Who Wants to Be a Mathematician Discussion
Go back to the Math Jam ArchiveAoPS instructor David Patrick will discuss the 2019-20 Who Wants to Be a Mathematician Championship contest. WWTBAM is a free(!) contest sponsored by the American Mathematical Society in which high school students compete for cash and prizes. Round 1 of qualifying for the contest will take place September 9-26, 2019, so this is your chance to learn about the contest, the type of problems that appear, and how to participate. 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 Denver in January 2020.
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Facilitator: Dave Patrick
DPatrick
2019-09-04 19:30:25
Time to get started!
Time to get started!
DPatrick
2019-09-04 19:30:30
Welcome to the 2019-20 Who Wants to Be a Mathematician Informational Math Jam!
Welcome to the 2019-20 Who Wants to Be a Mathematician Informational Math Jam!
DPatrick
2019-09-04 19:30:40
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 15 years, and I've written or co-written a few of our textbooks.
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 15 years, and I've written or co-written a few of our textbooks.
DPatrick
2019-09-04 19:30:49
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):
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
2019-09-04 19:30:53
DPatrick
2019-09-04 19:30:56
Photo Credit: Maria Melin, copyright 1999 ABC Television.
Photo Credit: Maria Melin, copyright 1999 ABC Television.
eed7573
2019-09-04 19:31:16
how much did you win
how much did you win
DPatrick
2019-09-04 19:31:24
Sadly, I didn't win the million bucks, but I did win enough to buy a new car.
Sadly, I didn't win the million bucks, but I did win enough to buy a new car.
DPatrick
2019-09-04 19:31:40
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.
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
2019-09-04 19:31:50
The classroom is moderated, meaning that students 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.
The classroom is moderated, meaning that students 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
2019-09-04 19:32:04
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.
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
2019-09-04 19:32:23
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.
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
2019-09-04 19:32:35
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!
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
2019-09-04 19:33:00
Tonight we'll be discussing the 2019-20 Who Wants to Be a Mathematician Championship contest.
Tonight we'll be discussing the 2019-20 Who Wants to Be a Mathematician Championship contest.
DPatrick
2019-09-04 19:33:09
We'll spend a few minutes discussing how the contest works, and then I'll go over some problems from last year's contest.
We'll spend a few minutes discussing how the contest works, and then I'll go over some problems from last year's contest.
DPatrick
2019-09-04 19:33:35
Who Wants to Be a Mathematician (or WWTBAM for short) is conducted by the American Mathematical Society (AMS).
Who Wants to Be a Mathematician (or WWTBAM for short) is conducted by the American Mathematical Society (AMS).
DPatrick
2019-09-04 19:33:41
DPatrick
2019-09-04 19:33:52
The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
DPatrick
2019-09-04 19:34:04
The AMS is one of the two big professional mathematics organizations in the U.S. (The other is the Mathematical Association of America, or MAA, which as you may know runs the AMC/AIME/USA(J)MO contests.)
The AMS is one of the two big professional mathematics organizations in the U.S. (The other is the Mathematical Association of America, or MAA, which as you may know runs the AMC/AIME/USA(J)MO contests.)
DPatrick
2019-09-04 19:34:17
Joining us tonight from the AMS is the co-creator and host of WWTBAM, Mike Breen (mikebreen).
Joining us tonight from the AMS is the co-creator and host of WWTBAM, Mike Breen (mikebreen).
mikebreen
2019-09-04 19:34:20
Good job with the logo and mission statement. Thanks.
Good job with the logo and mission statement. Thanks.
mikebreen
2019-09-04 19:34:32
Hello, everyone.
Hello, everyone.
DPatrick
2019-09-04 19:34:41
Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill Butterworth began Who Wants to Be a Mathematician for the American Mathematical Society in 2001, and the first national game was in 2010.
Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill Butterworth began Who Wants to Be a Mathematician for the American Mathematical Society in 2001, and the first national game was in 2010.
DPatrick
2019-09-04 19:34:56
Mike has been a contestant 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.
Mike has been a contestant 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.
DPatrick
2019-09-04 19:35:16
Bill Butterworth (TPiR) is with us tonight too. 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.
Bill Butterworth (TPiR) is with us tonight too. 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.
TPiR
2019-09-04 19:35:23
Welcome everyone
Welcome everyone
DPatrick
2019-09-04 19:35:28
Can you guess what Bill's username stands for?
Can you guess what Bill's username stands for?
AOP2004
2019-09-04 19:35:56
The Price is Right
The Price is Right
Ilikeapos
2019-09-04 19:35:56
The price is right
The price is right
DPatrick
2019-09-04 19:36:04
Correct! 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.)
Correct! 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
2019-09-04 19:36:14
Bill shares a life-long interest in game shows with colleague Mike Breen, with whom he works as the not-so-lovely assistant on Who Wants to Be a Mathematician.
Bill shares a life-long interest in game shows with colleague Mike Breen, with whom he works as the not-so-lovely assistant on Who Wants to Be a Mathematician.
DPatrick
2019-09-04 19:36:25
So between the three of us, we have a lot of game show experience!
So between the three of us, we have a lot of game show experience!
DPatrick
2019-09-04 19:36:42
Assisting for us tonight is Brent Nicklas (brentnicklas), who as far as I know hasn't been on any game shows.
Assisting for us tonight is Brent Nicklas (brentnicklas), who as far as I know hasn't been on any game shows.
DPatrick
2019-09-04 19:36:54
Brent first joined AoPS as an assistant in 2013. He received bachelors degrees from Michigan Tech in both mathematics and physics with concentrations in secondary education in 2014, then earned a masters degree in statistics in 2016. When he's not doing math, he enjoys drawing comic sketches, playing board games, cooking, riding his bicycle, and swing dancing.
