Who Wants to Be a Mathematician Discussion
Go back to the Math Jam ArchiveAoPS instructor David Patrick will discuss the 2019 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 10-26, 2018, 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 Baltimore in January 2019.
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Facilitator: Dave Patrick
DPatrick
2018-09-06 19:19:16
Welcome to the 2018-19 Who Wants to Be a Mathematician Informational Math Jam! We'll get started at 7:30 PM Eastern / 4:30 PM Pacific.
Welcome to the 2018-19 Who Wants to Be a Mathematician Informational Math Jam! We'll get started at 7:30 PM Eastern / 4:30 PM Pacific.
DPatrick
2018-09-06 19:21:26
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
2018-09-06 19:22:00
DPatrick
2018-09-06 19:24:19
Some of you are already asking questions. We'll get to your questions once we get started in about 6 minutes.
Some of you are already asking questions. We'll get to your questions once we get started in about 6 minutes.
DPatrick
2018-09-06 19:30:07
Hello and welcome to the 2018-19 Who Wants to Be a Mathematician Informational Math Jam!
Hello and welcome to the 2018-19 Who Wants to Be a Mathematician Informational Math Jam!
DPatrick
2018-09-06 19:30:16
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.
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
2018-09-06 19:30:26
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
2018-09-06 19:30:32
DPatrick
2018-09-06 19:30:38
Photo Credit: Maria Melin, copyright 1999 ABC Television.
Photo Credit: Maria Melin, copyright 1999 ABC Television.
Master_AoPs_Confirmed
2018-09-06 19:30:58
How much money did you get
How much money did you get
DPatrick
2018-09-06 19:31:02
Well, I didn't win the million bucks, but I did win enough to buy a new car.
Well, I didn't win the million bucks, but I did win enough to buy a new car.
DPatrick
2018-09-06 19:31:14
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
2018-09-06 19:31:29
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 moderators, 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 moderators, who may choose to share your comments with the room.
DPatrick
2018-09-06 19:31:43
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick
2018-09-06 19:31:53
Tonight we'll be discussing the 2018-19 Who Wants to Be a Mathematician Championship contest.
Tonight we'll be discussing the 2018-19 Who Wants to Be a Mathematician Championship contest.
DPatrick
2018-09-06 19:32:08
Who Wants to Be a Mathematician (or WWTBAM for short) 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.
Who Wants to Be a Mathematician (or WWTBAM for short) 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
2018-09-06 19:32:20
Joining us tonight is the co-creator and host of WWTBAM, Mike Breen (mikebreen).
Joining us tonight is the co-creator and host of WWTBAM, Mike Breen (mikebreen).
mikebreen
2018-09-06 19:32:31
Hello, everyone!
Hello, everyone!
DPatrick
2018-09-06 19:32:35
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
2018-09-06 19:32:51
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
2018-09-06 19:33:13
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
2018-09-06 19:33:19
Hi everyone!
Hi everyone!
DPatrick
2018-09-06 19:33:28
Can you guess what Bill's username stands for?
Can you guess what Bill's username stands for?
rocker27
2018-09-06 19:33:48
The Price is Right
The Price is Right
DPatrick
2018-09-06 19:33:54
Yes! 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.)
Yes! 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.)
TPiR
2018-09-06 19:34:00
Well done
Well done
DPatrick
2018-09-06 19:34:14
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
2018-09-06 19:34:22
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
2018-09-06 19:34:41
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. And a cool picture of Mike looking sharp:
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. And a cool picture of Mike looking sharp:
DPatrick
2018-09-06 19:34:50
DPatrick
2018-09-06 19:35:23
The 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 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
2018-09-06 19:35:46
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
2018-09-06 19:36:01
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!
mikebreen
2018-09-06 19:36:18
Yes, thanks Dave and AoPS.
Yes, thanks Dave and AoPS.
DPatrick
2018-09-06 19:36:23
To have your school sign up, ask your school or teacher to write to paoffice@ams.org with the subject line "WWTBAM Championship". Your school will then receive instructions for how to administer the qualifying test. Round 1 of qualifying will take place September 10-26 -- your school can administer the qualifying test on any day during this period.
