Transcript for the Math Jam "AoPS Classes Math Jam" on May 30.

Math Jam hosted by nsato (Naoki Sato ).

nsato (19:29:35)
Hello, and welcome to the AoPS Summer 2007 Classes Math Jam!

nsato (19:30:02)
Tonight, we will be discussing the MATHCOUNTS Problem Series, the AMC 10 Problem Series, and the Introduction to Number Theory class.

nsato (19:30:37)
We'll go through some sample problems in each class. If you have any questions, we'll take them after all three classes have been discussed.

nsato (19:31:05)
First, we'll discuss the MATHCOUNTS Problem Series.

nsato (19:31:11)
MATHCOUNTS

nsato (19:31:23)
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.

nsato (19:31:32)
The MATHCOUNTS Problem Series will be taught by Ashley Ahlin, a Westinghouse Science Talent Search winner. The course begins Monday, June 11, and classes are Monday nights from 7:30 to 9 PM ET (4:30 to 6 PM PT).

nsato (19:31:39)
Here is an excerpt of some of the curriculum and problems covered during the course.

nsato (19:31:48)


nsato (19:32:01)
Where can we start with this problem?

Anaxerzia (19:32:11)
we can find the primes

nsato (19:32:31)
What can we say about p and p + 1?

Tsui Yang (19:32:38)
even and odd

Anaxerzia (19:32:16)
wich are obviously 2 aand 3

MechaSnoopy (19:32:39)
they are 2 and 3

funmath (19:32:42)


plumbst (19:32:49)
theyre two and 3

cc1 (19:32:53)
they have to be 2 and 3

jama1 (19:33:00)
two and three?

emant777 (19:33:04)
onlu even prime is 2

nsato (19:33:27)
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.

nsato (19:33:35)
Now, how do we find the smallest composite number that is neither a multiple of 2 nor 3?

cc1 (19:34:09)
cant be even number

jama1 (19:34:19)
find numbers with factors that are neither 2 nor 3

nsato (19:34:50)
Right. And remember we're looking for a _composite_ number.

Anaxerzia (19:34:20)
the lowest prime number after them is 5, and the least composite number with 5 and not using 2 and 3 is 5 *5 = 25

cc1 (19:34:45)
25

MechaSnoopy (19:34:48)
25?

susie705 (19:34:48)
smallest composite odd number not divisible by 3: 25?

sri_ram@yahoo.com (19:35:04)
25

Cyrazeno (19:35:02)
25?

nsato (19:35:25)


nsato (19:35:35)


nsato (19:35:50)
How do we start?

nsato (19:36:15)
Can we identify one of the three numbers?

plumbst (19:36:01)
you know that 2 must be one of them

KingRoy (19:36:26)
2

jama1 (19:36:24)
one must be even so one is two

emant777 (19:36:39)
one even

GrantCBHS (19:36:42)
2 because three odds won't add to an even number

nsato (19:36:54)
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.

nsato (19:36:58)
So now what must we do?

cc1 (19:37:00)
find 2 primes that add up to 28

Anaxerzia (19:37:07)
find to primes that sum to 28

nsato (19:37:24)
We must find the largest possible product of two primes which have a sum of 28.

nsato (19:37:28)
How do we do it?

nsato (19:37:46)
We could list out the cases - what are our only options?

JANE CHUNG (19:37:17)
17 and 11

Anaxerzia (19:37:18)
11 & 17

jama1 (19:37:21)
find two primes that add to 28, 11 and 17

susie705 (19:37:48)
5 and 23

cc1 (19:37:54)
5, 23 or11,17

MechaSnoopy (19:38:04)
17,11 or 23, 5

nsato (19:38:29)
Our only options are 11 & 17, and 5 & 23.

nsato (19:38:33)
Which has a larger product?

plumbst (19:38:45)
11*17= 187

emant777 (19:38:47)
11 , 17

Tsui Yang (19:38:47)
11 and 17

funmath (19:38:49)
11 and 17

JANE CHUNG (19:38:52)
11 and 17

Anaxerzia (19:38:56)
11*17=187, 23*5=115

nsato (19:39:15)


nsato (19:39:19)
What is the answer to the problem?

