## 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 the 2016-2017 Who Wants to Be a Mathematician national contest. We will also be joined by Mike Breen and Bill Butterworth, the creators of the game. Mike is also the host of the national finals, to be held in Atlanta in January 2017.

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#### Facilitator: Dave Patrick

DPatrick
2016-11-09 19:30:44

**Welcome to the 2016-17***Who Wants to Be a Mathematician*Round 2 Math Jam!
DPatrick
2016-11-09 19:30:55

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 12 years, and I've written or co-written a few of our textbooks. I also once was a contestant on ABC's

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 12 years, and I've written or co-written a few of our textbooks. I also once was a contestant on ABC's

*Who Wants to Be a Millionaire*back before I started working at AoPS, when Regis was still the host. (No, I didn't win the million bucks, but I did win enough to buy a new car.)
DPatrick
2016-11-09 19:31:17

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
2016-11-09 19:31:31

The classroom is

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
2016-11-09 19:31:41

This helps keep the session organized and on track. This also means that only

This helps keep the session organized and on track. This also means that only

**well-written**comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick
2016-11-09 19:31:57

Also, we won't be going through all 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 all 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
2016-11-09 19:32:13

Assisting with the Math Jam tonight is Josh Morris (

Assisting with the Math Jam tonight is Josh Morris (

**QuantumCat**).
QuantumCat
2016-11-09 19:32:24

Hey everyone!

Hey everyone!

DPatrick
2016-11-09 19:32:26

Josh, after getting his Bachelor's and Master's in Physics from UT El Paso and teaching high school physics, decided he loved science and working with students so much he went back to school to work pursue a career in medicine. While he's applying to medical school and going through the interview process, he's helping out here at Art of Problem Solving and keeping his math skills sharp. In his spare time, he loves cooking, running, reading and is (still) slowly teaching himself how to play the piano.

Josh, after getting his Bachelor's and Master's in Physics from UT El Paso and teaching high school physics, decided he loved science and working with students so much he went back to school to work pursue a career in medicine. While he's applying to medical school and going through the interview process, he's helping out here at Art of Problem Solving and keeping his math skills sharp. In his spare time, he loves cooking, running, reading and is (still) slowly teaching himself how to play the piano.

DPatrick
2016-11-09 19:32:41

Josh can chat with you if you are having trouble, but again, due to the size of tonight's session, we may not be able to get to every question.

Josh can chat with you if you are having trouble, but again, due to the size of tonight's session, we may not be able to get to every question.

DPatrick
2016-11-09 19:32:52

Also joining us tonight are the co-creators of WWTBAM, Mike Breen (

Also joining us tonight are the co-creators of WWTBAM, Mike Breen (

**mikebreen**) and Bill Butterworth (**TPiR**).
mikebreen
2016-11-09 19:33:05

Hello, everyone!

Hello, everyone!

TPiR
2016-11-09 19:33:16

Hello all. Good to be here.

Hello all. Good to be here.

DPatrick
2016-11-09 19:33:18

Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill began

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.
DPatrick
2016-11-09 19:33:41

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

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*. 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
2016-11-09 19:34:08

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!

mathpractice2011
2016-11-09 19:34:11

I have been on wheel of fortune too

I have been on wheel of fortune too

mikebreen
2016-11-09 19:34:17

Bill is in the Cubs' afterglow

Bill is in the Cubs' afterglow

DPatrick
2016-11-09 19:34:28

Indeed, as is levans (our webmaster)!

Indeed, as is levans (our webmaster)!

DPatrick
2016-11-09 19:34:45

*Who Wants to Be a Mathematician*is run by the American Mathematical Society (AMS). The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
DPatrick
2016-11-09 19:35:02

Round 2 of the national contest consisted of 10 questions, with a 15-minute time limit. So the problems are quick: you have an average of 1.5 minutes per question. (Compare that to the AMC 10/12 which has an average of 3 minutes per question, or the AIME which has an average of 12 minutes per question.)