To have your school sign up, ask your school or teacher to write to paoffice@ams.org with the subject line "WWTBAM Championship". Your school will then receive instructions for how to administer the qualifying test. Round 1 of qualifying will take place September 10-26 -- your school can administer the qualifying test on any day during this period.
DPatrick
2018-09-06 19:37:15
A lot of you are asking questions. Please hold your questions until the end, and then Mike and I will be happy to answer them.
A lot of you are asking questions. Please hold your questions until the end, and then Mike and I will be happy to answer them.
DPatrick
2018-09-06 19:37:47
I will answer one now, though, that a few people are asking:
I will answer one now, though, that a few people are asking:
thebossishere
2018-09-06 19:37:48
What if your middle school doesn't want you to participate?
What if your middle school doesn't want you to participate?
o99999
2018-09-06 19:37:48
How do you sign up as an individual if your school doesn't sign up?
How do you sign up as an individual if your school doesn't sign up?
DPatrick
2018-09-06 19:38:02
Again, 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.)
Again, 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
2018-09-06 19:38:36
You will need to ask your school teacher to sign up for you. Again, it is free, and your school will get a coupon good towards AoPS books just for signing up!
You will need to ask your school teacher to sign up for you. Again, it is free, and your school will get a coupon good towards AoPS books just for signing up!
DPatrick
2018-09-06 19:39:00
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
2018-09-06 19:39:31
Any student who scores at least 8 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.
Any student who scores at least 8 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.
math2435
2018-09-06 19:39:49
where is round 2
where is round 2
DPatrick
2018-09-06 19:39:59
Also at your school. Your school administers Rounds 1 and 2.
Also at your school. Your school administers Rounds 1 and 2.
DPatrick
2018-09-06 19:40:09
After Round 2 is complete, 12 students will be invited to compete in the Championship Finals, held live in Baltimore at the 2019 Joint Mathematics Meetings in January. Travel costs to and from Baltimore 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 Baltimore at the 2019 Joint Mathematics Meetings in January. Travel costs to and from Baltimore will be covered by the AMS.
DPatrick
2018-09-06 19:40:51
I just got some late-breaking news: this year the score needed to advance from Round 1 to Round 2 is 7, not 8.
I just got some late-breaking news: this year the score needed to advance from Round 1 to Round 2 is 7, not 8.
DPatrick
2018-09-06 19:41:19
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
2018-09-06 19:41:24
DPatrick
2018-09-06 19:41:40
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).
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).
DPatrick
2018-09-06 19:42:24
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
2018-09-06 19:42:43
If there is still a tie, then a winner for the region will be selected by the contest organizers.
If there is still a tie, then a winner for the region will be selected by the contest organizers.
Guam21
2018-09-06 19:42:56
Does Guam count as one of the regions? or is it too far of a territory to participate/not a state?
Does Guam count as one of the regions? or is it too far of a territory to participate/not a state?
DPatrick
2018-09-06 19:43:05
I assume Guam is in Region 9. Mike, is that the case?
I assume Guam is in Region 9. Mike, is that the case?
mikebreen
2018-09-06 19:43:12
Yes.
Yes.
palisade
2018-09-06 19:43:16
Why Baltimore?
Why Baltimore?
DPatrick
2018-09-06 19:43:32
Because Baltimore is the location of the finals. WWTBAM always has a local finalist for the crowd to root for!
Because Baltimore is the location of the finals. WWTBAM always has a local finalist for the crowd to root for!
DPatrick
2018-09-06 19:44:08
The Championship finals are held live in front of an audience at the Joint Mathematics Meetings in Baltimore, 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 in Baltimore, 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
2018-09-06 19:44:23
All of the finalists will receive prizes (in addition to the free trip to Baltimore!), and the champion will win $10,000.
All of the finalists will receive prizes (in addition to the free trip to Baltimore!), and the champion will win $10,000.
DPatrick
2018-09-06 19:45:14
Again, all this info is on the AMS website (I'll post the link again at the end), and we'll answer further questions about the contest rules at the end as well.
