Who Wants to Be a Mathematician, Round 2
Go back to the Math Jam ArchiveAoPS instructor David Patrick will discuss the problems on Round 2 of qualifying for the 2019 Who Wants to Be a Mathematician Championship. We will be joined by Mike Breen and Bill Butterworth, the creators of the game. Mike is also the host of the Championship finals, to be held in Baltimore in January 2019.
Copyright © 2024 AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this file with others.
Facilitator: Dave Patrick
DPatrick
2018-10-23 19:13:36
DPatrick
2018-10-23 19:30:12
I think we're ready to get started!
I think we're ready to get started!
DPatrick
2018-10-23 19:30:16
Welcome to the 2018-19 Who Wants to Be a Mathematician Championship Round 2 Math Jam!
Welcome to the 2018-19 Who Wants to Be a Mathematician Championship Round 2 Math Jam!
DPatrick
2018-10-23 19:30:25
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
2018-10-23 19:30:39
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
2018-10-23 19:30:42
DPatrick
2018-10-23 19:30:46
Photo Credit: Maria Melin, copyright 1999 ABC Television.
Photo Credit: Maria Melin, copyright 1999 ABC Television.
j3370
2018-10-23 19:30:58
How much did you win!
How much did you win!
DPatrick
2018-10-23 19:31:07
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
2018-10-23 19:31:19
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
2018-10-23 19:31:35
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
2018-10-23 19:31:49
This helps keep the class organized and on track. This also means that only well-written comments will be passed 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 passed into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick
2018-10-23 19:32:04
There are a lot of students here! As I said, only (a fraction of the) well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
There are a lot of students here! As I said, only (a fraction of the) well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
DPatrick
2018-10-23 19:32:22
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the material for every problem as we go. 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!
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the material for every problem as we go. Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We usually do in our classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!
DPatrick
2018-10-23 19:32:37
Joining us tonight are the co-creators of WWTBAM, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
Joining us tonight are the co-creators of WWTBAM, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
mikebreen
2018-10-23 19:32:41
Hello, everyone.
Hello, everyone.
DPatrick
2018-10-23 19:32:49
Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill began Who Wants to Be a Mathematician for the American Mathematical Society in 2001. The first national game was in 2010. Mike has been on Jeopardy! and Wheel of Fortune (if you want to know if he won lots of money on either show, note that he is still working for a living) and may be the only person ever to cut his hand on the wheel. Who Wants to Be a Mathematician has so far been much safer.
Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill began Who Wants to Be a Mathematician for the American Mathematical Society in 2001. The first national game was in 2010. Mike has been on Jeopardy! and Wheel of Fortune (if you want to know if he won lots of money on either show, note that he is still working for a living) and may be the only person ever to cut his hand on the wheel. Who Wants to Be a Mathematician has so far been much safer.
TPiR
2018-10-23 19:33:03
Hi everyone!
Hi everyone!
DPatrick
2018-10-23 19:33:08
Bill earned an undergraduate degree in mathematics from Santa Clara University and a Ph.D. from Northwestern University, and is currently an associate professor and associate chair of mathematics at DePaul University. He shares a life-long interest in game shows with colleague Mike Breen, with whom he works as the not-so-lovely assistant on the mathematics game show Who Wants to Be a Mathematician.
Bill earned an undergraduate degree in mathematics from Santa Clara University and a Ph.D. from Northwestern University, and is currently an associate professor and associate chair of mathematics at DePaul University. He shares a life-long interest in game shows with colleague Mike Breen, with whom he works as the not-so-lovely assistant on the mathematics game show Who Wants to Be a Mathematician.
DPatrick
2018-10-23 19:33:17
And do you know why TPiR is his username?
And do you know why TPiR is his username?
Mathisfun04
2018-10-23 19:33:37
the price is right?
the price is right?
KSS
2018-10-23 19:33:37
The Price is right?
The Price is right?
champion999
2018-10-23 19:33:37
the price is right!
the price is right!
alex7
2018-10-23 19:33:37
The Price is Right
The Price is Right
WLongtin
2018-10-23 19:33:37
the price is right!
the price is right!
DPatrick
2018-10-23 19:33:43
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.)
DPatrick
2018-10-23 19:33:55
As you can see, we have a lot of game show background here tonight!
As you can see, we have a lot of game show background here tonight!
DPatrick
2018-10-23 19:34:04
Finally, we have a very special guest here with us tonight. Ken Ono (KenOno) is a professor of mathematics at Emory University and is Vice President of the American Mathematical Society (AMS), which sponsors Who Wants to Be a Mathematician. The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
Finally, we have a very special guest here with us tonight. Ken Ono (KenOno) is a professor of mathematics at Emory University and is Vice President of the American Mathematical Society (AMS), which sponsors Who Wants to Be a Mathematician. The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
KenOno
2018-10-23 19:34:23
Hi there.
Hi there.
DPatrick
2018-10-23 19:34:32
Ken is also the Director of the "Spirit of Ramanujan" project, which AoPS is also a partner with. If you'd like to learn more about Spirit of Ramanujan, please attend our Math Jam next week, October 30, at this same time, where we'll talk about Ramanujan's life, his mathematics, and the Spirit of Ramanujan talent search program.
Ken is also the Director of the "Spirit of Ramanujan" project, which AoPS is also a partner with. If you'd like to learn more about Spirit of Ramanujan, please attend our Math Jam next week, October 30, at this same time, where we'll talk about Ramanujan's life, his mathematics, and the Spirit of Ramanujan talent search program.
DPatrick
2018-10-23 19:35:09
Or if you can't make the Math Jam next Tuesday, visit spiritoframanujan.com to learn more.
Or if you can't make the Math Jam next Tuesday, visit spiritoframanujan.com to learn more.
KSS
2018-10-23 19:35:17
There are a lot of cool people here today!
There are a lot of cool people here today!
DPatrick
2018-10-23 19:35:22
Wait, one more!
Wait, one more!
DPatrick
2018-10-23 19:35:28
We also have an assistant here to help out tonight: Zach Stein-Perlman (ZachSteinPerlman). Zach has loved mathematics, brainteasers, and logic puzzles for longer than he can remember. He joined AoPS in 2011, back in the days when you faxed in homework. In addition to math, Zach is interested in philosophy and government. In his free time, Zach likes reading fantasy novels, cuddling with his cats, and running half marathons.
We also have an assistant here to help out tonight: Zach Stein-Perlman (ZachSteinPerlman). Zach has loved mathematics, brainteasers, and logic puzzles for longer than he can remember. He joined AoPS in 2011, back in the days when you faxed in homework. In addition to math, Zach is interested in philosophy and government. In his free time, Zach likes reading fantasy novels, cuddling with his cats, and running half marathons.
DPatrick
2018-10-23 19:35:47
Zach can try to help you if you have a question or are having some other difficulty. He may open a private window with you to chat if needed.
Zach can try to help you if you have a question or are having some other difficulty. He may open a private window with you to chat if needed.
DPatrick
2018-10-23 19:36:10
Tonight we'll be talking about Round 2 of the 2018-19 WWTBAM Championship contest, which concluded yesterday. Round 2 was only open to students who scored 7 or better (out of 10) on Round 1, which was held in September.
Tonight we'll be talking about Round 2 of the 2018-19 WWTBAM Championship contest, which concluded yesterday. Round 2 was only open to students who scored 7 or better (out of 10) on Round 1, which was held in September.
DPatrick
2018-10-23 19:36:24
Round 2 consisted of 10 questions, with a 15-minute time limit. So the problems are quick: you have an average of 90 seconds per question. (But as we'll see as we work through the problems, some of them may not take you nearly that long.) No books, notes, calculators, or internet access was permitted during the contest.
Round 2 consisted of 10 questions, with a 15-minute time limit. So the problems are quick: you have an average of 90 seconds per question. (But as we'll see as we work through the problems, some of them may not take you nearly that long.) No books, notes, calculators, or internet access was permitted during the contest.
DPatrick
2018-10-23 19:36:45
We'll take a bit longer than 15 minutes tonight (probably about an hour), because we'll stop along the way to discuss each question. Please also remember that the purpose of this Math Jam is to work through the solutions to the problems, and not to merely present the answers. "Working through the solutions" often includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer.
We'll take a bit longer than 15 minutes tonight (probably about an hour), because we'll stop along the way to discuss each question. Please also remember that the purpose of this Math Jam is to work through the solutions to the problems, and not to merely present the answers. "Working through the solutions" often includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer.
DPatrick
2018-10-23 19:37:03
And here is problem #1:
And here is problem #1:
DPatrick
2018-10-23 19:37:07
1. Find the remainder when $2019^{2019} + 2020^{2019}$ is divided by $7$.
1. Find the remainder when $2019^{2019} + 2020^{2019}$ is divided by $7$.
DPatrick
2018-10-23 19:37:14
(Most of the questions are short-answer.)
(Most of the questions are short-answer.)
Mathisfun04
2018-10-23 19:37:40
we should start with mods
we should start with mods
motorfinn
2018-10-23 19:37:40
Does this have something to do with using Modular Arithmetic?
Does this have something to do with using Modular Arithmetic?
DPatrick
2018-10-23 19:37:49
Good idea. We can use modular arithmetic and work modulo $7$. (If you haven't seen this before, basically we're just going to keep track of remainders).
Good idea. We can use modular arithmetic and work modulo $7$. (If you haven't seen this before, basically we're just going to keep track of remainders).
DPatrick
2018-10-23 19:38:09
In particular, what are $2019$ and $2020$ modulo $7$? (Meaning, what are their remainders when we divide by $7$?)
In particular, what are $2019$ and $2020$ modulo $7$? (Meaning, what are their remainders when we divide by $7$?)
motorfinn
2018-10-23 19:38:35
2019 mod 7 is equal to 3 I believe
2019 mod 7 is equal to 3 I believe
asdf334
2018-10-23 19:38:35
3 and 4
3 and 4
RuiyangWu
2018-10-23 19:38:35
2019 modulo 7 is 3, 2020 modulo 7 is 4
2019 modulo 7 is 3, 2020 modulo 7 is 4
lkarhat
2018-10-23 19:38:35
3 and 4
3 and 4
motorfinn
2018-10-23 19:38:35
And similarly, 2020 is 7 mod 4.
And similarly, 2020 is 7 mod 4.
JotatD
2018-10-23 19:38:35
3 and 4
3 and 4
Mathisfun04
2018-10-23 19:38:35
3 and 4 mod 7 respectively
3 and 4 mod 7 respectively
DPatrick
2018-10-23 19:38:41
Right. $2019 = 288 \cdot 7 + 3$, so $2019 \equiv 3 \pmod{7}$.
Right. $2019 = 288 \cdot 7 + 3$, so $2019 \equiv 3 \pmod{7}$.
DPatrick
2018-10-23 19:38:49
And thus $2020 \equiv 4 \pmod{7}$. (It's just one more.)
And thus $2020 \equiv 4 \pmod{7}$. (It's just one more.)
DPatrick
2018-10-23 19:38:53
So what we're trying to compute is the residue $3^{2019} + 4^{2019} \pmod{7}$.
So what we're trying to compute is the residue $3^{2019} + 4^{2019} \pmod{7}$.
sbudaraj
2018-10-23 19:39:14
now we use euler's
now we use euler's
motorfinn
2018-10-23 19:39:14
Do we find a pattern with the residues?
Do we find a pattern with the residues?