Round 2 of the national contest consisted of 10 questions, with a 15-minute time limit. So the problems are quick: you have an average of 1.5 minutes per question. (Compare that to the AMC 10/12 which has an average of 3 minutes per question, or the AIME which has an average of 12 minutes per question.)

DPatrick
2016-11-09 19:35:42

We'll take a bit longer than 15 minutes tonight, because we'll stop along the way to discuss each question. Please also remember that the purpose of this Math Jam is to work through the

We'll take a bit longer than 15 minutes tonight, because we'll stop along the way to discuss each question. Please also remember that the purpose of this Math Jam is to work through the

**solutions**to the problems, and not to merely present the answers. "Working through the solutions" often includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step.
DPatrick
2016-11-09 19:36:02

Several of the questions have interesting sidetracks, so we'll also stop and view some of the scenery along the way.

Several of the questions have interesting sidetracks, so we'll also stop and view some of the scenery along the way.

mathpractice2011
2016-11-09 19:36:07

How long is this session

How long is this session

DPatrick
2016-11-09 19:36:10

Probably about an hour.

Probably about an hour.

DPatrick
2016-11-09 19:36:29

Anyway, let's get started.

Anyway, let's get started.

DPatrick
2016-11-09 19:36:33

1. What is the sum of the roots of the polynomial $x^2 - 12x + 4$?

1. What is the sum of the roots of the polynomial $x^2 - 12x + 4$?

declanhiller
2016-11-09 19:37:24

would we use vieta's theorem?

would we use vieta's theorem?

Jorvis
2016-11-09 19:37:24

Use vieta's formula -> -(-12) = 12

Use vieta's formula -> -(-12) = 12

mc0131
2016-11-09 19:37:24

Vietta's

Vietta's

EpicRuler101
2016-11-09 19:37:24

vietas

vietas

chocolatemushrooms
2016-11-09 19:37:24

12 by vieta's

12 by vieta's

zetaintegrator
2016-11-09 19:37:24

vieta's rule

vieta's rule

NewtonFTW
2016-11-09 19:37:24

The sum of the roots of any quadratic equation are -b/a

The sum of the roots of any quadratic equation are -b/a

DPatrick
2016-11-09 19:37:41

Right! The fastest solution is to use one of Vieta's Formulas, which tell us that the sum of the roots of $ax^2 + bx + c = 0$ is $-\dfrac{b}{a}$.

Right! The fastest solution is to use one of Vieta's Formulas, which tell us that the sum of the roots of $ax^2 + bx + c = 0$ is $-\dfrac{b}{a}$.

DPatrick
2016-11-09 19:37:52

Is this equation, $b = -12$ and $a = 1$, so we get that the answer is $\boxed{12}$.

Is this equation, $b = -12$ and $a = 1$, so we get that the answer is $\boxed{12}$.

DPatrick
2016-11-09 19:38:22

We can see why this is true from the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

We can see why this is true from the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

DPatrick
2016-11-09 19:38:59

When we add the two roots, the square-root part cancels (one + with one -), and what's left is $2\left(\dfrac{-b}{2a}\right)$, or just $-\dfrac{b}{a}$.

When we add the two roots, the square-root part cancels (one + with one -), and what's left is $2\left(\dfrac{-b}{2a}\right)$, or just $-\dfrac{b}{a}$.

NewtonFTW
2016-11-09 19:39:07

So I'm guessing it might be useful to know the product formula as well, c/a?

So I'm guessing it might be useful to know the product formula as well, c/a?

EpicRuler101
2016-11-09 19:39:07

I also heard something about c/a, what would that be?

I also heard something about c/a, what would that be?

DPatrick
2016-11-09 19:39:21

Indeed, if you want the product of the two roots, that's $\dfrac{c}{a}$.

Indeed, if you want the product of the two roots, that's $\dfrac{c}{a}$.

DPatrick
2016-11-09 19:39:47

Yet another approach is to complete the square:

\[

(x^2 - 12x + 36) = 32

\]

so $(x-6)^2 = 32$.

Yet another approach is to complete the square:

\[

(x^2 - 12x + 36) = 32

\]

so $(x-6)^2 = 32$.

