| Transcript
for the Math
Jam "AoPS Classes Math Jam"
on May 29. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk19:29:12
Hello, and welcome to an Art of Problem Solving Math Jam. Today we'll be discussing the MATHCOUNTS Problem Series, Introduction to Algebra, and Introduction to Counting & Probability. We will go through a couple example problems from each class, and discuss both what these classes cover and how they work.
rrusczyk19:29:23
My name is Richard Rusczyk. I founded Art of Problem Solving and have written several Art of Problem Solving textbooks.
bwu19:29:38
Have you made the USAMO?
rrusczyk19:29:40
Three times, and I was a winner in 1989 :)
rrusczyk19:29:47
Before we get started I would like to take a moment to explain our Virtual Classroom to those who have not previously participated in a Math Jam or one of our online classes.
rrusczyk19:29:56
The classroom is moderated: students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read. Also, only moderators can enter into private chats with other people in the classroom.
rrusczyk19:30:31
Note that it is not possible for the instructor to personally respond to every comment that you submit -- please do not take it personally if your comment is not posted or responded to! I will try to respond to all questions to the extent that I can. I will let you know when to start asking questions about the classes
rrusczyk19:30:57
The virtual classroom is LaTeX enabled. LaTeX allows users to make nice equations and other math expressions. If you would like to learn how to write in LaTeX, click on the tab on the left side panel of our site and there is a tutorial and reference guide there.
rrusczyk19:31:23
Using LaTeX in the virtual classroom is slightly different than using it on the message board or in a LaTeX editor. If anything you type up in a post uses LaTeX, then you must use a semicolon (;) to begin your post. For example, if you type
rrusczyk19:31:29
rrusczyk19:31:45
This message will look like this when posted in the classroom:
rrusczyk19:31:48
rrusczyk19:31:54
Just remember, if your post uses LaTeX, use the semicolon (;) to begin your post!
lzhao19:31:56
Is LaTeX necessary for the class?
rrusczyk19:32:14
Not at all! In fact, most of our younger students don't bother learning it until they are older.
rrusczyk19:32:22
It is not necessary for any aspect of the class.
rrusczyk19:32:37
In this Math Jam, I will briefly describe a course, then go through a few example problems. Then, I will hold a question-and-answer session about that class.
isabella229619:33:04
How many example problems will there be?
rrusczyk19:33:09
2-5 for each class.
rrusczyk19:33:17
Let's get started with the first class we'll discuss tonight:
rrusczyk19:33:27
MATHCOUNTS Problem Series
rrusczyk19:33:41
The MATHCOUNTS Problem Series is a 12 week course designed to cover one by one the areas of problem solving covered by the MATHCOUNTS competition. Topics covered including methods of counting, probability, algebraic techniques, word problems, number theory, and more. This class will also help you with the AMC 8.
rrusczyk19:34:06
We are offering two different sections of the MATHCOUNTS class this summer. One section meets on Monday nights, 7:30 - 9:00 PM Eastern, and runs from June 9 through August 25. The other section meets on Thursday nights, 7:30 - 9:00 PM Eastern, and runs from June 12 through August 28.
isabella229619:34:35
Specifically, what kind of topics will be covered?
rrusczyk19:34:41
There are more details on this page:
nikki19:35:00
what's amc8?
rrusczyk19:35:13
See www.unl.edu/amc for more information about the AMC 8
isabella229619:35:28
American Mathematics Compitetion for grade 8 and under
rrusczyk19:35:29
This class is a Problem Series class, meaning that the major focus of the class will be working through various MATHCOUNTS problems. Although there will be weekly problem sets for each class posted on the message board, students do not submit their homeworks to be graded, and there is no personalized instructor feedback on the solutions. (However, the instructors will be monitoring the message board to answer questions and comment occasionally on the students' discussions there.)
wsjradha19:36:04
what level of MathCounts will this class help prepare for?
rrusczyk19:36:06
This class is primarily aimed at more experienced MATHCOUNTS students. There will be questions at difficulty ranging from Chapter through hard National.
rrusczyk19:36:18
This fall, we will introduce a beginner's MATHCOUNTS class.
basketball30019:36:27
will there be homework?
nikki19:36:27
where does one post homework?
rrusczyk19:36:55
For the MATHCOUNTS class, the "homework" will not be turned in - you will post answers to extra questions we give on the message board.
rrusczyk19:37:07
You post there just like you normally do on the message board.
bwu19:37:09
Will this help MathCounts participants advance to Nationals and do well there?
rrusczyk19:37:11
Yes.
rayli111119:37:21
is any there audio involve
rrusczyk19:37:29
There is no audio in the classroom.
rrusczyk19:37:35
Here is a webpage that describes why:
hluzader19:38:07
What is the math level of a student to be able to take Mathcounts beginner?
rrusczyk19:38:08
We don't yet have a MATHCOUNTS beginner class. When we do, we will expect students to at least be ready for pre-Algebra or enrolled in it.
bluegirl10119:38:33
do we have to do the hw
rrusczyk19:38:36
You don't *have* to do anything, but you won't learn much if you don't do any problems!
hluzader19:38:54
when will it be avaible? roughly?
rrusczyk19:38:55
This fall (the Intro MATHCOUNTS class, which will be grades 5-8)
basketball30019:38:57
is it okay to miss a few classes
rrusczyk19:39:34
Yes - There is a full transcript of *everying* that happens in the main classroom. You can then read everything you miss later.
rrusczyk19:39:42
This is part of why we do text only classes.
joslyn19:39:50
Can I get a transcript of this MathJam session ?
rrusczyk19:39:59
Yes - it will be in the Math Jams section of the community.
erictao20050019:40:03
can you give us a few examples of the problems we will cover?
rayli111119:40:03
can we start doing problems now?
rrusczyk19:40:06
Good idea!!
rrusczyk19:40:15
Let's do some problems, and then I will take more questions.
rrusczyk19:40:20
The instructor for this course is Josh Zucker. As a student (a while ago), Josh was invited to train with the US math team, was a member of the US physics team, and was a top-10 scorer on the Putnam exam (the most prestigious college math contest). More recently, he served as a MATHCOUNTS question writer, so he knows the contest very well!
rrusczyk19:40:53
Here is an excerpt of some of the curriculum and problems covered during the course.
rrusczyk19:41:05
rrusczyk19:41:18
Where can we start with this problem?
Urc19:41:48
One has to be 2
Lolligal10119:41:48
we know that one is odd and one is even so they have to be 2 and 3
natiator19:41:48
the only primes 1 away are 2 and 3
rrusczyk19:41:59
Since either p or p + 1 is even, one of them must be even, so one of them must be 2, the only even prime. Since 1 isn’t prime, the other one must be 3.
Now, how do we find the smallest composite number that is neither a multiple of 2 nor 3?
rrusczyk19:42:36
I see a lot of you saying 5. What's wrong with that?
19654219:42:38
What is a composite number?
rrusczyk19:42:52
A composite number is a number that has positive divisors besides itself and 1.
Visser Three19:42:59
5 is a prime number.
lzhao19:42:59
5 is not a composite number
FantasyLover19:42:59
5 is not composite number.
bl.html19:42:59
5 is prime
rrusczyk19:43:12
Yep, so what is the smallest *composite* number that is not divisible by 2 or 3?
aggarwal_219:43:34
25
footballer19:43:34
I tihnk the answer is 25
vjnmath19:43:34
25
footballer19:43:34
25
bl.html19:43:34
25 = 5 * 5
rrusczyk19:43:40
rrusczyk19:44:10
Let's try another problem about prime numbers.
rrusczyk19:44:13
rrusczyk19:44:25
How do we start?
