Transcript for the Math Jam "MOEMS Teachers Math Jam" on Feb 13.

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

rrusczyk (19:29:42)
Greetings and welcome to a MOEMS Teachers Math Jam!

rrusczyk (19:29:51)
My name is Richard Rusczyk, and I work here at Art of Problem Solving.

rrusczyk (19:29:54)
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 (19:29:59)
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 (19:30:02)
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php

rrusczyk (19:30:06)
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 (19:30:16)
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 (19:30:22)
Now, on with the show!

rrusczyk (19:30:27)
I'll now turn things over to your instructor for today, Jim Matthews.

JimMath-MOEMS (19:30:37)
Welcome to tonight's MATH JAM I see
some people who were here last week.
Please let me know if you are a
teacher,a student, a parent or
neither. If you are a teacher or
student, what grade level?

JimMath-MOEMS (19:30:56)
Also, where are you geographically
located tonight. Also, please let
know if you are a PICO. Thanks.
(city, state, country)?

JimMath-MOEMS (19:33:24)
so we have folks from east and west coasts and lots of places inbetween including hong kong

JimMath-MOEMS (19:33:47)
PROBLEM_1:
In the following addition problem,
each letter stands for a digit and
different letters stand for
different digits.
What do H, A and O stand for?
How do you know?
HA + HA + HA + HA = OH

RichKal-MOEMS (19:34:39)
PICO =Person In Charge of the Olympiads.

JimMath-MOEMS (19:36:18)
When you get an answer please submit it and try to explain how you got it.

greencheeze92 (19:35:33)
h is 2

JimMath-MOEMS (19:37:00)
why couldn't the value of H be 1?

suzifrizzle (19:36:47)
H has to be either 1 or 2 since its sum is a one-digit number. H = 2, A = 3, O = 9 A has to add up to a number ending in H. There is nothing divisible by 4 that ends in 1, so H has to be 2.

greencheeze92 (19:36:56)
thats because h has to be 1 or 2 for it to be a 2 digit number, and if its 1, then 4 numbers add up to 1, which doesn't happen

JimMath-MOEMS (19:38:03)
ok - so H is 2

KingRoy (19:35:36)
23+23+23+23=92

LindaKBrewer (19:35:59)
H=2, A=3, 0=9

nitish (19:37:22)
h= 2 a=3 o= 9

suzifrizzle (19:38:33)
the sum of the ones column must be 12, therefore A = 3

JimMath-MOEMS (19:39:17)
you're nailing this one

Glenda (19:39:08)
h=2, a=3, o=9

Dr. E (19:38:00)
Looks like A = 3, H = 2 and O = 9...H<2 to avoid carry, A = 3 to get carry and placement of 2 for H.

JimMath-MOEMS (19:40:01)
any questions on this one are your ready for the next problem?

JimMath-MOEMS (19:40:58)
PROBLEM_2:
In the following multiplication
problem, each letter stands for a
digit and different letters stand
for different digits.
What do A, B, C and D stand for?
How do you know?
ABC x C = DBC

greencheeze92 (19:41:20)
c has to be 1, 5, or 6

JimMath-MOEMS (19:41:49)
Could C be 0?

RichKal-MOEMS (19:41:52)
When I wanted to whisper, it shouted out. Now I want to raise my hand, so it whispered to RR!

KingRoy (19:41:52)
C=5, 6

LindaKBrewer (19:42:16)
C can be 5

JimMath-MOEMS (19:42:57)
Why can't C be 0 or 1?

greencheeze92 (19:42:50)
no because that would be a 1 digit number, c cant be 1 because then d is equal to a

KingRoy (19:43:13)
a*0=0

KingRoy (19:43:24)
a*1=a

LindaKBrewer (19:43:40)
It can't be 1 because A may not be the same as D

Glenda (19:43:47)
because something times 0 can't be a three digit number

LindaKBrewer (19:43:09)
A=1, B=2, C=5, D=6

nitish (19:43:51)
a= 1 b= 2 c= 5 d= 7

JimMath-MOEMS (19:45:37)
will both of these work?