plumbst (19:39:38)
374

MechaSnoopy (19:39:38)
374

funmath (19:39:39)
374

Anaxerzia (19:39:48)
187*2= 374

Cyrazeno (19:39:49)
374

nsato (19:40:05)


nsato (19:40:27)
Another point to keep in mind: Remember to answer what the problem is actually asking for.

nsato (19:40:39)
We'll do one more problem.

nsato (19:40:48)


nsato (19:41:07)
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.

nsato (19:41:17)
First, how many positive integral divisors does 792 have?

funmath (19:41:33)


nsato (19:42:00)


funmath (19:41:51)


ChrisA (19:41:58)
24

nsato (19:42:25)


nsato (19:42:55)
The even divisors have a positive exponent for the prime 2. How many such possibilities does that leave for a?

cc1 (19:43:17)
3

harbinger (19:43:18)
3

nsato (19:43:35)
For even divisors, a can be 1, 2, or 3. How many total even divisors are there?

Anaxerzia (19:43:08)
18

JANE CHUNG (19:43:56)
3*2*2

funmath (19:43:59)


emant777 (19:44:01)
3x3x2=18

harbinger (19:44:05)
18, i think

nsato (19:44:17)


nsato (19:44:25)
So 18 is the final answer.

nsato (19:44:50)
I'll now turn things over to Dave Patrick, who is the instructor for the AMC 10 Problem Series.

DPatrick (19:45:01)
Hi.

DPatrick (19:45:03)
The AMC 10 class starts on June 6, and meets every Wednesday from 7:30-9:00 PM Eastern. The class meets for 12 weeks and ends on August 29. The course is designed to cover a large portion of the curriculum tested on the AMC 10 exam.

DPatrick (19:45:16)
This class is a Problem Series class, meaning that the major focus of the class will be working through various AMC 10 or AMC 12 problems. Although there will be weekly problem sets for each class, students do not submit their homeworks to be graded, and there is no personalized instructor feedback.

DPatrick (19:45:23)
(As a result, these classes are somewhat less expensive that our regular subject classes.)

DPatrick (19:45:47)
The following are excerpts of a couple of the areas of problem solving covered in the AMC 10 Problem Series. (Don't worry if you don't know all the terminology -- it'll all be covered in the class!)

sri_ram@yahoo.com (19:45:45)
is there a difference between amc 10 and amc 12?

DPatrick (19:46:08)
Yes -- the AMC 12 is more difficult (in particular, it includes trigonometry).

DPatrick (19:46:19)
The AMC 10 is only for students in grades 10 and below.

DPatrick (19:46:34)
Here's a sample problem from the AMC 10 class:

DPatrick (19:46:40)


DPatrick (19:47:06)
Recall that arithmetic sequences involve common differences that are constant. Constants are our friends and we should remember how useful they can be.

DPatrick (19:47:12)
How can we use the given information to find the number of sides of the convex polygon?

funmath (19:47:40)
the total of the angles are 180(n-2)

JANE CHUNG (19:47:42)
160-5. (160-5)-5. and keep repeating

DPatrick (19:47:56)
Good -- those are the two key facts that we'll need.

DPatrick (19:48:16)
The largest angle is 160 and then the angles decrease by 5.

DPatrick (19:48:24)


DPatrick (19:48:36)
Note that the last value should be 160 - 5(n-1), not 160 - 5n because we started with 160.

DPatrick (19:48:49)
The angles are in arithmetic progression and so we can find their sum.
Conveniently, we also have a formula for the sum of the interior angles of a convex n-sided polygon: 180(n - 2).

DPatrick (19:49:02)
What is the sum of our arithmetic series?

Anaxerzia (19:48:29)
triangluar numbers

Anaxerzia (19:48:35)
mitliplied by 5

DPatrick (19:49:30)
Indeed...

DPatrick (19:49:41)


DPatrick (19:50:19)
The sum 1+2+...+(n-1) at the end is called a [b]triangular number[/b], and there is a nice formula for it!

DPatrick (19:50:41)
What is it?

GrantCBHS (19:50:51)
I thought triangular numbers were 1,3,6,10 etc...

DPatrick (19:51:09)
Indeed they are! 1 = 1, 3 = 1+2, 6 = 1+2+3, and so on.

funmath (19:51:12)


DPatrick (19:51:36)
Correct. But we only have 1+2+...+(n-1).

harbinger (19:50:55)
(n(n-1))/2

DPatrick (19:51:49)
Good.