DPatrick
2016-11-09 19:39:59

When we graph this, we get a parabola that ``bottoms out'' at $x=6$, so the two roots average to 6, and thus their sum is 12.

When we graph this, we get a parabola that ``bottoms out'' at $x=6$, so the two roots average to 6, and thus their sum is 12.

DPatrick
2016-11-09 19:40:18

That was our warmup...

That was our warmup...

DPatrick
2016-11-09 19:40:22

2. What is $\tan^{-1}\left(\cos\left(\sin\left(\cos\left(\dfrac{\pi}{2}\right)\right)\right)\right)$?

2. What is $\tan^{-1}\left(\cos\left(\sin\left(\cos\left(\dfrac{\pi}{2}\right)\right)\right)\right)$?

DPatrick
2016-11-09 19:40:45

What's the general approach here?

What's the general approach here?

mc0131
2016-11-09 19:41:18

You can just calculate each step

You can just calculate each step

verum4one
2016-11-09 19:41:18

Just work from the inside out.

Just work from the inside out.

Mathaddict11
2016-11-09 19:41:18

work from the inside out

work from the inside out

Juliaxyzz
2016-11-09 19:41:18

work from inside out

work from inside out

Jillian
2016-11-09 19:41:18

Is it solve for the innermost parantheses and work your way out?

Is it solve for the innermost parantheses and work your way out?

Liopleurodon
2016-11-09 19:41:18

find each value inside of the brackets

find each value inside of the brackets

MathNerd0120
2016-11-09 19:41:18

work from inside the parentheses?

work from inside the parentheses?

DPatrick
2016-11-09 19:41:27

Yes. We can work from the inside out.

Yes. We can work from the inside out.

vonnan
2016-11-09 19:41:34

cos(\pi/2)=0

cos(\pi/2)=0

DPatrick
2016-11-09 19:41:41

To start, $\cos\left(\dfrac{\pi}{2}\right)$ is 0, so our expression is $\tan^{-1}(\cos(\sin(0)))$.

To start, $\cos\left(\dfrac{\pi}{2}\right)$ is 0, so our expression is $\tan^{-1}(\cos(\sin(0)))$.

verum4one
2016-11-09 19:41:53

sin(0)=0

sin(0)=0

nvpthemvp
2016-11-09 19:41:59

sin of 0 =0

sin of 0 =0

DPatrick
2016-11-09 19:42:03

And then $\sin(0)$ is 0, so our expression is $\tan^{-1}(\cos(0))$.

And then $\sin(0)$ is 0, so our expression is $\tan^{-1}(\cos(0))$.

declanhiller
2016-11-09 19:42:22

cos(0)=1

cos(0)=1

EpicDragonSlayr
2016-11-09 19:42:22

cos 0 = 1

cos 0 = 1

DPatrick
2016-11-09 19:42:27

And then $\cos(0)$ is 1, so our expression is $\tan^{-1}(1)$.

And then $\cos(0)$ is 1, so our expression is $\tan^{-1}(1)$.

verum4one
2016-11-09 19:42:52

$\tan ^{-1}(1)=\pi/4$

$\tan ^{-1}(1)=\pi/4$

EpicDragonSlayr
2016-11-09 19:42:52

pi/4

pi/4

mc0131
2016-11-09 19:42:52

pi/4

pi/4

DPatrick
2016-11-09 19:42:56

And finally, $\tan^{-1}(1)$ is $\boxed{\dfrac{\pi}{4}}$.

And finally, $\tan^{-1}(1)$ is $\boxed{\dfrac{\pi}{4}}$.

DPatrick
2016-11-09 19:43:31

So this problem was mainly an exercise in being careful with your trig functions. (And obviously if you haven't seen trig functions like this yet, there's not much chance of solving this.)

So this problem was mainly an exercise in being careful with your trig functions. (And obviously if you haven't seen trig functions like this yet, there's not much chance of solving this.)