FantasyLover19:44:58
one of them should be 2
rrusczyk19:45:05
Why?
ashstorm19:45:27
one of the primes has to be '2' because all other primes are odd... and 3 odd numbers added together would make an odd
deltaren19:45:27
3 odd numbers can't add up to 30
FantasyLover19:45:27
because if there is 3 odd numbers, the result is odd
rrusczyk19:45:37
No matter what, one of the primes must be 2. If we have 3 numbers with an even sum, at least one of them is even. There's only one positive even prime, and that's 2.
rrusczyk19:45:43
So now what has the problem become?
seaturtle19:46:00
so then we need 28 between 2 primes
4thgraderstar19:46:00
2 nums sum to 28
Lolligal10119:46:00
two primes whose sum is 28
Visser Three19:46:00
Two numbers that add up to 28.
deltaren19:46:00
Find the largest product of numbers with a sum of 28
rrusczyk19:46:03
We must find the largest possible product of two primes which have a sum of 28.
rrusczyk19:46:12
How do we do it?
Urc19:46:43
There's only 11 and 17 or 5 and 23 I think
rrusczyk19:46:52
Our only options are 11 & 17, and 5 & 23.
rrusczyk19:46:55
Which has a larger product?
Visser Three19:47:01
11 and 17.
prithviramesh19:47:01
17 and 11
hluzader19:47:01
11 and 17
4thgraderstar19:47:01
11 and 17
rrusczyk19:47:08
rrusczyk19:47:14
What is the answer to the problem?
deltaren19:47:36
374
4thgraderstar19:47:36
374
littlesister19:47:36
374
ilovepi19:47:36
374
Jakey19:47:36
374
swimfish19:47:36
374
rrusczyk19:47:42
rrusczyk19:48:04
All right, those two problems are examples of relatively easy problems that we'll discuss.
rrusczyk19:48:16
Let's take a look at a couple harder problems we'll discuss in the MATHCOUNTS class.
hluzader19:48:27
which branch of math is this in?
rrusczyk19:48:31
Number Theory.
algebrafriek19:48:33
why do we multiply them together?
rrusczyk19:48:55
That's what the question asked for - the product of the numbers (last step should always be to read the question again and make sure you answer what is asked).
ilovepi19:48:59
it asks for the product of the 3 numbers
ajopajo19:48:59
because it says so in the problem
seaturtle19:48:59
because it's part of the problem
rrusczyk19:49:03
Exactly!
rrusczyk19:49:08
Next problem:
rrusczyk19:49:09
rrusczyk19:49:23
This is a tougher problem than most similar problems at MATHCOUNTS. In fact, I took the last question on a sprint round and added 'that are even'. But I wanted you all to take a shot at a problem that requires a little more thought than plugging into a formula.
bluegirl10119:49:29
what does integral mean
rrusczyk19:49:35
It means they are integers.
rrusczyk19:49:41
First, how many positive integral divisors does 792 have?
wsjradha19:49:51
prime factorize 792
Lolligal10119:49:51
should we prime factorize?
rrusczyk19:50:00
tigeralie19:50:40
what's prime factorizing?
rrusczyk19:50:50
Writing a number as a product of primes.
natiator19:50:57
it is a multiple of 2,3,and/or11
bwu19:50:57
multiple of a power of 2, 3, or 11
littlesister19:50:57
it is multiple of 2,3,11.
rrusczyk19:51:03
But could it be any multiple?
ajopajo19:51:26
only up to the power amount
rrusczyk19:51:44
Interesting. Let's explore this. Could a divisor of 792 have five 2's in its prime factorization?
FantasyLover19:51:57
no
ashstorm19:51:57
No.
natiator19:52:08
no we can only have 3
smiley99919:52:08
no
rrusczyk19:52:21
Right; 792 only has three 2's, so we can only have up to three 2's.
rrusczyk19:52:31
In this way, we can see what numbers can be divisors of 792:
natiator19:52:37
only a multiple that uses no more than 2^3,3^2,11
rrusczyk19:52:49
Exactly.
ABNoRMaLY_NoT_QuieT19:52:51
same with the 3s and 11s
basketball30019:52:51
same with 2 3s an 1 11
rrusczyk19:52:53
Good.
rrusczyk19:53:01
So, how do we count the divisors using this.
rrusczyk19:53:12
Here's the prime factorization again:
rrusczyk19:53:13
rrusczyk19:53:34
How many choices do we have for the power of 3 in a divisor of this number?
deltaren19:54:09
3: 0, 1, and 2
natiator19:54:09
3:0,1,2
Urc19:54:09
because there's 3^0, 3^1, and 3^2
littlesister19:54:09
3 choices. 1,3,9
Elphaba19:54:11
3 choices: 0, 1, or 2
rrusczyk19:54:17
Exactly.
rrusczyk19:54:47
We can continue in this way to count all the divisors of 792.
rrusczyk19:54:51
rrusczyk19:55:46
There are a couple ways to finish from here.
rrusczyk19:55:48
Here is one:
erictao20050019:55:52
24 - 6 odd divisors = 18 even divisors
deltaren19:55:52
there are 6 odd divisors (3x2), and you subtract that from 24
rrusczyk19:56:29
We can count the odd divisors by noting that an odd divisor has three choices for the power of 3, and two for the power of 11, so there are 3x2 = 6 odd divisors, leaving 24 - 6 = 18 even ones.
rrusczyk19:56:36
Does anyone see another way?
wsjradha19:56:47
2*3*3 = 18 as a can be 1,2 or 3 in an even
RussianRocket19:56:47
3x3x2= 18 possible choices
Jakey19:56:47
can't you just have 3 possibilities for 2 (exclude ;$2^0$) so we have 3x3x2=18
rrusczyk19:57:21
rrusczyk19:57:45
This is just like our previous reasoning, except now we only have 3 choices for the exponent of 2 instead of 4.
ABNoRMaLY_NoT_QuieT19:57:57
why?
rrusczyk19:58:13
Because we can't use 0 as the power of 2 if the divisor is even - there must be at least one 2!
Urc19:58:17
Was this on state or chapter?
rrusczyk19:58:22
This would be a National level problem.
rrusczyk19:58:30
As would, probably, our next problem:
rrusczyk19:58:36
rrusczyk19:58:49
Anyone want to find that number in base 10?
rrusczyk19:58:59
(No calculators!)
rrusczyk19:59:39
(Sorry, but I don't have time to explain base numbers in this Math Jam -- there is an explanation on the message board of the MATHCOUNTS class, though, and more thorough information elsewhere on the site and in our Introduction to Number Theory book)
rrusczyk19:59:58
I don't see anyone eager to convert to base 10.
rrusczyk20:00:12
(Without a calculator.)
rrusczyk20:00:43
Why might we think there's a way to do it without converting to base 10. How are the two bases in the problem related?
isabella229620:01:00
3*3=9
natiator20:01:00
9=3^2
rrusczyk20:01:05
Maybe this will help.
rrusczyk20:01:14
rrusczyk20:01:33
(As a note, this is what base 3 means -- it's just like base 10, except we use powers of 3 instead of powers of 10)
FantasyLover20:02:06
for example, 2*3^5 is 2*3*3^4, so its 6*3^4.
rrusczyk20:02:13
Interesting idea!
rrusczyk20:02:18
rrusczyk20:02:34
So, if we do this, what do we get for our base 9 number?