DPatrick
2018-10-23 19:39:30
Certainly we could use Fermat's Little Theorem, or experimentally determine, the two terms mod $7$ and work from there, but what's a simpler way?
Certainly we could use Fermat's Little Theorem, or experimentally determine, the two terms mod $7$ and work from there, but what's a simpler way?
champion999
2018-10-23 19:39:49
3 and -3
3 and -3
sbudaraj
2018-10-23 19:39:49
4 = (-3) mod 7
4 = (-3) mod 7
DPatrick
2018-10-23 19:39:54
Very nice.
Very nice.
DPatrick
2018-10-23 19:39:59
Note that $4 \equiv -3 \pmod{7}$!
Note that $4 \equiv -3 \pmod{7}$!
DPatrick
2018-10-23 19:40:07
So our quantity modulo $7$ is $3^{2019} + (-3)^{2019} \pmod{7}$.
So our quantity modulo $7$ is $3^{2019} + (-3)^{2019} \pmod{7}$.
RuiyangWu
2018-10-23 19:40:15
so is the answer 0, then?
so is the answer 0, then?
motorfinn
2018-10-23 19:40:15
Would they then sum to 0?
Would they then sum to 0?
sbudaraj
2018-10-23 19:40:22
WOAH OMG IT CANCELS.≥ its 0
WOAH OMG IT CANCELS.≥ its 0
Krypton36
2018-10-23 19:40:22
0
0
DPatrick
2018-10-23 19:40:26
And now we don't have to break a sweat at all!
And now we don't have to break a sweat at all!
DPatrick
2018-10-23 19:40:30
$3^{2019} + (-3)^{2019} \equiv 3^{2019} - 3^{2019} \equiv 0 \pmod{7}$.
$3^{2019} + (-3)^{2019} \equiv 3^{2019} - 3^{2019} \equiv 0 \pmod{7}$.
DPatrick
2018-10-23 19:40:37
So the answer is $\boxed{0}$.
So the answer is $\boxed{0}$.
Justin04
2018-10-23 19:40:55
That's very slick
That's very slick
DPatrick
2018-10-23 19:41:07
Indeed. I like it when I don't have to do a lot of computation. There's another quick way to solve the problem, using a little bit of algebra.
Indeed. I like it when I don't have to do a lot of computation. There's another quick way to solve the problem, using a little bit of algebra.
DPatrick
2018-10-23 19:41:12
How can we factor $a^n + b^n$ when we know $n$ is odd?
How can we factor $a^n + b^n$ when we know $n$ is odd?
DPatrick
2018-10-23 19:41:30
If we're not sure of the general case, we can try a small example.
If we're not sure of the general case, we can try a small example.
BBBBMMMMLLLLE
2018-10-23 19:41:40
a^3+b^3
a^3+b^3
asdf334
2018-10-23 19:41:42
a^3 + b^3?
a^3 + b^3?
DPatrick
2018-10-23 19:41:49
Sure -- How does $a^3 + b^3$ factor?
Sure -- How does $a^3 + b^3$ factor?
champion999
2018-10-23 19:42:08
(a+b)(a^2-ab+b^2)
(a+b)(a^2-ab+b^2)
asdf334
2018-10-23 19:42:08
(a + b)(a^2 - ab + b^2)
(a + b)(a^2 - ab + b^2)
Mathisfun04
2018-10-23 19:42:08
(a+b)(a^2-ab+b^2)
(a+b)(a^2-ab+b^2)
Puddles_Penguin
2018-10-23 19:42:08
(a+b)(a^2-ab+b^2)
(a+b)(a^2-ab+b^2)
dajeff
2018-10-23 19:42:08
$(a+b)(a^2-ab+b^2)$
$(a+b)(a^2-ab+b^2)$
DPatrick
2018-10-23 19:42:12
It factors as
$$a^3 + b^3 = (a+b)(a^2-ab+b^2).$$
It's a type of telescoping sum. (More about this in a later problem!)
It factors as
$$a^3 + b^3 = (a+b)(a^2-ab+b^2).$$
It's a type of telescoping sum. (More about this in a later problem!)
DPatrick
2018-10-23 19:42:22
And more generally,
$$a^n + b^n = (a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + \cdots - ab^{n-2} + b^{n-1}).$$
And more generally,
$$a^n + b^n = (a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + \cdots - ab^{n-2} + b^{n-1}).$$
DPatrick
2018-10-23 19:42:34
This only works when $n$ is odd (since we need it to end with a $+$ sign at $b^{n-1}$).
This only works when $n$ is odd (since we need it to end with a $+$ sign at $b^{n-1}$).
DPatrick
2018-10-23 19:43:00
So in particular, in our example:
$$2019^{2019} + 2020^{2019} = (2019+2020)(\text{some big number}).$$
So in particular, in our example:
$$2019^{2019} + 2020^{2019} = (2019+2020)(\text{some big number}).$$
champion999
2018-10-23 19:43:15
=0 b/c 2019+2020=0 mod 7
=0 b/c 2019+2020=0 mod 7
Mathisfun04
2018-10-23 19:43:17
we know 2019+2020=3+4=0mod7!
we know 2019+2020=3+4=0mod7!
DPatrick
2018-10-23 19:43:25
Right: that's good enough!
$2019 + 2020 = 4039$ is a multiple of $7$!. Specifically, $4039 = 7 \cdot 577$.
So the whole thing is a multiple of $7$. Hence the remainder is $0$.
Right: that's good enough!
$2019 + 2020 = 4039$ is a multiple of $7$!. Specifically, $4039 = 7 \cdot 577$.
So the whole thing is a multiple of $7$. Hence the remainder is $0$.
DPatrick
2018-10-23 19:43:46
On to #2:
On to #2:
DPatrick
2018-10-23 19:43:51
2. If $n$ is the fifth power of a positive integer, which of the following could be the number of positive integer divisors (or factors) of $n$? (Including $1$ and $n$)
$ $
(a) $451$ (b) $452$ (c) $453$ (d) $454$ (e) $455$
2. If $n$ is the fifth power of a positive integer, which of the following could be the number of positive integer divisors (or factors) of $n$? (Including $1$ and $n$)
$ $
(a) $451$ (b) $452$ (c) $453$ (d) $454$ (e) $455$
aops008
2018-10-23 19:44:25
Use prime factorization
Use prime factorization
asdf334
2018-10-23 19:44:43
so if n were prime, there would be 6
so if n were prime, there would be 6
DPatrick
2018-10-23 19:44:52
Right, primes are probably the key.
Right, primes are probably the key.
DPatrick
2018-10-23 19:45:10
And to slightly clarify what asdf334 said, I think you mean that if $n = p^5$ where $p$ is some prime, then there are $6$ factors.
And to slightly clarify what asdf334 said, I think you mean that if $n = p^5$ where $p$ is some prime, then there are $6$ factors.
DPatrick
2018-10-23 19:45:27
What's another easy sort of number to compute the number of factors of?
What's another easy sort of number to compute the number of factors of?
DPatrick
2018-10-23 19:45:44
Remember, the phrase "could be" in the problem means that we just have to find a single example that works.
Remember, the phrase "could be" in the problem means that we just have to find a single example that works.
PCampbell
2018-10-23 19:45:55
$p^{10}$
$p^{10}$
DPatrick
2018-10-23 19:46:09
Right, or more generally any 5th power of a prime power.
Right, or more generally any 5th power of a prime power.
Blackhawk
2018-10-23 19:46:25
2^450 has 451 divisors
2^450 has 451 divisors
DPatrick
2018-10-23 19:46:27
Aha!
Aha!
DPatrick
2018-10-23 19:46:37
If $p$ is prime, then $p^{5k} = (p^k)^5$ has $5k+1$ factors.
If $p$ is prime, then $p^{5k} = (p^k)^5$ has $5k+1$ factors.
DPatrick
2018-10-23 19:46:55
And as Blackhawk's example shows, $\boxed{(a)}$ $451$ clearly works when $k = 90$.
And as Blackhawk's example shows, $\boxed{(a)}$ $451$ clearly works when $k = 90$.
asdf334
2018-10-23 19:47:07
2^10 * 3^40 also has 451
2^10 * 3^40 also has 451
DPatrick
2018-10-23 19:47:15
Right: certainly there are lots of examples.
Right: certainly there are lots of examples.
DPatrick
2018-10-23 19:47:37
It requires just a little more work to show that the other answer choices can't happen, but it's not too difficult and I think I'll leave it for you to explore if you like.
It requires just a little more work to show that the other answer choices can't happen, but it's not too difficult and I think I'll leave it for you to explore if you like.
DPatrick
2018-10-23 19:48:12
And you can probably prove the more general statement that the number of divisors of an $n$th power must be equivalent to $1 \pmod{n}$.
And you can probably prove the more general statement that the number of divisors of an $n$th power must be equivalent to $1 \pmod{n}$.
DPatrick
2018-10-23 19:48:31
(The simplest example is that a perfect square always has an odd number of factors.)
(The simplest example is that a perfect square always has an odd number of factors.)
DPatrick
2018-10-23 19:48:40
Next up:
Next up:
DPatrick
2018-10-23 19:48:46
3. How many solutions are there to $\sin(4x) = \cos(2x)$ in $[0,2\pi]$? ($x$ in radians)
3. How many solutions are there to $\sin(4x) = \cos(2x)$ in $[0,2\pi]$? ($x$ in radians)
DPatrick
2018-10-23 19:49:01
There are a couple of ways we can approach this.
There are a couple of ways we can approach this.
aops008
2018-10-23 19:49:35
Double angle formula
Double angle formula
Blackhawk
2018-10-23 19:49:35
graph it
graph it
PCampbell
2018-10-23 19:49:44
Use the identity $\sin(2u)=2\sin(u)\cos(u)$, where $u=2x$.
Use the identity $\sin(2u)=2\sin(u)\cos(u)$, where $u=2x$.
DPatrick
2018-10-23 19:49:59
Those are the two ways that I thought of.
Those are the two ways that I thought of.
DPatrick
2018-10-23 19:50:14
I always like to try to answer questions like this by sketching a graph if it's easy enough.
I always like to try to answer questions like this by sketching a graph if it's easy enough.
DPatrick
2018-10-23 19:50:24
And I think this one probably *is* easy enough.
And I think this one probably *is* easy enough.
DPatrick
2018-10-23 19:50:30
DPatrick
2018-10-23 19:50:42
$y = \sin(4x)$ is in blue. Note it starts at $(0,0)$ and makes $4$ complete cycles between $x=0$ and $x=2\pi$.
$y = \sin(4x)$ is in blue. Note it starts at $(0,0)$ and makes $4$ complete cycles between $x=0$ and $x=2\pi$.
DPatrick
2018-10-23 19:50:47
$y = \cos(2x)$ is in red. Note it starts at $(0,1)$ and makes $2$ complete cycles between $x=0$ and $x=2\pi$.
$y = \cos(2x)$ is in red. Note it starts at $(0,1)$ and makes $2$ complete cycles between $x=0$ and $x=2\pi$.
DPatrick
2018-10-23 19:51:01
And your sketch doesn't have to be as accurate as mine to see what's going on.
And your sketch doesn't have to be as accurate as mine to see what's going on.
potato36
2018-10-23 19:51:04
So we simply count the interseciotns
So we simply count the interseciotns
aops008
2018-10-23 19:51:10
So it's 8
So it's 8
asdf334
2018-10-23 19:51:10
8 solutions!
