Transcript for the Math Jam "MOEMS Teachers Math Jam" on Nov 8.

Math Jam hosted by JimMath-MOEMS (Jim Matthews ).

rrusczyk (19:27:24)
Greetings and welcome to another MOEMS Math Jam.

rrusczyk (19:27:34)
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 (19:27:43)
For those of you who have showed up to see what AoPS classes are like, please view transcripts of other Math Jams from the past:

rrusczyk (19:27:47)
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php

rrusczyk (19:27:55)
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 (19:28:05)
Now, on with the show!

rrusczyk (19:28:10)
I'll now turn things over to your instructor for today, Jim Matthews

JimMath-MOEMS (19:28:12)
Hello everyone. I'm going to stay with a theme tonight, working from easier to harder problems.

JimMath-MOEMS (19:28:25)
Here's Problem 1

JimMath-MOEMS (19:29:15)
Pete has a penny, nickel dime, quarter and half dollar in his pocket. Four of these coins are taken out of the pocket and the sum of their values is calculated. How many sums are possible?

JimMath-MOEMS (19:29:43)
Try to send an explanation of your work with your answer.

gapteach (19:30:59)
5 sums are possible.

JimMath-MOEMS (19:31:47)
gapteach - how did you get this answer?

chaenny (19:31:46)
5 sums

MisterW (19:31:48)
iAGREE

chaenny (19:31:56)
1,2,3,4

MisterW (19:32:07)
91 CENTS - EACH COIN

JimMath-MOEMS (19:32:44)
what does 1,2, 3, 4 mean?

JimMath-MOEMS (19:33:28)
misterW - what does 91 cents mean? we want to know how many ways 4 coins can be taken out for different sums.

gapteach (19:32:05)
I make a chart like this

gapteach (19:32:29)
1 5 10 25 50

mathmom (19:30:53)
there are 5 coins, if 4 are chosen, that means one is not. There are 5 different choices of the one not chosen, and for each of these a distinct sum of the other coins. So, there are 5 sums.

JimMath-MOEMS (19:34:19)
nice explanation mathmom

gapteach (19:33:55)
Under each 4 coins I put an x leaving one blank each time. There are 5 possible combinations like this.

JimMath-MOEMS (19:34:58)
gapteach, you are doing the problem the same way as mathmom (basically) nice job

pamgoselin (19:32:57)
5 different ways of leaving one coin in the pocket

chaenny (19:33:40)
1,2,3,4,5 are 1st, 2nd, 3rd, 4th, 5th coin

chaenny (19:33:11)
1,2,3,4 1,2,3,5 1,2,4,5 1,3,4,5 2,3,4,5

mathmom (19:33:20)
you don't need to find the actual sums -- you just need to notice that if you add any 4, the sum will be different than if you added any other 4

RichKal-MOEMS (19:35:15)
The long way: 1+5+10+25; 1+5+10+50;

RichKal-MOEMS (19:31:02)
Jim - don't use my responses if someone else comes close... hold mine back. Rich

JimMath-MOEMS (19:36:29)
okay - any questions on the warm-up, problem 1?

mathmom (19:36:31)
what is a good way to help students approach this kind of question

JimMath-MOEMS (19:37:29)
any suggestions from the class on helping students approach problem 1?

gapteach (19:38:25)
In the beginning, let other students explain how they got it. After a while kids will begin to see use the chart plan for problems like these.

JimMath-MOEMS (19:38:49)
good mathematicians classify, organize, list...........so i would give a hint to students to try listing all possibilities in an organized way

RichKal-MOEMS (19:27:35)
Hi Jim & Richard! You can't hide from me!

RichKal-MOEMS (19:39:15)
Use manipulatives, 5 little disks, one for each coin. Let them act it out.

JimMath-MOEMS (19:40:08)
when i do coin problems with young students i actually bring enough real coins to let them use this works great (usually)

JimMath-MOEMS (19:41:03)
Problem 2 A purse contains 4 pennies, 2 nickels, 1 dime and 1 quarter. Different values can be obtained using one or more coins in the purse. How many different values?

JimMath-MOEMS (19:41:54)
This one is a little harder than Problem 1 but both are from old Math Olympiads for Elementary Level Students.

mathmom (19:42:16)
first I note that the only place where the same amount can be made in different ways is that 2 nickels equals 1 dime.