Glenda (19:44:50)
a=1, b=2=, c=5, d=6

greencheeze92 (19:45:33)
it cant be 6 because if it was, 6 times b would be something that ends in three, but thats not possible

suzifrizzle (19:45:57)
A = 1, B = 2, C=5, D=6. Csquared has to end in C, so it could only be 6 or 5. 6 doesn't work because the hundreds place would have to be a 2digit product. C*B must end in zero in order for there to be another B in the product

nitish (19:46:04)
my bad c=6

nitish (19:46:13)
i mean d=6

Glenda (19:46:14)
no, 125*5 doesn't equal 725

LindaKBrewer (19:46:16)
125 * 5 =625, so d=6

Dr. E (19:45:15)
125 * 5 = 625...C = 5 to get units digit to be same as C, B = 2 to allow multiplication by C and carry to equal 2, A = 1 to get correct digit without carry.

JimMath-MOEMS (19:47:26)
is 1,2,5, 6 the only solution?

Glenda (19:47:42)
I think so

JimMath-MOEMS (19:48:18)
if not, what is another solution? if it is unique, why?

greencheeze92 (19:49:06)
well, c has to be 5, no matter what

LindaKBrewer (19:48:49)
No, A=1, B=4, C=5, D=7 That should make nitish feel better.

Glenda (19:49:47)
No other solution ending in five works

JimMath-MOEMS (19:51:02)
glenda, what about linda's solution?

Glenda (19:51:21)
oops

Dr. E (19:51:35)
145 * 5 = 725...not a fit.

JimMath-MOEMS (19:52:22)
you were ok glenda. see dr. e's comment

suzifrizzle (19:51:03)
A could be zero, and then D would be 1, but we don't generally write numbers that way

JimMath-MOEMS (19:53:00)
that's right suzi

suzifrizzle (19:51:58)
B can't be 4 in the factor and 2 in the product

LindaKBrewer (19:52:22)
And A=1, B=8, C=5, D=9 Just work with even numbers for B.

LindaKBrewer (19:52:39)
OOPs!!!!!

JimMath-MOEMS (19:53:33)
good fast catch linda

greencheeze92 (19:51:25)
145 times 5 is 725, not 745

Glenda (19:53:53)
yeah, 145*5=725, not 745

nitish (19:54:05)
025*5= 125

RichKal-MOEMS (19:53:28)
MOEMS specifically says, ""No leading zeroes.""

JimMath-MOEMS (19:54:53)
but good mathematicians should make sure the problem is properly defined

suzifrizzle (19:54:45)
Oh, thanks for that MOEMS info

JimMath-MOEMS (19:55:34)
i didn't state this in the problem - my bad

JimMath-MOEMS (19:55:54)
there is another solution

JimMath-MOEMS (19:56:28)
a couple of people have found it but i'm waiting for others before i post

RichKal-MOEMS (19:57:19)
The key seems to be tied into multiples of 25.

Glenda (19:56:50)
It can't be any thing but 125*5 because any number ending in a five will always end in two five

JimMath-MOEMS (19:58:29)
glenda - are you saying that B will equal 5?

greencheeze92 (19:58:24)
150 times 5 is 650

greencheeze92 (19:58:31)
sorry

greencheeze92 (19:58:38)
I messed up

JimMath-MOEMS (19:59:13)
that's ok - fire away

Dr. E (19:49:10)
Looks like 175 * 5 = 875 so B = 7 is possible.

Dr. E (19:56:22)
Seems like the only other solution is 175 * 5, given the constraints imposed by carry, etc.

suzifrizzle (19:59:26)
A=1, B=7,C=5, D=8

LindaKBrewer (19:59:35)
A=1, B=7, C=5, C=8

Glenda (19:59:55)
a=1, b=7, c=5, d=8

nitish (19:59:56)
a= 1 b= 7 c= 5 d= 8

greencheeze92 (20:00:11)
175 times 5 is 875, aha!

KingRoy (20:00:02)
175*5=875

JimMath-MOEMS (20:00:40)
you were all ringing in at almost the same time!!!

JimMath-MOEMS (20:00:57)
looks like you got it. any questions or on to the next problem?

JimMath-MOEMS (20:01:34)
PROBLEM_3:
In the following addition problem,
each letter stands for a digit
and different letters stand for
different digits.
What do A, B and C stand for?
How do you know?
A + B + C = AB

LindaKBrewer (20:02:15)
A=1, B=3, C=9

greencheeze92 (20:02:18)
a and c have to add to ten, so maybe 4 and 6

KingRoy (20:02:46)
1+2+9=12

Dr. E (20:02:57)
A = 1, B = 5, C = 9 so 1 + 5 + 9 = 15 for starters

Glenda (20:03:09)
a has to be one or two

suzifrizzle (20:03:25)
There's no way to make a sum in the 20s if A = 2, so A must be 1. B+C has to be 9 or greater.

greencheeze92 (20:02:57)
no, wait A is 1, c is 9

LindaKBrewer (20:03:52)
A=1, B can be any integer from 1-9, and C=9

nitish (20:04:12)
a=1 b= 6 c= 9

Glenda (20:04:35)
a= 1, b=4, c= nine

suzifrizzle (20:04:46)
A=1, B=0, C=9

JimMath-MOEMS (20:05:16)
wait a second. how many answers are there for this one?

greencheeze92 (20:05:15)
b can be 0, right?