Again, all this info is on the AMS website (I'll post the link again at the end), and we'll answer further questions about the contest rules at the end as well.
DPatrick
2018-09-06 19:45:30
But next up, 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.
But next up, 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
2018-09-06 19:45:44
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=444
Round 2: https://artofproblemsolving.com/school/mathjams-transcripts?id=450
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=444
Round 2: https://artofproblemsolving.com/school/mathjams-transcripts?id=450
DPatrick
2018-09-06 19:45:52
First, here's #1 from last year's Round 1:
First, here's #1 from last year's Round 1:
DPatrick
2018-09-06 19:45:56
1. Find the $y$-intercept ($y$-coordinate only) of the line whose equation is $2x + 5y = 7$.
1. Find the $y$-intercept ($y$-coordinate only) of the line whose equation is $2x + 5y = 7$.
DPatrick
2018-09-06 19:46:24
Some of you got this more or less right away:
Some of you got this more or less right away:
sasb
2018-09-06 19:46:27
7/5
7/5
rubikscube2010
2018-09-06 19:46:27
7/5
7/5
sam1222
2018-09-06 19:46:27
7/5
7/5
ComicPuzzle
2018-09-06 19:46:27
7/5
7/5
Bubble3
2018-09-06 19:46:27
7/5
7/5
math2435
2018-09-06 19:46:27
7/5
7/5
thebossishere
2018-09-06 19:46:27
7/5
7/5
potato369
2018-09-06 19:46:27
7/5
7/5
DPatrick
2018-09-06 19:46:31
How'd you get it so fast?
How'd you get it so fast?
sasb
2018-09-06 19:46:54
Plug in x=0
Plug in x=0
rubikscube2010
2018-09-06 19:46:54
just let x=0
just let x=0
FateMonster500
2018-09-06 19:46:54
plug in 0 for x
plug in 0 for x
math2435
2018-09-06 19:46:54
because x is 0 and I just plug in zero and get 5y = 7 and y = 7/5
because x is 0 and I just plug in zero and get 5y = 7 and y = 7/5
DPatrick
2018-09-06 19:47:10
Good! The point we want is where the line intersects the $y$-axis.
Good! The point we want is where the line intersects the $y$-axis.
DPatrick
2018-09-06 19:47:20
But what do all the points on the $y$-axis have in common? They all have $x=0$.
But what do all the points on the $y$-axis have in common? They all have $x=0$.
DPatrick
2018-09-06 19:47:30
So we can just plug in $x=0$ into our equation and solve for $y$.
So we can just plug in $x=0$ into our equation and solve for $y$.
DPatrick
2018-09-06 19:47:40
This gives $5y = 7$, or $y = \boxed{\dfrac75}$.
This gives $5y = 7$, or $y = \boxed{\dfrac75}$.
DPatrick
2018-09-06 19:47:58
So that was pretty fast. The first 2 or 3 problems are typically like this: more or less straightforward computation. And nearly 85% of the participants last year got this correct. This is the warm-up.
So that was pretty fast. The first 2 or 3 problems are typically like this: more or less straightforward computation. And nearly 85% of the participants last year got this correct. This is the warm-up.
DPatrick
2018-09-06 19:48:27
Next is #6. It's not merely straightforward computation; it requires a bit more thought to do efficiently.
Next is #6. It's not merely straightforward computation; it requires a bit more thought to do efficiently.
DPatrick
2018-09-06 19:48:32
6. For which one of the following choices for $m$ are the base $m$ numbers $25_m$ and $27_m$ both prime?
a. 8 b. 9 c. 11 d. 12 e. 13
6. For which one of the following choices for $m$ are the base $m$ numbers $25_m$ and $27_m$ both prime?
a. 8 b. 9 c. 11 d. 12 e. 13
DPatrick
2018-09-06 19:49:13
So what do we notice that might help?
So what do we notice that might help?
proshi
2018-09-06 19:49:37
twin primes
twin primes
bao2022
2018-09-06 19:49:37
twin primes
twin primes
DPatrick
2018-09-06 19:49:51
Good -- these are twin primes, meaning that they're primes that differ by 2.