RussianRocket20:02:50
2612
theone14285720:02:50
2612
wsjradha20:02:55
2612
Visser Three20:02:55
2612.
rrusczyk20:03:02
rrusczyk20:03:25
If you found this problem very hard, that's because it's one of the harder ones we'll cover in the MATHCOUNTS Problem Series.
rrusczyk20:03:38
If you don't know anything about base numbers, then there are several ways you can learn:
rrusczyk20:03:52
1) There will be a message board discussion of them at the start of the MATHCOUNTS Problem Series.
rrusczyk20:04:05
2) There is a thorough discussion of them in the Introduction to Number Theory textbook.
rrusczyk20:04:24
3) Ask on the message board, and many people will help.
Urc20:04:33
is that number theory, or counting and probability?
rrusczyk20:04:41
This is number theory, indeed.
isabella229620:04:42
What would it be in base 10?
rrusczyk20:04:50
Is see that some of you have calculated it:
deltaren20:04:51
isn't it 1955 in base 10?
Franklin24520:04:57
Do you have a syllabus so the student can review the basic math before each class?
rrusczyk20:05:23
There is a listing of topics in the Course Introduction document, but the topics are very broad, so it will be tough to review *everything* before class.
bluegirl10120:05:29
what's a syllabus?
rrusczyk20:05:35
A list of subjects covered.
nikki20:05:43
where is the message board
rrusczyk20:06:14
There is a link to the class message board from the Course Homepage. You can also find it on the main Forum page if you scroll down. It will be active starting next week.
deltaren20:06:21
are higher level courses faster-paced?
rrusczyk20:06:24
Yes.
bwu20:06:26
Is the message board open to everyone at all times?
rrusczyk20:06:44
The class message board is open to all enrolled students at all times, and you can ask questions there at any time.
smiley99920:06:47
was the previous problem for middle school or high school?
rrusczyk20:07:01
That is a very hard MATHCOUNTS problem. Many high school students would have a hard time.
basketball30020:07:10
Are there any other books to help us with mathcounts
rrusczyk20:07:26
Our Introduction Series of texts, and our Volume 1 are good for MATHCOUNTS preparation.
isabella229620:07:37
I found it kind of complicated to solve that problem. Is there an easier way to convert bases?
rrusczyk20:07:55
You can learn more about this via the methods I mentioned earlier.
sda133720:07:57
Hi, what do you mean by "high school"? Grades 8-11? Sorry I'm from an Asian country.
rrusczyk20:08:03
Typically it means grades 9-12
lzhao20:08:06
When is the next problem?
rrusczyk20:08:13
After all the questions about this class.
basketball30020:08:15
where do we get the transcrips
rrusczyk20:08:23
Transcripts for the classes will be on the class homepage.
RussianRocket20:08:29
Is there a special class from AMC8 or AMC10
rrusczyk20:08:41
This class will help with AMC 8. We have a separate class for the AMC 10.
theone14285720:08:42
I did not see the intermediate NT book in the bookstore
rrusczyk20:08:52
There is not one; only an Introduction to Number Theory.
bwu20:08:56
What should we do, besides review books, to succeed in MathCounts?
rrusczyk20:09:02
Work on lots of problems!
prithviramesh20:09:05
How much problems are in the homework?
rrusczyk20:09:15
Usually we will give around a dozen extra problems to discuss.
theone14285720:09:18
do you have intermediate number theory class
rrusczyk20:09:25
Yes. It will be offered in the fall.
Jakey20:09:26
what's the maximum age/requirement for a student to join mathcounts?
rrusczyk20:09:43
I think it is grades 6-8 only. I don't know about age requirements.
nikki20:09:47
for the course, are the problems going to be similar to the earlier ones or the later ones?
rrusczyk20:10:00
There will be some of both. Most will be somewhere in-between in difficulty.
Urc20:10:11
Would amc 10 be like a follow up to this?
Elphaba20:10:11
What is the difference between AMC 8 and 10 and 12?
rrusczyk20:10:27
Difficulty: AMC 8 is up through 8th grade. AMC 10 up through 10th, AMC 12 up through 12th.
mathlete32520:10:28
The problems are harder
rrusczyk20:10:30
Exactly.
hluzader20:10:33
in class, will you discuss mostly homeworkproblems?
rrusczyk20:10:44
Not at all - the homework will only be discussed on the message board.
ilovepi20:10:49
Would you let us know where the problem that you taught will be found in the book?
rrusczyk20:10:59
There is not a textbook for the MATHCOUNTS class.
Ace120:11:02
will you also be covering material from intro algebra tonite? I thought that was what the email said.....
rrusczyk20:11:10
Yes, when I finish with questions here.
God_of_Math_220:11:21
Where can we get the transcripts
rrusczyk20:11:32
Class transcripts will be on the class homepage.
smiley99920:11:34
there's a high school mathcounts, right?
rrusczyk20:11:36
No.
RussianRocket20:11:38
Will we be tested on the material we cover?
rrusczyk20:11:41
No.
Jakey20:11:48
are there some mathcounts national winners who end up as IMO representatives?
rrusczyk20:11:50
Many
isabella229620:11:55
Is there anywhere where I could find MATHCOUNTS problems from previous years?
rrusczyk20:11:59
Check the MATHCOUNTS website.
prithviramesh20:12:01
How much longer will it take to finish the question to go on to the intro to algebra section?
rrusczyk20:12:05
About 5 minutes.
seaturtle20:12:06
if we're only in mathcounts should we stay for the others?
rrusczyk20:12:15
Sure - they're similar in difficulty.
algebrafriek20:12:19
what is IMO?
rrusczyk20:12:26
High school international math olympiad
iYOA20:12:27
how about if we write problems, would that help?
rrusczyk20:12:34
That is an excellent way to learn math.
vjnmath20:12:36
Will the course also improve our speed?
rrusczyk20:12:41
If you work on the problems, yes.
God_of_Math_220:12:51
What is the maximum class size for the mathcounts class/
rrusczyk20:13:04
No maximum. If many students enroll, we split the class and run two sessions.
basketball30020:13:16
will we work on Count down
rrusczyk20:13:25
Yes - one day will be a big Countdown round.
prithviramesh20:13:27
Is there a test at the end of the course?
rrusczyk20:13:29
No
hluzader20:13:38
so you meet withus once a week, ifwe miss a class we can look at the book?
rrusczyk20:13:52
We meet once a week, and if you miss a class, you can access the class transcript on the class homepage.
Franklin24520:13:54
Roughly how much time will the homework take on average?
rrusczyk20:14:04
For this class, typically 30-90 minutes.