8 solutions!
DPatrick
2018-10-23 19:51:26
Indeed, I count $\boxed{8}$ intersection points.
Indeed, I count $\boxed{8}$ intersection points.
aops008
2018-10-23 19:51:34
How do we do it with double angle formula
How do we do it with double angle formula
DPatrick
2018-10-23 19:51:40
Good question -- let's see.
Good question -- let's see.
DPatrick
2018-10-23 19:51:50
We use the trig identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$.
We use the trig identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$.
DPatrick
2018-10-23 19:51:56
Then our equation becomes
$$2\sin(2x)\cos(2x) = \cos(2x).$$
Then our equation becomes
$$2\sin(2x)\cos(2x) = \cos(2x).$$
DPatrick
2018-10-23 19:52:12
So there are two types of solutions to this: what are they?
So there are two types of solutions to this: what are they?
kdep
2018-10-23 19:52:43
factor out cos(2x)
factor out cos(2x)
geniusofart
2018-10-23 19:52:43
one is cos(2x)=0
one is cos(2x)=0
DPatrick
2018-10-23 19:52:52
Right. One solution is $\cos(2x) = 0$.
Right. One solution is $\cos(2x) = 0$.
geniusofart
2018-10-23 19:52:57
and the other is cos(2x) is cancel from both sides
and the other is cos(2x) is cancel from both sides
WLongtin
2018-10-23 19:52:59
cos(2x) = 0 or sin(2x) = 1/2
cos(2x) = 0 or sin(2x) = 1/2
asdf334
2018-10-23 19:53:02
sin(2x) = 1/2
sin(2x) = 1/2
DPatrick
2018-10-23 19:53:20
Exactly: if the $\cos$ term is nonzero, we divide by it and get $2\sin(2x) = 1$, so $\sin(2x) = \frac12$ is the other set of solutions.
Exactly: if the $\cos$ term is nonzero, we divide by it and get $2\sin(2x) = 1$, so $\sin(2x) = \frac12$ is the other set of solutions.
DottedCaculator
2018-10-23 19:53:36
4 solutions for each
4 solutions for each
DPatrick
2018-10-23 19:54:23
Exactly.
Exactly.
DPatrick
2018-10-23 19:54:31
$\cos(2x) = 0$ has two solutions with $0 \le 2x \le \pi$: either $2x = \frac\pi2$ or $2x = \frac{3\pi}{2}$.
$\cos(2x) = 0$ has two solutions with $0 \le 2x \le \pi$: either $2x = \frac\pi2$ or $2x = \frac{3\pi}{2}$.
DPatrick
2018-10-23 19:54:47
So it has four solutions with $0 \le 2x \le 2\pi$.
So it has four solutions with $0 \le 2x \le 2\pi$.
DPatrick
2018-10-23 19:55:02
$\sin(2x) = \frac12$ has two solutions with $0 \le 2x \le \pi$: either $2x = \frac\pi6$ or $2x = \frac{5\pi}{6}$. So again, four with $0 \le 2x \le 2\pi$.
$\sin(2x) = \frac12$ has two solutions with $0 \le 2x \le \pi$: either $2x = \frac\pi6$ or $2x = \frac{5\pi}{6}$. So again, four with $0 \le 2x \le 2\pi$.
motorfinn
2018-10-23 19:55:10
So we achieve our same end result of 8
So we achieve our same end result of 8
DPatrick
2018-10-23 19:55:20
Right, that's $4+4 = \boxed{8}$ solutions total.
Right, that's $4+4 = \boxed{8}$ solutions total.
DPatrick
2018-10-23 19:55:56
The first three problems were all roughly equal in difficulty: about 1/3 of students taking Round 2 got each of them correct.
The first three problems were all roughly equal in difficulty: about 1/3 of students taking Round 2 got each of them correct.
DPatrick
2018-10-23 19:56:15
The next two problems (4 and 5) were actually the easiest: about half got #4, which is...
The next two problems (4 and 5) were actually the easiest: about half got #4, which is...
DPatrick
2018-10-23 19:56:23
4. What is the next term in this geometric sequence: $\sqrt2$, $\sqrt[3]{2}$, $\sqrt[6]{2}$ ?
4. What is the next term in this geometric sequence: $\sqrt2$, $\sqrt[3]{2}$, $\sqrt[6]{2}$ ?
DPatrick
2018-10-23 19:56:44
This may be hard to read on your screen: the second term is the cube-root of $2$, and the third term is the sixth-root of $2$.
This may be hard to read on your screen: the second term is the cube-root of $2$, and the third term is the sixth-root of $2$.
DPatrick
2018-10-23 19:57:08
And since they're so hard to read, and to work with, what's a better way to write them to make them easier to work with?
And since they're so hard to read, and to work with, what's a better way to write them to make them easier to work with?
motorfinn
2018-10-23 19:57:23
Does it have something to do with raising to fractional powers?
Does it have something to do with raising to fractional powers?
geniusofart
2018-10-23 19:57:23
fractional powers
fractional powers
asdf334
2018-10-23 19:57:23
make it into exponents
make it into exponents
moab33
2018-10-23 19:57:26
$2^\frac{1}{2}, 2^\frac{1}{3}…$
$2^\frac{1}{2}, 2^\frac{1}{3}…$
DPatrick
2018-10-23 19:57:29
We can write them as fractional powers of $2$:
$$ 2^{\frac12},\, 2^{\frac13},\, 2^{\frac16},\, \underline{\phantom{2^{\frac{x}{y}}}}$$
We can write them as fractional powers of $2$:
$$ 2^{\frac12},\, 2^{\frac13},\, 2^{\frac16},\, \underline{\phantom{2^{\frac{x}{y}}}}$$
DPatrick
2018-10-23 19:57:38
What goes in the blank?
What goes in the blank?
KnightatNight
2018-10-23 19:57:58
1
1
spartacle
2018-10-23 19:57:58
1
1
WLongtin
2018-10-23 19:57:58
1
1
Puddles_Penguin
2018-10-23 19:57:58
1
1
smart_person_2000IQ
2018-10-23 19:57:58
1
1
pad
2018-10-23 19:57:58
1
1
Puddles_Penguin
2018-10-23 19:57:58
Is it 1?
Is it 1?
pad
2018-10-23 19:57:58
2^0=1
2^0=1
asdf334
2018-10-23 19:57:58
1.
1.
PCampbell
2018-10-23 19:57:58
$2^{\frac{0}{6}}=1$
$2^{\frac{0}{6}}=1$
DPatrick
2018-10-23 19:58:02
Right!
Right!
DPatrick
2018-10-23 19:58:23
If the powers of $2$ form a geometric sequence, then the exponents must form an arithmetic sequence!
If the powers of $2$ form a geometric sequence, then the exponents must form an arithmetic sequence!
DPatrick
2018-10-23 19:58:43
Or you could note that the common ratio of the geometric sequence is $2^{\frac16}$ -- specifically, we're multiplying by $2^{-\frac16}$ to get from one term to the next.
Or you could note that the common ratio of the geometric sequence is $2^{\frac16}$ -- specifically, we're multiplying by $2^{-\frac16}$ to get from one term to the next.
DPatrick
2018-10-23 19:58:55
...which decreases the exponent by $-\frac16$ each time.
...which decreases the exponent by $-\frac16$ each time.
DPatrick
2018-10-23 19:59:04
So the next term is $2^0$, or $\boxed{1}$.
So the next term is $2^0$, or $\boxed{1}$.
DPatrick
2018-10-23 19:59:40
On to #5, which turned out to be the easiest problem of the set.
On to #5, which turned out to be the easiest problem of the set.
DPatrick
2018-10-23 19:59:46
5. If $a$ is a positive real number, in how many points can the graphs of $y=ax$ and $y=x^2+1$ intersect?
$ $
(a) $0$ only (b) $0$ or $1$ only (c) $0$ or $2$ only (d) $1$ or $2$ only (e) $0$, $1$, or $2$
5. If $a$ is a positive real number, in how many points can the graphs of $y=ax$ and $y=x^2+1$ intersect?
$ $
(a) $0$ only (b) $0$ or $1$ only (c) $0$ or $2$ only (d) $1$ or $2$ only (e) $0$, $1$, or $2$
asdf334
2018-10-23 20:00:33
ax = x^2 + 1
ax = x^2 + 1
champion999
2018-10-23 20:00:33
Graph it!
Graph it!
cai40
2018-10-23 20:00:33
We should draw the graph of y=x^2+1
We should draw the graph of y=x^2+1
DPatrick
2018-10-23 20:00:45
Indeed, just like the trig problem, we can solve this using algebra or by sketching graphs.
Indeed, just like the trig problem, we can solve this using algebra or by sketching graphs.
DPatrick
2018-10-23 20:00:49
Let's do the algebra first.
Let's do the algebra first.
DPatrick
2018-10-23 20:00:57
We're trying to count the number of $x$'s such that $ax = x^2 + 1$.
We're trying to count the number of $x$'s such that $ax = x^2 + 1$.
DPatrick
2018-10-23 20:01:03
This rearranges to $x^2 - ax + 1 = 0$.
This rearranges to $x^2 - ax + 1 = 0$.
DPatrick
2018-10-23 20:01:14
How do we determine how many (real) solutions this has?
How do we determine how many (real) solutions this has?
champion999
2018-10-23 20:01:26
Look at the discriminant
Look at the discriminant
asdf334
2018-10-23 20:01:26
use quadratic formula
use quadratic formula
sbudaraj
2018-10-23 20:01:26
TAKE THE DETERMINANT
TAKE THE DETERMINANT
nawakre
2018-10-23 20:01:26
quadratic formula
quadratic formula
PCampbell
2018-10-23 20:01:26
with quadratic formula
with quadratic formula
dajeff
2018-10-23 20:01:26
discriminant
discriminant
DPatrick
2018-10-23 20:01:38
We can write the solutions explicitly using the quadratic formula.
We can write the solutions explicitly using the quadratic formula.
DPatrick
2018-10-23 20:01:48
$x = \dfrac{a \pm \sqrt{a^2-4}}{2}$
$x = \dfrac{a \pm \sqrt{a^2-4}}{2}$
DPatrick
2018-10-23 20:02:08
And as many of you pointed out, it's the discriminant $a^2 - 4$ that determines how many solutions there are.
And as many of you pointed out, it's the discriminant $a^2 - 4$ that determines how many solutions there are.
Assassino9931
2018-10-23 20:02:27
x^2 - ax + 1 = 0 has discriminant a^2 - 4 which can be both negative, 0 and positive
x^2 - ax + 1 = 0 has discriminant a^2 - 4 which can be both negative, 0 and positive
novus677
2018-10-23 20:02:41
This is a quadratic and has 0,1, or 2 solutions
This is a quadratic and has 0,1, or 2 solutions
kdep
2018-10-23 20:02:45
discriminant must be greater than zero for it to be real
discriminant must be greater than zero for it to be real
DPatrick
2018-10-23 20:02:50
If $a^2 - 4 < 0$, then there are no solutions.
If $a^2 - 4 = 0$, then there is 1 solution.
If $a^2 - 4 > 0$, then there are 2 solutions.
If $a^2 - 4 < 0$, then there are no solutions.
If $a^2 - 4 = 0$, then there is 1 solution.
If $a^2 - 4 > 0$, then there are 2 solutions.