Brent first joined AoPS as an assistant in 2013. He received bachelors degrees from Michigan Tech in both mathematics and physics with concentrations in secondary education in 2014, then earned a masters degree in statistics in 2016. When he's not doing math, he enjoys drawing comic sketches, playing board games, cooking, riding his bicycle, and swing dancing.
brentnicklas
2019-09-04 19:37:05
Greetings! I have not been on any game shows, but I'm excited to be here!
Greetings! I have not been on any game shows, but I'm excited to be here!
DPatrick
2019-09-04 19:37:15
We'll start with an overview of how the contest works. All of the information I'm about to tell you, and more, is on the WWTBAM web site at http://ams.org/wwtbam . There are also many years' worth of past contests on that website.
We'll start with an overview of how the contest works. All of the information I'm about to tell you, and more, is on the WWTBAM web site at http://ams.org/wwtbam . There are also many years' worth of past contests on that website.
DPatrick
2019-09-04 19:37:38
The WWTBAM Championship contest is open to students in the U.S., Canada, and the United Kingdom. Although the contest is designed for high school students, middle school students are eligible to participate too. Topics up to and including precalculus are included on the contest, so in terms of mathematical content, WWTBAM is similar to the AMC 12 contest.
The WWTBAM Championship contest is open to students in the U.S., Canada, and the United Kingdom. Although the contest is designed for high school students, middle school students are eligible to participate too. Topics up to and including precalculus are included on the contest, so in terms of mathematical content, WWTBAM is similar to the AMC 12 contest.
DPatrick
2019-09-04 19:37:58
Your school needs to sign up for the contest. You cannot sign up as an individual. But a homeschool counts as school too, so if you're homeschooled, your parent or other homeschool teacher can sign up.
Your school needs to sign up for the contest. You cannot sign up as an individual. But a homeschool counts as school too, so if you're homeschooled, your parent or other homeschool teacher can sign up.
DPatrick
2019-09-04 19:38:12
The good news is that the contest is free to sign up for. The even better news is that every school that signs up will also get an AoPS coupon.
The good news is that the contest is free to sign up for. The even better news is that every school that signs up will also get an AoPS coupon.
DPatrick
2019-09-04 19:38:25
To sign up, your school (not any individual student) should do one of the following:
1. Go to https://wwtbam.mobius.net/instructor/ and fill out the registration form.
2. Send an email to paoffice@ams.org with the subject line "WWTBAM Championship".
To sign up, your school (not any individual student) should do one of the following:
1. Go to https://wwtbam.mobius.net/instructor/ and fill out the registration form.
2. Send an email to paoffice@ams.org with the subject line "WWTBAM Championship".
Professor-Mom
2019-09-04 19:38:57
My parent can sign me up if my mom is my homeschool teacher and she has no other students?
My parent can sign me up if my mom is my homeschool teacher and she has no other students?
DPatrick
2019-09-04 19:39:04
Yes, that's right.
Yes, that's right.
DPatrick
2019-09-04 19:39:17
Your school will then receive instructions for how to administer the qualifying test. Round 1 of qualifying will take place September 9-26 -- your school can administer the qualifying test on any day during this period.
Your school will then receive instructions for how to administer the qualifying test. Round 1 of qualifying will take place September 9-26 -- your school can administer the qualifying test on any day during this period.
npratyush
2019-09-04 19:39:42
Can middle school sign up for this test?
Can middle school sign up for this test?
DPatrick
2019-09-04 19:39:54
Yes, although the contest is designed for high school, middle schools may sign up too.
Yes, although the contest is designed for high school, middle schools may sign up too.
vedadehhc
2019-09-04 19:40:01
Must the school sign up before Sept. 9, or can they sign up during the testing perion
Must the school sign up before Sept. 9, or can they sign up during the testing perion
DPatrick
2019-09-04 19:40:13
I believe you can sign up all the way until the end of the period, but I'm not 100% sure.
I believe you can sign up all the way until the end of the period, but I'm not 100% sure.
DPatrick
2019-09-04 19:40:53
A lot of you are asking more specific questions -- please hold those until the end and Mike and I will try to answer them, or you can probably find the info you need on the website.
A lot of you are asking more specific questions -- please hold those until the end and Mike and I will try to answer them, or you can probably find the info you need on the website.
DPatrick
2019-09-04 19:41:11
Qualifying Round 1 of the national contest will consist 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 you'll see a bit later when we look at some problems from last year's contest, some of the problems may not take you nearly that long.) No books, notes, calculators, or internet access is permitted during the test.
Qualifying Round 1 of the national contest will consist 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 you'll see a bit later when we look at some problems from last year's contest, some of the problems may not take you nearly that long.) No books, notes, calculators, or internet access is permitted during the test.
DPatrick
2019-09-04 19:41:31
Any student who scores at least 7 out of 10 correct on Round 1 will be invited to participate in Round 2, which will be held in October. The format for Round 2 is the same: 10 questions, 15 minutes, except that Round 2 is short-answer instead of multiple-choice.
Any student who scores at least 7 out of 10 correct on Round 1 will be invited to participate in Round 2, which will be held in October. The format for Round 2 is the same: 10 questions, 15 minutes, except that Round 2 is short-answer instead of multiple-choice.
DPatrick
2019-09-04 19:41:48
After Round 2 is complete, 12 students will be invited to compete in the Championship Finals, held live in Denver at the 2020 Joint Mathematics Meetings in January. Travel costs to and from Denver will be covered by the AMS.
After Round 2 is complete, 12 students will be invited to compete in the Championship Finals, held live in Denver at the 2020 Joint Mathematics Meetings in January. Travel costs to and from Denver will be covered by the AMS.