DPatrick (19:51:58)
1+2+...+(n-1) = n(n-1)/2.

DPatrick (19:52:25)


DPatrick (19:52:45)


DPatrick (19:52:58)
How do we solve this?

Tsui Yang (19:53:11)
distribute

JANE CHUNG (19:53:12)
multiply the equation by 2

DPatrick (19:53:51)
We can multiply by 2 to get rid of the fraction, and then multiply this out and collect terms.

DPatrick (19:54:06)


DPatrick (19:54:22)
What are the solutions to this quadratic?

cc1 (19:54:49)
-16 and 9

DPatrick (19:55:00)
We can solve the quadratic in a number of ways including factoring it into
(n - 9)(n + 16) = 0, so n = 9 or -16.

DPatrick (19:55:10)
So what is the answer to the original problem?

jama1 (19:55:11)
so its 9?

emant777 (19:55:17)
9 sides

cc1 (19:55:17)
9

DPatrick (19:55:27)
We know that a polygon cannot have a negative number of sides and so we have our answer, n = 9. (A).

DPatrick (19:55:43)
This problem shows us that we can often draw upon information not directly mentioned in the problem to help us create an equation to solve. Of course, we also needed to sum an arithmetic series to give us the second side of this equation.

DPatrick (19:56:03)


DPatrick (19:56:33)
How can we count this?

funmath (19:56:33)
there are 8 vertices

DPatrick (19:56:49)
Right, we start with the fact that a cube has 8 vertices.

Anaxerzia (19:56:49)
We are basically choosing two points from 8.

Anaxerzia (19:56:37)
its 8C2 = 28

plumbst (19:57:03)
how many ways to choose 2 of the 8 vertices

cc1 (19:57:05)
8 combination2

DPatrick (19:57:14)


DPatrick (19:57:32)
Again, if you don't know what this notation means, don't worry, because we're going to cover it in the class.

DPatrick (19:57:57)
Understanding combinations can directly lead us to quick solutions to some AMC problems. This one was relatively easy but the next one, although it uses similar concepts, is more difficult:

DPatrick (19:58:03)


DPatrick (19:58:36)
How can we count these numbers?

funmath (19:58:16)
casework

DPatrick (19:58:52)
What are the cases?

funmath (19:59:23)
increasing and decreasing

susie705 (19:59:35)
three digits in increasing order and three digits in decreasing order

DPatrick (19:59:50)
OK...we can break this problem into two cases. We can count the three digit numbers with increasing and decreasing digits separately.

DPatrick (19:59:55)
Let's begin with three digit numbers with increasing digits. How many are there going to be?

DPatrick (20:00:55)
Here's a hint. Suppose we have three different digits such as 2, 4, and 7. In how many ways can I arrange them to form a 3-digit number with increasing digits?

plumbst (20:01:09)
1

funmath (20:01:12)
only one

harbinger (20:01:12)
1

Nerd_of_the_Ages (20:01:14)
1

DPatrick (20:01:23)
Right, there's only one way: as 247.

DPatrick (20:01:36)
How does this help us count all of the 3-digit numbers with increasing digits?

sri_ram@yahoo.com (20:01:36)
9*8*7/3

DPatrick (20:02:15)
Close...you're just missing a punctuation mark.

Anaxerzia (20:02:12)
9C3

plumbst (20:02:13)
youre choosing 3 digits out of 9 since 0 cant be one of them

DPatrick (20:02:21)
Exactly.

DPatrick (20:02:36)


funmath (20:02:32)
aren't increasing and decreasing symmetric?

DPatrick (20:02:55)
Almost...how is decreasing slightly different?

MichaelFaraday (20:03:01)
we can use 0

plumbst (20:03:04)
you can use the 0's

Nerd_of_the_Ages (20:03:05)
you can use 0

DPatrick (20:03:16)
Right! For the decreasing ones, we're allowed to have a 0.

DPatrick (20:03:25)


DPatrick (20:03:43)
So what's the final answer?

Jzilla_2 (20:03:47)
204

choidavid91 (20:03:54)
120+84=204

DPatrick (20:04:01)
We now add the total numbers from each case:
84 + 120 = 204, so our answer is (C).

