Math Jams

Who Wants to Be a Mathematician, Round 1

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AoPS instructor David Patrick will discuss the problems on Round 1 of qualifying for the 2019 Who Wants to Be a Mathematician Championship. We will be joined by Mike Breen and Bill Butterworth, the creators of the game. Mike is also the host of the Championship finals, to be held in Baltimore in January 2019.

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Facilitator: Dave Patrick

DPatrick 2018-09-27 19:30:11
Welcome to the 2018-19 Who Wants to Be a Mathematician Championship Round 1 Math Jam!
DPatrick 2018-09-27 19:30:23
I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens of AoPS classes over the past 14 years, and I've written or co-written a few of our textbooks.
DPatrick 2018-09-27 19:30:36
I also once was a contestant on ABC's Who Wants to Be a Millionaire back before I started working at AoPS, way back when Regis Philbin was still the host. Here's a picture (I'm on the left, Regis is on the right):
DPatrick 2018-09-27 19:30:39
https://www.washington.edu/alumni/columns/march00/images/patrick.jpg
DPatrick 2018-09-27 19:30:43
Photo Credit: Maria Melin, copyright 1999 ABC Television..
Bubble3 2018-09-27 19:30:59
You won a car right
DPatrick 2018-09-27 19:31:07
Basically, yes: I didn't win the million bucks, but I did win enough to buy a new car.
DPatrick 2018-09-27 19:31:19
Of course, cars were cheaper back in 1999.
DPatrick 2018-09-27 19:31:35
Before we get started I would like to take a moment to explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.
DPatrick 2018-09-27 19:31:52
The classroom is moderated, meaning that participants can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
DPatrick 2018-09-27 19:32:17
This helps keep the session 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.
DPatrick 2018-09-27 19:32:34
There are a lot of students here! Only a fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
DPatrick 2018-09-27 19:32:48
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the material for every problem as we go.
DPatrick 2018-09-27 19:33:01
Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We usually do in our classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!
DPatrick 2018-09-27 19:33:15
We have an assistant here to help out tonight: David Stoner, or applepi2000. David is a senior at Harvard and longtime AoPSer. He plans to concentrate in theoretical mathematics, and he also enjoys studying computer science and philosophy. In his high school years, David enjoyed immersing himself in the mathematical community. He won the USAJMO in 2012, and won the USAMO in 2013, 2014, and 2015, achieving a 42 on the 2015 exam. He was a member of the winning United States team for the 2013 RMM and 2015 IMO competitions, earning a gold medal at each contest. Outside of mathematics, David enjoys playing blitz and classical chess, solving Nikoli puzzles, and eating apples.
DPatrick 2018-09-27 19:34:16
David can try to help you if you have a question or are having some other difficulty. He may open a private window with you to chat if he's able, but again it's pretty busy tonight so please understand if he doesn't get to you quickly (or at all).
DPatrick 2018-09-27 19:34:39
Also joining us tonight are the co-creators of the WWTBAM contest, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
DPatrick 2018-09-27 19:34:51
Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill began Who Wants to Be a Mathematician for the American Mathematical Society in 2001. The first national game was in 2010. Mike has been on Jeopardy! and Wheel of Fortune (if you want to know if he won lots of money on either show, note that he is still working for a living) and may be the only person ever to cut his hand on the wheel. Who Wants to Be a Mathematician has so far been much safer.
mikebreen 2018-09-27 19:34:57
Hello, everyone. Happy to be here with you and Dave.
TPiR 2018-09-27 19:35:01
Hello, everyone!
DPatrick 2018-09-27 19:35:13
Bill earned an undergraduate degree in mathematics from Santa Clara University and a Ph.D. from Northwestern University, and is currently an associate professor and associate chair of mathematics at DePaul University. He shares a life-long interest in game shows with colleague Mike Breen, with whom he works as the not-so-lovely assistant on the mathematics game show Who Wants to Be a Mathematician.
DPatrick 2018-09-27 19:35:22
Can you guess why Bill's username is TPiR?
shesa 2018-09-27 19:35:29
does your name have anything to do with the price is right
Mathisfun04 2018-09-27 19:35:33
the price is right?
DPatrick 2018-09-27 19:35:40
Yes indeed! In addition to authoring articles and presenting talks related to game-show mathematics, Bill served as mathematics consultant to the CBS television show The Price is Right from 1997 to 2009. (Hence, his username.)
DPatrick 2018-09-27 19:35:56
As you can see, we have a lot of game show background here tonight!
DPatrick 2018-09-27 19:36:06
On to the contest info:
DPatrick 2018-09-27 19:36:15
Who Wants to Be a Mathematician is run by the American Mathematical Society (AMS). The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
DPatrick 2018-09-27 19:36:27
Tonight we'll be talking about Round 1 of the Championship contest, which concluded yesterday.
DPatrick 2018-09-27 19:36:46