Good -- these are twin primes, meaning that they're primes that differ by 2.
DPatrick
2018-09-06 19:50:09
Does that help narrow the answer choices?
Does that help narrow the answer choices?
DPatrick
2018-09-06 19:50:45
What do we know about the odd number that immediately comes before a pair of twin primes? That is, what do we know about $23_m$?
What do we know about the odd number that immediately comes before a pair of twin primes? That is, what do we know about $23_m$?
coolak
2018-09-06 19:51:10
it's not a prime
it's not a prime
rubikscube2010
2018-09-06 19:51:10
It is composite?
It is composite?
DPatrick
2018-09-06 19:51:16
Right, but specifically, how do we know that?
Right, but specifically, how do we know that?
DPatrick
2018-09-06 19:51:29
Why can't three odd numbers in a row be prime?
Why can't three odd numbers in a row be prime?
Collatz__conjecture
2018-09-06 19:51:42
It is a multiple of 3
It is a multiple of 3
potato369
2018-09-06 19:51:42
muktiple of 3
muktiple of 3
casi
2018-09-06 19:51:42
multiples of 3
multiples of 3
User360702
2018-09-06 19:51:42
oh cuz one has to be divisible by 3
oh cuz one has to be divisible by 3
Ondhu
2018-09-06 19:51:42
it has to be a multiple of 3
it has to be a multiple of 3
coolak
2018-09-06 19:51:42
because one must be divisible by 3
because one must be divisible by 3
DPatrick
2018-09-06 19:51:56
Right. I actually lied a bit: 3, 5, 7 are three consecutive odd numbers that are prime.
Right. I actually lied a bit: 3, 5, 7 are three consecutive odd numbers that are prime.
DPatrick
2018-09-06 19:52:13
But if I give you any three consecutive odd numbers, one of them must be a multiple of 3.
But if I give you any three consecutive odd numbers, one of them must be a multiple of 3.
DPatrick
2018-09-06 19:52:30
So in particular, if $25_m$ and $27_m$ are prime, then $23_m$ is a multiple of 3.
So in particular, if $25_m$ and $27_m$ are prime, then $23_m$ is a multiple of 3.
DPatrick
2018-09-06 19:52:34
What does that tell us about $m$?
What does that tell us about $m$?
coolak
2018-09-06 19:52:48
m is a multiple of 3
m is a multiple of 3
rubikscube2010
2018-09-06 19:52:48
m is divisible by 3
m is divisible by 3
Ondhu
2018-09-06 19:52:48
it is divisible by 3
it is divisible by 3
thebossishere
2018-09-06 19:52:48
IT has to be either 9 or 12
IT has to be either 9 or 12
DPatrick
2018-09-06 19:53:00
Right. This means $23_m$, or $2m+3$, must be a multiple of 3, so $m$ itself must be a multiple of 3.
Right. This means $23_m$, or $2m+3$, must be a multiple of 3, so $m$ itself must be a multiple of 3.
DPatrick
2018-09-06 19:53:05
So the answer must be (b) 9 or (d) 12.
So the answer must be (b) 9 or (d) 12.
DPatrick
2018-09-06 19:53:24
Which is correct?
Which is correct?
MananV
2018-09-06 19:53:37
d
d
sasb
2018-09-06 19:53:37
D
D
KimmyT
2018-09-06 19:53:37
d
d
Supermathking
2018-09-06 19:53:37
$12$ is
$12$ is
Bubble3
2018-09-06 19:53:37
12
12
BobNot
2018-09-06 19:53:37
d
d
snow_monkey
2018-09-06 19:53:37
12
12
DPatrick
2018-09-06 19:53:42
$m=9$ doesn't work because $27_9 = 25$ isn't prime.
$m=9$ doesn't work because $27_9 = 25$ isn't prime.
DPatrick
2018-09-06 19:53:54
So the answer must be $\boxed{\text{(d) } 12}$. We don't even need to check further.
So the answer must be $\boxed{\text{(d) } 12}$. We don't even need to check further.
DPatrick
2018-09-06 19:54:08
But if we do, we can see that $25_{12} = 29$ and $27_{12} = 31$ are both prime.