DPatrick
2016-11-09 19:43:59

Next is one of the hallmarks of WWTBAM, which is that one of the questions usually involves mathematical history:

Next is one of the hallmarks of WWTBAM, which is that one of the questions usually involves mathematical history:

DPatrick
2016-11-09 19:44:03

3. What is the last name (family name) of the famous British mathematician who worked with Ramanujan and wrote

3. What is the last name (family name) of the famous British mathematician who worked with Ramanujan and wrote

*A Mathematician's Apology*?
Eugenis
2016-11-09 19:44:16

david patrick?

david patrick?

DPatrick
2016-11-09 19:44:31

I'm a little too young, by about a century I'm afraid.

I'm a little too young, by about a century I'm afraid.

bogstop320
2016-11-09 19:44:45

Hardy

Hardy

Mathgirl03
2016-11-09 19:44:45

Hardy

Hardy

Mathaddict11
2016-11-09 19:44:45

Hardy

Hardy

ngundotra
2016-11-09 19:44:45

Hardy

Hardy

POKEMON123
2016-11-09 19:44:45

G. H. Hardy

G. H. Hardy

declanhiller
2016-11-09 19:44:45

hardy

hardy

DPatrick
2016-11-09 19:44:51

Correct -- the mathematician in question is G. H.

Correct -- the mathematician in question is G. H.

**Hardy**.
DPatrick
2016-11-09 19:44:57

DPatrick
2016-11-09 19:45:09

And the person mentioned by name in the question is the Indian mathematician Srinivasa Ramanujan.

And the person mentioned by name in the question is the Indian mathematician Srinivasa Ramanujan.

DPatrick
2016-11-09 19:45:19

mikebreen
2016-11-09 19:45:30

We thought we might take advantage of the movie The Man Who Knew Infinity.

We thought we might take advantage of the movie The Man Who Knew Infinity.

DPatrick
2016-11-09 19:45:40

Indeed, the story of Ramanujan and Hardy was recently dramatized in the movie

Indeed, the story of Ramanujan and Hardy was recently dramatized in the movie

*The Man Who Knew Infinity*, starring Dev Patel as Ramanujan and Jeremy Irons as Hardy.
Mathaddict11
2016-11-09 19:45:53

AoPS books teach us this, which turns out to be surprisingly helpful for WWTBAM

AoPS books teach us this, which turns out to be surprisingly helpful for WWTBAM

DPatrick
2016-11-09 19:46:08

Yes, we like to sprinkle mathematical history and anecdotes into our books from time to time!

Yes, we like to sprinkle mathematical history and anecdotes into our books from time to time!

DPatrick
2016-11-09 19:46:20

As some of you know, Ramanujan was a math prodigy who grew up in India and was essentially self-taught. In 1913 at age 25, having studied Hardy's work extensively, Ramanujan wrote to Hardy and started a correspondence. Hardy immediately saw Ramanujan's mathematical talent and arrange for a position for him at the University of Cambridge in England where Hardy was a professor.

As some of you know, Ramanujan was a math prodigy who grew up in India and was essentially self-taught. In 1913 at age 25, having studied Hardy's work extensively, Ramanujan wrote to Hardy and started a correspondence. Hardy immediately saw Ramanujan's mathematical talent and arrange for a position for him at the University of Cambridge in England where Hardy was a professor.

DPatrick
2016-11-09 19:46:42

Ramanujan traveled to England in 1914 and studied with Hardy for the next several years. This ultimately lead to Ramanujan's election to the Royal Society of London, an extremely prestigious award. Unfortunately, Ramanujan was in very poor health at the time, and he returned to India in 1919, only to die there just a year later at age 32.

Ramanujan traveled to England in 1914 and studied with Hardy for the next several years. This ultimately lead to Ramanujan's election to the Royal Society of London, an extremely prestigious award. Unfortunately, Ramanujan was in very poor health at the time, and he returned to India in 1919, only to die there just a year later at age 32.