Round 1 consisted of 10 questions, with a 15-minute time limit. So the problems are quick: you have an average of 90 seconds per question. (But as we'll see as we work through the problems, some of them shouldn't take you nearly that long.) No books, notes, calculators, or internet access were permitted during the contest.
DPatrick 2018-09-27 19:37:00
We'll take a bit longer than 15 minutes tonight, because we'll stop along the way to discuss each question. Please also remember that the purpose of this Math Jam is to work through the solutions to the problems, and not to merely present the answers. "Working through the solutions" often includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step.
DPatrick 2018-09-27 19:37:31
Some of the questions may have interesting sidetracks, so we may also stop and view some of the scenery along the way.
DPatrick 2018-09-27 19:37:59
This year, the qualifying round had roughly 5200 participants, of which roughly 1850 advanced to Round 2. (More about that later.)
DPatrick 2018-09-27 19:38:15
So let's get started on the problems!
DPatrick 2018-09-27 19:38:22
1. What is the ones (units) digit of $2019^2 - 2018^2$?
$ $
(a) $1$ (b) $3$ (c) $5$ (d) $7$
DPatrick 2018-09-27 19:38:30
(All of the Round 1 questions are multiple-choice.)
DPatrick 2018-09-27 19:38:38
Reminder: please do not simply post your answer. Instead, please post suggestion(s) as to how to solve the problem.
mathman629 2018-09-27 19:39:03
just use difference of squares
shesa 2018-09-27 19:39:03
we only care about the units digit, so let's just look at the units digit of 9^2 and 8^2
Allen31415 2018-09-27 19:39:03
look at the last digit: you get 9^2-8^2
Orangecounty2 2018-09-27 19:39:03
difference of squares
amuthup 2018-09-27 19:39:03
difference of squares
WalkerTesla 2018-09-27 19:39:03
$a^{2}-b^{2} = (a-b)(a+b)$
aussie_math_kid 2018-09-27 19:39:03
9 squared minus 8 squared
DPatrick 2018-09-27 19:39:12
Good -- there are basically two ways to approach it.
DPatrick 2018-09-27 19:39:25
(Possibly more ways too -- but two that I though of, and apparently most of you did too.)
DPatrick 2018-09-27 19:39:36
One is to note that only the units digits matter.
DPatrick 2018-09-27 19:39:50
That is, the units digit of $2019^2$ is the same as the units digit of $9^2$, which is $81$. So its units digit is $1$.
DPatrick 2018-09-27 19:40:04
And similarly the units digit of $2018^2$ is the same as the units digit of $8^2$, which is $64$. So its units digit is $4$.
DPatrick 2018-09-27 19:40:08
And when we subtract a number ending in $4$ from a number ending in $1$, what units digit do we get?
hobbit16 2018-09-27 19:40:29
7
jcao0715 2018-09-27 19:40:29
7
GoodInMathEverytime 2018-09-27 19:40:29
7
drakecap 2018-09-27 19:40:29
7
Apurple 2018-09-27 19:40:29
7
mutinykids 2018-09-27 19:40:29
7
DPatrick 2018-09-27 19:40:32
We get, for example, $11-4 = 7$. So the units digit of the difference is $7$. $\boxed{\text{(d)}}$
DPatrick 2018-09-27 19:40:47
You could also be a little more fancy and write $1 - 4 \equiv -3 \equiv 7 \pmod{10}$. (If you're familiar with modular arithmetic, that is.)
DPatrick 2018-09-27 19:41:05
But also, as many of you mentioned, we can use the difference-of-squares factorization instead.
flyhawkeye 2018-09-27 19:41:19
$2019^2-2018^2=(2019+2018)(2019-2018)=4037$, so the answer is $7$.
RootThreeOverTwo 2018-09-27 19:41:19
(2019+2018)(2019-2018)
DPatrick 2018-09-27 19:41:27
Right. We get $2019^2 - 2018^2 = (2019 + 2018)(2019 - 2018)$.
DPatrick 2018-09-27 19:41:41
This is especially convenient because the second factor is just $1$.
DPatrick 2018-09-27 19:41:48
So the difference is equal to $2019 + 2018 = 4037$, and has units digit $7$.
DPatrick 2018-09-27 19:42:02
Next up:
DPatrick 2018-09-27 19:42:06
2. What is the area in the first quadrant bounded by the graph of $x + 2y = 4$ and the $x$- and $y$-axes?
$ $
(a) $1$ (b) $2$ (c) $4$ (d) $8$
phoenix9 2018-09-27 19:42:36
I just drew a quick graph of it
khina 2018-09-27 19:42:36
just draw a triangle
a1b2 2018-09-27 19:42:36
It forms a triangle
phoenix9 2018-09-27 19:42:36
the y and x intercpt
DPatrick 2018-09-27 19:42:58
Good idea -- let's just draw it. The graph is a line.
DPatrick 2018-09-27 19:43:03
And we care about where it intersects the axes.
Orangestripe 2018-09-27 19:43:17
You could find the x and y intercepts by substituting zero for each of the variables respectively.
DPatrick 2018-09-27 19:43:29
Excellent idea.
DPatrick 2018-09-27 19:43:38
For example, the line intersects the $x$-axis where $y=0$.
DPatrick 2018-09-27 19:43:55
So we just substitute $y=0$ into the equation for the line. This gives $x=4$, so the point $(4,0)$ is the intersection point.
DPatrick 2018-09-27 19:44:11
And the intersection with the $y$-axis is?
kcbhatraju 2018-09-27 19:44:32
(0, 2) for the y intercept
bibilutza 2018-09-27 19:44:32
where x=0
thea72 2018-09-27 19:44:32
(0,2)
DPatrick 2018-09-27 19:44:38
It intersects the $y$-axis where $x=0$.
DPatrick 2018-09-27 19:44:43
This gives $2y = 4$, so the point $(0,2)$ is the intersection point.
DPatrick 2018-09-27 19:44:51
So now we can make a quick sketch:
DPatrick 2018-09-27 19:44:54
ASnerd 2018-09-27 19:45:16
2*4/2=4
Scipio1 2018-09-27 19:45:16
2x4/2=4
CoolKidWiz 2018-09-27 19:45:16
Then we use 1/2 * h * b
Orangecounty2 2018-09-27 19:45:19
now its 4*2/2 = 4
IlinoisMathlete 2018-09-27 19:45:19
1/2*b*h= (4*2)/2 which is 4
WalkerTesla 2018-09-27 19:45:19
$\dfrac{2\cdot 4}{2} = 4$
DPatrick 2018-09-27 19:45:26
Yes. This is a right triangle with legs of length $2$ and $4$.
DPatrick 2018-09-27 19:45:31
So its area is $\frac12(2)(4) = 4$. $\boxed{\text{(c)}}$
DPatrick 2018-09-27 19:46:11