MisterW (19:43:58)
EIGHT TOTALS 25 + 10 = 35 25 + 15 = 40 25 + 5 + 5 = 45 +1 +1 + 1 + 1 = 46 47 48 49

marsbake (19:44:54)
I am looking at how you can combine the pennies with the other coins, so you can make 1, 5+1, 10 +1, etc. with your coins

JimMath-MOEMS (19:45:06)
MisterW - can you get 3 cents?

marsbake (19:45:31)
Remember you can use only one coin, two coins, etc.

mathmom (19:44:03)
and because there are 4 pennies, you can make all the ""in between"" amounts, that is all the numbers less than 5, all the numbers betwen 5 and 10 etc

Julie Farr (19:45:31)
Sure, 3 pennies

JimMath-MOEMS (19:46:10)
Everyone - i like the way you are all playing off of each other during this jam - keep it up

chaenny (19:46:03)
I don't know why my responses are taking very long to see on the screen after I press enter.

JimMath-MOEMS (19:47:12)
everyone - i'm controlling when the responses are posted. almost like a real classroom! it's to help the flow of our discussion

mathmom (19:45:33)
I think you can make every value up to 49 cents

mathmom (19:45:58)
So, I would say you can make 49 different totals

chaenny (19:47:06)
There are 49 values because you can get anywhere from $.01 to $.49

marsbake (19:47:39)
I think you can make all the combinations up to 49 cents, so there are 49 ways.

gapteach (19:47:13)
I am looking at marsbake's way but then adding another one to each of those possibilities, then another 1 and another, then another 5 to ech of those, never duplicating a complete combo.

pamgoselin (19:48:01)
i agree...49 cents = 49 ways

mathmom (19:48:06)
I think the thing I noticed at first, about being able to make 10 cents two ways isn't relevant in this case, though I think that that kind of thing does become relevant sometimes when you are looking for how many distinct sums... in other similar problems

marsbake (19:48:38)
I find it's a great strategy when you can use a pattern to try to look for all the possible ways to do something.

Julie Farr (19:48:43)
Yes, I agree with 49 ways, 1cent to 49 cents

MisterW (19:48:54)
7 COINS X 7 = 49

marsbake (19:49:07)
1

marsbake (19:49:28)
oops...I was trying to show a pattern and hit return instead. Sorry

JimMath-MOEMS (19:49:29)
aah, misterW, what if all of your coins were nickels?

MisterW (19:49:42)
7 COINS X (6 + 1)

mathmom (19:49:43)
there's 8 coins, not 7 ;-)

RichKal-MOEMS (19:50:06)
Consider separately the quarters, dimes/nickels, pennies. Quarters allow 2 choices (0, 25); nickel/dimes allow 5 choices (0, 5, 10, 15, 20); pennies allow 5 choices (0, 1, 2, 3, 4). Then 2x5x5 = 50 ways, but 0cents is not allowed: 49 ways.

MisterW (19:50:12)
i THOUGHT THERE WERE ONLY 2 NICKELS ...

marsbake (19:50:32)
Can you explain what you are doing, MisterW?

pamgoselin (19:50:30)
all coins nickels = 7 values

mathmom (19:51:01)
I was originally trying to do something like what Rich just posted, but got stymied by the nickels/dimes

MisterW (19:51:21)
pAIRING EACH OF 4 DIFFERENT TYPES OF COINS WITH MULTIPLE AND/OR PAIRINGS OF THE COINS THAT REMAIN

JimMath-MOEMS (19:52:49)
MisterW - we'll get back to that if we have time but let's move on to Problem 3

mathmom (19:52:35)
Rich, if there were a third nickel, you would run into ""overlap"" with your quarters. Then how would you approach it?

JimMath-MOEMS (19:53:35)
I have four 3 cent stamps and three 5 cent stamps. How many different amounts of postage can I make?

RichKal-MOEMS (19:53:29)
Very carefully, mathmom

mathmom (19:54:04)
thanks rich :)

pamgoselin (19:54:16)
how many stamps can you use?

JimMath-MOEMS (19:55:02)
at least one and at most all 7

chaenny (19:55:44)
13, combined multiples of 3 and 5 up to 27

JimMath-MOEMS (19:56:57)
chaenny - what do you mean by combined multiples?

mathmom (19:56:48)
in this case, I'd use Richard's approach from problem #2. you have 5 choices of how many 3's to use (0 through 4) and 4 choices of how many 5's to use. That's 20 different combinations. Since you don't have enough 3-cent stamps to make a multiple of 5 (you'd need 5) so I think that you're safe from having to worry about getting the same value 2 different ways. So, I think the answer is 20.