DPatrick
2019-09-04 19:42:02
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:
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
2019-09-04 19:42:06
DPatrick
2019-09-04 19:42:16
The other two contestants will be the top scorer in the United Kingdom and the top scorer in the Denver metro area (or the second-highest scorer, if the highest scorer in Region 8 happens to be from Denver).
The other two contestants will be the top scorer in the United Kingdom and the top scorer in the Denver metro area (or the second-highest scorer, if the highest scorer in Region 8 happens to be from Denver).
DPatrick
2019-09-04 19:42:35
Since with only 10 questions there are likely to be ties for the top scorer, there will be tiebreakers if necessary. The first tiebreaker will be question #10 of Round 2: this question will be an "estimation" question that will be hard to get an exact answer for, but for which it is reasonable to try to estimate (or guess).
Since with only 10 questions there are likely to be ties for the top scorer, there will be tiebreakers if necessary. The first tiebreaker will be question #10 of Round 2: this question will be an "estimation" question that will be hard to get an exact answer for, but for which it is reasonable to try to estimate (or guess).
DPatrick
2019-09-04 19:43:07
For example, last year's tiebreaker question #10 was:
For example, last year's tiebreaker question #10 was:
DPatrick
2019-09-04 19:43:11
10. A partition of a positive integer $n$ is a way of writing $n$ as a sum of positive integers. For example, the number $3$ has three partitions: $3$ (the number itself is a partition with one addend), $2+1$, and $1+1+1$ (the order of the addends doesn't matter). How many partitions does $101$ have?
10. A partition of a positive integer $n$ is a way of writing $n$ as a sum of positive integers. For example, the number $3$ has three partitions: $3$ (the number itself is a partition with one addend), $2+1$, and $1+1+1$ (the order of the addends doesn't matter). How many partitions does $101$ have?
DPatrick
2019-09-04 19:43:26
This is essentially impossible to compute exactly by hand, but one can make an educated guess.
This is essentially impossible to compute exactly by hand, but one can make an educated guess.
DPatrick
2019-09-04 19:43:57
As it turns out, the actual answer is $\boxed{214{,}481{,}126}$.
As it turns out, the actual answer is $\boxed{214{,}481{,}126}$.
DPatrick
2019-09-04 19:44:25
If there is still a tie (which is unlikely), then a winner for the region will be selected by the contest organizers.
If there is still a tie (which is unlikely), then a winner for the region will be selected by the contest organizers.
Gershwin_Fan
2019-09-04 19:44:30
How did you figure it out??
How did you figure it out??
reitracks
2019-09-04 19:44:30
how did you work that out?
how did you work that out?
DPatrick
2019-09-04 19:44:50
You can take a look at last year's Math Jam transcript to see how we approached it.
You can take a look at last year's Math Jam transcript to see how we approached it.
DPatrick
2019-09-04 19:44:56
Indeed, if you want to see all the problems from both qualifying tests from last year, along with worked-out solutions, you can read the Math Jam transcripts from last year:
Round 1: https://artofproblemsolving.com/school/mathjams-transcripts?id=470
Round 2: https://artofproblemsolving.com/school/mathjams-transcripts?id=471
Indeed, if you want to see all the problems from both qualifying tests from last year, along with worked-out solutions, you can read the Math Jam transcripts from last year:
Round 1: https://artofproblemsolving.com/school/mathjams-transcripts?id=470
Round 2: https://artofproblemsolving.com/school/mathjams-transcripts?id=471
DPatrick
2019-09-04 19:45:16
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.
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.
DPatrick
2019-09-04 19:45:27
All of the finalists will receive prizes (in addition to the free trip to Denver!), and the champion will win $10,000.
All of the finalists will receive prizes (in addition to the free trip to Denver!), and the champion will win $10,000.
t3sp
2019-09-04 19:45:45
will we be figuring out problems like that later?
will we be figuring out problems like that later?
t3sp
2019-09-04 19:45:45
or just from round 1?
or just from round 1?
DPatrick
2019-09-04 19:45:55
We'll be doing that right now! To give you some idea of the content and difficulty level of the contest, I'd like to present three of the problems from last year's Round 1 Qualifying Test.
We'll be doing that right now! To give you some idea of the content and difficulty level of the contest, I'd like to present three of the problems from last year's Round 1 Qualifying Test.
DPatrick
2019-09-04 19:46:10
Here is last year's #1:
Here is last year's #1:
DPatrick
2019-09-04 19:46:14
1. What is the ones (units) digit of $2019^2 - 2018^2$?
$ $
(a) $1$ (b) $3$ (c) $5$ (d) $7$
1. What is the ones (units) digit of $2019^2 - 2018^2$?
$ $
(a) $1$ (b) $3$ (c) $5$ (d) $7$
DPatrick
2019-09-04 19:46:23
Reminder: all the Round 1 questions are multiple-choice.
Reminder: all the Round 1 questions are multiple-choice.
DPatrick
2019-09-04 19:46:37
Please do not simply post your answer. Instead, please post suggestions as to how to solve the problem.
Please do not simply post your answer. Instead, please post suggestions as to how to solve the problem.
PiMath12345
2019-09-04 19:47:04
Modular arithmetic, modulo 10
Modular arithmetic, modulo 10
aopspractice
2019-09-04 19:47:04
We observe the units digit of both squares
We observe the units digit of both squares
RadioActive1729
2019-09-04 19:47:04
First you can find the unit digit.
First you can find the unit digit.