DPatrick
2018-10-23 20:02:59
But these can all happen!
But these can all happen!
DPatrick
2018-10-23 20:03:11
If $a < 2$, then there are no solutions.
If $a = 2$, then there is 1 solution.
If $a > 2$, then there are 2 solutions.
If $a < 2$, then there are no solutions.
If $a = 2$, then there is 1 solution.
If $a > 2$, then there are 2 solutions.
JotatD
2018-10-23 20:03:20
0,1,2
0,1,2
Hermain
2018-10-23 20:03:20
so it must be e because it's the only one with 3 options
so it must be e because it's the only one with 3 options
DottedCaculator
2018-10-23 20:03:20
The answer is (e)
The answer is (e)
DPatrick
2018-10-23 20:03:23
So the answer is $\boxed{\text{(e)}}$.
So the answer is $\boxed{\text{(e)}}$.
DPatrick
2018-10-23 20:03:40
And we could also solve this by sketching graphs.
And we could also solve this by sketching graphs.
DPatrick
2018-10-23 20:03:49
The graph of $y = x^2 + 1$ is an upwards-opening parabola with vertex at $(0,1)$:
The graph of $y = x^2 + 1$ is an upwards-opening parabola with vertex at $(0,1)$:
DPatrick
2018-10-23 20:03:53
DPatrick
2018-10-23 20:04:04
The graph of $y = ax$ (with $a > 0$) is a line through the origin $(0,0)$ with positive slope.
The graph of $y = ax$ (with $a > 0$) is a line through the origin $(0,0)$ with positive slope.
DPatrick
2018-10-23 20:04:25
Is it clear that such a line could intersect in $0$, $1$, or $2$ points?
Is it clear that such a line could intersect in $0$, $1$, or $2$ points?
DPatrick
2018-10-23 20:04:40
If $a$ is small, then the line will miss the parabola completely:
If $a$ is small, then the line will miss the parabola completely:
DPatrick
2018-10-23 20:04:44
DPatrick
2018-10-23 20:04:53
If $a$ is big, then we'll cross the parabola twice:
If $a$ is big, then we'll cross the parabola twice:
DPatrick
2018-10-23 20:04:57
DPatrick
2018-10-23 20:05:06
And can we glance and touch the parabola exactly once!
And can we glance and touch the parabola exactly once!
JotatD
2018-10-23 20:05:11
1 WHEN IS TANGENT
1 WHEN IS TANGENT
motorfinn
2018-10-23 20:05:31
Yes, if we make it tangent
Yes, if we make it tangent
KSS
2018-10-23 20:05:31
When tangent
When tangent
DPatrick
2018-10-23 20:05:32
Yes. Without doing any algebra, you can think of rotating the red line from the first image counterclockwise. At some point it will exactly touch the parabola.
Yes. Without doing any algebra, you can think of rotating the red line from the first image counterclockwise. At some point it will exactly touch the parabola.
DPatrick
2018-10-23 20:05:37
DPatrick
2018-10-23 20:05:47
So the answer is $\boxed{\text{(e)}}$.
So the answer is $\boxed{\text{(e)}}$.
DPatrick
2018-10-23 20:06:11
Now the problems get a bit harder. Here's #6:
Now the problems get a bit harder. Here's #6:
DPatrick
2018-10-23 20:06:16
6. Express $\displaystyle\sum_{n=1}^{10}\frac{2}{n(n+2)} - \sum_{n=1}^{10}\frac{1}{n(n+1)}$ as a fraction in lowest terms.
6. Express $\displaystyle\sum_{n=1}^{10}\frac{2}{n(n+2)} - \sum_{n=1}^{10}\frac{1}{n(n+1)}$ as a fraction in lowest terms.
DPatrick
2018-10-23 20:06:36
That $\displaystyle\sum$ symbol might be confusing.
That $\displaystyle\sum$ symbol might be confusing.
DPatrick
2018-10-23 20:06:43
It means "sum." In particular, $\displaystyle\sum_{n=1}^{10}$ means the sum of all the terms where we plug in $n=1$, $n=2$, $n=3$, all the way up to $n=10$.
It means "sum." In particular, $\displaystyle\sum_{n=1}^{10}$ means the sum of all the terms where we plug in $n=1$, $n=2$, $n=3$, all the way up to $n=10$.
DPatrick
2018-10-23 20:06:51
That is,
\[
\sum_{n=1}^{10}\frac{2}{n(n+2)} = \frac{2}{1(3)} + \frac{2}{2(4)} + \frac{2}{3(5)} + \cdots + \frac{2}{10(12)}.
\]
That is,
\[
\sum_{n=1}^{10}\frac{2}{n(n+2)} = \frac{2}{1(3)} + \frac{2}{2(4)} + \frac{2}{3(5)} + \cdots + \frac{2}{10(12)}.
\]
DPatrick
2018-10-23 20:07:00
Wow, that seems like a pain to add up.
Wow, that seems like a pain to add up.
novus677
2018-10-23 20:07:07
partial fraction decomposition and telescoping
partial fraction decomposition and telescoping
arjvik
2018-10-23 20:07:07
PFD?
PFD?
DPatrick
2018-10-23 20:07:18
Fortunately, there's a little trick that helps with this, called partial fraction decomposition.
Fortunately, there's a little trick that helps with this, called partial fraction decomposition.
DPatrick
2018-10-23 20:07:51
Basically, we want to write $\dfrac{2}{n(n+2)}$ as a sum or difference of two separate fractions, one with denominator $n$ and one with denominator $n+2$.
Basically, we want to write $\dfrac{2}{n(n+2)}$ as a sum or difference of two separate fractions, one with denominator $n$ and one with denominator $n+2$.
DPatrick
2018-10-23 20:08:05
It's a little bit of guess-and-check to figure it out.
It's a little bit of guess-and-check to figure it out.
novus677
2018-10-23 20:08:22
$\frac{1}{n}-\frac{1}{n+2}$
$\frac{1}{n}-\frac{1}{n+2}$
DPatrick
2018-10-23 20:08:29
That's right:
$$ \frac{2}{n(n+2)} = \frac{1}{n} - \frac{1}{n+2} $$
That's right:
$$ \frac{2}{n(n+2)} = \frac{1}{n} - \frac{1}{n+2} $$
DPatrick
2018-10-23 20:08:50
You can put it over a common denominator to check, the $n$'s will cancel out and we'll just have $2$ in the numerator.
You can put it over a common denominator to check, the $n$'s will cancel out and we'll just have $2$ in the numerator.
DPatrick
2018-10-23 20:09:03
So let's write it out for our first sum:
$$
\sum_{n=1}^{10} \frac{2}{n(n+2)} = \sum_{n=1}^{10}\left(\frac{1}{n} - \frac{1}{n+2}\right)
$$
So let's write it out for our first sum:
$$
\sum_{n=1}^{10} \frac{2}{n(n+2)} = \sum_{n=1}^{10}\left(\frac{1}{n} - \frac{1}{n+2}\right)
$$
DPatrick
2018-10-23 20:09:11
It's not clear that this helps...
It's not clear that this helps...
DPatrick
2018-10-23 20:09:30
But if we write it without the $\displaystyle\sum$ notation it's a lot easier to see what's going on.
But if we write it without the $\displaystyle\sum$ notation it's a lot easier to see what's going on.
DPatrick
2018-10-23 20:09:37
$$
\left(\frac11-\frac13\right) + \left(\frac12-\frac14\right) + \left(\frac13-\frac15\right) + \left(\frac14 - \frac16\right) + \cdots
$$
$$
\left(\frac11-\frac13\right) + \left(\frac12-\frac14\right) + \left(\frac13-\frac15\right) + \left(\frac14 - \frac16\right) + \cdots
$$
DPatrick
2018-10-23 20:10:06
Notice anything interesting?
Notice anything interesting?
mathlogician
2018-10-23 20:10:18
It telescopes
It telescopes
PCampbell
2018-10-23 20:10:18
Term cancellation
Term cancellation
nawakre
2018-10-23 20:10:18
some fractions cancel out
some fractions cancel out
sbudaraj
2018-10-23 20:10:18
TELESCOPE
TELESCOPE
asdf334
2018-10-23 20:10:18
it cancels
it cancels
phoenix9
2018-10-23 20:10:20
they cancel out!!!
they cancel out!!!
harsha12345
2018-10-23 20:10:24
most terms cancel ouy\
most terms cancel ouy\
asdf334
2018-10-23 20:10:27
you are left with 1/1, -1/11, 1/2, and -1/12?
you are left with 1/1, -1/11, 1/2, and -1/12?
DPatrick
2018-10-23 20:10:30
The $-\dfrac13$ in the first term will cancel with the $\dfrac13$ in the third term!
The $-\dfrac13$ in the first term will cancel with the $\dfrac13$ in the third term!
DPatrick
2018-10-23 20:10:36
The $-\dfrac14$ in the second term will cancel with the $\dfrac14$ in the fourth term!
The $-\dfrac14$ in the second term will cancel with the $\dfrac14$ in the fourth term!
DPatrick
2018-10-23 20:10:40
And so on!
And so on!
DPatrick
2018-10-23 20:10:43
The only terms that don't cancel are $\dfrac11 + \dfrac12 - \dfrac{1}{11} - \dfrac{1}{12}$.
The only terms that don't cancel are $\dfrac11 + \dfrac12 - \dfrac{1}{11} - \dfrac{1}{12}$.
DPatrick
2018-10-23 20:11:00
Let's leave it like this for now -- remember this is only the first sum in our expression. Some of these might cancel further with terms from the second sum.
Let's leave it like this for now -- remember this is only the first sum in our expression. Some of these might cancel further with terms from the second sum.
motorfinn
2018-10-23 20:11:03
Can we do the same for the second?
Can we do the same for the second?
WLongtin
2018-10-23 20:11:03
so do the same for the second one
so do the same for the second one
DPatrick
2018-10-23 20:11:14
So now we try to compute $\displaystyle\sum_{n=1}^{10} \frac{1}{n(n+1)}$. We can use the same sort of idea.
So now we try to compute $\displaystyle\sum_{n=1}^{10} \frac{1}{n(n+1)}$. We can use the same sort of idea.
DPatrick
2018-10-23 20:11:31
This one is a little simpler:
$$ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.$$
This one is a little simpler:
$$ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.$$
DPatrick
2018-10-23 20:11:41
So the sum becomes:
$$ \left(\frac11 - \frac12\right) + \left(\frac12 - \frac13\right) + \left(\frac13 - \frac14\right) + \cdots$$
So the sum becomes:
$$ \left(\frac11 - \frac12\right) + \left(\frac12 - \frac13\right) + \left(\frac13 - \frac14\right) + \cdots$$
kcbhatraju
2018-10-23 20:11:45
Telescoping sum for second as well!
Telescoping sum for second as well!
asdf334
2018-10-23 20:12:00
1/1 and -1/11
1/1 and -1/11
JotatD
2018-10-23 20:12:00
another cancelation
another cancelation
DottedCaculator
2018-10-23 20:12:00
1/1-1/11
1/1-1/11
cai40
2018-10-23 20:12:09
we end up wwith 1/1-1/11
we end up wwith 1/1-1/11
DPatrick
2018-10-23 20:12:10
All the terms cancel except $\dfrac11 - \dfrac{1}{11}$.
All the terms cancel except $\dfrac11 - \dfrac{1}{11}$.