JimMath-MOEMS (19:58:23)
almost right on the answer mathmom but you still have a little error -

Julie Farr (19:57:05)
3,5,6,8,9,10,11,12,13,14,16,17,18,19,21,24,27

JimMath-MOEMS (19:58:48)
julie - i think you're missing some

chaenny (19:57:53)
mult of 3 -3,6,9,12,15,18,21,24,27 and of 5- 5,10,20,25

JimMath-MOEMS (19:59:11)
so are you chaenny

pamgoselin (19:58:17)
I got 19...there are four choices by all 3 cent stamps and 3 choices by all 5 cent stamps = 7 options. Using 1 -3 cent stamps with all combos of 5 cent stamps gives me another 3 options. Using 2 -3 cent stamps gives another 3, Using 3 - 3 cent stamps gives another 3, using 4 -3 cent stamps gives another 3...7 + 12 = 19

marsbake (19:59:36)
But then you could do multiples of 8

mathmom (19:58:38)
19, becaues 0 of each is not allowed

pamgoselin (19:58:56)
0-4 is 5 options

pamgoselin (19:59:16)
chaenny...there are only three 5 cent stamps

MisterW (19:59:19)
1 2 3 4 5 10 (2 X5) 15 (3x5) 6 (5 +1) 7 ( 5+2) 8 (5+3) 9(5 +4) 11 ... 12 ... 13 14 16 17 18 19

chaenny (19:59:29)
25!

chaenny (19:59:50)
you start at $.03 and end at $.27

gapteach (20:00:09)
Using a chart I get 3,6,9,12,17,22,27,5,10,15,18,21,24

pamgoselin (19:58:31)
20 -1 ...zero doesn't count

MisterW (19:56:35)
i GOT 19 COMIBINATIONS MAKING AN ORGANIZED LIST

JimMath-MOEMS (20:01:43)
misterW sent this last response and i asked him to provide his list. he's got it but with a little time it could be cleaned up a bit

chaenny (20:00:30)
I thought there wher 4 3-cent stamps

mathmom (20:00:56)
gapteach, misterW how did you organize your lists

marsbake (20:02:06)
I listed 3, 6, 9, 12 on one side, then 5, 10, 15 on the other, made a chart with the numbers from 1-27 and began crossing off combos of the two multiple combinations

Julie Farr (20:01:58)
Mister W, how do you get 4 cents?

JimMath-MOEMS (20:03:06)
So I think you can get 3, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 24, and 27

JimMath-MOEMS (20:03:27)
can not get 1, 2, 4, 7, 20, 23, 25, 26

MisterW (20:03:02)
MY MISTAKE ... I was thinking 4 1 cent stamps

chaenny (20:03:05)
I got 19 also - 3,5,6,8,9,10,11,12,13,14,15,18,19,20,21,24,25,27

mathmom (20:03:14)
you can't make 1 or 2 either. Is it easy to list all of the numbers you can't get?

JimMath-MOEMS (20:04:36)
your collective work suggests changing the problem to 5 three centers and 3 five centers. then there is at least one value that can be done two different ways.

Julie Farr (20:04:04)
How do you get 22 cents?

Julie Farr (20:04:20)
oophs, never mind!!

marsbake (20:05:06)
10 +12

gapteach (20:04:59)
I made a chart with 3,3,3,3,5,5,5 across the top. First I put an x under the first 3, Then an x under 3 and 3. Then an x under 3,3,3 and so on until I exhausted those combos. next I started with an x under the 5. then an x under 5, 5, then under 555, then under 5553 etc until I hit all the combos.

JimMath-MOEMS (20:05:52)
gapteach has a cool and different way to approach this one.

JimMath-MOEMS (20:06:17)
ready for problem 5

marsbake (20:06:26)
That's a great strategy, gapteach. I think I saw something similar in a book entitled Fostering Algebraic Thinking by Mark Driscoll....it's great!

Julie Farr (20:06:24)
sure

marsbake (20:06:56)
We're rolling, JimMath!

JimMath-MOEMS (20:07:24)
Suppose we modify problem 4 so that you have unlimited 2 cent and unlimited 4 cent stamps. what is the largest postage value you cannot make?

mathmom (20:08:05)
in addition to the 3's and 5's or just the 2's and 4's??

marsbake (20:08:36)
I don't think you can ever make an odd number if you are just using even/mult. of 4 stamps.