JimMath-MOEMS (20:05:31)
yep

Glenda (20:05:27)
if a is one and c is nine, b can be any number for 2 through8

KingRoy (20:05:17)
B=2-8 and 0 because it can't be 1 or 9 because A=1 and C=9

LindaKBrewer (20:05:44)
Oops! Sure B can be 0

nitish (20:05:47)
10 solutions

suzifrizzle (20:05:56)
I like Linda's answer

greencheeze92 (20:05:24)
ten

KingRoy (20:03:37)
B=2, 3, 4, 5, 6, 7, 8

greencheeze92 (20:06:19)
oops

greencheeze92 (20:06:25)
seven

Glenda (20:06:52)
nine 0+2+3+4+5+6+7+8

greencheeze92 (20:06:55)
oops again, eight

KingRoy (20:06:47)
8 answers possible

LindaKBrewer (20:07:13)
It says that each letter stands for a different digit, so I'm with KingRoy. All Hail!

JimMath-MOEMS (20:07:35)
all right - i think you have it

Dr. E (20:07:27)
Looks like A =1 to get correct carry. B can be any digit besides 1 and 9

JimMath-MOEMS (20:08:16)
ok - next one coming up

JimMath-MOEMS (20:08:31)
PROBLEM_4:
In the following addition problem,
each letter stands for a digit and
different letters stand for different
digits. What do A, B and C stand for?
How do you know?
AA + BB + CC = BAC

greencheeze92 (20:08:54)
well, a and c add to ten

Glenda (20:10:26)
b can only be one

LindaKBrewer (20:10:26)
A + B needs add to 10

Zak! (20:11:01)
I think a + b = 10 since the last digit of a + b + c = c

suvisa (20:11:20)
a and b add to 10 -- b=1 a=9?

greencheeze92 (20:11:52)
since b is 1, a has to be 9

Glenda (20:12:12)
can't have a+c= ten, aa+bb would always be 110

Zak! (20:12:31)
Then C = 8 and it checks, but why does b have to be 1?

greencheeze92 (20:10:48)
b is 1

KingRoy (20:12:04)
B=1 A=9 C=8

greencheeze92 (20:10:32)
and then a and b and c and 1 add to b, so a and c add to 9

suzifrizzle (20:12:48)
A=9, B=1, C=8

suzifrizzle (20:13:16)
You can't get B to be a 2 in the sum if it is also a 2 in the addend

nitish (20:14:03)
a= 9 b= 1 c=8

Dr. E (20:14:13)
A = 9, B = 1, C = 8 as stated by others.

JimMath-MOEMS (20:14:27)
Why?

suzifrizzle (20:14:27)
Now is Jim going to tell us there are 7 more ways to solve this?

greencheeze92 (20:14:33)
i agree with KingRoy

Zak! (20:14:36)
Thank you suzifrizzle, that is right.

JimMath-MOEMS (20:14:52)
Not this time

Glenda (20:14:50)
because the numbers that go up to over 200 can't contain a 2, also, no leading zeroes

suzifrizzle (20:14:56)
Yay!

Zak! (20:15:11)
B = 1 as stated above.

Zak! (20:15:19)
Then A = 9 since A+B =10

RichKal-MOEMS (20:13:05)
Units column: A+B=10. B is 1 or 2. If B = 2, then A = 8 and 88+22+CC =28C, which is tooooo large. So B = 1.

LindaKBrewer (20:15:24)
Since 99 + 11 =110, Then 110 + 11C =190 + C. If that makes sense.

RichKal-MOEMS (20:15:21)
Then B = 1 and A = 9, so 99+11+CC = 19C. So C=8. Check: 99+11+88= 198

suzifrizzle (20:15:56)
C has to be 8 so that with the 1 that is carried, it will make a 10 so that the sum ends in a 9 for A

LindaKBrewer (20:16:02)
Solve that little equation, and you get C=80

LindaKBrewer (20:16:22)
Oops, I mean 8.

JimMath-MOEMS (20:16:49)
my postings got a little separated but you've made your points.