But if we do, we can see that $25_{12} = 29$ and $27_{12} = 31$ are both prime.
DPatrick
2018-09-06 19:54:37
Of course, you could have just checked them all. But that's a lot slower. And part of the contest is figuring out an efficient way to quickly answer some of the problems!
Of course, you could have just checked them all. But that's a lot slower. And part of the contest is figuring out an efficient way to quickly answer some of the problems!
DPatrick
2018-09-06 19:54:54
Finally, let's look at #10 from last year. The last two or three problems will usually require some additional insight, and may even require a little estimation or guesswork.
Finally, let's look at #10 from last year. The last two or three problems will usually require some additional insight, and may even require a little estimation or guesswork.
DPatrick
2018-09-06 19:54:59
10. What is the largest prime factor of $2^{14} + 1$.
10. What is the largest prime factor of $2^{14} + 1$.
mathlogician
2018-09-06 19:55:34
Sum of squares
Sum of squares
DPatrick
2018-09-06 19:55:44
Indeed, this is a sum of two perfect squares -- does that help?
Indeed, this is a sum of two perfect squares -- does that help?
DPatrick
2018-09-06 19:55:54
I can easily factor a difference of two squares, but what can we do with a sum?
I can easily factor a difference of two squares, but what can we do with a sum?
UnstoppableGoddess
2018-09-06 19:56:06
Could I technically find 2^14 and go from there
Could I technically find 2^14 and go from there
DPatrick
2018-09-06 19:56:12
Sure, let's take a look.
Sure, let's take a look.
DPatrick
2018-09-06 19:56:17
You may know that $2^{10} = 1024$. (This is a really useful fact to memorize.) So $2^{14} = 2^{10} \cdot 2^4 = 1024 \cdot 16$.
You may know that $2^{10} = 1024$. (This is a really useful fact to memorize.) So $2^{14} = 2^{10} \cdot 2^4 = 1024 \cdot 16$.
Ondhu
2018-09-06 19:56:31
16384
16384
DPatrick
2018-09-06 19:56:34
This works out to $16384$.
This works out to $16384$.
DPatrick
2018-09-06 19:56:37
So the number we are given is $16385$.
So the number we are given is $16385$.
jason168
2018-09-06 19:56:42
just try to factor 16385?
just try to factor 16385?
veeruvasireddy
2018-09-06 19:56:48
5
5
Bubble3
2018-09-06 19:56:48
div by 5
div by 5
FateMonster500
2018-09-06 19:56:53
3277
3277
DPatrick
2018-09-06 19:56:57
Clearly 5 is a prime factor, and we can divide to get $16385 = 5 \cdot 3277$.
Clearly 5 is a prime factor, and we can divide to get $16385 = 5 \cdot 3277$.
DPatrick
2018-09-06 19:57:11
So now we would have to decide if $3277$ is prime, and if not try to factor it further.
So now we would have to decide if $3277$ is prime, and if not try to factor it further.
DPatrick
2018-09-06 19:57:23
This is probably doable but might take a while.
This is probably doable but might take a while.
DPatrick
2018-09-06 19:57:36
So let's go back to the observation that what we started with is a sum of two perfect squares.
So let's go back to the observation that what we started with is a sum of two perfect squares.
rocker27
2018-09-06 19:57:40
I thought a sum of squares wasn’t factorable.
I thought a sum of squares wasn’t factorable.
DPatrick
2018-09-06 19:57:56
True: in general it's not (unless we use imaginary numbers, but that sounds scary).
True: in general it's not (unless we use imaginary numbers, but that sounds scary).
Bubble3
2018-09-06 19:58:04
Yes: a^2 + b^2 = (a+b)^2 - 2ab
Yes: a^2 + b^2 = (a+b)^2 - 2ab
rubikscube2010
2018-09-06 19:58:04
Add and subtract 2*2^7 gives a difference of squares
Add and subtract 2*2^7 gives a difference of squares
DPatrick
2018-09-06 19:58:14
Good idea: there's always the identity $x^2 + y^2 = (x+y)^2 - 2xy$.
