses 79, and you will see the number of tables of 8 fit the criteria on the graph.

marsbake (20:12:45)
I'm getting a few numerical answers, but no explanations...can you send your strategies?

vishalarul (20:10:47)
7

Rocket95 (20:12:51)
35/3=11 R2
9 then 7.
It is 7

JordanV (20:13:10)
9 because it keeps rotating the last digit between 9 7 1 3

vishalarul (20:13:37)
Notice the pattern 3^1 ends in 3, 3^2 ends in 9, 3^3 ends in 7, and 3^4 ends in 1. Continuing the pattern, we can find that 3^35 ends in 7.

frostmourne (20:13:00)
i think its seven, because there is a pattern of 3,9,7,1,3,9,7,1, and so on, so after 32 of them, then it would be a one, so the 33rd is a 3, the 34th a 9, and the 35th a 7

marsbake (20:15:28)
You are certainly looking for patterns, and finding an organized way of solving the problem. Great job!

eobrien19 (20:15:14)
Since there are four numbers that repeat in a cycle, are we looking for the remainder after we divide 35 by four? Very cool!

RichKal-MOEMS (20:15:25)
In a MULTIPLICATION string, the key is the 1, since the next number times 1 will start the cycle over.

marsbake (20:17:03)
Since we have only about 10 minutes left, let's solve another problem: 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?

JordanV (20:18:44)
168= 2*2*2*3*7

marsbake (20:19:23)
Why does this strategy work?

saktamsase (20:20:07)
find the least common multiple of 6,7 and 8

marsbake (20:21:04)
What if a student doesn't know about LCM's? Would there be another way to find the answer?

eobrien19 (20:18:11)
We could list out the multiples of 6 and of 7 and of 8 until we have a winner.

Rocket95 (20:21:55)
it might take a while though

eobrien19 (20:22:03)
6 x 7 = 42; we could list out the multiples of 42 until we get a number that is divisible by 8.

marsbake (20:22:55)
Why would we use multiples of 42?

eobrien19 (20:23:34)
6 x 7 is part of our answer, since we are looking for groups of 6 and of 7.

marsbake (20:24:19)
Out time is about done. I'm going to leave you with a problem to explore on your own.

marsbake (20:24:51)
At a new middle school, there are exactly 1000 students and 1000 lockers. The lockers are numbered in order from 1 to 1000. On April Fool's Day the students played the following prank. The first student to enter the building opened every locker. The second student closed every locker that had an even number. The third student changed every third locker, closing those that were open and opening those that were closed. The fourth student changed every fourth locker, opening those that were closed and closing those that were open, and so on. After all 1000 students passed through the locker room, which lockers were open?

marsbake (20:25:48)
Let me suggest you try "acting this out" by putting the locker numbers on cards. I've also had my students play the lockers. You'll be surprised at what you get!

marsbake (20:26:42)
I'll also suggest you try a simpler problem if you are stuck!

marsbake (20:28:28)
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.