And on to #3. (By the way, numbers 1-3 were the three easiest problems on the contest, in the sense that they had the highest percentages of correct answers.)
DPatrick 2018-09-27 19:46:15
3. Let $f(x) = x^2 + 7$. What is $f(f(2))$?
$ $
(a) $11$ (b) $107$ (c) $121$ (d) $128$
DPatrick 2018-09-27 19:46:26
How do we evaluate this?
abdeet 2018-09-27 19:46:43
f(2)=11
mutinykids 2018-09-27 19:46:43
substitute 2 into the equation first
GoodInMathEverytime 2018-09-27 19:46:43
plug x as 2
hobbit16 2018-09-27 19:46:43
First find f(2) and then plug that value into f again
Panda_dino 2018-09-27 19:46:43
substitute x for 2
kevalshah2005 2018-09-27 19:46:43
find out f(2) then find out f() of that.
mathman629 2018-09-27 19:46:46
Plug in 2, and then plug in the result of f(2) into the function again. That's the answer
jcao0715 2018-09-27 19:46:50
substitue 2 into the function and substitute that value into the function again
DPatrick 2018-09-27 19:47:03
Exactly. First we find $f(2)$, and then whatever that is, we plug it back into $f$ to get the final answer.
DPatrick 2018-09-27 19:47:11
So we start with $f(2) = 2^2 + 7 = 4 + 7 = 11$.
DPatrick 2018-09-27 19:47:25
So $f(f(2)) = f(11)$...which is what?
Robot7620_2 2018-09-27 19:47:43
128
Bubble3 2018-09-27 19:47:43
128
Mathdude102 2018-09-27 19:47:43
128
o99999 2018-09-27 19:47:43
128
Apurple 2018-09-27 19:47:43
11^2+7=128
flyhawkeye 2018-09-27 19:47:43
$11^2+7=128$
DPatrick 2018-09-27 19:47:46
$f(11) = 11^2 + 7 = 121 + 7 = 128$. $\boxed{\text{(d)}}$
DPatrick 2018-09-27 19:48:08
This problem turned out to be the easiest of all ten problems on the contest.
DPatrick 2018-09-27 19:48:31
Now they start to get a little harder.
DPatrick 2018-09-27 19:48:35
4. The radius of a sphere is $6$ cm. What is the sphere's volume divided by its surface area (ignore units)?
$ $
(a) $1.5$ (b) $2$ (c) $2.5$ (d) $3$
DPatrick 2018-09-27 19:49:01
So I think either you know these formulas or you don't. They're very hard to derive in real time.
a1b2 2018-09-27 19:49:42
Use the volume and SA formula
flyhawkeye 2018-09-27 19:49:42
Use $V=\frac{4}{3}\pi r^3$ and $SA=4\pi r^2$
phoenix9 2018-09-27 19:49:42
Find the volume and SA of the sphere using their formulas
hobbit16 2018-09-27 19:49:42
Find the volume as 4/3 pi r^3 and surface area as 4 pi r^2 and ratio them.
asdf334 2018-09-27 19:49:42
(4/3 * pi * r^3) / (4 * pi * r^2)
Scipio1 2018-09-27 19:49:42
find the volume and surface area with formulas, 4/3pi•r^3, and 4pi•r^2?
drakecap 2018-09-27 19:49:42
4/3 pi r^2 / 4 pi r^2 = 1/3 r
DPatrick 2018-09-27 19:49:55
Right -- let me work through these.
DPatrick 2018-09-27 19:50:07
The volume of a sphere is $\dfrac43 \pi r^3$, where $r$ is the radius.
DPatrick 2018-09-27 19:50:14
And the surface area is $4 \pi r^2$.
DPatrick 2018-09-27 19:50:27
So when we divide volume by surface area, we get $$\dfrac{\frac43 \pi r^3}{4 \pi r^2}.$$
kevalshah2005 2018-09-27 19:50:38
pi cancels out
abank 2018-09-27 19:50:40
r/3
DPatrick 2018-09-27 19:50:48
Hurray! This simplifies to $\dfrac{r}{3}$. (The $\pi$s cancel, which is good, since none of the answer choices contain $\pi$.)
Bubble3 2018-09-27 19:50:58
Substitute $6$ for $r$
Orangestripe 2018-09-27 19:50:58
6/3=2
bibilutza 2018-09-27 19:51:02
6/3=2
FreedomLantern 2018-09-27 19:51:02
r/3=6/3=2
krazykid548 2018-09-27 19:51:02
Thus the answer is 2!
DPatrick 2018-09-27 19:51:11
And when $r=6$, the final answer is just $\dfrac63 = 2$. $\boxed{\text{(b)}}$
o99999 2018-09-27 19:51:19
How do you derive the surface area of a sphere formula?
DPatrick 2018-09-27 19:51:26
Well, it's pretty hard.
DPatrick 2018-09-27 19:51:38
But there is a little historical gadget that may make them easier to remember.
DPatrick 2018-09-27 19:51:45
The Greek mathematician Archimedes first discovered these formulas over 2000 years ago!
DPatrick 2018-09-27 19:52:05
His very clever idea was to inscribe the sphere inside a cylinder. This makes the cylinder have radius $r$ and height $2r$. This is my very, very crude picture of this:
DPatrick 2018-09-27 19:52:10
DPatrick 2018-09-27 19:52:41
His further very very very clever observation was that both the volume and the surface area of the sphere are exactly $\frac23$ that of the surrounding cylinder. This requires quite a bit more work to see.
DPatrick 2018-09-27 19:53:15
But the nice thing is that the volume and surface area formulas for the cylinder are a lot easier.
DPatrick 2018-09-27 19:53:26
The volume of the cylinder is $V = bh$ where $b = \pi r^2$ is the area of the base and $h = 2r$ is the height. So $V = 2\pi r^3$.
DPatrick 2018-09-27 19:53:42
So the volume of the sphere is $\frac23$ of this, or $\frac23(2\pi r^3) = \frac43\pi r^3$.
DPatrick 2018-09-27 19:54:01
And, the surface area of the cylinder is twice the base area ($\pi r^2$) plus the vertical area, which is the height ($2r$) times the circumference $(2\pi r)$.
DPatrick 2018-09-27 19:54:12
Adding this up gives $S = 2(\pi r^2) + (2r)(2\pi r) = 6\pi r^2$.
DPatrick 2018-09-27 19:54:20
So the surface area of the sphere is $\frac23(6 \pi r^2) = 4 \pi r^2$.
phoenix9 2018-09-27 19:54:30
how do we know its 2/3
DPatrick 2018-09-27 19:54:52
Well, it turns our Archimedes was a really clever guy! He was able to figure it out, over 2000 years ago!
mathleticguyyy 2018-09-27 19:54:55
Why isn't it 1/3 like with cones
DPatrick 2018-09-27 19:55:02
These facts are definitely related.
DPatrick 2018-09-27 19:55:31
I encourage you to look it up sometime!
DPatrick 2018-09-27 19:55:48
But right now, let's move on with the contest problems.
DPatrick 2018-09-27 19:55:58