DPatrick
2018-10-23 20:12:16
So our entire expression is $$\left(\dfrac11 + \dfrac12 - \dfrac{1}{11} - \dfrac{1}{12}\right) - \left(\dfrac11 - \dfrac{1}{11}\right).$$
So our entire expression is $$\left(\dfrac11 + \dfrac12 - \dfrac{1}{11} - \dfrac{1}{12}\right) - \left(\dfrac11 - \dfrac{1}{11}\right).$$
DPatrick
2018-10-23 20:12:33
That's great -- even more cancellation!
That's great -- even more cancellation!
asdf334
2018-10-23 20:12:38
when you subtract it from the first term, you get $\frac{1}{2}-\frac{1}{12}$
when you subtract it from the first term, you get $\frac{1}{2}-\frac{1}{12}$
kdep
2018-10-23 20:12:38
1/2 - 1/12
1/2 - 1/12
motorfinn
2018-10-23 20:12:41
1/2-1/12
1/2-1/12
motorfinn
2018-10-23 20:12:41
so 5/12
so 5/12
PCampbell
2018-10-23 20:12:41
so $\frac{1}{2} - \frac{1}{12}$
so $\frac{1}{2} - \frac{1}{12}$
DPatrick
2018-10-23 20:12:44
What's left is just $\dfrac12 - \dfrac{1}{12}$.
What's left is just $\dfrac12 - \dfrac{1}{12}$.
DPatrick
2018-10-23 20:12:52
And this is just $\dfrac{6}{12} - \dfrac{1}{12} = \boxed{\dfrac{5}{12}}$.
And this is just $\dfrac{6}{12} - \dfrac{1}{12} = \boxed{\dfrac{5}{12}}$.
DPatrick
2018-10-23 20:13:09
This tactic of writing each term of a sum as a difference of two terms, in order to get mass cancellation, is called a telescoping sum.
This tactic of writing each term of a sum as a difference of two terms, in order to get mass cancellation, is called a telescoping sum.
DPatrick
2018-10-23 20:13:34
On to #7:
On to #7:
DPatrick
2018-10-23 20:13:38
7. The regular octagon pictured has side length $5$. What is the nearest integer to the length of $AB$?
7. The regular octagon pictured has side length $5$. What is the nearest integer to the length of $AB$?
DPatrick
2018-10-23 20:13:41
DPatrick
2018-10-23 20:13:57
Hmmm...any ideas?
Hmmm...any ideas?
RuiyangWu
2018-10-23 20:14:22
draw triangles
draw triangles
KSS
2018-10-23 20:14:22
Make some triangles?
Make some triangles?
asdf334
2018-10-23 20:14:22
make a right triangle
make a right triangle
cai40
2018-10-23 20:14:22
PYTHAGOREAN THEOROM?
PYTHAGOREAN THEOROM?
asdf334
2018-10-23 20:14:22
with AB as hypotenuse
with AB as hypotenuse
sbudaraj
2018-10-23 20:14:22
Draw right triangle CAB
Draw right triangle CAB
RuiyangWu
2018-10-23 20:14:22
right triangle hypotenuse
right triangle hypotenuse
motorfinn
2018-10-23 20:14:22
We could use Pythagorean Theorem
We could use Pythagorean Theorem
DPatrick
2018-10-23 20:14:28
One idea is that $AB$ is the hypotenuse of the green right triangle:
One idea is that $AB$ is the hypotenuse of the green right triangle:
DPatrick
2018-10-23 20:14:30
DPatrick
2018-10-23 20:14:40
One leg is easy: $AX = 5$.
One leg is easy: $AX = 5$.
DPatrick
2018-10-23 20:14:44
But what is $XB$?
But what is $XB$?
WLongtin
2018-10-23 20:15:24
yes and also a small 45 45 90 triangle around each vertex
yes and also a small 45 45 90 triangle around each vertex
Vexaria
2018-10-23 20:15:24
make another 45 45 90 triangle with side 5 as hypt
make another 45 45 90 triangle with side 5 as hypt
amuthup
2018-10-23 20:15:24
draw a square around the octagon
draw a square around the octagon
PCampbell
2018-10-23 20:15:24
Connect the top vertices to $XB$ such that the connections are perpendicular
Connect the top vertices to $XB$ such that the connections are perpendicular
DPatrick
2018-10-23 20:15:39
There's a couple of ways to do this, but I think they're essentially the same work.
There's a couple of ways to do this, but I think they're essentially the same work.
DPatrick
2018-10-23 20:15:48
I drew a couple of perpendiculars:
I drew a couple of perpendiculars:
DPatrick
2018-10-23 20:15:52
DPatrick
2018-10-23 20:15:59
Now $XB$ is broken up into 3 segments.
Now $XB$ is broken up into 3 segments.
DPatrick
2018-10-23 20:16:06
The middle segment has the same length as $YZ$, which is $5$.
The middle segment has the same length as $YZ$, which is $5$.
DPatrick
2018-10-23 20:16:31
And the two outside segments are each legs of a 45-45-90 triangle with hypotenuse $5$.
And the two outside segments are each legs of a 45-45-90 triangle with hypotenuse $5$.
skrublord420
2018-10-23 20:16:39
Project other vertices onto $XB$ to find that $XB = 5/\sqrt{2} + 5 + 5/\sqrt{2} = 5 + 5\sqrt{2}.$
Project other vertices onto $XB$ to find that $XB = 5/\sqrt{2} + 5 + 5/\sqrt{2} = 5 + 5\sqrt{2}.$
WLongtin
2018-10-23 20:16:39
5 + 2 * 5/root(2)
5 + 2 * 5/root(2)
motorfinn
2018-10-23 20:16:39
5+5sqrt2
5+5sqrt2
DPatrick
2018-10-23 20:16:47
Right. Each is $\frac{5}{\sqrt2}$, and together they are $2 \cdot \frac{5}{\sqrt2} = 5\sqrt2$.
Right. Each is $\frac{5}{\sqrt2}$, and together they are $2 \cdot \frac{5}{\sqrt2} = 5\sqrt2$.
DPatrick
2018-10-23 20:16:51
So combined, $XB = 5 + 5\sqrt2$.
So combined, $XB = 5 + 5\sqrt2$.
DPatrick
2018-10-23 20:17:04
This might be more convenient as $XB = 5(1 + \sqrt2$).
This might be more convenient as $XB = 5(1 + \sqrt2$).
motorfinn
2018-10-23 20:17:08
Now we use pythagorean theorem
Now we use pythagorean theorem
harsha12345
2018-10-23 20:17:08
now pythagorean theorem
now pythagorean theorem
DPatrick
2018-10-23 20:17:14
Right. $AB$ is the hypotenuse of a right triangle with legs $5$ and $5(1+\sqrt2)$.
Right. $AB$ is the hypotenuse of a right triangle with legs $5$ and $5(1+\sqrt2)$.
DPatrick
2018-10-23 20:17:29
Therefore, $AB = 5\sqrt{1^2 + (1+\sqrt2)^2}$.
Therefore, $AB = 5\sqrt{1^2 + (1+\sqrt2)^2}$.
DPatrick
2018-10-23 20:17:41
This simplifies to $AB = 5\sqrt{1 + (1 + 2\sqrt2 + 2)} = 5\sqrt{4 + 2\sqrt2}$.
This simplifies to $AB = 5\sqrt{1 + (1 + 2\sqrt2 + 2)} = 5\sqrt{4 + 2\sqrt2}$.
DPatrick
2018-10-23 20:17:45
Yikes. What integer is this closest to?
Yikes. What integer is this closest to?
WLongtin
2018-10-23 20:18:03
AB^2 is very close to 169
AB^2 is very close to 169
DPatrick
2018-10-23 20:18:18
Indeed, it's hard to tell straight away, but it's easier to work with if we square it.
Indeed, it's hard to tell straight away, but it's easier to work with if we square it.
DPatrick
2018-10-23 20:18:23
$AB^2 = 25(4 + 2\sqrt2) = 100 + 50\sqrt2$.
$AB^2 = 25(4 + 2\sqrt2) = 100 + 50\sqrt2$.
DPatrick
2018-10-23 20:18:37
Now we can use the fact that $\sqrt2 \approx 1.4$. (Hopefully that approximation is good enough.) This gives $AB^2 \approx 100 + 50(1.4) = 170$.
Now we can use the fact that $\sqrt2 \approx 1.4$. (Hopefully that approximation is good enough.) This gives $AB^2 \approx 100 + 50(1.4) = 170$.
bibilutza
2018-10-23 20:18:51
13
13
DottedCaculator
2018-10-23 20:18:51
13
13
JotatD
2018-10-23 20:18:51
13 then?
13 then?
qwertykeyboard
2018-10-23 20:18:51
so AB is close to 13
so AB is close to 13
DPatrick
2018-10-23 20:18:58
That is just about $13^2 = 169$, so $AB \approx \boxed{13}$ looks like the answer.
That is just about $13^2 = 169$, so $AB \approx \boxed{13}$ looks like the answer.
DPatrick
2018-10-23 20:19:09
If we wanted to be extra-careful, we could show that $(12.5)^2 < AB^2 < (13.5)^2$. But I'll skip that for now.
If we wanted to be extra-careful, we could show that $(12.5)^2 < AB^2 < (13.5)^2$. But I'll skip that for now.
mikebreen
2018-10-23 20:19:15
There's a connection with the Fibonacci sequence.
There's a connection with the Fibonacci sequence.
DPatrick
2018-10-23 20:19:36
Interesting...I didn't know that!
Interesting...I didn't know that!
mikebreen
2018-10-23 20:19:57
If the side length was 8, AB is close to 21, ... We couldn't figure out why, though.
If the side length was 8, AB is close to 21, ... We couldn't figure out why, though.
DPatrick
2018-10-23 20:20:08
Something to think about...
Something to think about...
DPatrick
2018-10-23 20:20:32
On to #8:
On to #8:
DPatrick
2018-10-23 20:20:37
8. The real polynomial $p(x) = x^3 - x^2 + bx + c$ has $2+i$ as one of its roots. What is the sum of the $x$- and $y$-intercepts of the graph of $y = p(x)$? (no $b$ or $c$ in your answer)
8. The real polynomial $p(x) = x^3 - x^2 + bx + c$ has $2+i$ as one of its roots. What is the sum of the $x$- and $y$-intercepts of the graph of $y = p(x)$? (no $b$ or $c$ in your answer)
DPatrick
2018-10-23 20:21:00
In case you don't know, $i$ is the imaginary number with the property that $i^2 = -1$. (If you haven't worked much with imaginary numbers before, the solution to this problem may be hard to follow.)
In case you don't know, $i$ is the imaginary number with the property that $i^2 = -1$. (If you haven't worked much with imaginary numbers before, the solution to this problem may be hard to follow.)
DPatrick
2018-10-23 20:21:23
What do we know about complex (non-real) solutions to real polynomials?
What do we know about complex (non-real) solutions to real polynomials?
skonar
2018-10-23 20:21:34
2-i is also a root
2-i is also a root
motorfinn
2018-10-23 20:21:34
If 2+i is a root, 2-i is a root
If 2+i is a root, 2-i is a root
kdep
2018-10-23 20:21:34
2-i is also a root
2-i is also a root
novus677
2018-10-23 20:21:34
$2-i$ is also a root
$2-i$ is also a root
WLongtin
2018-10-23 20:21:34
so $2-i$ is a root as well
so $2-i$ is a root as well
champion999
2018-10-23 20:21:34
they occur in conjugate pairs
they occur in conjugate pairs
bibilutza
2018-10-23 20:21:34
you must have the conjugate $2-1$
you must have the conjugate $2-1$
Vexaria
2018-10-23 20:21:39
2-i is also a root
2-i is also a root
crazi4math001
2018-10-23 20:21:39
The other root is the conjugate, or $(2-i)$.