DPatrick
2016-11-09 19:46:56

The most famous anecdote regarding Ramanujan and Hardy is told by Hardy in his biography of Ramanujan:

The most famous anecdote regarding Ramanujan and Hardy is told by Hardy in his biography of Ramanujan:

*I remember once going to see him when he was lying ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."*
DPatrick
2016-11-09 19:47:36

Meanwhile, Hardy himself was also a fellow of the Royal Society of London, and a prolific mathematician in his own right, and (other than his mentorship of Ramanujan) is probably best known for his book

Meanwhile, Hardy himself was also a fellow of the Royal Society of London, and a prolific mathematician in his own right, and (other than his mentorship of Ramanujan) is probably best known for his book

*A Mathematician's Apology*, which he wrote later in life (in 1940, when he was in his 60s), which was his attempt to explain what it means to be a mathematician.
DPatrick
2016-11-09 19:47:50

Note: the photos and the stories are from the terrific math history website

Note: the photos and the stories are from the terrific math history website

*MacTutor History of Mathematics*at www-history.mcs.st-andrews.ac.uk/, run by the University of St. Andrews in Scotland.
NewtonFTW
2016-11-09 19:47:57

Will there be history of math at the national level?

Will there be history of math at the national level?

DPatrick
2016-11-09 19:48:13

There almost always is a math history question at the national finals.

There almost always is a math history question at the national finals.

DPatrick
2016-11-09 19:48:36

Anyway, on to the next problem:

Anyway, on to the next problem:

DPatrick
2016-11-09 19:48:41

4. A machine makes parts, but $1\%$ of the parts are defective. If an inspector selects 100 parts at random, which of the following intervals below contains the probability that none of the 100 parts are defective?

(a) $[0,0.2)$ (b) $[0.2, 0.4)$ (c) $[0.4,0.6)$ (d) $[0.6,0.8)$ (e) $[0.8,1.0]$

4. A machine makes parts, but $1\%$ of the parts are defective. If an inspector selects 100 parts at random, which of the following intervals below contains the probability that none of the 100 parts are defective?

(a) $[0,0.2)$ (b) $[0.2, 0.4)$ (c) $[0.4,0.6)$ (d) $[0.6,0.8)$ (e) $[0.8,1.0]$

Wave-Particle
2016-11-09 19:49:32

There is a $0.99$ chance it isn't defective

There is a $0.99$ chance it isn't defective

DPatrick
2016-11-09 19:49:51

Right, that's where we can start: given any individual part, the probability it isn't defective is 0.99.

Right, that's where we can start: given any individual part, the probability it isn't defective is 0.99.

NewtonFTW
2016-11-09 19:49:55

And that's for each part checked

And that's for each part checked

bogstop320
2016-11-09 19:49:57

(99/100)^100

(99/100)^100

DPatrick
2016-11-09 19:50:11

For none of the parts to be defective, they all have to be working. Each is working with probability $0.99$, so the probability that all 100 are working is $(0.99)^{100}$.

For none of the parts to be defective, they all have to be working. Each is working with probability $0.99$, so the probability that all 100 are working is $(0.99)^{100}$.

DPatrick
2016-11-09 19:50:35

So the question becomes: which of those answer choices contains $(0.99)^{100}$. (Need I mention that you're not allowed to use a computing device?)

So the question becomes: which of those answer choices contains $(0.99)^{100}$. (Need I mention that you're not allowed to use a computing device?)

Math1331Math
2016-11-09 19:50:50

Rewrite as $1-.01$

Rewrite as $1-.01$

DPatrick
2016-11-09 19:51:00

Good idea: why does that help?

Good idea: why does that help?

atmchallenge
2016-11-09 19:51:13

Let's try binomial theorem

Let's try binomial theorem

fractal161
2016-11-09 19:51:13

Binomial Theorem!!

Binomial Theorem!!

DPatrick
2016-11-09 19:51:21

Yes, we can use the Binomial Theorem to estimate:

\begin{align*}

(0.99)^{100} &= (1 - 0.01)^{100} \\

&= 1 - \binom{100}{1}(0.01) + \binom{100}{2}(0.01)^2 - \binom{100}{3}(0.01)^3 + \cdots - \binom{100}{99}(0.01)^{99} + \binom{100}{100}(0.01)^{100}.