5. Which of the following is the closest integer to the cube of $\tan\left(\frac\pi3\right)$ ($\frac\pi3$ is in radians, not degrees)?
$ $
(a) $5$ (b) $6$ (c) $7$ (d) $8$
DPatrick 2018-09-27 19:56:13
First, what is $\tan\left(\frac\pi3\right)$?
amuthup 2018-09-27 19:56:25
30-60-90 triangles
Robot7620_2 2018-09-27 19:56:25
pi = 180 deg
asdf334 2018-09-27 19:56:25
tan(60)
Robot7620_2 2018-09-27 19:56:25
tan 60
ASnerd 2018-09-27 19:56:27
tan(60)
DPatrick 2018-09-27 19:56:35
True, you might prefer to think of $\frac\pi3$ radians in terms of degrees.
DPatrick 2018-09-27 19:56:39
The key is to remember that $\pi$ radians is $180^\circ$.
DPatrick 2018-09-27 19:56:44
So $\frac\pi3$ radians is $60^\circ$.
spookyflame2 2018-09-27 19:56:53
sqrt3
hobbit16 2018-09-27 19:56:53
root 3
Orangecounty2 2018-09-27 19:56:53
tan(60) = sqrt(3)
ParadigmShift 2018-09-27 19:56:53
$\sqrt{3}$
Dseg123 2018-09-27 19:56:53
sqrt(3)
Catgirl1234 2018-09-27 19:56:53
tan(60)
krazykid548 2018-09-27 19:56:55
You can draw a 30-60-90 triangle (because pi/3 is 60) and find the ratio of the opposite over adjacent sides
DPatrick 2018-09-27 19:57:03
That's it: $\tan\left(\frac\pi3\right) = \sqrt3$. Think of a 30-60-90 triangle with legs $1$ and $\sqrt3$ and hypotenuse $2$. The tangent is the ratio of the two legs.
DPatrick 2018-09-27 19:57:08
DPatrick 2018-09-27 19:57:31
So our quantity is $(\sqrt3)^3$.
DPatrick 2018-09-27 19:57:37
What's probably a better way to write that?
DatBOIIIIIIII 2018-09-27 19:57:53
3rt3
krazykid548 2018-09-27 19:57:53
Then to cube that, you do 3 * sqrt(3)
jcao0715 2018-09-27 19:57:53
3sqrt3
proshi 2018-09-27 19:57:53
3root 3
r_dat_13 2018-09-27 19:57:53
3sqrt(3)
Apurple 2018-09-27 19:57:53
3sqrt3
kootrapali 2018-09-27 19:57:53
$3\sqrt{3}$
abank 2018-09-27 19:57:53
3sqrt3
Panda_dino 2018-09-27 19:57:53
3 * root3
DPatrick 2018-09-27 19:58:06
Yeah, I'd probably write $(\sqrt3)^3 = 3\sqrt3$.
DPatrick 2018-09-27 19:58:18
So we want the integer closest to $3\sqrt3$.
kootrapali 2018-09-27 19:58:38
$3\cdot1.7=5.1$
coolak 2018-09-27 19:58:38
sqrt 3 is approximately 1.7
Dseg123 2018-09-27 19:58:38
about 3 * 1.7 = 5.1, answer is 5?
hobbit16 2018-09-27 19:58:38
which is 5 since root 3 is 1.73 approximately
DPatrick 2018-09-27 19:58:49
Right. You may know that $\sqrt3 = 1.7\ldots$.
DPatrick 2018-09-27 19:59:24
You can approximate to withing a tenth by noting that $1.7^2 = 2.89$ and $1.8^2 = 3.24$.
DPatrick 2018-09-27 19:59:57
So $3\sqrt3$ is between $5.1$ and $5.4$. And that's closest to $5$.
DPatrick 2018-09-27 20:00:02
So the answer is $5$. $\boxed{\text{(a)}}$
ChickenAgent2227-_- 2018-09-27 20:00:10
or we write it as sqrt27 and see that 5^2=25 and 6^2=36
coolak 2018-09-27 20:00:10
if you didnt, you could think about sqrt 27
kevalshah2005 2018-09-27 20:00:16
you can convert it to sqrt 27 which is a little greater than 5
DPatrick 2018-09-27 20:00:40
Certainly, you can also use $3\sqrt3 = \sqrt{27}$, and then $\sqrt{25} < \sqrt{27} < \sqrt{36}$.
DPatrick 2018-09-27 20:00:55
...and is a lot closer to $\sqrt{25}$.
DPatrick 2018-09-27 20:01:09
On to #6:
DPatrick 2018-09-27 20:01:13
6. Let $n$ represent a positive integer greater than $1$. The number of points of the intersection of the graphs $y = x^n$ and $y = n^x$ is
$ $
(a) always odd
(b) always even
(c) odd when $n$ is even and even when $n$ is odd
(d) even when $n$ is even and odd when $n$ is odd
Orangestripe 2018-09-27 20:01:34
I just sketched a graph of n=2 and n=3
Scipio1 2018-09-27 20:01:34
try it out
Apurple 2018-09-27 20:01:34
pick sample integers to try
phoenix9 2018-09-27 20:01:34
guess and check?
DPatrick 2018-09-27 20:01:42
Right: the answer choices help us out here.
DPatrick 2018-09-27 20:01:46
Specifically, if we can figure out what happens when $n=2$ and $n=3$, we'll know which answer is correct.
DPatrick 2018-09-27 20:02:00
So let's first look at $n=2$. We're comparing $y = x^2$ and $y = 2^x$.
DPatrick 2018-09-27 20:02:18
How many solutions does it have, and how can we be reasonably sure?
krazykid548 2018-09-27 20:02:47
They intersect twice, at x = 2 and x = 4
DPatrick 2018-09-27 20:02:53
Certainly there are at least 2 solutions.
DPatrick 2018-09-27 20:03:01
$x=2$ gives us $2^2 = 2^2$, which I hope is true.
DPatrick 2018-09-27 20:03:12
And $x=4$ gives us $4^2 = 2^4$, which is also true.
Mathisfun04 2018-09-27 20:03:24
after that, the graph of 2^x outgrows x^2, so they don't meet again
krazykid548 2018-09-27 20:03:40
Because 2^x is an function which increases at a faster rate than x^2 after x = 4, they never intersect again.
DPatrick 2018-09-27 20:03:44
True...
a1b2 2018-09-27 20:03:52
They intersect once at a negative x
mathleticguyyy 2018-09-27 20:03:52
what about negative x?
spartacle 2018-09-27 20:03:52
but wait they do meet again for negative x
DPatrick 2018-09-27 20:04:02
Hmmm...maybe we should sketch the graphs.
DPatrick 2018-09-27 20:04:14
What does the graph of $y = x^2$ look like?
shesa 2018-09-27 20:04:32
parabola
Allen31415 2018-09-27 20:04:32
a parabola
jcao0715 2018-09-27 20:04:32
parabola
mathman629 2018-09-27 20:04:32
parabola
ParadigmShift 2018-09-27 20:04:32
Upwards opening parabola
DPatrick 2018-09-27 20:04:36
It's an upwards-opening parabola with vertex at $(0,0)$.
DPatrick 2018-09-27 20:04:40
DPatrick 2018-09-27 20:04:46
What does the graph of the $y = 2^x$ look like?
Dseg123 2018-09-27 20:05:00
exponential curve
hobbit16 2018-09-27 20:05:00
exponential curve
DPatrick 2018-09-27 20:05:17
Yeah, it's a little hard to describe in words.
DPatrick 2018-09-27 20:05:24
It's steadily increasing.