The other root is the conjugate, or $(2-i)$.
DPatrick
2018-10-23 20:21:44
They come in conjugate pairs! That is, if $2+i$ is a root, then $2-i$ must also be a root.
They come in conjugate pairs! That is, if $2+i$ is a root, then $2-i$ must also be a root.
DPatrick
2018-10-23 20:22:04
The polynomial is a cubic. We know $2 \pm i$ are two of the roots. Can we determine the third root?
The polynomial is a cubic. We know $2 \pm i$ are two of the roots. Can we determine the third root?
champion999
2018-10-23 20:22:26
vieta gives us the third root
vieta gives us the third root
crazi4math001
2018-10-23 20:22:26
Vieta's?
Vieta's?
KSS
2018-10-23 20:22:28
Can we use vieta for it
Can we use vieta for it
DPatrick
2018-10-23 20:22:34
Yes -- can you elaborate?
Yes -- can you elaborate?
DPatrick
2018-10-23 20:22:52
What Vieta formula specifically is useful here?
What Vieta formula specifically is useful here?
novus677
2018-10-23 20:23:17
the sum of the roots is 1
the sum of the roots is 1
Puddles_Penguin
2018-10-23 20:23:17
Sum is 1
Sum is 1
DPatrick
2018-10-23 20:23:29
Right!
Right!
DPatrick
2018-10-23 20:24:00
One of Vieta's formulas tells us that the sum of the roots of monic polynomial is $-1$ times the coefficient of the $x^{n-1}$ term.
One of Vieta's formulas tells us that the sum of the roots of monic polynomial is $-1$ times the coefficient of the $x^{n-1}$ term.
DPatrick
2018-10-23 20:24:21
"monic" means the polynomial starts with an $x^n$ with coefficient $1$.
"monic" means the polynomial starts with an $x^n$ with coefficient $1$.
DPatrick
2018-10-23 20:24:31
So in this example, since the coefficient of $x^2$ is $-1$, we know that the sum of the roots is $1$.
So in this example, since the coefficient of $x^2$ is $-1$, we know that the sum of the roots is $1$.
asdf334
2018-10-23 20:24:41
3rd root is -3
3rd root is -3
skonar
2018-10-23 20:24:41
$2+i+2-i=4$ so other root is $-3$
$2+i+2-i=4$ so other root is $-3$
motorfinn
2018-10-23 20:24:41
So the third root is -3
So the third root is -3
JotatD
2018-10-23 20:24:41
-3 is the third root
-3 is the third root
DPatrick
2018-10-23 20:24:56
Thus, if $2 \pm i$ are two of the roots, the third root must be $-3$ so that $(2+i) + (2-i) + (-3) = 1$.
Thus, if $2 \pm i$ are two of the roots, the third root must be $-3$ so that $(2+i) + (2-i) + (-3) = 1$.
Puddles_Penguin
2018-10-23 20:25:05
So -3 is x-intercept
So -3 is x-intercept
DPatrick
2018-10-23 20:25:15
Right, now we know the $x$-intercept! The $x$-intercept is where the graph crosses the $x$-axis, so it's where $y = 0$.
Right, now we know the $x$-intercept! The $x$-intercept is where the graph crosses the $x$-axis, so it's where $y = 0$.
DPatrick
2018-10-23 20:25:26
But the only real root is $x=-3$. So $(-3,0)$ is the $x$-intercept.
But the only real root is $x=-3$. So $(-3,0)$ is the $x$-intercept.
motorfinn
2018-10-23 20:25:36
We want the sum of the values
We want the sum of the values
motorfinn
2018-10-23 20:25:36
For both x, and y intercepts
For both x, and y intercepts
DPatrick
2018-10-23 20:25:43
Right, we still need to find the $y$-intercept.
Right, we still need to find the $y$-intercept.
DPatrick
2018-10-23 20:25:49
How do we do that?
How do we do that?
Vexaria
2018-10-23 20:26:10
use prod of roots, which equals -c
use prod of roots, which equals -c
skonar
2018-10-23 20:26:10
we need c bc p(0)=c
we need c bc p(0)=c
Puddles_Penguin
2018-10-23 20:26:15
Maybe you can use vietas to find c
Maybe you can use vietas to find c
WLongtin
2018-10-23 20:26:15
we need $c$
we need $c$
DPatrick
2018-10-23 20:26:32
If we plug in $x=0$, we get $y = p(0) = c$. So $(0,c)$ is the $y$-intercept.
If we plug in $x=0$, we get $y = p(0) = c$. So $(0,c)$ is the $y$-intercept.
DPatrick
2018-10-23 20:26:48
And we can find $c$ by using Vieta's formula for the product of the roots!
And we can find $c$ by using Vieta's formula for the product of the roots!
exal
2018-10-23 20:26:53
-c = product of roots
-c = product of roots
DPatrick
2018-10-23 20:27:01
Right. $c$ is the negative of the product of the roots! (It's negative because the polynomial has odd degree.)
Right. $c$ is the negative of the product of the roots! (It's negative because the polynomial has odd degree.)
DPatrick
2018-10-23 20:27:10
That is, $c = -(2+i)(2-i)(-3)$.
That is, $c = -(2+i)(2-i)(-3)$.
skonar
2018-10-23 20:27:23
c=15
c=15
JotatD
2018-10-23 20:27:23
15!
15!
DPatrick
2018-10-23 20:27:28
And $(2+i)(2-i) = 4 - i^2 = 4 - (-1) = 5$.
And $(2+i)(2-i) = 4 - i^2 = 4 - (-1) = 5$.
DPatrick
2018-10-23 20:27:32
So $c = -(5)(-3) = 15$. Therefore, $(0,15)$ is the $y$-intercept.
So $c = -(5)(-3) = 15$. Therefore, $(0,15)$ is the $y$-intercept.
DottedCaculator
2018-10-23 20:27:45
The answer is 12.
The answer is 12.
buzhidao
2018-10-23 20:27:45
So our final answer is 12.
So our final answer is 12.
mathChamp06
2018-10-23 20:27:47
12 is the answer then
12 is the answer then
DPatrick
2018-10-23 20:27:50
And the final answer is thus $(-3) + 15 = \boxed{12}$.
And the final answer is thus $(-3) + 15 = \boxed{12}$.
DPatrick
2018-10-23 20:28:10
If you haven't seen Vieta's Formulas before, they are well worth learning. They come up in a lot of polynomial problems.
If you haven't seen Vieta's Formulas before, they are well worth learning. They come up in a lot of polynomial problems.
DPatrick
2018-10-23 20:28:27
But you have to be really really careful with the signs!
But you have to be really really careful with the signs!
DPatrick
2018-10-23 20:28:55
On to #9, which was the hardest of the regular problems. (#10 was not a "regular" problem as we'll soon see.)
On to #9, which was the hardest of the regular problems. (#10 was not a "regular" problem as we'll soon see.)
DPatrick
2018-10-23 20:29:00
9. You roll two fair six-sided dice and your opponent rolls one die. What is the probability that the maximum of the numbers shown on the top faces of your two dice is strictly higher that the number shown on the top face of your opponent's die?
9. You roll two fair six-sided dice and your opponent rolls one die. What is the probability that the maximum of the numbers shown on the top faces of your two dice is strictly higher that the number shown on the top face of your opponent's die?
DPatrick
2018-10-23 20:29:22
Anybody recognize what board game this scenario is from?
Anybody recognize what board game this scenario is from?
PCampbell
2018-10-23 20:29:37
Risk
Risk
nawakre
2018-10-23 20:29:37
risk?
risk?
DPatrick
2018-10-23 20:29:41
If you've ever played the board game Risk, you may recognize this as what happens in the game when 2 armies attack 1 army.
If you've ever played the board game Risk, you may recognize this as what happens in the game when 2 armies attack 1 army.
DPatrick
2018-10-23 20:29:59
Let's call a roll where the maximum of my two dice is greater than my opponent's die a winner, and all other rolls losers.
Let's call a roll where the maximum of my two dice is greater than my opponent's die a winner, and all other rolls losers.
DPatrick
2018-10-23 20:30:12
We want the probability of a winner.
We want the probability of a winner.
DPatrick
2018-10-23 20:30:22
First, how many rolls in total are there?
First, how many rolls in total are there?
Vexaria
2018-10-23 20:30:41
6^3 = 216
6^3 = 216
happyyouni
2018-10-23 20:30:41
216
216
DPatrick
2018-10-23 20:30:45
Each of the three dice has $6$ possible outcomes, so there are $6^3 = 216$ possible rolls.
Each of the three dice has $6$ possible outcomes, so there are $6^3 = 216$ possible rolls.
DPatrick
2018-10-23 20:30:52
Is it easier to count winners or losers?
Is it easier to count winners or losers?
BestAOPS
2018-10-23 20:31:04
losers?
losers?
crazi4math001
2018-10-23 20:31:04
Losers
Losers
DPatrick
2018-10-23 20:31:11
To count winners, we have to count rolls where my first die is greater than my opponent's die OR my second die is greater than my opponent's die.
To count winners, we have to count rolls where my first die is greater than my opponent's die OR my second die is greater than my opponent's die.
DPatrick
2018-10-23 20:31:14
To count losers, we have to count rolls where my first die is less than or equal to my opponent's die AND my second die is less than or equal to my opponent's die.
To count losers, we have to count rolls where my first die is less than or equal to my opponent's die AND my second die is less than or equal to my opponent's die.
DPatrick
2018-10-23 20:31:27
Usually, counting "AND" conditions is easier. We can often count "OR" using casework, but if they cases overlap (and they do in this problem) it can get a little tricky.
Usually, counting "AND" conditions is easier. We can often count "OR" using casework, but if they cases overlap (and they do in this problem) it can get a little tricky.
DPatrick
2018-10-23 20:31:37
So let's count losers. How can we count them?
So let's count losers. How can we count them?
sbudaraj
2018-10-23 20:31:45
y not just split up into 6 cases and then it becomes easy
y not just split up into 6 cases and then it becomes easy
motorfinn
2018-10-23 20:32:01
Casework
Casework
DPatrick
2018-10-23 20:32:02
Indeed, we can do casework based on what my opponent rolls.
Indeed, we can do casework based on what my opponent rolls.
DPatrick
2018-10-23 20:32:07
For example, if my opponent rolls a $1$, how many of my rolls are losers?
For example, if my opponent rolls a $1$, how many of my rolls are losers?
sbudaraj
2018-10-23 20:32:27
1
1
Owen1204
2018-10-23 20:32:27
1
1
bibilutza
2018-10-23 20:32:27
1
1
DPatrick
2018-10-23 20:32:30
I lose only if I roll a $1$ on both my dice. So only $1$ roll is a loser.
I lose only if I roll a $1$ on both my dice. So only $1$ roll is a loser.
DPatrick
2018-10-23 20:32:35
How about if my opponent rolls a $2$?