Urc20:14:08
How would you do CD with a huge amount of people?
rrusczyk20:14:11
You'll see :)
mathlete32520:14:14
What is Countdown?
rrusczyk20:14:21
Try the For The Win game on our site.
joslyn20:14:45
Is it okay to miss a class?
rrusczyk20:14:48
Yes.
rrusczyk20:14:56
Let's move on to the Introduction to Algebra class.
rrusczyk20:15:06
Then, I'll take more questions after talking about that class for a bit.
bluegirl10120:15:12
about what time does this class end?
rrusczyk20:15:21
Probably in about 45 minutes, hopefully
prithviramesh20:15:24
If we don't understand, may we ask you to repeat the explanation
rrusczyk20:15:25
Yes.
rrusczyk20:15:47
In the regular classes, there will be extra instructors to work 1-on-1 with questions.
rrusczyk20:16:02
Introduction to Algebra
rrusczyk20:16:06
Introduction to Algebra is intended to cover all the fundamental concepts of Algebra, including the following:
rrusczyk20:16:09
expressions
equations
quadratics and other polynomials
complex numbers
graphing
functions
sequences and series
exponents and logarithms
rrusczyk20:16:28
In terms of what schools usually call Algebra 1 and Algebra 2, our course covers
rrusczyk20:16:32
* essentially all the algebraic topics in Algebra 1
* most of the algebraic topics in Algebra 2
* certain advanced topics (like telescoping sums and piecewise functions)
nikki20:16:39
i dont know what some of these are
rrusczyk20:16:47
That's OK - that's why you take these classes :)
rrusczyk20:16:57
The reason I referred to 'algebraic topics' is that schools often teach the basics of geometry and counting in their algebra classes. Instead of teaching those topics in our algebra classes, we teach them in our Geometry and Counting & Probability classes, as part of a deeper exploration of those subjects.
chili.peppers20:17:19
will we go into more depth than at school?
rrusczyk20:17:22
Much more.
hluzader20:17:27
how good should I be in math? I'm in 5th, will go to 6th, I'm good but not super.
rrusczyk20:17:50
If you aren't sure what class to take, email me your math background at classes@artofproblemsolving.com and I will advise you.
ABNoRMaLY_NoT_QuieT20:17:57
wow. if you're in this room, you're already amazing
rrusczyk20:18:01
That is probably true.
rubik_220:18:03
What's the level of material covered for Intro Algebra
rrusczyk20:18:28
It starts at Algebra 1 and runs through much of Algebra 2, but is much more challenging than typical school Algebra 1 or 2.
rrusczyk20:18:32
Introduction to Algebra, like all our classes, emphasizes problem solving and conceptual understanding rather than rote memorization. So in addition to teaching students how to manipulate equations, we teach them why the techniques are logically sound, and we talk about general problem solving strategies. Our class meetings are largely interactive, meaning that most of the time is spent solving problems. As much as possible, the students do the solving; the teacher only guides them along and provides useful hints.
rrusczyk20:19:01
There are a couple major goals of Introduction to Algebra:
rrusczyk20:19:04
Students should develop the ability to translate a situation (which might be a real-life situation, a puzzle, etc.) into the abstract language of equations, and then manipulate the equations to develop insight into the original problem. This skill is the key that opens the door to numerous other fields of study, like higher math, chemistry, physics, and engineering.
nikki20:19:30
will the teachers explain if students dont understand?
rrusczyk20:19:52
Yes - in the regular classes, we have extra assistants to take questions, or the main instructor will handle them in the main room.
wsjradha20:19:56
Is there a large overlap between the Intro to Algebra and Intermediate Algebra curriculums?
rrusczyk20:20:24
Not much. We aren't like a regular school - we don't spend 50% of every course reviewing the last course.
rrusczyk20:20:28
The real world is the taking-off point which motivates many mathematical concepts. Once one begins to study those concepts, many new questions arise: Do all equations have solutions? Is every number a fraction? Can we invent a number whose square is negative? We delve into some of these questions so that students gain an appreciation for the structure of mathematics, the ability to think abstractly, and the confidence to tackle very difficult questions.
chili.peppers20:20:43
will we go in order according to our books?
rrusczyk20:20:45
Most of the time, yes.
ABNoRMaLY_NoT_QuieT20:20:53
will the problems be from the book?
rrusczyk20:20:55
Some will, some won't
bwu20:20:59
What does Internediate Algebra cover?
rrusczyk20:21:16
Basically all of the non-trig precalculus that's not in the Introduction to Algebra clas.
hluzader20:21:19
Do you cover the whole book in three months?
rrusczyk20:21:27
Six months.
rrusczyk20:21:42
This is a 24 week class.
rrusczyk20:21:58
It starts on June 11 and runs through the middle of November.
ABNoRMaLY_NoT_QuieT20:22:06
which one?
ABNoRMaLY_NoT_QuieT20:22:06
intermediate or intro?
rrusczyk20:22:12
Both are 24 weeks.
rrusczyk20:22:28
Let's look at a couple sample problems, and then I'll take more questions.
rrusczyk20:22:34
What's the solution to the following equation?
rrusczyk20:22:37
rrusczyk20:23:08
I see a lot of you answering -1.
rrusczyk20:23:16
What did you do as your first step?
ashstorm20:23:22
is the -2x under the radical?
rrusczyk20:23:24
No.
rrusczyk20:23:52
People have suggested this:
rrusczyk20:23:55
rrusczyk20:23:58
After all, the square root of the square of a number seems to be the original number. For example,
rrusczyk20:24:00
rrusczyk20:24:05
Let's try this and see where it goes. What's our new equation?
Visser Three20:24:22
x - 2x = 1.
seaturtle20:24:26
x-2x=1
footballer20:24:27
x-2x=1
rrusczyk20:24:31
rrusczyk20:24:33
Now what?
FantasyLover20:24:48
-x=1
FantasyLover20:24:48
-x=1, so x=-1.
4thgraderstar_220:24:48
-x=1
Visser Three20:24:48
- x = 1.
rrusczyk20:24:50
We can combine like terms.
rrusczyk20:24:54
rrusczyk20:24:56
And then we get:
sda133720:25:04
x-2x=1, -x=1, x=-1
seaturtle20:25:04
x=-1
lzhao20:25:04
or x = -1
rrusczyk20:25:08
rrusczyk20:25:15
gives us x = -1.
rrusczyk20:25:20
Alright, now, it's always a good idea to check your work by plugging your solution into the original equation. Let's do that. If we plug x = -1 into
rrusczyk20:25:22
rrusczyk20:25:33
what do we get on the left hand side?
bwu20:26:29
3
bpgbcg20:26:29
3
chili.peppers20:26:29
1+2
bl.html20:26:29
3
rrusczyk20:26:40
rrusczyk20:26:48
Uhoh! Is x = -1 a solution?
bwu20:26:51
3 = 1??
Jakey20:26:56
3 is not equal to 1
rrusczyk20:27:10
Exactly. If we put in -1, we get 3=1, which is obviously not true!!!
rrusczyk20:27:15
So, -1 is not a solution.
rrusczyk20:27:20
But there *does* exist a solution. What is it?
ilovepi20:27:38
x= -1/3
EOTiger20:27:38
so x is -1/3
deltaren20:27:38
-1/3
FantasyLover20:27:38
-1/3
mathgenius20:27:38
doesn't -1/3 work?
rrusczyk20:27:42
Check for yourself that x = -1/3 works.
rrusczyk20:27:49
We missed this solution. So somewhere we must have made a faulty assumption. Where?
deltaren20:28:23
that the sqrt of the square of x is x
bsalleman20:28:23
square root of x^2 = x
ABNoRMaLY_NoT_QuieT20:28:23
that the square root of x squared = x?
rrusczyk20:28:27
It was wrong to assume that
rrusczyk20:28:30
rrusczyk20:28:36
We can only make this assumption if we know that x is positive. If x is negative, what does
rrusczyk20:28:39
rrusczyk20:28:41
come out to?