Good idea: there's always the identity $x^2 + y^2 = (x+y)^2 - 2xy$.
DPatrick
2018-09-06 19:58:21
In this case, that gives us $2^{14} + 1 = (2^7+1)^2 - 2 \cdot 2^7$.
In this case, that gives us $2^{14} + 1 = (2^7+1)^2 - 2 \cdot 2^7$.
Bubble3
2018-09-06 19:58:31
factor into (2^7 + 1)^2 - 2^8
factor into (2^7 + 1)^2 - 2^8
sasb
2018-09-06 19:58:37
DIfference of squares
DIfference of squares
DPatrick
2018-09-06 19:58:39
That's $(2^7+1)^2 - 2^8$, which is $(2^7+1)^2 - (2^4)^2$. It's a difference of two perfect squares!
That's $(2^7+1)^2 - 2^8$, which is $(2^7+1)^2 - (2^4)^2$. It's a difference of two perfect squares!
DPatrick
2018-09-06 19:58:51
So our number factors as $(2^7+1+2^4)(2^7+1-2^4)$.
So our number factors as $(2^7+1+2^4)(2^7+1-2^4)$.
sasb
2018-09-06 19:59:17
113 * 145
113 * 145
rubikscube2010
2018-09-06 19:59:17
is it 113
is it 113
Bubble3
2018-09-06 19:59:17
113
113
potato369
2018-09-06 19:59:20
145 and 113
145 and 113
a000
2018-09-06 19:59:20
$145*113$
$145*113$
DPatrick
2018-09-06 19:59:26
Right, this is $145 \cdot 113$.
Right, this is $145 \cdot 113$.
DPatrick
2018-09-06 19:59:30
So after factoring out the 5 as before, we get $5 \cdot 29 \cdot 113$.
So after factoring out the 5 as before, we get $5 \cdot 29 \cdot 113$.
thebossishere
2018-09-06 19:59:42
113
113
proshi
2018-09-06 19:59:42
113
113
ComicPuzzle
2018-09-06 19:59:42
113
113
shemao
2018-09-06 19:59:42
113 is prime
113 is prime
DPatrick
2018-09-06 19:59:50
Yes, $113$ is prime! (This is easy to check: since $11^2 = 121 > 113$, we only need to check divisibility by 2, 3, 5, and 7, and none of them work.)
Yes, $113$ is prime! (This is easy to check: since $11^2 = 121 > 113$, we only need to check divisibility by 2, 3, 5, and 7, and none of them work.)
DPatrick
2018-09-06 19:59:58
So we have the prime factorization $2^{14} + 1 = 5 \cdot 29 \cdot 113$, and thus $\boxed{113}$ is the largest prime factor.
So we have the prime factorization $2^{14} + 1 = 5 \cdot 29 \cdot 113$, and thus $\boxed{113}$ is the largest prime factor.
DPatrick
2018-09-06 20:00:15
This is also a special case of the Sophie Germain Identity, which is
\[ 4x^4 + y^4 = (2x^2+y^2)^2 - 4x^2y^2 = (2x^2+y^2+2xy)(2x^2+y^2-2xy).\]
This is also a special case of the Sophie Germain Identity, which is
\[ 4x^4 + y^4 = (2x^2+y^2)^2 - 4x^2y^2 = (2x^2+y^2+2xy)(2x^2+y^2-2xy).\]
DPatrick
2018-09-06 20:00:27
In our problem, $x = 2^3$ and $y = 1$.
In our problem, $x = 2^3$ and $y = 1$.
DPatrick
2018-09-06 20:01:05
So that was #10 from last year's Round 1, and if I remember correctly only about 20% or so of the participants solved it.
So that was #10 from last year's Round 1, and if I remember correctly only about 20% or so of the participants solved it.
DPatrick
2018-09-06 20:01:27
You can find lots of past years' Qualifying Tests on the AMS's website at http://ams.org/wwtbam, along with all of the registration information for the contest.
You can find lots of past years' Qualifying Tests on the AMS's website at http://ams.org/wwtbam, along with all of the registration information for the contest.