It's always positive.



It crosses the $y$-axis at $y = 2^0 = 1$; that is, the point $(0,1)$.
DPatrick 2018-09-27 20:05:41
(In fact, all of the functions $y = n^x$ where $n > 1$ is a positive integer have these properties. They all look about the same.)
DPatrick 2018-09-27 20:05:47
shesa 2018-09-27 20:05:56
they meet at one point on the negative side, because the parabola crosses to (0,0)!!
DPatrick 2018-09-27 20:06:09
Right: they have to meet on the $x<0$ side as well!
DPatrick 2018-09-27 20:06:34
My graph wasn't big enough to show the $(4,16)$ intersection point, but it's to the right of what I drew.
DPatrick 2018-09-27 20:06:53
And as we mentioned, after that point, the red (exponential) curve is growing way faster than the blue (polynomial) curve, so the red will stay above the blue and they won't intersect again.
Orangestripe 2018-09-27 20:07:01
So there are 3 solutions to that system
DPatrick 2018-09-27 20:07:17
Right: there are $3$ intersection points of $y=2^x$ and $y=x^2$: one negative and two positive ($x=2$ and $x=4$).
CoolKidWiz 2018-09-27 20:07:25
always an odd number
DPatrick 2018-09-27 20:07:31
So when $n$ is even, the answer must be odd. We haven't proved this for any $n > 2$, of course, but the answer choices lock us in.
DPatrick 2018-09-27 20:07:42
So the answer is going to be either (a) or (c).
DPatrick 2018-09-27 20:07:53
How about when $n$ is odd? In particular, what about $y = 3^x$ and $y = x^3$?
Mathisfun04 2018-09-27 20:08:28
intersect once at x=3
mathleticguyyy 2018-09-27 20:08:28
the negative part is not going to intersect
DPatrick 2018-09-27 20:08:40
We definitely get an intersection at $x=3$.
DPatrick 2018-09-27 20:08:46
Since $3^3 = 3^3$
DPatrick 2018-09-27 20:08:59
Let me try to sketch these:
DPatrick 2018-09-27 20:09:04
a1b2 2018-09-27 20:09:26
there will be another positive intersection because it is not tangent
Abra_Kadabra 2018-09-27 20:09:26
they intersect twice, right?
DPatrick 2018-09-27 20:09:38
They do intersect twice!
DPatrick 2018-09-27 20:09:55
Because the red (exponential) has to end up above the blue (polynomial), since the red grows faster.
DPatrick 2018-09-27 20:10:08
The red starts above the blue and ends above the blue.
DPatrick 2018-09-27 20:10:31
So if they cross once, the red goes below the blue. They have to cross a second time for the red to get back on top!
IcyEmerald05 2018-09-27 20:11:09
so the answer is c
kootrapali 2018-09-27 20:11:09
The answer is c
bibilutza 2018-09-27 20:11:09
(c) odd when $n$ is even and even when $n$ is odd
DPatrick 2018-09-27 20:11:18
Right! $n=3$ has 2 intersections, which is even.
DPatrick 2018-09-27 20:11:32
So when $n$ is odd, we get an even number of solutions.
DPatrick 2018-09-27 20:11:34
Therefore, the final answer is $\boxed{\text{(c)}}$.
Orangestripe 2018-09-27 20:11:55
How do you find the solutions algebraically (without a graph)?
DPatrick 2018-09-27 20:12:08
I don't think we can. Exponentials are really hard to work with.
DPatrick 2018-09-27 20:12:42
OK, that was tricky, so next is an easier one for us to catch our breath before the end.
DPatrick 2018-09-27 20:12:49
7. On a flat surface, a bug walks 1 foot north, 2 feet west, 3 feet south, 4 feet east, 5 feet north, and 6 feet west. It then walks straight back to its original starting point. How far did the bug walk total, in feet?
$ $
(a) $21$ (b) $24$ (c) $26$ (d) $21 + \sqrt5$
Apurple 2018-09-27 20:13:09
draw it out
shesa 2018-09-27 20:13:09
just draw this out on a graph!
Scipio1 2018-09-27 20:13:09
draw a diagram
DPatrick 2018-09-27 20:13:18