EOTiger20:29:14
sqrt(x^2) = -x if x itself is negative
bwu20:29:14
-x
wsjradha20:29:14
-x
FantasyLover20:29:14
-x
eggylv99920:29:14
-x
EOTiger20:29:14
-x
rrusczyk20:29:18
rrusczyk20:29:24
Let's try solving the equation again, this time assuming x is negative.
rrusczyk20:29:30
rrusczyk20:29:36
What do we get when we assume x is negative?
bpgbcg20:29:54
-3x=1
FantasyLover20:29:54
then it's -3x=1
4thgraderstar_220:29:54
-3x=1,x=-1/3
seaturtle20:29:54
-3x=1
smiley99920:29:54
-3x=1
b557msa20:29:58
-x-2x=1 then x = -1/3
rrusczyk20:30:07
First, we simplify the square root:
rrusczyk20:30:10
rrusczyk20:30:12
This reduces to
rrusczyk20:30:14
rrusczyk20:30:23
rrusczyk20:30:29
Now we've found the solution.
rrusczyk20:30:32
This problem illustrates the importance of not making any hidden assumptions when you attempt to solve an equation. You must keep *all* possible solutions under consideration rather than assuming the solution is positive.
God_of_Math20:30:48
Can we do another one?
rrusczyk20:30:50
We'll do one more problem.
nikki20:31:21
how would u know if x is positive or negative?
rrusczyk20:31:46
We first tried assuming x is positive, when we let sqrt(x^2) be x, but we got a negative result. So, there were no solutions in this case.
rrusczyk20:31:57
Then we tried x being negative, and we found a negative solution.
rrusczyk20:32:12
Sometimes, you can find valid solutions in both cases. Or in neither case!
rrusczyk20:32:33
Bottle A contains 10% alcohol. Bottle B contains 20% alcohol. In what ratio do I need to mix liquid from the two bottles if I want to create a liquid that is 16% alcohol?
rrusczyk20:32:59
Where do we start?
rrusczyk20:33:20
I see a lot of you giving answers with no explanation... I need explanations.
FantasyLover20:33:23
make an equation
rrusczyk20:33:38
It's a word problem, so our goal will be to convert it to an equation.
rrusczyk20:33:41
How do we do it?
isabella229620:33:56
let a = amount from bottle a
isabella229620:33:56
let b = amount from bottle b
rrusczyk20:34:04
Exactly! We assign variables.
rrusczyk20:34:11
rrusczyk20:34:39
How can we build an equation now? What is an expression for the amount of alcohol we want in the end?
rrusczyk20:35:44
I see many of you jumping way ahead.
sda133720:35:52
0.16(a+b)?
isabella229620:35:52
0.16(a+b)?
rrusczyk20:35:57
16% of (a + b) = amount of alcohol in our mixture
rrusczyk20:36:12
We can write 16% of (a+b) as 0.16(a+b).
rrusczyk20:36:38
What must this equal?
F-86K Sabre Dog20:36:48
.1a+.2b=.16(a+b)
deltaren20:36:48
so .1a + .2b = .16 (a+b)
hluzader20:36:48
so 0.1 a + 0.2 b = .16(a+b)
sda133720:37:03
0.1a+0.1b
bpgbcg20:37:03
0.1a+0.2b
rrusczyk20:37:28
Exactly, we know we have 0.1a alcohol from A and 0.2 alcohol from B, for a total of 0.1a + 0.2b.
rrusczyk20:37:45
This must equal the total amount of alcohol in the mixture, so:
isabella229620:37:48
Then we have 0.1a + 0.2b = 0.16(a+b)
rrusczyk20:37:54
We have an equation now!
rrusczyk20:37:57
.16(a+b) = .10a + .20b
rrusczyk20:38:23
How can we simplify this? What might we do about the decimals?
footballer20:38:29
You multiply the .16 by 100
rrusczyk20:38:47
We can multiply by 100 to get an equation without decimals:
rrusczyk20:38:49
16a + 16b = 10a + 20b
rrusczyk20:38:51
Now what?
nikki20:39:07
wouldn't that change the equation though?
rrusczyk20:39:32
Yes, but it changes it to an equation that is still true! (Good question). Multiplying both sides of an equation by 100 gives us a new equation that is still true.
God_of_Math20:39:39
6a = 4b
wsjradha20:39:39
6a=4b
smiley99920:39:39
combine like terms
rrusczyk20:39:53
We subtract 10a from both sides and subtract 16b from both sides, and we have:
rrusczyk20:39:57
6a = 4b
rrusczyk20:40:03
So, what is the ratio of a to b?
FantasyLover20:40:11
a:b=2:3
bwu20:40:11
2/3
lzhao20:40:11
2:3
rrusczyk20:40:23
We divide both sides by 6 and by b, and we get
rrusczyk20:40:27
a/b = 4/6 = 2/3.
rrusczyk20:40:35
So, we have the ratio of a to b, which is what we wanted!
rrusczyk20:41:37
On your own, see if you can find an intuitive solution to this problem that doesn't use algebra (we won't discuss it tonight because we still have another class to discuss!)
EOTiger20:41:42
will this be retaught in the real class
rrusczyk20:41:44
Yes.
ABNoRMaLY_NoT_QuieT20:41:45
wait, why did we want the ratio again?
rrusczyk20:41:51
That is what the question asked for.
isabella229620:41:53
Can we do another one?
rrusczyk20:42:01
We'll do some counting questions in a few minutes.
rrusczyk20:42:10
Here's a little more information about the class.
rrusczyk20:42:27
The course will meet for 24 weeks on Wednesdays, starting June 11, at 7:30 PM Eastern / 4:30 PM Pacific. Each class is 90 minutes, and each is 7:30 - 9 PM ET (4:30 - 6 PM PT)
rrusczyk20:42:33
I'll be teaching this course; you can find a short biography for me and all the other AoPS teachers here:
rrusczyk20:42:40
This course will use a textbook in conjunction with the course: our own Introduction to Algebra book. The material covered in the textbook is roughly equivalent to the material covered in the course. You can see the table of contents and some excerpts from the book here:
rrusczyk20:43:03
The book is required for the course. Students will be able to read additional material that complements the lectures, and will have access to a large number of practice problems at varying levels of difficulty. We recommend that students read the corresponding chapter(s) in the book before each lecture, and attempt some of that chapter's Review and Challenge Problems after each lecture. We also expect to spend some class time answering students' questions about problems from the textbook.
rrusczyk20:43:27
The homework for the class consists of weekly problems that will be posted to the class message board -- for these problems, you do not turn your solutions in, but you may post them to the message board if you like. The class also has 4 Challenge Sets for which you should write up your full solutions and submit them. You will receive thorough feedback for your work on these Challenge Sets that will comment both on your mathematical accuracy and how well you write solutions.
nikki20:43:45
what was the first one we discussed?
rrusczyk20:43:47
The first class we discussed was the MATHCOUNTS Problem Series.
isabella229620:43:49
What grade levels do you recommend this class for?
rrusczyk20:44:10
We've had students from grades 5-9 in this course.
rrusczyk20:44:17
The main thing is that you be ready for algebra.
rrusczyk20:44:27
Try the pre-test on the class description page in the Online School.
vallabh_suresh20:44:33
Will we ever talk about the golden ratio in this class?
rrusczyk20:44:41
Probably :)
nikki20:44:45
i think this one is easier
rrusczyk20:44:54
Yes, this class is easier than the MATHCOUNTS Problem Series.
joslyn20:45:02
Can our parents help us with homework??
rrusczyk20:45:05
Certainly.
mathlete32520:45:08
Are we going to be able to use Aops grades to get into more advanced math programs at school?
rrusczyk20:45:15
We typically don't give grades.
rrusczyk20:45:27
We can if your school needs them.
andrewjjiang20:45:44
Will we know if we are correct?
rrusczyk20:45:50
On the homework, yes.