DPatrick
2018-09-06 20:02:00
Once again: To have your school sign up, ask your school or teacher to write to paoffice@ams.org with the subject line "WWTBAM Championship". Your school will then receive instructions for how to administer the qualifying test. Round 1 of qualifying will take place September 10-26 -- your school can administer the qualifying test on any day during this period, and it'll only take 15 minutes.
Once again: To have your school sign up, ask your school or teacher to write to paoffice@ams.org with the subject line "WWTBAM Championship". Your school will then receive instructions for how to administer the qualifying test. Round 1 of qualifying will take place September 10-26 -- your school can administer the qualifying test on any day during this period, and it'll only take 15 minutes.
DPatrick
2018-09-06 20:02:29
At this point, I'd like to open the discussion for questions about the contest.
At this point, I'd like to open the discussion for questions about the contest.
sasb
2018-09-06 20:02:33
Is too late for a school to register?
Is too late for a school to register?
DPatrick
2018-09-06 20:02:49
Not at all. I believe you can register all the way up to the final day. (Though I don't recommend waiting that long.)
Not at all. I believe you can register all the way up to the final day. (Though I don't recommend waiting that long.)
mikebreen
2018-09-06 20:02:49
No, you can register until the contest closes on the 26th
No, you can register until the contest closes on the 26th
thebossishere
2018-09-06 20:02:57
Can our parents enter us as a homeschool
Can our parents enter us as a homeschool
DPatrick
2018-09-06 20:03:09
Yes. Have your parent email the AMS at the address I listed above.
Yes. Have your parent email the AMS at the address I listed above.
Master_AoPs_Confirmed
2018-09-06 20:03:26
what rewards do other contestants get
what rewards do other contestants get
Ondhu
2018-09-06 20:03:26
are there prizes even if you don't make it to the championships
are there prizes even if you don't make it to the championships
DPatrick
2018-09-06 20:03:51
I don't believe that there are prized except for the 12 finalists. But your school will receive an AoPS coupon good for AoPS books.
I don't believe that there are prized except for the 12 finalists. But your school will receive an AoPS coupon good for AoPS books.
sub_math
2018-09-06 20:04:11
what amc level is the contest around? Like AMC8 AMC10 or what
what amc level is the contest around? Like AMC8 AMC10 or what
shemao
2018-09-06 20:04:11
how hard are the questions in terms of the AMC/AIME level contests
how hard are the questions in terms of the AMC/AIME level contests
DPatrick
2018-09-06 20:04:16
I would say AMC 12 generally.
I would say AMC 12 generally.
ComicPuzzle
2018-09-06 20:04:35
Is a perfect score on round two practically necessary to get in?
Is a perfect score on round two practically necessary to get in?
ComicPuzzle
2018-09-06 20:04:35
On average what is the score to go into the championships?
On average what is the score to go into the championships?
DPatrick
2018-09-06 20:04:51
I will defer to Mike on this. And I think in the UK the process is somewhat different as well.
I will defer to Mike on this. And I think in the UK the process is somewhat different as well.
mikebreen
2018-09-06 20:05:01
9 of 10 on Round 2 in the US and Canada
9 of 10 on Round 2 in the US and Canada
math2435
2018-09-06 20:05:16
Can I enter even If I am in 6th grade
Can I enter even If I am in 6th grade
Bubble3
2018-09-06 20:05:16
can a 5th grader participate
can a 5th grader participate
mikebreen
2018-09-06 20:05:18
In the UK, the top 4-6 scorers compete in a live final to get to the US
In the UK, the top 4-6 scorers compete in a live final to get to the US
DPatrick
2018-09-06 20:05:36
You can, but be advised that the math on the contest may be up through precalculus level.
You can, but be advised that the math on the contest may be up through precalculus level.
mikebreen
2018-09-06 20:05:37
Yes. No lower age limit
Yes. No lower age limit
proshi
2018-09-06 20:05:43
what is the time limt?
what is the time limt?
DPatrick
2018-09-06 20:05:48
10 problems, 15 minutes.
