Transcript for the Math Jam "MOEMS Teachers Math Jam" on Sep 27.

Math Jam hosted by marsbake (Marshalyn Baker ).

marsbake (19:23:38)
Hello to all,

marsbake (19:24:21)
I'm Marshalyn Baker and I will be your moderator for this evening's hour session. I will post problems and ask for your solutions and the strategies you use to solve the problems.

marsbake (19:24:32)
Here's the first problem:

marsbake (19:25:32)
If 3 smiles = 10 grins and 6 grins = 9 laughs, how many laughs equal 2 smiles?

marsbake (19:27:48)
Welcome to those who have joined me. Try the smiles problem and I'll share your solutions and strategies.

pamgoselin (19:27:21)
10 laughs...I set up it up as a rate problem since that is what we are working on in one of my classes right now. I want to look at this from a student's perspective. I first checked my units to make sure my problem was set up right then I did the calculations by reducing.

marsbake (19:30:48)
Can you take others through this with the steps that you used?

jasnik (19:28:13)
2smiles=10 laughs because 3:10 and 6:9 scaled up is 9S:45L

eobrien19 (19:33:14)
Since 6 grins = 9 laughs, 2 grins = 3 laughs.

marsbake (19:33:22)
This looks like a great way to set up the proportion. What if a student doesn't follow this? How would you use questioning to help set up the path for this?

marsbake (19:34:31)
eobrien...where do you go after the 6g=9laughs, 2g=3 laughs?

eobrien19 (19:35:21)
Comparing 3 smiles=10 grins to 2 grins = 3 laughs becomes an easy proportion

eobrien19 (19:36:15)
Then 3 smiles = 15 laughs can be reduced to 1 smile = 5 laughs

eobrien19 (19:36:31)
Then 2 smiles = 10 laughs

eobrien19 (19:36:44)
How is that?

marsbake (19:37:14)
Great...do you think that all students would think to solve this problem with proportional reasoning? How else might you solve it?

marsbake (19:40:13)
Great strategies...pictures, equations, proportional reasoning, this problem is rich with so many strategies.

marsbake (19:40:26)
Ready to leave this one?

marsbake (19:44:14)
Here's Problem #2: A certain test of 12 questions is graded by giving 10 points for each correct answer and then by deducting 5 points for each incorrect answer. David answers all 12 questions and scores a total of 75 points. How many wrong answers does he have?

marsbake (19:45:35)
Are there other strategies to use besides guess and check?

marsbake (19:45:59)
Even when guessing, what did you start with? Was there an efficient way to guess and check?

aznphatso (19:46:34)
you can start with 120, you lose 15 pts for every incorrect answer

marsbake (19:47:23)
Why do you lose 15 points for every incorrect answer?

marsbake (19:48:32)
So with algebra and equations, you are actually setting up a system of equations:

marsbake (19:49:09)
10a + 5b = 75 and a + b = 12

marsbake (19:49:44)
If an student didn't understand how to work with those, how would you help them? Can you relate that to the context of the problem?

marsbake (19:51:38)
What if I'm working with students who haven't had algebra? What would you do to steer them into thinking algebraically, but not formally with the algebraic equation?

marsbake (19:53:27)
Is there a way to set up this problem in terms of one variable? Or one unknown (if the students don't know what a variable is?)

eobrien19 (19:53:57)
How about creating a table, with rows for "Correct, Deductions and Score"?

marsbake (19:54:07)
Can you elaborate on the candy idea?

marsbake (19:55:11)
Is there a way for patterns to come from the table that can then help the students see a "rule"?

jasnik (19:55:46)
welllll.....have candy A be q's Right and candy B is q's Wrong.........move them around (but always have a total of 12) then on a paper w/ a table write out the scores

jasnik (19:56:18)
remove one Q right u must balance that and add one Q wrong

pamgoselin (19:56:08)
75 = 10a - 5(12-a) where a = # right

marsbake (20:00:26)
So we are linking using pictorial representations to the table, by making a table and moving the candy from right answers to wrong answers and showing a decrease of 15 (10 gained from the correct answer and 5 lost for wrong)?

marsbake (20:01:10)
Pamgoselin, can you explain your equation? How would you relate this to objects and the table?

cwoodwell (20:03:02)
If you did not want to use candy, couldn't you just lable table headings right, wrong and total points, then 12 0 120, 11 1 105, 10 2 90, 9 3 75 You must substract 15 each time you get one wrong starting from 120

pamgoselin (20:03:28)
5 (12-a) is how many points would be lost; 10a is how many points would be gained

marsbake (20:05:11)
I don't know about you, but my students have a lot of trouble showing 2 unknowns as one in terms of the other such as you've done with a total of 12, one unknown is a and the other is 12-a. Any suggestions?

ch1n353ch3s54a1l (20:06:08)
aren't you assuming they don't know algebra?

cwoodwell (20:06:44)
Thank you - kids can see a pattern (one of the strategies used to solve problems

pamgoselin (20:06:50)
Start with something easy: true/false questions for example that are only worth one point apiece. Either it is right or wrong...points or no points

marsbake (20:08:13)
That's a great suggestion to start with a simpler problem and build up to this problem. Are we ready for another problem?