JimMath-MOEMS (20:17:03)
i think everyone is OK

JimMath-MOEMS (20:17:39)
based on our time left tonight i'll change gears a little for the last problem

JimMath-MOEMS (20:18:00)
In the Game of 24 you are given
4 digits between 1 and 9 and you
must form an expression equal to
24 using each of the digits exactly
one time along with any combination
of operators +, -, x, and /
and any number of parentheses.

JimMath-MOEMS (20:18:24)
Hint:
Be careful about the order of
operations.

JimMath-MOEMS (20:18:41)
For example,
given 3, 6, 7 and 8
3 + 6 + 7 + 8
is a good answer.

JimMath-MOEMS (20:18:58)
SO HERE IS THE PROBLEM

JimMath-MOEMS (20:19:10)
Find an expression
for 1, 7, 3, and 2
the year George Washington
was born.

suvisa (20:19:32)
7*3+2+1

KingRoy (20:19:34)
1+7*3+2

nitish (20:19:43)
7*3=21+1+2=24

greencheeze92 (20:19:45)
(7+2-1)(3)

LindaKBrewer (20:19:51)
7*3 +1+2

Glenda (20:20:02)
Got it. 7*3+1+2=24

RichKal-MOEMS (20:20:11)
1+7*3+2, (in order!)

Dr. E (20:20:13)
Unclear...7*3 + 1 + 2 = 24...is that what you want?

JimMath-MOEMS (20:20:32)
Yes- you are all getting it.

JimMath-MOEMS (20:20:50)
OK now try to make 24 with 8, 8, 3 and 3

KingRoy (20:21:23)
Can you square?

JimMath-MOEMS (20:21:58)
nope - look back at the rules

JimMath-MOEMS (20:22:11)
+ - * and / with ()

Glenda (20:22:13)
only one off. darn!

JimMath-MOEMS (20:23:52)
A hush has fallen over the crowd!!

suvisa (20:23:46)
ha ha i have 22, 23, and 25 -- too bad they dont count :(

suvisa (20:24:04)
is integer division ok?

suvisa (20:24:24)
if so 8*3+(3/8)

JimMath-MOEMS (20:25:00)
no integer division. your solution is 24.375

greencheeze92 (20:25:23)
any hints?

JimMath-MOEMS (20:25:38)
yes. there is a solution

JimMath-MOEMS (20:26:55)
and there is only one way to do it

Glenda (20:27:29)
still one off. confusing!

Glenda (20:28:13)
can we say 83 or 38?

JimMath-MOEMS (20:29:25)
i think so but for this one it won't help

JimMath-MOEMS (20:29:55)
we getting to the end of our time so i'll post the solution for the first person who gets it.

JimMath-MOEMS (20:31:11)
ok another hing

JimMath-MOEMS (20:31:18)
oops - hint

JimMath-MOEMS (20:31:30)
8 divided by what is 24?

greencheeze92 (20:31:46)
1\3

JimMath-MOEMS (20:32:04)
good - can you go from there?

Glenda (20:32:18)
how do we get to 1/3

JimMath-MOEMS (20:32:33)
how do you get to 1/2?

JimMath-MOEMS (20:32:43)
whoops - 1/3

suzifrizzle (20:32:39)
without using up the 8

Glenda (20:32:58)
yes.

JimMath-MOEMS (20:33:13)
you do have 2 8s

Zak! (20:33:16)
1/3=(8/8)/3

JimMath-MOEMS (20:33:50)
so now what zak? you just have a 3 left

Glenda (20:33:52)
but that uses both eights

JimMath-MOEMS (20:34:34)
someone has an answer - do you want to see it?

suzifrizzle (20:34:42)
Yes, I have to go back to work now

greencheeze92 (20:34:45)
yes-please

Glenda (20:34:51)
yes yes yes yes yes!

LindaKBrewer (20:35:05)
yes, time for bed.

JimMath-MOEMS (20:35:24)
ok - good one kingroy

KingRoy (20:34:58)
8/(3-8/3)=8/(1/3)=24

greencheeze92 (20:35:47)
ohhhhh!

suvisa (20:35:52)
cool

RichKal-MOEMS (20:35:52)
WOW!!

nitish (20:36:01)
sweet

LindaKBrewer (20:36:07)
All Hail, once again!

Glenda (20:36:14)
wow. congrats! awesome4

JimMath-MOEMS (20:36:38)
You guys were all great tonight. Thanks.

JimMath-MOEMS (20:37:03)
There will be other math jams based on math olympiads coming up.