Franklin24520:45:53
Is it possible to change the registration from mathcout to this class?
rrusczyk20:46:06
Yes - you should have your parents call during our office hours to sort that out.
Elphaba20:46:09
What if you've covered algebra 1 already but not algebra 2?
rrusczyk20:46:35
This is a tough question, and depends on how good your algebra 1 class was. I recommend you look at the post-test for this class to decide if you need this class.
vallabh_suresh20:46:40
What if you still don't understand something after a class? Can we get extra help from somewhere?
rrusczyk20:46:50
You can ask on the class message board, and an instructor will help you.
hluzader20:46:52
do you grade them or we check ourselves?
rrusczyk20:47:09
For the Challenge Sets, we check them. For the message board problems, you do.
bwu20:47:11
What would you recommend to take, if you have taken Algebra 1 and Algebra 2?
rrusczyk20:47:19
Intermediate Algebra.
ABNoRMaLY_NoT_QuieT20:47:21
how do we hand in hw?
rrusczyk20:47:28
There will be instructions on the homework.
isabella229620:47:31
Do we turn in homework and have it graded?
rrusczyk20:47:42
For the Challenge Sets, you turn it in and we give feedback.
prithviramesh20:47:44
What happens if we go on vacation?
rrusczyk20:48:01
You can read the class transcripts and catch up on everything in the class.
mathgenius20:48:12
can we go on to the probability class?
rrusczyk20:48:19
Let's do so, then I'll take more questions.
rrusczyk20:48:23
Introduction to Counting & Probability
rrusczyk20:48:27
In the Introduction to Counting and Probability class, we cover basic and intermediate counting concepts, including casework, multiplication, permutations, combinations, Pascal's triangle, probability, combinatorial identities, and the Binomial Theorem.
rrusczyk20:48:34
The main emphasis of this class is learning how to take an organized approach to counting, and understanding that nearly all of counting is learning when to use the basic arithmetic operations division, multiplication, addition, and subtraction (and of course why to use them when you use them).
rrusczyk20:48:41
Students completing this course, who work most of the problems, should come out of the course knowing how to tackle any MATHCOUNTS counting problems, most AMC 10 and AMC 12 counting problems, and even some AIME counting problems. The concepts in this course are also crucial to understanding computer science.
rrusczyk20:49:04
We'll now take a look at a couple of sample problems from the course, which highlight some of the tactics we'll investigate in this class.
rrusczyk20:49:33
rrusczyk20:50:00
I see a lot of you saying 7!
rrusczyk20:50:05
What is that? And why is it the answer?
ashstorm20:50:07
are each of the girls and boys distinct? or is each kid of the two genders the same?
rrusczyk20:50:19
The kids are distinct (I should hope :) )
ilovepi20:50:48
the first seat has 7 possible people
ilovepi20:50:50
the second 6, then 5, then 4, 3 , 2 ,1
bwu20:50:52
The first seat has 7 choices, the second has 6, 3rd, 5........and the 7th seat has 1 choice left.
isabella229620:50:54
It's a permutations problem. We have 7 choices for the first student, 6 for the second, and so on. So we do: 7*6*5...=5,040.
rrusczyk20:50:59
Good.
rrusczyk20:51:07
This is a straightforward application of multiplication: there are 7 students who could sit in the first seat. For each of these choices we make for the first seat, there are six ways to choose a student for the next seat, so there are 7 x 6 ways to seat the first two students.
rrusczyk20:51:16
Continuing in this vein, for each of these 7 x 6 ways to seat the first two students, there are 5 ways to pick a student for the third seat. Thus, there are 7 x 6 x 5 ways to seat the first three students.
rrusczyk20:51:25
We keep going like this: there are 4 ways to seat the fourth student, 3 ways to seat the fifth, 2 ways to seat the sixth, and one way to seat the last student. This gives us 7 x 6 x 5 x 4 x 3 x 2 x 1 ways to seat all the students.
rrusczyk20:51:34
We run into products like 7 x 6 x 5 x 4 x 3 x 2 x 1 so much in mathematics that we have a symbol and a name for it. We write 7 x 6 x 5 x 4 x 3 x 2 x 1 = 7! and we call this 'seven factorial'.
rrusczyk20:51:49
Similarly,
rrusczyk20:51:52
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.
rrusczyk20:51:57
That problem was pretty simple. Let's put a wrinkle in it. Suppose we must have a girl in the first chair and a girl in the last chair. Then how many seatings are there?
rrusczyk20:52:09
We'll first look at a wrong solution:
rrusczyk20:52:12
What's wrong with this answer:
rrusczyk20:52:20
There are 4 ways to choose the girl for the first chair. After that, we have 6 students left for the next chair, then 5 for the next, and so on, giving us a total of:
rrusczyk20:52:23
4 x 6 x 5 x 4 x 3 x 2 x 1 seatings.
rrusczyk20:52:32
What's wrong with that?
erictao20050020:52:51
you cant be sure that the last one is a girl
Visser Three20:52:51
All the girls could be seated already.
bwu20:52:51
The last seat must be a girl
wsjradha20:52:54
it doesn't guarantee a girl in the last seat
rrusczyk20:52:59
The problem here is the last chair - we must have a girl in that last chair, but our approach above definitely does not guarantee this. We might end up with a boy left at the end, which would violate the problem.
rrusczyk20:53:06
What should we do to solve the problem?
deltaren20:53:52
can you repeat the problem? I just came in
ilovepi20:53:52
the first seat could have 4 possibilities, for there are foru girls
rrusczyk20:54:07
Yes, the first seat could have 4 possibilities for the girls. Then what?
RussianRocket20:54:10
Maybe make sure that the first and last seat is occupied by a girl first
ilovepi20:54:19
the last seat has three possibilities, because one girl already sat down on the first seat
Wickedestjr20:54:20
There are four possible girls for the left most seat, and three for the right most seat, so that means that there is five factorial combinations for the five remaining seats so it is 4 x 3 x 5!
erictao20050020:54:22
you can pick two girls to sit on the ends, so the first chair has 4 selections, the last has 3, and then it si 5*4*3*2*1
rrusczyk20:54:33
We can think to ourselves 'How would we seat the kids according to these restrictions if we had to make up a seating ourselves?' Our answer is: we'd seat the girls at the ends first, so we make sure we satisfy that restriction.
rrusczyk20:54:40
As before, there are 4 ways to seat a girl in the first seat. Next we seat a second girl in the last seat - there are 3 girls left, so there are 3 choices. Now we have our restriction taken care of. We can then seat the rest of the students as before. There are 5 students left to choose one for the second chair, then 4 students for the third chair, and so on.
rrusczyk20:54:55
Thus, we have 4 x 3 ways to seat girls at each end, and for each of these seatings we have 5 x 4 x 3 x 2 x 1 ways to seat the rest of the students, for a total of:
bwu20:55:07
4*5*4*3*2*1*3
littlesister20:55:07
it is 4*3*5!
bpgbcg20:55:07
So 1440
Wickedestjr20:55:17
I think that the answer is 1440
nikki20:55:20
1440
rrusczyk20:55:23
4 x 3 x 5 x 4 x 3 x 2 x 1 = 1440
rrusczyk20:55:26
ways to seat the students such that there is a girl on either end.
rrusczyk20:55:30
This example brings up two important counting concepts.