\end{align*}

Yes, we can use the Binomial Theorem to estimate:

\begin{align*}

(0.99)^{100} &= (1 - 0.01)^{100} \\

&= 1 - \binom{100}{1}(0.01) + \binom{100}{2}(0.01)^2 - \binom{100}{3}(0.01)^3 + \cdots - \binom{100}{99}(0.01)^{99} + \binom{100}{100}(0.01)^{100}.

\end{align*}

DPatrick
2016-11-09 19:51:37

What do we notice about this?

What do we notice about this?

yuvi18
2016-11-09 19:51:59

they cancel eachother out

they cancel eachother out

DPatrick
2016-11-09 19:52:03

The first two terms cancel out.

The first two terms cancel out.

Wave-Particle
2016-11-09 19:52:16

Some parts are very very small

Some parts are very very small

atmchallenge
2016-11-09 19:52:16

They begin to get small pretty fast

They begin to get small pretty fast

Math1331Math
2016-11-09 19:52:16

Eventually the .01^x power gets so small it doesn't have an impact

Eventually the .01^x power gets so small it doesn't have an impact

DPatrick
2016-11-09 19:52:29

Well, I'm not so sure. We've got all those 100's in the top of $\dbinom{100}{k}$ too.

Well, I'm not so sure. We've got all those 100's in the top of $\dbinom{100}{k}$ too.

fractal161
2016-11-09 19:53:01

All negated by each $\frac{1}{100}$ term, though.

All negated by each $\frac{1}{100}$ term, though.

DPatrick
2016-11-09 19:53:14

Well...let's take a close look. The third term is $\dfrac{100 \cdot 99}{2 \cdot 1}(0.01)^2$.

Well...let's take a close look. The third term is $\dfrac{100 \cdot 99}{2 \cdot 1}(0.01)^2$.

DPatrick
2016-11-09 19:53:43

The numerator of the $\dbinom{100}{2}$ part is just about canceled out by the $(0.01)^2$.

The numerator of the $\dbinom{100}{2}$ part is just about canceled out by the $(0.01)^2$.

Wave-Particle
2016-11-09 19:53:49

Like 0.495?

Like 0.495?

atmchallenge
2016-11-09 19:53:49

Which is .495

Which is .495

vonnan
2016-11-09 19:53:52

1/2

1/2

DPatrick
2016-11-09 19:54:04

It's exactly 0.495, but probably $\dfrac12$ is close enough.

It's exactly 0.495, but probably $\dfrac12$ is close enough.

DPatrick
2016-11-09 19:54:12

Does this sort of thing continue?

Does this sort of thing continue?

awin
2016-11-09 19:54:26

Yes

Yes

mathpractice2011
2016-11-09 19:54:26

yes

yes

MathNerd0120
2016-11-09 19:54:26

yes

yes

vonnan
2016-11-09 19:54:26

yes

yes

Liopleurodon
2016-11-09 19:54:26

yes

yes

yuvi18
2016-11-09 19:54:26

yes!

yes!

atmchallenge
2016-11-09 19:54:28

yes, then it is $1/3!$, then $1/4!$, and so on...

yes, then it is $1/3!$, then $1/4!$, and so on...

DPatrick
2016-11-09 19:54:34

In fact, in all of the first few terms of the form $\binom{100}{k}(0.01)^k$, we can write it as

\[

\frac{100 \cdot 99 \cdot \cdots \cdot (101-k)}{k!} \cdot \frac{1}{100^k}.

\]

In fact, in all of the first few terms of the form $\binom{100}{k}(0.01)^k$, we can write it as

\[

\frac{100 \cdot 99 \cdot \cdots \cdot (101-k)}{k!} \cdot \frac{1}{100^k}.

\]

DPatrick
2016-11-09 19:54:52

The $100^k$ term very nearly cancels out the numerator of this fraction.

The $100^k$ term very nearly cancels out the numerator of this fraction.

DPatrick
2016-11-09 19:55:10

So we can approximate the expansion as

\[

(0.99)^{100} \approx 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \cdots.