Right, this should be pretty straightforward to draw.
DPatrick 2018-09-27 20:13:34
It makes sense to think of the bug walking on the coordinate plane, starting at $(0,0)$. That way it'll be easy to tell where it ends up.
DPatrick 2018-09-27 20:13:37
DPatrick 2018-09-27 20:13:59
The segments parallel to the axes are easy: they total to $1+2+3+4+5+6 = 21$.
krazykid548 2018-09-27 20:14:11
The diagonal has lengths 3 and 4 so its length is 5
phoenix9 2018-09-27 20:14:11
use Pythagorean theorem
Scipio1 2018-09-27 20:14:11
3 4 5 triangle
GMbob 2018-09-27 20:14:11
it's a 3-4-5 right triangle to get back to where it started.
DPatrick 2018-09-27 20:14:18
Right. At the end it goes from the point $(-4,3)$ back to $(0,0)$.
DPatrick 2018-09-27 20:14:29
So it's the hypotenuse of a right triangle with legs $3$ and $4$!
DPatrick 2018-09-27 20:14:36
Thus the last step is length $5$.
Catgirl1234 2018-09-27 20:14:42
So the answer is C
kevalshah2005 2018-09-27 20:14:42
so the answer is 26
bibilutza 2018-09-27 20:14:42
last line is 5, so 21+5= c) $26$
DPatrick 2018-09-27 20:14:44
So the total length of the path is $21 + 5 = 26$. $\boxed{\text{(c)}}$
DPatrick 2018-09-27 20:15:03
On to #8:
DPatrick 2018-09-27 20:15:07
8. Double the sum of the first $1{,}111$ positive integers and subtract $1{,}111$. What is the result?
$ $
(a) $616{,}605$ (b) $1{,}234{,}321$ (c) $1{,}423{,}231$ (d) $1{,}432{,}231$
mathman629 2018-09-27 20:15:29
Use n(n+1)/2 for n=1111
asdf334 2018-09-27 20:15:29
triangular number so n(n+1)/2
IlinoisMathlete 2018-09-27 20:15:29
n(n+1)/2 = sum of numbers formula
DPatrick 2018-09-27 20:15:35
$1{,}111$ is a big number. I don't like big numbers.
DPatrick 2018-09-27 20:15:38
Let's call it $n$ instead.
DPatrick 2018-09-27 20:16:01
And as many of you have pointed out, the sum of the first $n$ positive integers is $\dfrac{n(n+1)}{2}$.
DPatrick 2018-09-27 20:16:13
(If you've never seen or proved this formula before, it's a fun exercise to do so!)
asdf334 2018-09-27 20:16:23
n(n+1)/2 but you double it so it becomes n(n+1)
DPatrick 2018-09-27 20:16:29
Right: double that sum is just $n(n+1)$.
DPatrick 2018-09-27 20:16:38
And when we subtract $n$, we get $n(n+1) - n$.
DPatrick 2018-09-27 20:16:46
And what does this simplify to?
spookyflame2 2018-09-27 20:16:59
n(n+1)-n=n^2
Robot7620_2 2018-09-27 20:16:59
factor
amuthup 2018-09-27 20:16:59
which is n^2
Robot7620_2 2018-09-27 20:16:59
n^2
CoolKidWiz 2018-09-27 20:16:59
n^2
hobbit16 2018-09-27 20:16:59
n^2
DPatrick 2018-09-27 20:17:04
We can simplify this by factoring out an $n$, to get $n((n+1) - 1)$.
DPatrick 2018-09-27 20:17:08
Hey, that's just $n(n) = n^2$!
DPatrick 2018-09-27 20:17:14
So the result we're looking for is just $1{,}111^2$.
DPatrick 2018-09-27 20:17:32
(And see how much easier that was to write instead of writing $1{,}111$ over and over! Don't use a big number until you have to!)
Irshad 2018-09-27 20:17:47
1234321
thebossishere 2018-09-27 20:17:47
1234321
Irshad 2018-09-27 20:17:47
its 1234321
aadinair 2018-09-27 20:17:47
1234321, so b
jcao0715 2018-09-27 20:17:47
1234321
DatBOIIIIIIII 2018-09-27 20:17:47
1234321
IcyEmerald05 2018-09-27 20:17:47
1234321 so b
thebossishere 2018-09-27 20:17:47
Which is 1234321
DPatrick 2018-09-27 20:17:53
This works out to $1{,}234{,}321$. $\boxed{\text{(b)}}$
kevalshah2005 2018-09-27 20:18:05
There's a cool trick: for 11x11, its 121 for 111x111 its 12321 and for 1111x1111 its 1234321
DPatrick 2018-09-27 20:18:09
Indeed, to quickly see why this is the answer, think about doing the multiplication the old-fashioned way:
DPatrick 2018-09-27 20:18:12
$$