Ancy
2019-09-04 19:47:04
only units digits matter. so problem simplifies to $9^2 - 8^2$
only units digits matter. so problem simplifies to $9^2 - 8^2$
Pumpkin66
2019-09-04 19:47:10
difference of squares
difference of squares
GammaZero
2019-09-04 19:47:10
difference of square
difference of square
Mathematician1010
2019-09-04 19:47:10
Difference of squares:a^2-b^2=(a+b)(a-b)
Difference of squares:a^2-b^2=(a+b)(a-b)
coolmath2017
2019-09-04 19:47:10
using difference of squares, we find (2019+2018)*1 so the units digit is 7
using difference of squares, we find (2019+2018)*1 so the units digit is 7
Professor-Mom
2019-09-04 19:47:10
difference of squares
difference of squares
DPatrick
2019-09-04 19:47:17
Right, there are a couple of ways to do this.
Right, there are a couple of ways to do this.
DPatrick
2019-09-04 19:47:36
We could observe that the units digit of $2019^2$ is the same as the units digit of $9^2$, which is $81$. So its units digit is $1$.
We could observe that 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
2019-09-04 19:47:57
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 $4$.
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 $4$.
Selseev
2019-09-04 19:48:21
d)7. This is the answer becaause 9 squared minus 8 squared is 81-64= 17. Therefore the units digit is 7
d)7. This is the answer becaause 9 squared minus 8 squared is 81-64= 17. Therefore the units digit is 7
DPatrick
2019-09-04 19:48:34
Right, or for example, $11-4 = 7$. So the units digit of the difference is $7$. $\boxed{\text{(d)}}$
Right, or for example, $11-4 = 7$. So the units digit of the difference is $7$. $\boxed{\text{(d)}}$
DPatrick
2019-09-04 19:48:53
Or, as many of you mentioned, we can use the difference-of-squares factorization!
Or, as many of you mentioned, we can use the difference-of-squares factorization!
DPatrick
2019-09-04 19:49:00
We get $2019^2 - 2018^2 = (2019 + 2018)(2019 - 2018)$.
We get $2019^2 - 2018^2 = (2019 + 2018)(2019 - 2018)$.
mikebreen
2019-09-04 19:49:03
Very good.
Very good.
DPatrick
2019-09-04 19:49:12
This is especially convenient because the second factor is just $1$.
This is especially convenient because the second factor is just $1$.
DPatrick
2019-09-04 19:49:19
So the difference is equal to $2019 + 2018 = 4037$, and has units digit $7$.
So the difference is equal to $2019 + 2018 = 4037$, and has units digit $7$.
DPatrick
2019-09-04 19:49:40
About 3/4 of the students taking Round 1 last year got this question correct.
About 3/4 of the students taking Round 1 last year got this question correct.
DPatrick
2019-09-04 19:49:57
Next is #5 from last year, which is "medium" difficulty:
Next is #5 from last year, which is "medium" difficulty:
DPatrick
2019-09-04 19:50:01
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) $8$
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
2019-09-04 19:50:25
Again, WWTBAM is a high-school level contest, so trigonometry problems are fair game. Middle school students are certainly permitted to participate if they wish, but advanced high school topics like intermediate algebra (including logarithms) and trigonometry will appear on the contest.
Again, WWTBAM is a high-school level contest, so trigonometry problems are fair game. Middle school students are certainly permitted to participate if they wish, but advanced high school topics like intermediate algebra (including logarithms) and trigonometry will appear on the contest.
DPatrick
2019-09-04 19:50:52
First, what is $\tan\left(\frac\pi3\right)$?
First, what is $\tan\left(\frac\pi3\right)$?
carzland
2019-09-04 19:51:10
Sqrt(3)
Sqrt(3)
prajna1225
2019-09-04 19:51:10
sqrt(3)
sqrt(3)
ctw0611
2019-09-04 19:51:10
sqrt(3)
sqrt(3)
Ancy
2019-09-04 19:51:10
$tan(\frac{pi}{3}) = \sqrt{3}$
$tan(\frac{pi}{3}) = \sqrt{3}$
reitracks
2019-09-04 19:51:10
$\sqrt{3}$
$\sqrt{3}$
chc237
2019-09-04 19:51:10
sqrt(3)
sqrt(3)
DPatrick
2019-09-04 19:51:16
Good!
Good!
mada123
2019-09-04 19:51:25
60 degrees
60 degrees
npratyush
2019-09-04 19:51:25
TAN 60?
TAN 60?
Mathematician1010
2019-09-04 19:51:25
tan(60) degrees
tan(60) degrees
DPatrick
2019-09-04 19:51:34
Certainly, you might prefer to think of $\frac\pi3$ radians in terms of degrees.
Certainly, you might prefer to think of $\frac\pi3$ radians in terms of degrees.
DPatrick
2019-09-04 19:51:43
The key is to remember that $\pi$ radians is $180^\circ$. So $\frac\pi3$ radians is $60^\circ$.
The key is to remember that $\pi$ radians is $180^\circ$. So $\frac\pi3$ radians is $60^\circ$.
DPatrick
2019-09-04 19:51:53
And then $\tan\left(\frac\pi3\right) = \sqrt3$. Think of a 30-60-90 triangle with legs $1$ and $\sqrt3$ and hypotenuse $2$. The tangent is the ratio of the two legs.
And then $\tan\left(\frac\pi3\right) = \sqrt3$. 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
2019-09-04 19:51:57
DPatrick
2019-09-04 19:52:16
So the quantity we want to round to the nearest integer is $(\sqrt3)^3$.
So the quantity we want to round to the nearest integer is $(\sqrt3)^3$.
Professor-Mom
2019-09-04 19:52:36
or $3\sqrt{3}$ ;)
or $3\sqrt{3}$ ;)
DPatrick
2019-09-04 19:52:44
Yeah, a better way to write that is $3\sqrt3$.
Yeah, a better way to write that is $3\sqrt3$.