\]

So we can approximate the expansion as

\[

(0.99)^{100} \approx 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \cdots.

\]

DPatrick
2016-11-09 19:55:29

As we noticed before, the first two terms cancel out. Just using the other three terms I've written, what do we get?

As we noticed before, the first two terms cancel out. Just using the other three terms I've written, what do we get?

Liopleurodon
2016-11-09 19:56:04

$\frac38$

$\frac38$

EpicRuler101
2016-11-09 19:56:04

1/2 - 1/6 + 1/24

1/2 - 1/6 + 1/24

thedoge
2016-11-09 19:56:04

3/8

3/8

NewtonFTW
2016-11-09 19:56:04

3/8

3/8

EulerMacaroni
2016-11-09 19:56:04

$\frac{3}{8}$

$\frac{3}{8}$

fractal161
2016-11-09 19:56:04

~$\frac{3}{8}$?

~$\frac{3}{8}$?

DPatrick
2016-11-09 19:56:08

$\dfrac12 - \dfrac16 + \dfrac{1}{24} = \dfrac{9}{24} = \dfrac38 = 0.375$

$\dfrac12 - \dfrac16 + \dfrac{1}{24} = \dfrac{9}{24} = \dfrac38 = 0.375$

DPatrick
2016-11-09 19:56:28

And the terms start to get really small after that, so the quantity won't change much.

And the terms start to get really small after that, so the quantity won't change much.

atmchallenge
2016-11-09 19:56:35

So b!

So b!

Wave-Particle
2016-11-09 19:56:35

So it's b?

So it's b?

DPatrick
2016-11-09 19:56:39

This is solidly inside the interval $[0.2,0.4)$ so the answer is $\boxed{(b)}$.

This is solidly inside the interval $[0.2,0.4)$ so the answer is $\boxed{(b)}$.

mikebreen
2016-11-09 19:56:49

Third-hardest question as it turns out.

Third-hardest question as it turns out.

DPatrick
2016-11-09 19:57:13

Indeed, #10 is the hardest (and you'll see why when we get there), the Hardy question was 2nd-hardest, and this was 3rd-hardest with around 36% correct.

Indeed, #10 is the hardest (and you'll see why when we get there), the Hardy question was 2nd-hardest, and this was 3rd-hardest with around 36% correct.

DPatrick
2016-11-09 19:57:34

Now if you've played with these sorts of expressions before, you might notice something familiar...

Now if you've played with these sorts of expressions before, you might notice something familiar...

Kpatton
2016-11-09 19:57:51

B, the probability is about 1/e

B, the probability is about 1/e

EulerMacaroni
2016-11-09 19:57:51

$$\bigg(\frac{99}{100}\bigg)^{100}=\bigg(1-\frac{1}{100}\bigg)^{100}\approx e^{-1}$$

$$\bigg(\frac{99}{100}\bigg)^{100}=\bigg(1-\frac{1}{100}\bigg)^{100}\approx e^{-1}$$

EulerMacaroni
2016-11-09 19:57:51

which is $\frac{1}{e}$

which is $\frac{1}{e}$

thedoge
2016-11-09 19:57:51

that looks like the formula for dearrangements...

that looks like the formula for dearrangements...

DPatrick
2016-11-09 19:58:13

Indeed, this is related to one of the definitions of $e$ as $e = \displaystyle\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$.

Indeed, this is related to one of the definitions of $e$ as $e = \displaystyle\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$.

DPatrick
2016-11-09 19:58:34

That is, as $n$ gets really big, the expression $(1 + 1/n)^n$ gets really close to the number $e$.

That is, as $n$ gets really big, the expression $(1 + 1/n)^n$ gets really close to the number $e$.

DPatrick
2016-11-09 19:58:41

This is the flip side: $\dfrac{1}{e} = \displaystyle\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^n$.

This is the flip side: $\dfrac{1}{e} = \displaystyle\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^n$.

DPatrick
2016-11-09 19:59:13

And in fact $\dfrac{1}{e} \approx 0.367879\ldots$ is very close to the exact answer of $(0.99)^{100} = 0.366032\ldots$.