\begin{array}{ccccccc}

&&&1&1&1&1 \\

&&\times&1&1&1&1 \\ \hline

&&&1&1&1&1 \\

&&1&1&1&1 \\

&1&1&1&1 \\

1&1&1&1 \\ \hline

1&2&3&4&3&2&1

\end{array}

$$
DPatrick 2018-09-27 20:18:42
OK, a couple of harder problems to finish.
DPatrick 2018-09-27 20:18:48
9. Suppose $\log_{10}8 = r$ and $\log_{10}9 = s$. What is $\log_{10}5$ in terms of $r$ and/or $s$?
$ $
(a) $\sqrt[3]{r} + \sqrt{s}$ (b) $\left(\sqrt[3]{r}\right)\left(\sqrt{s}\right)$ (c) $(r/3) + (s/2)$ (d) $1 - (r/3)$
DPatrick 2018-09-27 20:19:30
So here you might run into a roadblock. If you don't know what "log" means, I don't think there's much you can do.
mathman629 2018-09-27 20:19:39
Rewrite in exponential form?
DPatrick 2018-09-27 20:19:47
I almost always like to do this.
DPatrick 2018-09-27 20:20:05
We can rewrite the given information as $10^r = 8$ and $10^s = 9$.
DPatrick 2018-09-27 20:20:27
(That's what "log" means. $\log_{10} 8 = r$ by definition means $10^r = 8$.)
DPatrick 2018-09-27 20:20:38
And we're trying to solve for $x$ in $10^x = 5$.
DPatrick 2018-09-27 20:20:58
There's not much we can do with these except multiply them or take them to a power. How are we possibly going to get a $5$ that way, though? All I see are powers of $2$'s and $3$'s.
a1b2 2018-09-27 20:21:12
$2\times5=10$
DPatrick 2018-09-27 20:21:22
Hmmm...how does that help?
asdf334 2018-09-27 20:21:59
cube rt(8) = 2
DPatrick 2018-09-27 20:22:33
OK...so if we take the cube root of $10^r = 8$, we get $\left(10^r\right)^\frac13 = 2$.
DPatrick 2018-09-27 20:22:45
Or $10^{\frac{r}{3}} = 2$.
DPatrick 2018-09-27 20:22:48
And now what?
Stephenpiano 2018-09-27 20:22:57
divide by 10!
Catgirl1234 2018-09-27 20:23:08
divide by 10
hobbit16 2018-09-27 20:23:10
10/2 = 5 so 10/10^r/3 = 5
DPatrick 2018-09-27 20:23:32
Good idea! But how about we divide it into 10 instead, it's a little cleaner.
DPatrick 2018-09-27 20:23:47
We now have $\dfrac{10}{10^{\frac{r}{3}}} = \dfrac{10}{2} = 5$.
DPatrick 2018-09-27 20:23:53
Hey, that's what we want!
DPatrick 2018-09-27 20:24:19
And the fraction on the left simplifies to $10^{\left(1 - \frac{r}{3}\right)} = 5$.
phoenix9 2018-09-27 20:24:33
= x
aadinair 2018-09-27 20:24:36
(d)
Orangestripe 2018-09-27 20:24:36
So the answer is d
Apurple 2018-09-27 20:24:36
d
DPatrick 2018-09-27 20:24:37
So $10^{1 - (r/3)} = 5$, and thus $x = 1 - (r/3)$ is the solution. $\boxed{\text{(d)}}$
DPatrick 2018-09-27 20:25:00
(The inclusion of $s$ was just there to distract us!)
DPatrick 2018-09-27 20:25:27
Like I said, when I see logs, I usually try to convert to exponentials, which are more familiar territory for me.
bibilutza 2018-09-27 20:25:33
and last.....
DPatrick 2018-09-27 20:25:36
10. Which of the following is closest to the number of ordered pairs of points $(m,n)$, where $m$ and $n$ are both between $1$ and $100$ inclusive and relatively prime (their greatest common divisor is $1$)? [$(2,3)$ and $(3,2)$ count as two such points.]
$ $
(a) $5{,}000$ (b) $6{,}000$ (c) $7{,}000$ (d) $8{,}000$
DPatrick 2018-09-27 20:26:13
There are $100^2 = 10{,}000$ ordered pairs in total. How can we estimate how many of them are relatively prime?
kootrapali 2018-09-27 20:26:32
Subtract the cases with two evens
Apurple 2018-09-27 20:26:42
elimination of those that are not
DPatrick 2018-09-27 20:26:46
That's a really good place to start.
DPatrick 2018-09-27 20:27:05
If $(m,n)$ are both even, then we have to throw them out, as their gcd is at least $2$.
DPatrick 2018-09-27 20:27:15
How many pairs $(m,n)$ are both even?
bradleyguo 2018-09-27 20:27:28
50*50
AforApple 2018-09-27 20:27:28
2500
Mathisfun04 2018-09-27 20:27:28
2500
DPatrick 2018-09-27 20:27:46
Right $50 \cdot 50 = 2500$ of them. Or, exactly $\frac14$ of them.
DPatrick 2018-09-27 20:27:54
So we can eliminate $\frac14$ of our pairs, and we're left with $7500$.
DPatrick 2018-09-27 20:27:58
Now what?
asdf334 2018-09-27 20:28:10
you can just do two multiples of the same prime will be eliminated
StanDaMan 2018-09-27 20:28:10
eliminate the pairs that are both divisible by 3
bigmath 2018-09-27 20:28:10
now factors of 3 and ones that dont work
mathman629 2018-09-27 20:28:10
Now try multiples of 3
DPatrick 2018-09-27 20:28:19
Sure, let's just keep going! How many of the remaining pairs have both terms multiples of $3$?
DPatrick 2018-09-27 20:28:36
Remember: we don't need an exact count! We're only trying to get to the nearest 1000, after all.
mathleticguyyy 2018-09-27 20:28:56
1/9 of remaining
a1b2 2018-09-27 20:28:56
$\frac19$ of them
DPatrick 2018-09-27 20:29:04
Right. About $\frac19$ of them. (About $\frac13$ have $m$ a multiple of $3$ and about $\frac13$ have $n$ a multiple of $3$.)
DPatrick 2018-09-27 20:29:14
That should be close enough.
DPatrick 2018-09-27 20:29:21
So we throw those away, and what's left is about $\frac89 \cdot 7500$.
DPatrick 2018-09-27 20:29:29
This is about $6700$. And we've still got more to get rid of!
Orangestripe 2018-09-27 20:29:37
Then we would do multiples of 5
StanDaMan 2018-09-27 20:29:43
then do multiples of 5, 7, etc
mathman629 2018-09-27 20:29:43
now try the next prime, 5!
thebossishere 2018-09-27 20:29:43
try 5s
DPatrick 2018-09-27 20:29:57
Sure, let's keep going! About $\frac{1}{25}$ of them are both multiples of $5$. (About $\frac15$ have $m$ a multiple of $5$ and about $\frac15$ have $n$ a multiple of $5$.)
DPatrick 2018-09-27 20:30:12
So we throw those away, and what's left is about $\frac{24}{25} \cdot 6700$. This is about $6400$.
DPatrick 2018-09-27 20:30:24
We'll throw away another $150$ or so ($\frac{1}{49}$ of $6400$) when we look at the $7$'s. Down to about $6250$.
DPatrick 2018-09-27 20:30:30
We'll throw away another $50$ or so ($\frac{1}{121}$ of $6250$) when we look at the $11$'s. Down to about $6200$.
kevalshah2005 2018-09-27 20:30:39
If you keep going, it goes to 6000
krazykid548 2018-09-27 20:30:39
We will even more slowly approach 6000
DPatrick 2018-09-27 20:30:52
It sure looks like we'll probably end up somewhere around $6000$, maybe a little higher since there isn't much left to throw away.
DPatrick 2018-09-27 20:31:08
At least, it seems clear that $6000$ is the closest of the answer choices. $\boxed{\text{(b)}}$