TheGeometer
2019-09-04 19:53:01
3*1.7=5.1
3*1.7=5.1
applepear
2019-09-04 19:53:01
i think that rounds to 5
i think that rounds to 5
DPatrick
2019-09-04 19:53:32
Right: you may know that $\sqrt 3 \approx 1.7$ (and you can check that $(1.7)^2 = 2.89$).
Right: you may know that $\sqrt 3 \approx 1.7$ (and you can check that $(1.7)^2 = 2.89$).
DPatrick
2019-09-04 19:53:52
And certainly $1.7 < \sqrt3 < 1.8$.
And certainly $1.7 < \sqrt3 < 1.8$.
DPatrick
2019-09-04 19:54:03
So $5.1 < 3\sqrt3 < 5.4$.
So $5.1 < 3\sqrt3 < 5.4$.
RadioActive1729
2019-09-04 19:54:27
A
A
Professor-Mom
2019-09-04 19:54:33
(a)5
(a)5
anshrai
2019-09-04 19:54:33
a.) 5
a.) 5
DPatrick
2019-09-04 19:54:42
So the answer is $5$. $\boxed{\text{(a)}}$
So the answer is $5$. $\boxed{\text{(a)}}$
npratyush
2019-09-04 19:54:52
is a calculator allowed?
is a calculator allowed?
DPatrick
2019-09-04 19:55:16
No. (This problem would be pretty easy with a calculator!)
No. (This problem would be pretty easy with a calculator!)
DPatrick
2019-09-04 19:55:34
Slightly over half of the students taking the contest last year got this correct.
Slightly over half of the students taking the contest last year got this correct.
DPatrick
2019-09-04 19:55:52
Finally, here was #10 (the final problem) from last year's Round 1:
Finally, here was #10 (the final problem) from last year's Round 1:
DPatrick
2019-09-04 19:55:58
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) $8{,}000$
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
2019-09-04 19:56:40
So, there are $10{,}000$ ordered pairs in total. How can we count how many of them are relatively prime?
So, there are $10{,}000$ ordered pairs in total. How can we count how many of them are relatively prime?
carzland
2019-09-04 19:57:11
Only 25 possible prime factors to check
Only 25 possible prime factors to check
DPatrick
2019-09-04 19:57:23
True: there are 25 primes under $100$, but what does it mean "to check"?
True: there are 25 primes under $100$, but what does it mean "to check"?
t3sp
2019-09-04 19:57:29
by subtracting all those that are multiples of a prime factor
by subtracting all those that are multiples of a prime factor
DPatrick
2019-09-04 19:58:02
Right, but to be more precise: we want to subtract those $(m,n)$ where $m$ and $n$ are multiples of the same prime factor.
Right, but to be more precise: we want to subtract those $(m,n)$ where $m$ and $n$ are multiples of the same prime factor.
DPatrick
2019-09-04 19:58:34
Maybe we can count these. Or at least get an estimate that clearly rounds to one of the answer choices.
Maybe we can count these. Or at least get an estimate that clearly rounds to one of the answer choices.
DPatrick
2019-09-04 19:58:47
For instance: how many pairs $(m,n)$ are both multiples of $2$?
For instance: how many pairs $(m,n)$ are both multiples of $2$?
mada123
2019-09-04 19:59:09
well about a quarter have two multiples of 2
well about a quarter have two multiples of 2
edgymemelord
2019-09-04 19:59:09
2500
2500
prajna1225
2019-09-04 19:59:09
2500
2500
DPatrick
2019-09-04 19:59:20
Right: exactly $\frac14$ of them, or $2500$.
Right: exactly $\frac14$ of them, or $2500$.
DPatrick
2019-09-04 19:59:29
So we can eliminate $\frac14$ of our pairs, and we're left with $10000 - 2500 = 7500$.
So we can eliminate $\frac14$ of our pairs, and we're left with $10000 - 2500 = 7500$.
DPatrick
2019-09-04 19:59:38
How many of the remaining pairs have both terms multiples of $3$?
How many of the remaining pairs have both terms multiples of $3$?
KKD
2019-09-04 20:00:09
1/9 of the pairs
1/9 of the pairs
DPatrick
2019-09-04 20:00:19
Right: about $\frac19$ of them. (About $\frac13$ have $m$ a multiple of $3$ and about $\frac13$ have $n$ a multiple of $3$.)
Right: about $\frac19$ of them. (About $\frac13$ have $m$ a multiple of $3$ and about $\frac13$ have $n$ a multiple of $3$.)
DPatrick
2019-09-04 20:00:33
So we throw those away, and what's left is about $\frac89 \cdot 7500$.
So we throw those away, and what's left is about $\frac89 \cdot 7500$.
KKD
2019-09-04 20:00:44
but like arent we double counting like (6,12)
but like arent we double counting like (6,12)
carzland
2019-09-04 20:00:44
Aren't we overcounting?
Aren't we overcounting?
DPatrick
2019-09-04 20:00:52
That's a really good concern.
That's a really good concern.
DPatrick
2019-09-04 20:01:04
But we're not, because I'm only throwing away $\frac19$ of what's remaining.
But we're not, because I'm only throwing away $\frac19$ of what's remaining.
DPatrick
2019-09-04 20:01:23
So now we're down to about $6700$. And we've still got more to get rid of.
So now we're down to about $6700$. And we've still got more to get rid of.
DPatrick
2019-09-04 20:01:29
How many of the remaining pairs have both terms multiples of $5$?