JimMath-MOEMS (20:08:51)
thanks - good mathematicians try to clarify the problem statement before working on it. not in addition - just 2s and 4s

chaenny (20:08:48)
23?

JimMath-MOEMS (20:10:00)
marsbaker - hi - hold back cause i can't control your postings since you have moderator privledge

mathmom (20:09:12)
well, then there is no largest, because as marsbake said, you can't make any odd numbers

Julie Farr (20:10:04)
Why 23? why not 73?

pamgoselin (20:10:22)
infinite number of stamps = infinite number of values therefore two time infinity will be an even number, plus 1 to make it odd

Julie Farr (20:10:26)
agree

RichKal-MOEMS (20:09:34)
Then no odd number is possible, no matter how large.

mathmom (20:10:33)
you can make every even amount and no odd amount. Every multiple of an even number is still even

JimMath-MOEMS (20:11:11)
okay - so you can only make even amounts? no odd amounts?

RichKal-MOEMS (20:11:26)
If marsbake closes her moderator window (""CLOSE"" at bottom of that window), she can become a regular participant.

Julie Farr (20:11:30)
right

pamgoselin (20:11:31)
post is not clarification: answer is 2 x infinity + 1

mathmom (20:11:45)
right

JimMath-MOEMS (20:13:55)
suppose you have unlimited 2 centers and 3 centers. what is the largest postage amount you cannot make?

chaenny (20:13:15)
I've never seen a stamp that high :)

chaenny (20:14:19)
$.01

MisterW (20:14:48)
20 cents ?????

JimMath-MOEMS (20:15:20)
remember you're looking for the highest amount you CANNOT make

mathmom (20:15:07)
you can make 20 cents with a bunch of 2's

MisterW (20:15:23)
no largest = 20

JimMath-MOEMS (20:15:59)
can you make 22 cents?

marsbake (20:15:53)
20 cents could be 2 + 3 five times

pamgoselin (20:16:08)
I agree with chaenny...$0.01 because 2 and 3 are smallest prime and can be added to make all other primes in some fashion

mathmom (20:16:11)
I think chaenny is right

JimMath-MOEMS (20:16:52)
good - problem 7 what if you had 1 centers and 10 centers

MisterW (20:16:49)
I think I missed something about 2 cent and 4 cent stamps when i was working with pen and pencil ... =please ignore the ramblings of an old fool who has difficulty muti-tasking

mathmom (20:16:56)
you can make any even number using however many 2-cent stamps you need. YOu can make any odd number higher than 2 by replacing the last 2-center with a 3-center

JimMath-MOEMS (20:17:33)
keep rambling misterW - that's the fun of all of this

pamgoselin (20:17:33)
JimMath...how many stamps...what are we looking for?

JimMath-MOEMS (20:18:01)
unlimited 1s and 10s

pamgoselin (20:18:31)
then 9 cents if looking for largest value not possible

mathmom (20:17:28)
clearly for problem 7 you can make any value. Even if you only had 1-centers, you can make any value

mathmom (20:19:14)
pamgoselin, 9 cents is possible using nine 1-cent stamps

marsbake (20:19:18)
Couldn't you just use 1's to get 9?

pamgoselin (20:19:24)
oh yeah...oops

mathmom (20:18:43)
with unlimited stamps, any value that is a multiple of any other given value is unnecessary and does not change the answer

pamgoselin (20:19:34)
yeah...yeah...yeah

JimMath-MOEMS (20:19:56)
okay - grand finale problem for tonight

JimMath-MOEMS (20:20:53)
A fast food place serves chicken nuggets in containers that hold 6, 9 or 20 nuggets. They have an unlimited number of containers of each size. What's the largest order they cannot fill?

JimMath-MOEMS (20:21:27)
As a hint - Can you fill an order of 15?

Julie Farr (20:21:49)
Sure, 6 + 9

JimMath-MOEMS (20:22:15)
good - how about an order of 21?

pamgoselin (20:22:29)
9+12

mathmom (20:22:38)
I don't think so for 21

RichKal-MOEMS (20:22:45)
Where do they store the containers?