How about if my opponent rolls a $2$?
sbudaraj
2018-10-23 20:32:59
4
4
mathChamp06
2018-10-23 20:32:59
4
4
popcorn1
2018-10-23 20:32:59
4
4
cai40
2018-10-23 20:32:59
4
4
skonar
2018-10-23 20:32:59
4 rolls are losers
4 rolls are losers
piplupdragon
2018-10-23 20:32:59
4
4
DPatrick
2018-10-23 20:33:04
I lose only if I roll a $1$ or a $2$ on both my dice. So $2^2 = 4$ rolls are losers.
I lose only if I roll a $1$ or a $2$ on both my dice. So $2^2 = 4$ rolls are losers.
mathChamp06
2018-10-23 20:33:17
doesn't it just become the sum of the first 6 perfect squares?
doesn't it just become the sum of the first 6 perfect squares?
DPatrick
2018-10-23 20:33:20
Aha, good!
Aha, good!
popcorn1
2018-10-23 20:33:23
1,4,9,16,...
1,4,9,16,...
DPatrick
2018-10-23 20:33:27
If my opponent rolls $n$, then I lose if I roll $n$ or less on both dice. This happens in $n^2$ ways.
If my opponent rolls $n$, then I lose if I roll $n$ or less on both dice. This happens in $n^2$ ways.
DPatrick
2018-10-23 20:33:33
So the number of losers is $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2$.
So the number of losers is $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2$.
DPatrick
2018-10-23 20:33:49
(And note that if my opponent rolls a $6$, I'm doomed. All $36$ of my rolls are losers.)
(And note that if my opponent rolls a $6$, I'm doomed. All $36$ of my rolls are losers.)
DPatrick
2018-10-23 20:33:58
This is $1+4+9+16+25+36 = 91$ rolls that are losers.
This is $1+4+9+16+25+36 = 91$ rolls that are losers.
Puddles_Penguin
2018-10-23 20:34:14
Is it 125/216?
Is it 125/216?
skonar
2018-10-23 20:34:17
125/216
125/216
DPatrick
2018-10-23 20:34:24
Therefore, $216 - 91 = 125$ rolls are winners, and the probability of winning is $\boxed{\dfrac{125}{216}}$.
Therefore, $216 - 91 = 125$ rolls are winners, and the probability of winning is $\boxed{\dfrac{125}{216}}$.
popcorn1
2018-10-23 20:34:34
But thats 5^3/6^3
But thats 5^3/6^3
buzhidao
2018-10-23 20:34:34
125 is a nice number
125 is a nice number
motorfinn
2018-10-23 20:34:34
or (5/6)^3
or (5/6)^3
DPatrick
2018-10-23 20:34:49
Indeed, that's the first thing I noticed too. It's an odd coincidence that the answer is $\dfrac{5^3}{6^3}$. One might wonder if, given a $k$-sided die, the answer is always $\dfrac{(k-1)^3}{k^3}$.
Indeed, that's the first thing I noticed too. It's an odd coincidence that the answer is $\dfrac{5^3}{6^3}$. One might wonder if, given a $k$-sided die, the answer is always $\dfrac{(k-1)^3}{k^3}$.
DPatrick
2018-10-23 20:34:59
But it's not. It's just a coincidence.
But it's not. It's just a coincidence.
DPatrick
2018-10-23 20:35:09
(We checked.)
(We checked.)
DPatrick
2018-10-23 20:35:30
And finally, #10, which is a little bit different.
And finally, #10, which is a little bit different.
DPatrick
2018-10-23 20:35:35
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
2018-10-23 20:35:46
This is the tiebreaker question.
This is the tiebreaker question.
DPatrick
2018-10-23 20:35:53
No one is expected to get the exact answer.
No one is expected to get the exact answer.
DPatrick
2018-10-23 20:36:07
But this will break ties if two or more students tie on 1-9 in any region (more about that later).
But this will break ties if two or more students tie on 1-9 in any region (more about that later).
WLongtin
2018-10-23 20:36:20
ramanujan!
ramanujan!
Puddles_Penguin
2018-10-23 20:36:20
Ramanujan know this
Ramanujan know this
DPatrick
2018-10-23 20:36:42
Indeed, Ramanujan and his collaborator Hardy studied this question quite a bit and came up with a formula to approximate this.
Indeed, Ramanujan and his collaborator Hardy studied this question quite a bit and came up with a formula to approximate this.
DPatrick
2018-10-23 20:36:51
But first, any guesses? And any ideas how we might go about estimating this number?
But first, any guesses? And any ideas how we might go about estimating this number?
Hermain
2018-10-23 20:37:06
stars and bars?
stars and bars?
math-legend
2018-10-23 20:37:06
I got some crazy answer like $$2^101$$
I got some crazy answer like $$2^101$$
Owen1204
2018-10-23 20:37:06
wouldn't it just be 2^101?
wouldn't it just be 2^101?
skonar
2018-10-23 20:37:06
sticks and stones
sticks and stones
DPatrick
2018-10-23 20:37:11
Well...that's a related problem.
Well...that's a related problem.
DPatrick
2018-10-23 20:37:25
An easier problem is the number of ordered partitions. That is, in our example above $2+1$ and $1+2$ would count as different partitions of $3$ if we were counting ordered partitions.
An easier problem is the number of ordered partitions. That is, in our example above $2+1$ and $1+2$ would count as different partitions of $3$ if we were counting ordered partitions.
DPatrick
2018-10-23 20:37:44
This is a lot easier to count.
This is a lot easier to count.
DPatrick
2018-10-23 20:37:49
For example, suppose we want an ordered partition of $6$.
For example, suppose we want an ordered partition of $6$.
DPatrick
2018-10-23 20:37:55
We draw $6$ dots in a row:
;![[asy]<br /><br /><br />
unitsize(4mm);<br /><br /><br />
for (int i=0; i<6; ++i) {<br /><br /><br />
fill(shift(i,0)*scale(0.2)*unitcircle,black);<br /><br /><br />
}<br /><br /><br />
[/asy]](//latex.artofproblemsolving.com/4/2/e/42ea3da3fed3b621456e1f95f361433614c19ada.png)
We draw $6$ dots in a row:
;
![[asy]<br /><br /><br />
unitsize(4mm);<br /><br /><br />
for (int i=0; i<6; ++i) {<br /><br /><br />
fill(shift(i,0)*scale(0.2)*unitcircle,black);<br /><br /><br />
}<br /><br /><br />
[/asy]](http://latex.artofproblemsolving.com/4/2/e/42ea3da3fed3b621456e1f95f361433614c19ada.png)
DPatrick
2018-10-23 20:37:59
Then we put walls between the dots to break them up into groups.
Then we put walls between the dots to break them up into groups.
DPatrick
2018-10-23 20:38:07
For example:
;![[asy]<br /><br /><br />
unitsize(4mm);<br /><br /><br />
path p = ((0,-0.4)--(0,0.4));<br /><br /><br />
for (int i=0; i<6; ++i) {<br /><br /><br />
fill(shift(i,0)*scale(0.2)*unitcircle,black);<br /><br /><br />
}<br /><br /><br />
draw(shift(0.5,0)*p,black+2bp);<br /><br /><br />
draw(shift(3.5,0)*p,black+2bp);<br /><br /><br />
[/asy]](//latex.artofproblemsolving.com/9/b/6/9b6132253da6ec21e8aafc66732c6bf0eda2ab9e.png)
This corresponds to the ordered partition $1+3+2$.
For example:
;
![[asy]<br /><br /><br />
unitsize(4mm);<br /><br /><br />
path p = ((0,-0.4)--(0,0.4));<br /><br /><br />
for (int i=0; i<6; ++i) {<br /><br /><br />
fill(shift(i,0)*scale(0.2)*unitcircle,black);<br /><br /><br />
}<br /><br /><br />
draw(shift(0.5,0)*p,black+2bp);<br /><br /><br />
draw(shift(3.5,0)*p,black+2bp);<br /><br /><br />
[/asy]](http://latex.artofproblemsolving.com/9/b/6/9b6132253da6ec21e8aafc66732c6bf0eda2ab9e.png)
This corresponds to the ordered partition $1+3+2$.
DPatrick
2018-10-23 20:38:22
So, since there are 5 spaces between dots where we could put a wall, there are $2^5 = 32$ ordered partitions of $6$.
So, since there are 5 spaces between dots where we could put a wall, there are $2^5 = 32$ ordered partitions of $6$.
asdf334
2018-10-23 20:38:34
1+3+2 is the same as 1+2+3
1+3+2 is the same as 1+2+3
asdf334
2018-10-23 20:38:40
overcounting
overcounting
DPatrick
2018-10-23 20:38:45
Right: this is a substantial overcount. There are only eleven unordered partitions of $6$:
$6$
$5+1$
$4+2$
$4+1+1$
$3+3$
$3+2+1$
$3+1+1+1$
$2+2+2$
$2+2+1+1$
$2+1+1+1+1$
$1+1+1+1+1+1$
Right: this is a substantial overcount. There are only eleven unordered partitions of $6$:
$6$
$5+1$
$4+2$
$4+1+1$
$3+3$
$3+2+1$
$3+1+1+1$
$2+2+2$
$2+2+1+1$
$2+1+1+1+1$
$1+1+1+1+1+1$
DPatrick
2018-10-23 20:38:52
And we can see the overcount more clearly if we keep track of how many ways each unordered partition can be ordered
And we can see the overcount more clearly if we keep track of how many ways each unordered partition can be ordered
DPatrick
2018-10-23 20:38:56
$$\begin{array}{l|l}
\text{Unordered} & \text{# of orderings} \\ \hline
6 & 1 \\
5+1 & 2 \\
4+2 & 2 \\
4+1+1 & 3 \\
3+3 & 1 \\
3+2+1 & 6 \\
3+1+1+1 & 4 \\
2+2+2 & 1 \\
2+2+1+1 & 6 \\
2+1+1+1+1 & 5 \\
1+1+1+1+1+1 & 1
\end{array}$$
$$\begin{array}{l|l}
\text{Unordered} & \text{# of orderings} \\ \hline
6 & 1 \\
5+1 & 2 \\
4+2 & 2 \\
4+1+1 & 3 \\
3+3 & 1 \\
3+2+1 & 6 \\
3+1+1+1 & 4 \\
2+2+2 & 1 \\
2+2+1+1 & 6 \\
2+1+1+1+1 & 5 \\
1+1+1+1+1+1 & 1
\end{array}$$
DPatrick
2018-10-23 20:39:04
The numbers in the right column add up to $32$, the number of ordered partitions.
The numbers in the right column add up to $32$, the number of ordered partitions.
motorfinn
2018-10-23 20:39:14
Similarly, we have 100 spaces for 101 numbers
Similarly, we have 100 spaces for 101 numbers
DPatrick
2018-10-23 20:39:20
Indeed, if we line up $101$ dots in a row:
Indeed, if we line up $101$ dots in a row:
DPatrick
2018-10-23 20:39:24
DPatrick
2018-10-23 20:39:35
We can put walls into any of the $100$ slots between dots and get a partition:
We can put walls into any of the $100$ slots between dots and get a partition:
DPatrick
2018-10-23 20:39:47
DPatrick
2018-10-23 20:39:50
It's a little hard to tell, but this is the ordered partition $4 + 8 + 46 + 25 + 1 + 17$.
It's a little hard to tell, but this is the ordered partition $4 + 8 + 46 + 25 + 1 + 17$.