How many of the remaining pairs have both terms multiples of $5$?
prajna1225
2019-09-04 20:01:44
1/25
1/25
t3sp
2019-09-04 20:01:44
1/25
1/25
reitracks
2019-09-04 20:01:44
1/25
1/25
KKD
2019-09-04 20:01:44
1/25
1/25
mada123
2019-09-04 20:01:44
1/25
1/25
liyuxiao
2019-09-04 20:01:44
1/25 of them
1/25 of them
DPatrick
2019-09-04 20:01:52
Right: about $\frac{1}{25}$ of them. (About $\frac15$ have $m$ a multiple of $5$ and about $\frac15$ have $n$ a multiple of $5$.)
Right: about $\frac{1}{25}$ of them. (About $\frac15$ have $m$ a multiple of $5$ and about $\frac15$ have $n$ a multiple of $5$.)
DPatrick
2019-09-04 20:02:00
So we throw those away, and what's left is about $\frac{24}{25} \cdot 6700$.
So we throw those away, and what's left is about $\frac{24}{25} \cdot 6700$.
DPatrick
2019-09-04 20:02:08
This is about $6400$.
This is about $6400$.
DPatrick
2019-09-04 20:02:15
We're throwing away fewer and fewer each time.
We're throwing away fewer and fewer each time.
DPatrick
2019-09-04 20:02:22
We'll throw away another $150$ or so ($\frac{1}{49}$ of $6400$) when we look at the $7$'s. Down to about $6250$.
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
2019-09-04 20:02:28
We'll throw away another $50$ or so ($\frac{1}{121}$ of $6250$) when we look at the $11$'s. Down to about $6200$.
We'll throw away another $50$ or so ($\frac{1}{121}$ of $6250$) when we look at the $11$'s. Down to about $6200$.
reitracks
2019-09-04 20:02:33
guesstimate of around 6000
guesstimate of around 6000
DPatrick
2019-09-04 20:02:45
Yeah, it looks like we'll end up somewhere around $6000$, maybe a little higher since there isn't much left to throw away.
Yeah, it looks like we'll end up somewhere around $6000$, maybe a little higher since there isn't much left to throw away.
carzland
2019-09-04 20:02:50
Yeah...B
Yeah...B
Gershwin_Fan
2019-09-04 20:02:50
The answer is (b) 6000.
The answer is (b) 6000.
molocyxu
2019-09-04 20:02:50
So b)
So b)
DPatrick
2019-09-04 20:02:57
It seems clear that $6000$ is the closest of the answer choices. $\boxed{\text{(b)}}$
It seems clear that $6000$ is the closest of the answer choices. $\boxed{\text{(b)}}$
ctw0611
2019-09-04 20:03:09
what is a rigorous way to do this question
what is a rigorous way to do this question
DPatrick
2019-09-04 20:03:18
This actually is rigorous! If we did this calculation more carefully, we might even get the actual answer, which is $6{,}087$.
This actually is rigorous! If we did this calculation more carefully, we might even get the actual answer, which is $6{,}087$.
edgymemelord
2019-09-04 20:03:39
how do we do that in 90 seconds?
how do we do that in 90 seconds?
npratyush
2019-09-04 20:03:39
So this was solved in ~1.5min?
So this was solved in ~1.5min?
DPatrick
2019-09-04 20:03:54
Probably not...but then again, the early problems might not have actually taken 90 seconds each.
Probably not...but then again, the early problems might not have actually taken 90 seconds each.
DPatrick
2019-09-04 20:04:06
You might have gotten the $2019^2 - 2018^2$ problem much more quickly, for example.
You might have gotten the $2019^2 - 2018^2$ problem much more quickly, for example.
DPatrick
2019-09-04 20:04:20
And there's no penalty for guessing, so you can always make an educated guess.
And there's no penalty for guessing, so you can always make an educated guess.
DPatrick
2019-09-04 20:04:46
By the way, does anybody know the general result for the situation in problem #10?
By the way, does anybody know the general result for the situation in problem #10?
zeroman
2019-09-04 20:05:24
isnt it like probability 6/pi^2 or something
isnt it like probability 6/pi^2 or something
DPatrick
2019-09-04 20:05:28
Exactly!
Exactly!
DPatrick
2019-09-04 20:05:31
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.)
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.)
DPatrick
2019-09-04 20:05:42
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.
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.
PiMath12345
2019-09-04 20:05:58
Who proved it?
Who proved it?
DPatrick
2019-09-04 20:06:08
I think it's generally credited to Euler.
I think it's generally credited to Euler.
DPatrick
2019-09-04 20:06:37
And finally, here is a pretty picture I made. Every dot is a pair $(m,n)$ where $m$ and $n$ are relatively prime.
And finally, here is a pretty picture I made. Every dot is a pair $(m,n)$ where $m$ and $n$ are relatively prime.
DPatrick
2019-09-04 20:06:41
DPatrick
2019-09-04 20:06:48
If you count them (good luck!), you'll count $6087$ dots.
If you count them (good luck!), you'll count $6087$ dots.
DPatrick
2019-09-04 20:07:07
A little over 1/3 of students last year got this correct. (Keep in mind that there's no penalty for guessing, so even if everyone guessed you'd expect about 1/4 to get it right!)
A little over 1/3 of students last year got this correct. (Keep in mind that there's no penalty for guessing, so even if everyone guessed you'd expect about 1/4 to get it right!)
DPatrick
2019-09-04 20:07:32
Overall, about one-third of the students who participated in Round 1 scored 7 or higher, which qualified them to advance to Round 2.
Overall, about one-third of the students who participated in Round 1 scored 7 or higher, which qualified them to advance to Round 2.
DPatrick
2019-09-04 20:07:47
You can find lots of past years' Qualifying Tests on the AMS's website at http://ams.org/wwtbam
You can find lots of past years' Qualifying Tests on the AMS's website at http://ams.org/wwtbam
DPatrick
2019-09-04 20:08:20
And you can find our Math Jam transcripts from previous years' contests on the AoPS website (click on "Math Jams" in the "Online School" menu).