DPatrick (20:04:09)
The key to this problem was recognizing that the case for increasing digits is different from the case for decreasing digits. When solving combinatorial problems it is necessary to consider exactly how we are choosing members of a group. When the methods are different we must set up cases and calculate them separately.

DPatrick (20:04:26)
Here is the schedule of topics for this summer's AMC 10 Problem Series course:

DPatrick (20:04:33)
Week 1 (June 6): The Mechanics of Algebra
Week 2 (June 13): Word Problems
Week 3 (June 20): Patterns, Sequences, and Series
Week 4 (June 27): Functions and Graphs
No class July 4 (holiday)
Week 5 (July 11): Systems of Equations and Inequalities
Week 6 (July 18): Number Theory

DPatrick (20:04:37)
Week 7 (July 25): Counting Methods
Week 8 (August 1): Probability
Week 9 (August 8): The Mechanics of Geometry
Week 10 (August 15): Triangles
Week 11 (August 22): Circles
Week 12 (August 29): Polygons and Solids

DPatrick (20:05:11)
Now I'll turn things back over to Naoki, who will discuss the Introduction to Number Theory course. (We'll answer your questions at the end.)

nsato (20:05:41)
Introduction to Number Theory

nsato (20:05:47)
The Introduction to Number Theory Subject Class will be taught by Joshua Zucker. The course begins Thursday, June 7, and classes are Thursday nights from 7:30 to 9 PM ET (4:30 to 6 PM PT).

nsato (20:05:57)
Many students who already know how to solve MATHCOUNTS level problems about divisibility, base numbers, and divisor counting might think they have little need for this class, but many of the students who have made the most of this class were participants at national MATHCOUNTS and the AIME and found many of the problems discussed in class very challenging.

nsato (20:06:07)
The Introduction to Number Theory class covers divisibility problems, clever uses of prime factorization, base numbers, linear Diophantine equations, the Euclidean Algorithm, and covers the mechanics of modular arithmetic in thorough detail. There are also many other topics covered in this course that are rarely mentioned in math books at all such as the ways in which we can determine information about the number of digits in a repeating decimal expansion whether in base 10 or in some other number base.

nsato (20:06:31)
The topics covered in this course are crucial to an understanding of such areas as cryptography and computer science.

nsato (20:06:39)
The following mini-lesson is excerpted from one of the Number Theory classes. Pay attention to the reasons behind the way we discuss the material because some of the harder problems at the end might surprise you.

nsato (20:06:47)
COUNTING DIVISORS

nsato (20:06:53)
Once we know how to decompose numbers into their prime factorization we can begin to use this tool to solve other problems.

nsato (20:06:58)
One such problem is answering the question 'how many positive divisors does a particular integer have?' This kind of counting problem is common in number theory.

nsato (20:07:04)


nsato (20:07:18)
How can we describe any divisor of 200?

nsato (20:07:33)
Hint: Think prime factorization!

nsato (20:07:40)


plumbst (20:07:54)
4 is greater than 3

jama1 (20:08:15)
because 200 is not divisible by 16

choidavid91 (20:08:19)
it is impossible to make that number out of 2^3 and 5^2

nsato (20:08:32)


emant777 (20:07:32)
2^a x 5^b

MichaelFaraday (20:07:40)
as a multiple of ;$2^n x 5^y$

nsato (20:08:40)


nsato (20:08:49)
What are the possible values of each $a$ and $b$ for these divisors and why?

Anaxerzia (20:09:09)
Any divisor is in the form 2^a * 5^b, with A being 0,1,2,3 and B being 0,1,2

susie705 (20:09:23)
a can be 0, 1, 2, 3

harbinger (20:09:24)
0<=a<=3 0<=b<=2

cc1 (20:09:36)
0,1,2,3=a, 0,1,2=b

nsato (20:09:46)
We know that a can be 0, 1, 2, or 3 while b can be 0, 1, or 2 because the exponents to each prime of a divisor can be no larger than exponent to those primes in the factorization of 200.

nsato (20:09:57)
Now that you know what the possible values for a and b are, how can you use this to count the total number of divisors of 200?

nsato (20:10:03)
Consider this tree diagram:

nsato (20:10:17)


nsato (20:10:19)
http://www.artofproblemsolving.com/Classes/IntroNumbers/L3/exponenttree.gif