WLongtin
2018-10-23 20:40:02
but the over counting will be more severe
but the over counting will be more severe
motorfinn
2018-10-23 20:40:02
But that's overcounting
But that's overcounting
DPatrick
2018-10-23 20:40:06
Since there are $100$ slots, there are $2^{100}$ ordered partitions.
Since there are $100$ slots, there are $2^{100}$ ordered partitions.
DPatrick
2018-10-23 20:40:17
But this gives lots of different unordered partitions. For example, the partition above can be ordered in $6! = 720$ ways. That is, there are $720$ different ordered partitions that all correspond to the same unordered partition.
But this gives lots of different unordered partitions. For example, the partition above can be ordered in $6! = 720$ ways. That is, there are $720$ different ordered partitions that all correspond to the same unordered partition.
DPatrick
2018-10-23 20:40:24
And I don't know of a good way to account for all of the overcounting.
And I don't know of a good way to account for all of the overcounting.
DPatrick
2018-10-23 20:40:36
At least we have an upper bound. And $2^{100} = (2^{10})^{10} \approx (10^3)^{10} = 10^{30}$.
At least we have an upper bound. And $2^{100} = (2^{10})^{10} \approx (10^3)^{10} = 10^{30}$.
DPatrick
2018-10-23 20:40:45
But the overcounting is massive. For example, if I had a partition with $12$ different parts, then (assuming the parts are all different) it is overcounted by a factor of $12!$, which is about $500$ million.
But the overcounting is massive. For example, if I had a partition with $12$ different parts, then (assuming the parts are all different) it is overcounted by a factor of $12!$, which is about $500$ million.
DPatrick
2018-10-23 20:41:09
The best I could do was to try to figure out the "average" partition.
The best I could do was to try to figure out the "average" partition.
DPatrick
2018-10-23 20:41:33
I figured, using expected value, that the average ordered partition of $101$ has 26 1's, 12 2's, 6 3's, 3 4's, 2 5's, one unique higher
I figured, using expected value, that the average ordered partition of $101$ has 26 1's, 12 2's, 6 3's, 3 4's, 2 5's, one unique higher
DPatrick
2018-10-23 20:41:45
So the overcounting is by about $\dfrac{50!}{26!12!6!3!2!} \approx 10^{25}$.
So the overcounting is by about $\dfrac{50!}{26!12!6!3!2!} \approx 10^{25}$.
DPatrick
2018-10-23 20:42:07
So my guess would have been $10^{30} / 10^{25} = 10^5 = 100{,}000$.
So my guess would have been $10^{30} / 10^{25} = 10^5 = 100{,}000$.
DPatrick
2018-10-23 20:42:21
Turns out this is bogus and is way way too low.
Turns out this is bogus and is way way too low.
DPatrick
2018-10-23 20:42:33
Let's see what Ramanujan and Hardy did -- they're way smarter than me.
Let's see what Ramanujan and Hardy did -- they're way smarter than me.
DPatrick
2018-10-23 20:42:52
The Hardy-Ramanujan (1918) asymptotic formula is $$p(n) \sim \dfrac{1}{4n\sqrt3}\exp(c\sqrt{n}),$$ where $\exp(x) = e^x$ is the natural exponential (where $e = 2.71828\ldots$) and $c = \pi\sqrt{\frac23}$.
The Hardy-Ramanujan (1918) asymptotic formula is $$p(n) \sim \dfrac{1}{4n\sqrt3}\exp(c\sqrt{n}),$$ where $\exp(x) = e^x$ is the natural exponential (where $e = 2.71828\ldots$) and $c = \pi\sqrt{\frac23}$.
DPatrick
2018-10-23 20:43:22
As a reminder, if you'd like to learn more about Srinivasa Ramanujan and his mathematical achievements (much of which were in collaboration with G. H. Hardy), please attend our "Spirit of Ramanujan" Math Jam next Tuesday at this same time!
As a reminder, if you'd like to learn more about Srinivasa Ramanujan and his mathematical achievements (much of which were in collaboration with G. H. Hardy), please attend our "Spirit of Ramanujan" Math Jam next Tuesday at this same time!
spartacle
2018-10-23 20:43:27
not sure this is much better
not sure this is much better
DPatrick
2018-10-23 20:43:32
True, but it's workable at least.
True, but it's workable at least.
DPatrick
2018-10-23 20:43:42
Plugging in $n=101$ gives
$$ p(101) \approx \frac{1}{404\sqrt3}\exp\left(\pi\sqrt\frac{202}{3}\right).$$
Plugging in $n=101$ gives
$$ p(101) \approx \frac{1}{404\sqrt3}\exp\left(\pi\sqrt\frac{202}{3}\right).$$
DPatrick
2018-10-23 20:44:04
To approximate without a computer, I would note that $404\sqrt3 \approx 700$ and $\pi\sqrt\frac{202}{3} \approx 26$, so this is about $\dfrac{e^{26}}{700}$.
To approximate without a computer, I would note that $404\sqrt3 \approx 700$ and $\pi\sqrt\frac{202}{3} \approx 26$, so this is about $\dfrac{e^{26}}{700}$.
DPatrick
2018-10-23 20:44:17
And to approximate the exponential, note $2 < e < 3$ but is closer to $3$.
And to approximate the exponential, note $2 < e < 3$ but is closer to $3$.
DPatrick
2018-10-23 20:44:31
And then $2^{26} \approx (10^{3})^{2.6} \approx 10^8$ and $3^{26} \approx (3^2)^{13} \approx 10^{13}$, so maybe $e^{26} \approx 10^{11}$ is a decent approximation.
And then $2^{26} \approx (10^{3})^{2.6} \approx 10^8$ and $3^{26} \approx (3^2)^{13} \approx 10^{13}$, so maybe $e^{26} \approx 10^{11}$ is a decent approximation.
DPatrick
2018-10-23 20:44:44
Dividing by $700$ gives $1.4 \cdot 10^8 = 140{,}000{,}000$ as my approximation.
Dividing by $700$ gives $1.4 \cdot 10^8 = 140{,}000{,}000$ as my approximation.
DPatrick
2018-10-23 20:44:53
That's not terrible. The actual answer is $\boxed{214{,}481{,}126}$.
That's not terrible. The actual answer is $\boxed{214{,}481{,}126}$.
WLongtin
2018-10-23 20:45:09
maybe some kind of recursive formula giving partitions of $n$ in terms of partitions of $n-1$
maybe some kind of recursive formula giving partitions of $n$ in terms of partitions of $n-1$
DPatrick
2018-10-23 20:45:24
That's another approach. A simple recurrence is $p(n) = p(n-1) + \text{partitions of $n$ that don't use $1$}$.
That's another approach. A simple recurrence is $p(n) = p(n-1) + \text{partitions of $n$ that don't use $1$}$.
DPatrick
2018-10-23 20:45:31
But that last part is still tricky.
But that last part is still tricky.
motorfinn
2018-10-23 20:45:40
It would take a long time to derive such a formula
It would take a long time to derive such a formula
DPatrick
2018-10-23 20:45:53
Indeed. The mathematics behind the formula is not simple, to say the least.
Indeed. The mathematics behind the formula is not simple, to say the least.
WLongtin
2018-10-23 20:46:15
It was featured in the Ramanujan movie
It was featured in the Ramanujan movie
skonar
2018-10-23 20:46:31
The Man Who Knew Infinity
The Man Who Knew Infinity
DPatrick
2018-10-23 20:46:42
Anyway, that's it for the Round 2 problems!
Anyway, that's it for the Round 2 problems!
mikebreen
2018-10-23 20:46:46
No one got #10 correct. We were only looking for fairly good approximations. Within a factor of 10 is very good.
No one got #10 correct. We were only looking for fairly good approximations. Within a factor of 10 is very good.
DPatrick
2018-10-23 20:46:51
After the Round 2 scoring is complete, 12 students will be invited to compete in the Championship Finals, live in Baltimore at the 2019 Joint Mathematics Meetings on January 19. Travel costs to and from Baltimore will be covered by the AMS.
After the Round 2 scoring is complete, 12 students will be invited to compete in the Championship Finals, live in Baltimore at the 2019 Joint Mathematics Meetings on January 19. Travel costs to and from Baltimore will be covered by the AMS.
DPatrick
2018-10-23 20:47:02
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
2018-10-23 20:47:05
DPatrick
2018-10-23 20:47:18
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
2018-10-23 20:47:31
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
2018-10-23 20:47:47
And I'll be there too!
And I'll be there too!
DPatrick
2018-10-23 20:47:55
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
2018-10-23 20:48:30
You can learn more about WWTBAM on the AMS website at http://www.ams.org/wwtbam
You can learn more about WWTBAM on the AMS website at http://www.ams.org/wwtbam
mikebreen
2018-10-23 20:48:41
Emails to qualifiers probably going out tomorrow.
Emails to qualifiers probably going out tomorrow.
DPatrick
2018-10-23 20:49:06
Thanks for coming tonight. Hopefully I'll see many of you next week at our Spirit of Ramanujan Math Jam too!
Thanks for coming tonight. Hopefully I'll see many of you next week at our Spirit of Ramanujan Math Jam too!
asdf334
2018-10-23 20:49:15
how can i signup next year for the competition?
how can i signup next year for the competition?
DPatrick
2018-10-23 20:49:32
I don't think you can yet, but you can probably go to the AMS website I just linked to above to sign up to get on the mailing list.
I don't think you can yet, but you can probably go to the AMS website I just linked to above to sign up to get on the mailing list.
mikebreen
2018-10-23 20:49:39
Email paoffice with your teacher's name and email or ask your teacher to email us.
Email paoffice with your teacher's name and email or ask your teacher to email us.
mikebreen
2018-10-23 20:49:49
paoffice@ams.org
paoffice@ams.org
motorfinn
2018-10-23 20:49:58
What are the qualifications for taking this competition?
What are the qualifications for taking this competition?
DPatrick
2018-10-23 20:50:15
You just need to be a student (high school or lower) in the US, Canada, or the UK.
You just need to be a student (high school or lower) in the US, Canada, or the UK.
DPatrick
2018-10-23 20:51:02
Details are on the AMS website or your teacher can email the address that Mike just posted.
Details are on the AMS website or your teacher can email the address that Mike just posted.
DPatrick
2018-10-23 20:51:30
The AMS website will also have the livestream of the Finals in January. We'll announce it on AoPS too.
The AMS website will also have the livestream of the Finals in January. We'll announce it on AoPS too.
mathChamp06
2018-10-23 20:52:38
is this the end of the class then?
is this the end of the class then?
DPatrick
2018-10-23 20:53:02
I think so -- thanks to everyone for coming and to our special guests Mike, Bill, and Ken from the AMS for joining us. (Ken had to leave early.)
I think so -- thanks to everyone for coming and to our special guests Mike, Bill, and Ken from the AMS for joining us. (Ken had to leave early.)
DPatrick
2018-10-23 20:53:13
Ken will also be with us next week for Spirit of Ramanujan.
Ken will also be with us next week for Spirit of Ramanujan.
mikebreen
2018-10-23 20:53:21
Thanks, everyone. Great problem solving.
Thanks, everyone. Great problem solving.
TPiR
2018-10-23 20:53:26
Thanks everyone. Great job all the way around.
Thanks everyone. Great job all the way around.
DPatrick
2018-10-23 20:53:37
Have a good night!
Have a good night!
Copyright © 2024 AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this file with others.