marsbake (20:10:55)
All counting numbers are arranged in a triangular pattern as shown by the first four rows. What is the first number in the 13th row?

marsbake (20:11:05)
1

marsbake (20:11:07)
23

marsbake (20:11:29)
Sorry...I am having trouble getting my pictures on the screen. I'll try again

marsbake (20:11:36)
First row: 1

marsbake (20:11:45)
Second row:2 3 4

marsbake (20:11:56)
Third row: 5 6 7 8 9

marsbake (20:12:23)
Fourth row: 10 11 12 13 14 15 16

marsbake (20:13:06)
Arrange the numbers as a triangle so 1, 3, 7, and 13 are in the middle of the triangle of numbers.

cwoodwell (20:10:14)
Can I ask the group if in general, 4th graders are successful in this program?

eobrien19 (20:13:59)
Yes 4th graders can start to get involved

eobrien19 (20:14:21)
They won't be able to do the algebra at the level we have seen. . .

marsbake (20:14:31)
Yes, 4th graders are successful. I have found that practice with lots of problems in the classroom and then the discourse after the contest is over is soooo valuable. Students learn so much from one another.

eobrien19 (20:14:32)
But patterns move them along.

RichKal-MOEMS (20:16:00)
$th graders do well, but of course, the older the students are, the better they do

pamgoselin (20:16:13)
170...draw diagram, check for pattern, compute answer

marsbake (20:17:10)
The Olympiads can be anywhere from grades 2-3 up to grade 8. There are two divisions: elementary and middle. Our school system has some 3rd graders, 4th, 5th and 6th do the elementary problem contest. The more they do, the better they get.

RichKal-MOEMS (20:17:28)
On a problem, if 30% of 4th graders do a problem correctly, 35% of 5th might and 40% of 6th.

eobrien19 (20:17:38)
Are there any reasons why 170 is a logical answer to the triangle pattern problem?

jasnik (20:11:35)
ok

pamgoselin (20:17:28)
pattern is to add consecutive odd numbers to first number in line

RichKal-MOEMS (20:18:17)
The real key is that the first year is a students learning year, as they get the feel for what is asked.

marsbake (20:18:40)
Is 170 the answer?

pamgoselin (20:18:10)
I'm sure there are logical solutions...trying to figure it out....great brain tickling!

marsbake (20:19:22)
What do you notice about the last number in each line?

Mika (20:18:56)
if you add those consecutive odd numbers, you get 145

eobrien19 (20:20:25)
So we have an answer of 145 and an answer of 170. which do you think is correct?

marsbake (20:20:30)
How does adding the consecutive odd numbers and getting 145 help answer the question, what is the first number in the 13th row?

ch1n353ch3s54a1l (20:19:47)
squares of the line

pamgoselin (20:19:59)
it is a perfect square

pamgoselin (20:20:43)
perfect square of row it is on...1^2 = 1; 2^2 = 4; etc. 13^ = 169 though

marsbake (20:21:32)
We seem to have several different answers coming up....is there a way to try and prove which is correct and why others have gotten different results?

eobrien19 (20:22:18)
So perhaps looking at the front number of each row is not the most important part of this problem.

marsbake (20:22:23)
I have seen 155, 175, 145, 169...which one, if any would be correct?

Mika (20:22:18)
the perfect squares are at the end of each row, not at the begining

marsbake (20:23:20)
So if the perfect squares are at the end, what connection do they have to the row? Any where would you go for the 13th row? Should you check the 12th? Why?

marsbake (20:24:12)
Sorry...any should be and

ch1n353ch3s54a1l (20:23:55)
well the end of the twelth is 144

ch1n353ch3s54a1l (20:24:02)
and the first is 144+1

ch1n353ch3s54a1l (20:24:11)
first of the thirteenth row *

pamgoselin (20:21:13)
found my error...169 is right

Mika (20:23:58)
the last number in the 12th row is 144 (12^2) so the first one in the 13th row is 145

marsbake (20:25:47)
Great thinking about the row number and its connection to the square number at the end. Do you see any patterns with anything else in the triangle? Is there a way to extend this problem?

ch1n353ch3s54a1l (20:26:15)
i guess you could also look at the middle number

Zzarkc-20 (20:26:29)
Wait, if you take the row number and add what the row is minus 1, it shows the difference between the first and last number

marsbake (20:27:11)
What do you notice about the middle numbers?

pamgoselin (20:27:00)
you could draw it an a right triangle and look at area rather than an isos. triangle

marsbake (20:28:24)
That's an interesting connection...to geometry. Can you elaborate?

pamgoselin (20:27:39)
consecutive even numbers down the middle

ch1n353ch3s54a1l (20:27:47)
they in crease by two more each time

Zzarkc-20 (20:27:47)
The middle numbers follow the last digits of the increasing powers of 7.

marsbake (20:29:44)
Great pattern finding and interesting extensions.

pamgoselin (20:29:13)
please explain more about powers of 7

marsbake (20:30:07)
What do we see about the powers of 7?

marsbake (20:31:12)
That's about all the time we have for tonight. Thanks to all who have participated. It has been a wonderful experience for me to be your moderator. Happy problme

marsbake (20:31:31)
Happy problem solving...sorry about the misspelling (mistypo)

RichKal-MOEMS (20:31:56)
Thank you, Marsbake.