rrusczyk20:55:33
First, when dealing with a counting problem that has restrictions, it often pays to think about how you would create one possible arrangement yourself. Here, we realize that if we seated the students ourselves, we'd start with the girls on the ends. This brings us to our second important counting concept:
rrusczyk20:55:44
When dealing with restrictions, it usually helps to deal with the restrictions first. Here, we took care of the girls on the ends first since that was our restriction.
rrusczyk20:56:13
However, there are other clever ways for dealing with restrictions. Let's try one more:
rrusczyk20:56:19
We still have 7 students to seat in a row, but two of them, Ali and Brianna, refuse to sit next to each other. In how many ways can we seat the students now?
rrusczyk20:56:29
As before, we'll look at a wrong solution first:
rrusczyk20:56:32
There are 7 ways to seat Ali. We deal with the restriction first and realize that we can't seat Brianna in either of the seats next to Ali. Hence, Brianna has 4 choices. Then the next student has 5 choices, the one after that has 4 choices, and so on.
rrusczyk20:56:35
What's wrong?
ashstorm20:57:12
Ali could be on the end
bpgbcg20:57:12
If Ali is on the end Brianna has 5 choices
wsjradha20:57:12
what if Ali is on the end?
ashstorm20:57:12
If Ali was on the end, Brianna would have *5* seat choices
rrusczyk20:57:16
Uh-oh.
rrusczyk20:57:21
The problem here is that there are not always 2 seats next to Ali - sometimes he may be put at the end. Hence, sometimes Brianna will have 5 choices for her seat.
rrusczyk20:57:33
We could deal with this by using casework (and we'll discuss very important casework strategies in the course - these tricky casework problems are often the difference in proceeding to the next level in MATHCOUNTS/AMC), but there is a slicker approach. What else could we do?
bwu20:57:41
Find all the solutions and subtract the solutions with Ali and Brianna sitting next to each other.
Wickedestjr20:57:41
Find all the possible arrangements of the seven people, and then subtract the number of ways that Ali and Brianna can sit next to each other to find the compliment which would be the answer.
rrusczyk20:57:53
Interesting idea!
rrusczyk20:58:03
What makes this problem hard is the restriction that Ali and Brianna are not adjacent. We know there are 7! ways to seat the students without any restrictions.
rrusczyk20:58:10
Instead of counting our desired seatings directly, we count what we don't want and subtract.
wsjradha20:58:11
not compliment, complement
rrusczyk20:58:15
:) You are correct!
rrusczyk20:58:46
We know there are 7! ways without restrictions, so we will try to count those that violate our restriction that Ali and Brianna are separate. We'll then subtract these violators from our total.
rrusczyk20:58:50
In how many ways can we seat Ali and Brianna if they are together?
rrusczyk20:59:03
Again, we could seat Ali and then note that Brianna has . . . uh-oh. Brianna might have 1 or 2 choices. We don't want to do casework. What can we do with Ali and Brianna to easily count those cases in which they are together?
FantasyLover20:59:36
120*6*2
rrusczyk20:59:40
Why?
FantasyLover21:00:11
first, we put ali and brianna together, and there's 5 students left, so its 5!, which is 120.
rrusczyk21:00:17
And where does the 2 come from?
FantasyLover21:00:29
they can switch seats
wsjradha21:00:33
120*6*2 because 6 for the position of Ali and Brianna - 2 for the order of Ali and Brianna - 120 for the rest
littlesister21:00:35
Because you could flip them around
bwu21:00:35
Ali and brianna are interchangeable
rrusczyk21:00:40
Ali and Brianna are not the same person. They could be AliBrianna or BriannaAli. Thus, for each of our 6! seatings, there are 2 orders in which we can seat Ali and Brianna in their slot. Hence, there are 2 x 6! ways to seat the students such that Ali and Brianna are together.
ashstorm21:00:47
Count 'blocks'. They could sit like this: A B _ _ _ _ _, _ A B _ _ _ _, etc... for 6 blocks. Then 5! options for the other students, and you can reverse A and B. so 6 x 5! x 2
rrusczyk21:00:51
We could also have counted this with casework:
bpgbcg21:00:56
If Ali is on the end, 2 seats for Ali, each corresponding to 1 seat for Brianna, plus 5 seats for Ali in the middle times 2 choices for Brianna equals 12 choices*120 for the other five people=1440 violators
rrusczyk21:01:00
So, in how many ways can we seat them so that they are apart?
RussianRocket21:01:19
7!- 6x2x5! = 3600
FantasyLover21:01:19
5040-1440=3600
bpgbcg21:01:23
5040-1440=3600
rrusczyk21:01:31
There are 7! ways without restrictions, and 2 x 6! ways for them to be together. This leaves 7! - 2 x 6! ways for them to be apart.
rrusczyk21:01:39
7! - 2 x 6! = 7 x 6! - 2 x 6! = 5 x 6! = 5 x 720 = 3600.
rrusczyk21:01:43
This example brings up a couple more important tactics.
rrusczyk21:01:46
First, when it looks hard to count something directly, try counting the opposite of what you're asked for. We call this approach complementary counting, since 'complement' in dealing with groups of objects in mathematics roughly means 'opposite'. I also call this 'counting what you don't want'.
rrusczyk21:01:52
Second, when your restriction is that some of your items must remain together when putting them in a row, a useful tactic is to consider the items all together as a single item, as we did AliBrianna above. Then you separately consider how many ways you can order the items within the group.
rrusczyk21:02:03
These three basic examples show why it is pointless to memorize your way through counting - I can ask zillions of variations of the above questions. Instead of memorizing your way through each variation, you should learn when to add, when to subtract, when to multiply, and when to divide. Since you already know how to perform these operations, once you know when to do them, you know how to count!
rrusczyk21:02:33
The first of these three problems was considerably easier than most of the problems we will do in the course. The second and third are closer to the middle in difficulty, though they are still a little easier than the average problem.
rrusczyk21:02:43
In general in the course we will go through the ideas more gradually than we did here - each idea will be explored with gradually more difficult examples. Thus, the pace at which new ideas are introduced is slower than we did here (in which we introduced 4 general tactics in two problems!)
rrusczyk21:02:52
You can find more questions like those we cover in the course by checking out the Post Test for the course here:
erictao20050021:03:04
i know how to count: 1,2,3,4,5,6,7,8,9,10...
rrusczyk21:03:10
And that's where we start this class :)
rrusczyk21:03:15
The course will meet for 12 weeks on Tuesdays, 7:30 PM - 9 PM Eastern, starting June 10. The course ends August 26. Each class is 90 minutes.
rrusczyk21:03:22
The course will be taught by Ashley Ahlin. Ashley was the first female to win a medal at National MATHCOUNTS, placing 3rd in 1987. She also won 1st place in the Westinghouse Science Talent Search. Later, Ashley served on the MATHCOUNTS problem writing committee and spoke at the national awards banquet. She has taught at the high school and college levels, and in summer programs for all age levels. Ashley finished her Ph.D. in math at the University of Chicago in 2001.
rrusczyk21:03:36
This course will use a textbook in conjunction with the course: our own Introduction to Counting & Probability book. The material covered in the textbook is roughly equivalent to the material covered in the course. You can see the table of contents and some excerpts from the book here:
rrusczyk21:03:55
The book is required for the course. Students will be able to read additional material that complements the lectures, and will have access to a large number of practice problems at varying levels of difficulty. We are recommending that students read the corresponding chapter(s) in the book before each lecture, and attempt some of that chapter's Review and Challenge Problems after each lecture. We also expect to spend some class time answering students' questions about problems from the textbook.
rrusczyk21:04:13
The homework for the class consists of weekly problems that will be posted to the class message board -- for these problems, you do not turn your solutions in, however you may post them to the message board if you like. The class also has two longer Challenge Sets -- for which you should write up your full solutions and submit them. These solutions will be read, and you will receive detailed feedback.
rrusczyk21:04:33
Are there any questions?