And you can find our Math Jam transcripts from previous years' contests on the AoPS website (click on "Math Jams" in the "Online School" menu).
DPatrick
2019-09-04 20:08:32
A reminder: if your school would like to participate (and it's free! and your school gets an AoPS coupon just for signing up!), a teacher at your school should do one of the following:
1. Go to https://wwtbam.mobius.net/instructor/ and fill out the registration form.
2. Send an email to paoffice@ams.org with the subject line "WWTBAM Championship".
A reminder: if your school would like to participate (and it's free! and your school gets an AoPS coupon just for signing up!), a teacher at your school should do one of the following:
1. Go to https://wwtbam.mobius.net/instructor/ and fill out the registration form.
2. Send an email to paoffice@ams.org with the subject line "WWTBAM Championship".
DPatrick
2019-09-04 20:09:04
At this point, Mike and I can take questions about the contest, if you have any.
At this point, Mike and I can take questions about the contest, if you have any.
audbear
2019-09-04 20:09:29
Is there a limit for how many students can compete from one school?
Is there a limit for how many students can compete from one school?
mikebreen
2019-09-04 20:09:32
No
No
DPatrick
2019-09-04 20:09:40
No, as long as all the students take it at the same time, of course.
No, as long as all the students take it at the same time, of course.
reitracks
2019-09-04 20:09:46
Can UK students apply?
Can UK students apply?
DPatrick
2019-09-04 20:09:54
Yes, but I believe the UK procedures are slightly different.
Yes, but I believe the UK procedures are slightly different.
mikebreen
2019-09-04 20:10:07
Yes, but the same in Rounds 1 and 2
Yes, but the same in Rounds 1 and 2
DPatrick
2019-09-04 20:10:07
You'll probably want to write to paoffice@ams.org for more info.
You'll probably want to write to paoffice@ams.org for more info.
Puddles_Penguin
2019-09-04 20:10:17
has 2019 already started?
has 2019 already started?
mikebreen
2019-09-04 20:10:27
Not quite. It starts Monday.
Not quite. It starts Monday.
coolmath2017
2019-09-04 20:10:37
is it possible for individual students to sign up(without a school)?
is it possible for individual students to sign up(without a school)?
DPatrick
2019-09-04 20:10:42
No. You have to sign up via a school.
No. You have to sign up via a school.
winstonfan
2019-09-04 20:11:11
If the student take it this year, can the student take it next year?
If the student take it this year, can the student take it next year?
DPatrick
2019-09-04 20:11:14
Yes.
Yes.
gr8phul1
2019-09-04 20:11:17
can homeschool students apply
can homeschool students apply
DPatrick
2019-09-04 20:11:27
Yes, but your teacher or homeschooling parent should apply.
Yes, but your teacher or homeschooling parent should apply.
Gershwin_Fan
2019-09-04 20:11:38
What are the prizes from 2nd place to 10th place???
What are the prizes from 2nd place to 10th place???
DPatrick
2019-09-04 20:11:46
I'm not sure, but the trip to Denver is included.
I'm not sure, but the trip to Denver is included.
math_lover111
2019-09-04 20:12:11
could any of you clarify whether we can register after Sep. 9th? i.e. any day up to Sep. 26 for this year?
could any of you clarify whether we can register after Sep. 9th? i.e. any day up to Sep. 26 for this year?
npratyush
2019-09-04 20:12:11
It starts monday. What if the school hasnt signed up yet? Is there a way to start it like next week?
It starts monday. What if the school hasnt signed up yet? Is there a way to start it like next week?
DPatrick
2019-09-04 20:12:29
I believe you can register right up to the end of the Round 1 period.
I believe you can register right up to the end of the Round 1 period.
mikebreen
2019-09-04 20:12:40
Yes, Dave is absolutely right.
Yes, Dave is absolutely right.
DPatrick
2019-09-04 20:12:46
I'd probably register as soon as possible though (you don't have to take the contest the day you register), just in case.
I'd probably register as soon as possible though (you don't have to take the contest the day you register), just in case.
Puddles_Penguin
2019-09-04 20:12:48
Can you register now?
Can you register now?
DPatrick
2019-09-04 20:12:56
Yes, the website is open now.
Yes, the website is open now.
mada123
2019-09-04 20:13:10
Is the second round the same format as the first?
Is the second round the same format as the first?
DPatrick
2019-09-04 20:13:18
Yes, except it's short answer instead of multiple choice.
Yes, except it's short answer instead of multiple choice.
mikebreen
2019-09-04 20:13:26
And harder questions.
And harder questions.
akailash
2019-09-04 20:13:32
Can private coaching institutes like Russian School of Math register?
Can private coaching institutes like Russian School of Math register?
DPatrick
2019-09-04 20:13:46
I don't believe so, but I'm not sure. I'll defer to Mike on this.
I don't believe so, but I'm not sure. I'll defer to Mike on this.
mikebreen
2019-09-04 20:14:39
Send paoffice@ams.org some details about the school and we'll decide.
Send paoffice@ams.org some details about the school and we'll decide.
DPatrick
2019-09-04 20:14:58
To go back to the question about registering in the UK, here's the website: https://goodthinkingsociety.org/who-wants-to-be-a-mathematician-is-back-2/
To go back to the question about registering in the UK, here's the website: https://goodthinkingsociety.org/who-wants-to-be-a-mathematician-is-back-2/
PiMath12345
2019-09-04 20:15:27
Can you take the exam at other schools?
Can you take the exam at other schools?
fqokjfioqjfoiqw
2019-09-04 20:15:27
Can my school register when I’ll be the only one doing it?