10 problems, 15 minutes.
mikebreen
2018-09-06 20:05:50
15 minutes on each round
15 minutes on each round
Apurple
2018-09-06 20:06:00
how much do you have to get right to pass round 1?
how much do you have to get right to pass round 1?
DPatrick
2018-09-06 20:06:03
7 out of 10
7 out of 10
ADMathNoob
2018-09-06 20:06:16
Is it true that this contest also touches on math history?
Is it true that this contest also touches on math history?
mikebreen
2018-09-06 20:06:29
It used to. NOt this year.
It used to. NOt this year.
smartmonkey888
2018-09-06 20:06:41
what is the championship round like?
what is the championship round like?
DPatrick
2018-09-06 20:06:51
It is the finalists competing on stage in front of a live audience!
It is the finalists competing on stage in front of a live audience!
DPatrick
2018-09-06 20:06:59
You can watch videos of past years on the AMS website.
You can watch videos of past years on the AMS website.
mikebreen
2018-09-06 20:07:01
Very exciting.
Very exciting.
MananV
2018-09-06 20:07:18
How many people enter
How many people enter
DPatrick
2018-09-06 20:07:25
I think about 3,000 last year, is that right Mike?
I think about 3,000 last year, is that right Mike?
mikebreen
2018-09-06 20:07:36
Yes, roughly. We hope more enter this year.
Yes, roughly. We hope more enter this year.
ComicPuzzle
2018-09-06 20:07:52
when will the championship round be?
when will the championship round be?
mikebreen
2018-09-06 20:08:02
On Jan. 19 in Baltimore
On Jan. 19 in Baltimore
shemao
2018-09-06 20:08:15
do they factor in your round 1 score when deciding who goes to nationals
do they factor in your round 1 score when deciding who goes to nationals
doctordairy
2018-09-06 20:08:15
Is it a cumulative score to get to round three or is it just based on your round two score?
Is it a cumulative score to get to round three or is it just based on your round two score?
DPatrick
2018-09-06 20:08:20
I believe only the Round 2 score is considered.
I believe only the Round 2 score is considered.
mikebreen
2018-09-06 20:08:28
Right.Not cumulative
Right.Not cumulative
DPatrick
2018-09-06 20:09:17
We are at the point where we're getting a lot of repeats of questions that we've already answered.
We are at the point where we're getting a lot of repeats of questions that we've already answered.
DPatrick
2018-09-06 20:09:34
So I think I will bring the session to an end at this point.
So I think I will bring the session to an end at this point.
DPatrick
2018-09-06 20:09:42
I'll repeat the most important informatioN:
I'll repeat the most important informatioN:
DPatrick
2018-09-06 20:09:48
To have your school sign up, ask your school or teacher to write to paoffice@ams.org with the subject line "WWTBAM Championship". Your school will then receive instructions for how to administer the qualifying test. Round 1 of qualifying will take place September 10-26 -- your school can administer the qualifying test on any day during this period.
To have your school sign up, ask your school or teacher to write to paoffice@ams.org with the subject line "WWTBAM Championship". Your school will then receive instructions for how to administer the qualifying test. Round 1 of qualifying will take place September 10-26 -- your school can administer the qualifying test on any day during this period.
DPatrick
2018-09-06 20:10:11
And again, 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.
And again, 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.
UnstoppableGoddess
2018-09-06 20:10:26
How much does it cost
How much does it cost
mikebreen
2018-09-06 20:10:31
Free
Free
DPatrick
2018-09-06 20:10:36
And I'll say this again too because it's important: it's free!!
And I'll say this again too because it's important: it's free!!
DPatrick
2018-09-06 20:10:46
Thanks for participating tonight, and good luck on WWTBAM!
Thanks for participating tonight, and good luck on WWTBAM!
DPatrick
2018-09-06 20:10:53
Please join us again for a discussion of Qualifying Round 1 on Thursday, September 27, 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 Tuesday, October 23.
Please join us again for a discussion of Qualifying Round 1 on Thursday, September 27, 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 Tuesday, October 23.
mikebreen
2018-09-06 20:10:53
Yes, good luck everyone.
Yes, good luck everyone.
DPatrick
2018-09-06 20:12:13
Good night!
Good night!
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