Northboy21:04:35
will u be going back to the intro to algebra anymore?
rrusczyk21:04:46
You can ask questions about that if you want - no more math problems tonight!
prithviramesh21:05:03
What will be the first topic covered in intro to algebra?
rrusczyk21:05:20
The basics: properties of addition and multiplication and order of operations.
wsjradha21:05:22
Will the challenge set questions be available to people who aren't in the Intro to Counting class?
rrusczyk21:05:33
No. Only enrolled students get the Challenge Sets.
erictao20050021:05:36
so am i taking all the cources?
rrusczyk21:06:03
I don't know what classes you are taking - click on the My Classes link when you are logged in. That will tell you what classes you are in.
Northboy21:06:05
will we have a chalenge set every week to turn in the next week?
rrusczyk21:06:32
No - you have 6 weeks for the Challenge Sets. There are other weekly problems that you don't turn in (you talk about them on the message board).
Barry Field21:06:36
should we read the appropriate chapter in the algebra course before each class?
rrusczyk21:06:54
Yes. Same goes for the counting class. The students who read the book do much better than those who don't.
littlesister21:06:57
Will there be answer keys for the homework? How will we correct it?
rrusczyk21:07:05
Yes. We will correct the Challenge Set problems.
wsjradha21:07:07
Are there challenge sets in the AIME A class?
rrusczyk21:07:10
No.
Wickedestjr21:07:11
So will I be getting the Challenge Set?
rrusczyk21:07:16
If you are enrolled in the class..
bwu21:07:18
Is there a MathCounts book for the class?
rrusczyk21:07:20
No.
joslyn21:07:22
Where do we get a challenge set?
rrusczyk21:07:33
On the class home page. The first one will be up after the second class.
RussianRocket21:07:35
Are the assingments graded?
rrusczyk21:07:56
No formal grade unless you need it. Instead, we give something more valuable - thorough feedback on your solutions.
Wickedestjr21:07:59
How do you enroll?
rrusczyk21:08:21
See that link.
hamsterdilly21:08:23
how do you get to the message board again??
rrusczyk21:08:42
The message board is not yet available for the classes, but there will be a link for it a few days before classes start.
prithviramesh21:08:45
When do we begin getting homework?
rrusczyk21:08:52
In the first class.
Ace121:08:55
will the intro to algebra class include instruction on the TI-84?
rrusczyk21:09:11
No. Our classes are not calculator classes.
prithviramesh21:09:14
Is that today?
rrusczyk21:09:16
No.
rrusczyk21:09:29
The three classes we talk about today will not start until the week after next.
ABNoRMaLY_NoT_QuieT21:09:31
I know that before class, we are supposed to learn the material by ourselves, how well should we prepare?
rrusczyk21:09:43
You should read the text as described in the course information document.
Wickedestjr21:09:45
Does it cost money to enroll?
rrusczyk21:09:52
Yes. See the link I posted for the fees.
littlesister21:09:56
How do we correct the homework?
rrusczyk21:10:08
We will give you the solutions to the message board problems that you can compare to your work.
rrusczyk21:10:16
For the Challenge Sets, we do the correcting.
sda133721:10:18
Do you accept Paypal?
rrusczyk21:10:20
Yes.
LightningStreak21:10:22
DO WE HAVE ANY TEST
rrusczyk21:10:38
The Challenge Sets are essentially a 6-week take home test. There are no timed tests.
F-86K Sabre Dog21:10:40
what is the range of questions
rrusczyk21:10:57
From easy review to very hard! No one gets every problem right on the Challenge Sets.
vallabh_suresh21:10:59
So will we be meeting next week? If not when will the classes start?
rrusczyk21:11:06
The class schedule is here:
prithviramesh21:11:11
Do we have a FINAL EXAM?
rrusczyk21:11:21
There are no 'Exams'. There are only problem sets.
Ace121:11:23
will you be doing another demo class before the november classes start?
rrusczyk21:11:26
Yes.
Wickedestjr21:11:28
Is there anything that we can do without having to enroll?
rrusczyk21:11:33
Only Math Jams like this one.
littlesister21:11:35
Will the counting and probability class help us in MathCounts or other contests?
rrusczyk21:12:11
Absolutely. The coach of the National MATHCOUNTS championship team 3 of the last 4 years said that he spends around half his coaching time on counting. And guess what book he'll be using next year :)
prithviramesh21:12:13
Can I bring problems to class?
rrusczyk21:12:27
We prefer you post them on the message board, where we'll have more time to discuss them thoroughly.
aggarwal_221:12:29
do you know when the next math jam like this is
rrusczyk21:12:41
Here is a schedule:
rrusczyk21:12:51
I'll be updating it throughout the summer.
joslyn21:12:53
Can our parents help us on the Challenge Sets??
rrusczyk21:13:05
Yes, but please note that on the Challenge Set when you turn it in.
Northboy21:13:07
how long after the homework is assigned will the answer be on the message board?
rrusczyk21:13:21
Each week, the solutions to the previous week's problems will be on the message baord.
gafrasa21:13:24
Approx. how much time will the homework take each week for the Intro to Algebra course?
rrusczyk21:13:42
You should spend 4-7 hours a week on Introduction to Algebra. This includes reading, homeworks, class time, etc.
hamsterdilly21:13:44
will part of our homework be on the book??
rrusczyk21:13:48
Yes.
prithviramesh21:13:50
What if we don't do our homework?
rrusczyk21:13:54
Then you will learn very little.
Northboy21:13:56
can u access the message board at any time?
rrusczyk21:14:10
Yes, once we activate the board (usally a few days before the class starts)
littlesister21:14:14
How much time will the homework take us weekly on average?
rrusczyk21:14:30
For algebra and counting you should spend 4-7 hours a week on everything as described above.
ashstorm21:14:32
For the AIME, which book would be most useful?
rrusczyk21:14:43
Our Intermediate series of textbooks or Volume 2.
F-86K Sabre Dog21:14:45
what is a message board
rrusczyk21:14:53
It's the screen you go to when you log in.
wsjradha21:14:55
How much homework on the AIME A class?
rrusczyk21:15:04
About 10 problems a week. Some are quite hard!
prithviramesh21:15:23
What if we turn it in late?
rrusczyk21:15:47
You need to ask your teacher for permission ahead of time. If you ask nicely and your teacher is a good mood, you might get a few extra days :)
wsjradha21:15:50
How long will that take on average (AIME A)?
rrusczyk21:16:05
If you take the class seriously, you should be spending 3-5 hours on the class each week.
rrusczyk21:16:48
That's it for the Math Jam tonight. If you have any more questions, you can email me at classes@artofproblemsolving.com!
rrusczyk21:16:58
We look forward to working with those of you who will be in the classes!
vallabh_suresh21:17:09
If we don't get a chance to attend 1 or 2 of the classes, where can we find someone who can teach us what was taught in the class we missed
rrusczyk21:17:24
You can read the transcripts for the classes, then post questions on the class message board if you have any.