Anaxerzia (20:10:05)
4*3=12

harbinger (20:10:31)
12

nsato (20:10:39)
Each combination of values for a and b represents one divisor that we can construct for the number 200. Since we have 4 values for a to choose from and 3 values for b to choose from, we have 4 x 3 = 12 total divisors.

susie705 (20:10:48)
there are 4 values for a to choose from and 3 values for b to choose from

nsato (20:11:02)
Indeed we can see that this is true:
1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, and 200 are all the positive divisors of 200.

nsato (20:11:09)
So, why is it that we multiply the number of possible values for the exponent of each prime together?

nsato (20:11:41)
The reason we simply multiply the numbers of values for the exponents together is because we can select the values for each exponent independently from the values of the other exponent(s).

nsato (20:11:49)


nsato (20:12:02)


nsato (20:12:17)
In general what can we say about the number of positive divisors of an integer n with a prime factorization

nsato (20:12:20)


MichaelFaraday (20:13:00)
the number of different values of each e multiplied together

harbinger (20:13:02)
(e1+1)*(e2+1)...(em+1)

plumbst (20:13:19)
add 1 to each exponent and multiply them

nsato (20:13:25)


nsato (20:13:41)


nsato (20:13:50)


nsato (20:14:01)
To make this method more clear, we shall now work through some exercises.

nsato (20:14:08)
Exercise: How many positive divisors does 60 have?

nsato (20:14:14)
First, what is the prime factorization of 60?

susie705 (20:14:28)
2 x 2 x 3 x 5

Anaxerzia (20:14:30)
2^2*3*5

jsun_100 (20:14:30)
2^2x3x5

funmath (20:14:34)


plumbst (20:14:35)
2^2*3^1*5^1

Cyrazeno (20:14:37)
2^2*3*5

nsato (20:14:53)


nsato (20:15:00)


nsato (20:15:08)


LHM (20:15:19)
3,2,2

plumbst (20:15:23)
3 ; 2 ; 2

ChrisA (20:15:24)
3,2,2

susie705 (20:15:33)
e1 has 3, e2 and e3 each have 2

nsato (20:15:48)


nsato (20:15:53)
Given the number of possibilities for each exponent, how do we use this to determine the total number of divisors of 60?

sri_ram@yahoo.com (20:14:44)
12

funmath (20:15:00)


chessmaster (20:15:03)
2*3*2 or 12 factors

Anaxerzia (20:15:18)
3*2*2 = 12

emant777 (20:15:19)
(2+1)(1+1)(1+1)=12

jsun_100 (20:15:32)
so there are 12 divsors of 60?

MichaelFaraday (20:16:03)
multiply them together

choidavid91 (20:16:14)
3*2*2=12

nsato (20:16:40)


nsato (20:16:44)
As the class continues we begin to discuss more difficult questions such as 'How many of the divisors of 360 are even?' and 'How many of the divisors of 9800 are perfect squares?'

nsato (20:16:54)
Here are a few more kinds or problems that we will be tackling in class:

nsato (20:17:00)


nsato (20:17:06)


nsato (20:17:12)


nsato (20:17:19)


nsato (20:17:28)


nsato (20:17:34)


jsun_100 (20:17:19)
whats the Diophantine equation thing?

nsato (20:17:45)
It means we want integer solutions.

nsato (20:18:02)
One more word about this number theory class. If you do not know modular arithmetic well enough to use it on most any problem up to AMC-12 level, this is a class that you would benefit from taking.

nsato (20:18:35)
Now we'll take questions about the classes.

GrantCBHS (19:35:39)
Hi, I'm interested in enrolling my little brother in this course. He's just entering 6th grade and has only basic prealgebra skills (no algebra I or geometry, or at least very minimal). Would this course be recommended, and if he is enrolled and finds it too diffucult, are we able to drop him out and get a refund?

nsato (20:19:40)
That should be fine, because the course starts at a very basic level. You can drop the course until the third class, and receive a full refund. All of our classes have the same drop policy.

Anaxerzia (20:04:59)
Will taking the AMC 10 help for National MathCounts level problems?