DPatrick 2018-09-27 20:31:32
If we did this calculation more carefully, we might even get the actual answer, which is $6{,}087$.
DPatrick 2018-09-27 20:31:43
Does anybody know the general result?
DPatrick 2018-09-27 20:32:10
That is, if I replaced $100$ with a larger number $N$, do you happen to know the formula?
Mathisfun04 2018-09-27 20:32:17
it is 10000 * 6/pi^2?
DPatrick 2018-09-27 20:32:27
Excellent! I'm glad someone has seen this!
mathman629 2018-09-27 20:32:35
$10000\cdot\tfrac{6}{\pi^2}\approx\boxed{6000}$
DPatrick 2018-09-27 20:32:37
It turns out that the number of pairs $(m,n)$ of relatively prime positive integers where $m$ and $n$ are less than or equal to $N$ is approximately $\dfrac{6}{\pi^2} \cdot N^2$. (This requires some calculus to prove.)
CoolKidWiz 2018-09-27 20:32:50
how did pi get in there?
DPatrick 2018-09-27 20:32:57
That's where the calculus comes in.
DPatrick 2018-09-27 20:33:04
And indeed, when $N = 100$, this formula gives $6079$ (rounded to the nearest integer), which is remarkably close. It's only off by $8$, which is within $0.13\%$ of the actual answer.
DPatrick 2018-09-27 20:33:21
And the formula gets even better as $N$ increases.
DPatrick 2018-09-27 20:33:28
When $N=1000$, the formula gives $607{,}927$ and the actual answer is $608{,}383$. The formula is only off by $0.075\%$.
spartacle 2018-09-27 20:33:38
It comes from 1/1 + 1/4 + 1/9 + ...
DPatrick 2018-09-27 20:33:59
Indeed, it is related to the famous Riemann Zeta Function, which in turn is related to the (probably) still-unsolved Riemann Hypothesis.
DPatrick 2018-09-27 20:34:13
And finally, here is a pretty picture I made. Every dot is a pair $(m,n)$ where $m$ and $n$ are relatively prime.
DPatrick 2018-09-27 20:34:17
DPatrick 2018-09-27 20:34:30
If you count them (good luck!), you'll count $6087$ dots.
asdf334 2018-09-27 20:34:52
what is that diagnoal line? is it when they are the same?
DPatrick 2018-09-27 20:35:12
Yes, the white diagonal are the points $(m,m)$ that have gcd $m$, so they're not colored in (except when $m=1$).
Catgirl1234 2018-09-27 20:35:23
Is the graph following a set pattern?
mikebreen 2018-09-27 20:35:32
Symmetric?
DPatrick 2018-09-27 20:35:50
Nobody really knows! (If we did, then we could probably solve the Riemann Hypothesis and we'd really be a millionaire!)
DPatrick 2018-09-27 20:36:16
So those were the 10 problems on Round 1 of this year's Championship contest!
DPatrick 2018-09-27 20:36:33
If you participated in the contest earlier this month and got 7 or more correct, congratulations! -- you're moving on to Round 2. Your teacher should have your invitation (or should get it soon). Round 2 will be held in October. And come back here for our Round 2 Math Jam on Tuesday, October 23 at 7:30 pm ET / 4:30 pm PT.
Bubble3 2018-09-27 20:36:43
What was the mean score?
DPatrick 2018-09-27 20:36:48
It was about 5.5.
bibilutza 2018-09-27 20:36:52
round 3 is Baltimore, right?
DPatrick 2018-09-27 20:36:58
Yes: after Round 2 is complete, 12 students will be invited to compete in the Championship Finals, live in Baltimore at the 2019 Joint Mathematics Meetings on January 19. Travel costs to and from Baltimore will be covered by the AMS.
DPatrick 2018-09-27 20:37:09
Here's how the 12 finalists will be determined: 10 of the 12 will be the top scorer from Round 2 in each of the following regions:
DPatrick 2018-09-27 20:37:13
http://www.ams.org/images/wwtbam-map-us-canada.jpg
DPatrick 2018-09-27 20:37:26
The other two contestants will be the top scorer in the United Kingdom and the top scorer in the Baltimore metro area (or, I suppose, the second-highest scorer, if the highest scorer in Region 2 happens to be from Baltimore).
Orangestripe 2018-09-27 20:37:32
What if there is a tie?
DPatrick 2018-09-27 20:37:55
It's unlikely: problem #10 on Round 2 will ask you to estimate a quantity. The closest answer will break ties.
USA 2018-09-27 20:38:08
what's the scoring method for round 2?
DPatrick 2018-09-27 20:38:35
1 point per question, just like Round 1. Questions will be short-answer, though, not multiple-choice (except where multiple choice is more convenient for some reason).
mikebreen 2018-09-27 20:38:42
Next round one starts next September
DPatrick 2018-09-27 20:39:19
Right: if you didn't participate in Round 1 earlier this month, you can't participate further in 2018-19. Only those who advanced in Round 1 will get to take Round 2.
DPatrick 2018-09-27 20:39:33
The Championship Finals are held live in front of an audience at the Joint Mathematics Meetings, and are also live streamed on the web. (You can watch the archives of past years' finals on the WWTBAM website.) Contestants will compete directly against each other in semi-final rounds, with the semi-final winners advancing to a Jeopardy!-style buzz-in final round to determine a champion.
m22pi7 2018-09-27 20:39:54
How many questions are in round 2?
DPatrick 2018-09-27 20:39:58
10 problems, same as Round 1.
AwesomeDude86 2018-09-27 20:40:01
Are round one and two scores accumulated?
DPatrick 2018-09-27 20:40:12
No -- the highest scorer on Round 2 in each region advances to the finals.
DPatrick 2018-09-27 20:40:24
...with ties broken by the best estimate on #10, if necessary.
mikebreen 2018-09-27 20:41:16
Usually junior or senior, but no lower age limit
DPatrick 2018-09-27 20:41:38
If you have questions about how the contest is administered, and to see if maybe one of the many regional contests is coming your way, please visit the AMS website.
DPatrick 2018-09-27 20:42:26
And to answer the question that I accidentally deleted: yes, Puerto Ricans and all overseas US residents can participate!
TPiR 2018-09-27 20:42:31
Thanks, everyone. Great work!
UnstoppableGoddess 2018-09-27 20:42:44
How much does it cost
DPatrick 2018-09-27 20:42:47
It's free!
mikebreen 2018-09-27 20:42:52
Yes, thanks Dave and all the AoPS fans.
DPatrick 2018-09-27 20:43:12
That's it for tonight's session -- thanks for coming!
DPatrick 2018-09-27 20:43:30
Please join us again on October 23 when we discuss the Round 2 problems!
DPatrick 2018-09-27 20:43:32
Good night!

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