Can my school register when I’ll be the only one doing it?
DPatrick
2019-09-04 20:15:45
Other schools: I don't think so, but you can write to the AMS and explain your situation.
Other schools: I don't think so, but you can write to the AMS and explain your situation.
DPatrick
2019-09-04 20:15:59
There's no minimum per school. You can register even if only 1 student is participating.
There's no minimum per school. You can register even if only 1 student is participating.
DPatrick
2019-09-04 20:16:08
And to reiterate: the contest is free!
And to reiterate: the contest is free!
RadioActive1729
2019-09-04 20:16:28
What happens if the teacher won`t sign up like in my area.
What happens if the teacher won`t sign up like in my area.
Mathcounts333
2019-09-04 20:16:28
Can any student attending a school take it if their school registers?
Can any student attending a school take it if their school registers?
mikebreen
2019-09-04 20:16:37
At other schools, yes.
At other schools, yes.
mikebreen
2019-09-04 20:16:44
Any student: yes.
Any student: yes.
mingxias
2019-09-04 20:17:03
when is the deadline?
when is the deadline?
DPatrick
2019-09-04 20:17:10
September 26 is the end of Round 1.
September 26 is the end of Round 1.
fqokjfioqjfoiqw
2019-09-04 20:17:19
I’m in middle school i can still participate correct?
I’m in middle school i can still participate correct?
DPatrick
2019-09-04 20:17:21
Yes.
Yes.
flashomatic
2019-09-04 20:17:55
Can a math circle register?
Can a math circle register?
mikebreen
2019-09-04 20:18:05
Yes.
Yes.
DPatrick
2019-09-04 20:19:13
The questions are starting to get really specific, so I think we'll end the general Q&A here.
The questions are starting to get really specific, so I think we'll end the general Q&A here.
DPatrick
2019-09-04 20:19:29
If you still have questions, you can visit the website at http://www.ams.org/wwtbam, or write to paoffice@ams.org
If you still have questions, you can visit the website at http://www.ams.org/wwtbam, or write to paoffice@ams.org
DPatrick
2019-09-04 20:19:49
Please join us again for a discussion of Qualifying Round 1 on Monday, September 30, at 7:30 PM Eastern / 4:30 PM Pacific. We'll go through all 10 problems from Round 1. And there will be a Qualifying Round 2 discussion on Monday, October 14, after Round 2 concludes.
Please join us again for a discussion of Qualifying Round 1 on Monday, September 30, at 7:30 PM Eastern / 4:30 PM Pacific. We'll go through all 10 problems from Round 1. And there will be a Qualifying Round 2 discussion on Monday, October 14, after Round 2 concludes.
DPatrick
2019-09-04 20:19:59
Thanks for participating tonight, and good luck on WWTBAM!
Thanks for participating tonight, and good luck on WWTBAM!
mikebreen
2019-09-04 20:20:27
Yes, nice to chat with you. Good luck, everyone!
Yes, nice to chat with you. Good luck, everyone!
TPiR
2019-09-04 20:20:46
Goodnight everyone. Thank you!
Goodnight everyone. Thank you!
brentnicklas
2019-09-04 20:21:08
Good luck everyone!
Good luck everyone!
audbear
2019-09-04 20:21:33
thank you again! but how can i share this information (ie this chat) with my math club officers?
thank you again! but how can i share this information (ie this chat) with my math club officers?
DPatrick
2019-09-04 20:21:55
We'll be posting the transcript of the session on the AoPS website later this evening.
We'll be posting the transcript of the session on the AoPS website later this evening.
DPatrick
2019-09-04 20:22:09
You'll find it at https://artofproblemsolving.com/school/mathjams-transcripts
You'll find it at https://artofproblemsolving.com/school/mathjams-transcripts
asdf334
2019-09-04 20:24:45
when will transcript be open
when will transcript be open
DPatrick
2019-09-04 20:24:55
Hopefully by 9 PM (Eastern) tonight.
Hopefully by 9 PM (Eastern) tonight.
eminentabyss
2019-09-04 20:26:23
A huge thank you to Dave, Mike, Bill, and everyone who attended the math jam tonight.
A huge thank you to Dave, Mike, Bill, and everyone who attended the math jam tonight.
eminentabyss
2019-09-04 20:26:26
This room will be closing shortly!
This room will be closing shortly!
Professor-Mom
2019-09-04 20:27:24
Thanks!
Thanks!
Professor-Mom
2019-09-04 20:27:24
Goodnight!
Goodnight!
bidb2019
2019-09-04 20:27:24
good night and Thanks!
good night and Thanks!
aopspractice
2019-09-04 20:27:24
Thank you
Thank you
aopspractice
2019-09-04 20:27:24
Goodnight
Goodnight
aopspractice
2019-09-04 20:27:27
Bye
Bye
Stbcosmos
2019-09-04 20:27:28
Thank you!
Thank you!
Mathcounts333
2019-09-04 20:27:31
Thanks!
Thanks!
Professor-Mom
2019-09-04 20:27:35
Thanks for taking your time to do this mathjam!
Thanks for taking your time to do this mathjam!
epowerpii
2019-09-04 20:27:47
thx
thx
epowerpii
2019-09-04 20:27:47
bye
bye
motorfinn
2019-09-04 20:27:47
Tyty
Tyty
audbear
2019-09-04 20:27:47
Wonderful thank you!
Wonderful thank you!
mingxias
2019-09-04 20:27:47
Thanks you!
Thanks you!
mathapple101
2019-09-04 20:27:47
Bye!
Bye!
chc237
2019-09-04 20:27:47
Thanks so much!
Thanks so much!
applepear
2019-09-04 20:27:50
thank you
thank you
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