JimMath-MOEMS (20:23:15)
they have a very very tall place

RichKal-MOEMS (20:23:00)
9+6+6

Julie Farr (20:23:15)
No. but that's not the largest. You can't make 68, or 87 . . . I think I'm confused about how there is a largest . . .

mathmom (20:23:28)
ok, they can make 21

RichKal-MOEMS (20:23:32)
Heaven!

marsbake (20:23:27)
21 can be 9 + 6 + 6

mathmom (20:22:20)
you can't fill an order of 19, but I don't think that's the highest

JimMath-MOEMS (20:24:23)
that's right - an order of 19 cannot be filled

JimMath-MOEMS (20:25:44)
julie - for 87 they can do 20, 20, 20, 9, 9, 9

mathmom (20:24:24)
Julie, I think there is a largest that they can NOT make because after a certain point, you can always rearrange the existing numbers and make any total

mathmom (20:24:44)
how about 22?

pamgoselin (20:25:09)
1081 can't be filled

RichKal-MOEMS (20:25:17)
Would it help to see what can be made?

mathmom (20:25:25)
we need to find a pattern by which we can make all nubers over a certain size, to be sure that when we find something that cannot be made, it is really the largest

pamgoselin (20:26:03)
we seem to be heading back to 20 x 6 x 9 x infinity + 1

gapteach (20:26:06)
how about 22?

Julie Farr (20:26:22)
ok some large prime number . . which I don't know

marsbake (20:26:37)
You can do 68, Julie, with 20 + 6 + 6 + 6 + 6+ 9 +6 +9

mathmom (20:26:48)
some primes can be made by adding composites

JimMath-MOEMS (20:27:27)
and some non-primes (composites) cannot be done. like 16

gapteach (20:27:20)
I 'm listing multiples of 6 then of 9 then of 20 and then making the combinations of these

marsbake (20:27:33)
I was trying to get a list of multiples of different combinations. Then I could cross those off from a hundred chart.

pamgoselin (20:27:51)
so 5 nuggets then?

RichKal-MOEMS (20:28:10)
All multiples of 3 above 3 can be made. Since 20 is 2 more than a multiple of 3, all remainder 2s above 23 can be made. Since 40 is 1 more than a multiple of 3, all remainder -1s above 43 can be made. So I would think 41.

gapteach (20:28:25)
With my list I then draw lines for possible combinations. (The lines look kind of like a fan.)

mathmom (20:28:11)
only odd multiples of 9 are interesting, the evens are also multiples of 6

JimMath-MOEMS (20:29:09)
you're coming up with lots of nice theorems or lemmas - great

JimMath-MOEMS (20:29:42)
i'll give you a hint since we're running out of time - the largest that you CANNOT make is less than 60

mathmom (20:30:04)
all remainder 2's above 23 can be made because.... you could add 20 to a multiple of 3, right?

pamgoselin (20:30:15)
thanks JimMath...gotta go!

RichKal-MOEMS (20:30:53)
Oops, I meant 43 is the largest that can't be made with 6s, 9s, and 20s.

mathmom (20:31:03)
so similarly you could make all remainder 2's above 40 by adding 20 + 20 to a multiple of 3

Julie Farr (20:31:27)
thanks. look forward to the next session

mathmom (20:31:45)
so I see what Rich is doing....

marsbake (20:31:55)
I working backwards and know you can make 59, 58, 57 and 56.

JimMath-MOEMS (20:33:30)
if you work forward from 6 on you will find that you can't make 43 but you can make 44, 45, 46, 47, 48 and 49. once you make six in a row you can always add a container of 6 onto each of these giving you the next 6 consecutive integers. this can be done indefinitely

RichKal-MOEMS (20:32:37)
Great problem!

marsbake (20:32:49)
I just made 55 and 54

mathmom (20:32:50)
oops, I meant remainder 1's above 40, but it's really remainder 1's 46 and above that we can make

mathmom (20:33:43)
so the highest thing we can't make is something that leaves a remainder of 1 when divided by 3, and is less than 46, so 43.

JimMath-MOEMS (20:34:49)
of course, like we were doing with the stamp problem we can change the 6, 9 and 20 to other sizes! this is called the frobenius problem in some math circles. you can google frobenius problem for more info

mathmom (20:33:59)
there's the trick, making 6 in a row,

Julie Farr (20:34:43)
Thanks for the explanation, Jim

mathmom (20:35:02)
thanks, Rich, I like the way you worked that out

JimMath-MOEMS (20:35:43)
hope you liked the theme for tonight's jam. you guys were great. thanks.