Anaxerzia (20:05:15)
AMC 10 course i mean

nsato (20:21:17)
The AMC 10 class, not surprisingly, is geared towards the AMC 10 contest, so there will be slightly different subjects, and we go a little deeper into the concepts involved. As I understand, MATHCOUNTS is about solving problems quickly, so the two contests emphasize different skills. So I would say you might get some benefit, but not a great amount.

JANE CHUNG (20:18:39)
Is this class transcript going to be able for a download?

plumbst (20:20:36)
how do i get the class transcript?

nsato (20:22:07)
Transcripts for all classes are provided. They are usually available within a day of the class.

choidavid91 (20:18:45)
Is it recommended to have the AOPS intro to number theory book?

nsato (20:22:40)
Very much, because the book is based on the course.

harbinger (20:19:29)
AMC 10: How do these sessions operate?

nsato (20:23:57)
The classes will be like the math jam tonight, taking part in a virtual classroom. Also, we will be asking problems and taking up their solutions, again just like we have done in the math jam.

LHM (20:19:30)
if you already do lots of Mathcounts past problems, does the Mathcounts class help?

nsato (20:25:36)
It depends on how comfortable you are at solving MathCounts problems. Even if you do lots, it also helps to know the underlying principles of what's going on in the problems, and that's what we teach in the class.

JANE CHUNG (20:20:21)
How long will the homework take? and will the answers to the homework be thoroughly explained?

nsato (20:26:52)
It depends on the student (and by homework, I assume you mean the two exams). Some put a lot of effort into them, and others not so much. The students that put the most time and effort tend to learn the most as well. We will provide full solutions after the solutions from the students have been collected.

Nerd_of_the_Ages (20:20:25)
Is mathcounts of AMC 10 harder

nsato (20:27:50)
MathCounts is for students in grades 6 to 8, so that would make the AMC 10 more challenging.

GrantCBHS (20:20:25)
What would be the main differences between taking the AMC 12 course or the AMC 10 course be?

nsato (20:30:07)
Mainly the difficulty level. The topics covered are about the same (e.g. algebra, geometry, etc.), but the AMC 12 class will cover the topics at a greater level.

susie705 (20:21:10)
Can instructors and tutors enroll in these courses?

nsato (20:30:31)
Yes, the classes are open to anyone who is interested in taking them.

cc1 (20:21:54)
why dont you have an AMC8 course

nsato (20:31:54)
Good question, and I don't have a good answer. I think it's because the AMC 8 contest was introduced only very recently, after we set up our classes. Or it could be that the MathCounts class would cover more or less the same thing.

harbinger (20:22:27)
Is the AMC 10 class going to deal with any Trig?

nsato (20:32:37)
No, there's no trig. We do cover geometry (i.e. circles and triangles).

tennismom (20:22:51)
what is the best class for a 7th grader who has completed Algebra 1 and is going to start beginning Mu Alpha Theta?

nsato (20:33:20)
I would recommend the MathCounts class.

Anaxerzia (20:22:58)
When will the other Introductory level courses be available?

nsato (20:34:24)
We schedule our classes semester by semester, but we always have at least one introductory level class each semester. I couldn't tell you which one it will be for this fall.

JANE CHUNG (20:24:25)
about how many people are in a class?

nsato (20:34:46)
It varies a lot - on average about 60, and about 20 to 30 will show up in the classroom.

choidavid91 (20:31:28)
This is more of a question about AOPS classes in general --are calculators allowed to be used? Also are calculators allowed for the AMC?

nsato (20:36:17)
You may certainly use a calculator in the class, or whatever tools you want to. (The exception might be MathCounts classes where we simulate the contest.) I believe that calculators will no longer be allowed on the AMC contests, starting this year? Maybe next year.

jsun_100 (20:31:31)
what do you mean by exams?

nsato (20:36:35)
In the subject classes, there is a midterm and a final.

JANE CHUNG (20:33:25)
For a 8th grader who is currently taking Algebra 2, (next year Pre-Calc), do you recommend the AMC 10 or AMC 12.

nsato (20:37:47)
I would recommend the AMC10. Even though he (she?) has taken some advanced classes, there are still probably some gaps that would be nicely filled in by the AMC 10.

nsato (20:38:03)
Any other questions?

nsato (20:39:00)
Then that's it for tonight's math jam. Thanks for coming!