And in fact $\dfrac{1}{e} \approx 0.367879\ldots$ is very close to the exact answer of $(0.99)^{100} = 0.366032\ldots$.

DPatrick
2016-11-09 20:00:15

And as thedoge also mentioned, this is also very closely related to the combinatorics problem of

And as thedoge also mentioned, this is also very closely related to the combinatorics problem of

**derangements**: if $n$ people all drop their keys in a hat, and each randomly selects a key, the probability that no one draws his or her original key is about $1/e$.
DPatrick
2016-11-09 20:00:37

Let's move on:

Let's move on:

DPatrick
2016-11-09 20:00:41

5. Find the remainder when $18^6 + 87^6$ is divided by $7$.

5. Find the remainder when $18^6 + 87^6$ is divided by $7$.

Wave-Particle
2016-11-09 20:01:11

Modular arithmetic

Modular arithmetic

Math1331Math
2016-11-09 20:01:11

Reduce the large numbers mod7

Reduce the large numbers mod7

DPatrick
2016-11-09 20:01:20

Since all we care about is the remainder when divided by 7, we can replace the numbers that we are exponentiating with their remainders upon division by 7.

Since all we care about is the remainder when divided by 7, we can replace the numbers that we are exponentiating with their remainders upon division by 7.

DPatrick
2016-11-09 20:01:35

We have $18 = 2 \cdot 7 + 4$ and $87 = 12 \cdot 7 + 3$, so the problem becomes:

We have $18 = 2 \cdot 7 + 4$ and $87 = 12 \cdot 7 + 3$, so the problem becomes:

DPatrick
2016-11-09 20:01:40

5. Find the remainder when $4^6 + 3^6$ is divided by $7$.

5. Find the remainder when $4^6 + 3^6$ is divided by $7$.

Math1331Math
2016-11-09 20:01:58

not too hard to compute to finish

not too hard to compute to finish

DPatrick
2016-11-09 20:02:09

Right, one option is to just compute to finish.

Right, one option is to just compute to finish.

DPatrick
2016-11-09 20:02:22

$4^6 = 2^{12} = 4096$ and $3^6 = 729$, so $4^6 + 3^6 = 4096 + 729 = 4825$.

$4^6 = 2^{12} = 4096$ and $3^6 = 729$, so $4^6 + 3^6 = 4096 + 729 = 4825$.

DPatrick
2016-11-09 20:02:28

Then $4825 = 689 \cdot 7 + 2$, so the remainder when $4825$ is divided by $7$ is $\boxed{2}$.

Then $4825 = 689 \cdot 7 + 2$, so the remainder when $4825$ is divided by $7$ is $\boxed{2}$.

EpicDragonSlayr
2016-11-09 20:02:38

find power patterns

find power patterns

Jorvis
2016-11-09 20:02:38

4^6 = 4^3 * 4^3, and 4^3 = 1 mod 7

4^6 = 4^3 * 4^3, and 4^3 = 1 mod 7

DPatrick
2016-11-09 20:02:47

Right, a slightly more sophisticated option is to look for a pattern among powers of 4 and powers of 3.

\begin{align*}

4^1 &\equiv 4 \pmod{7} \\

4^2 &\equiv 16 \equiv 2 \pmod{7} \\

4^3 &\equiv 8 \equiv 1 \pmod{7} \\

4^4 &\equiv 4 \pmod{7}

\end{align*}

Right, a slightly more sophisticated option is to look for a pattern among powers of 4 and powers of 3.

\begin{align*}

4^1 &\equiv 4 \pmod{7} \\

4^2 &\equiv 16 \equiv 2 \pmod{7} \\

4^3 &\equiv 8 \equiv 1 \pmod{7} \\

4^4 &\equiv 4 \pmod{7}

\end{align*}

DPatrick
2016-11-09 20:03:09

(If you're not familiar with the notation, (mod 7) basically just means the remainder when we divide by 7.)

(If you're not familiar with the notation, (mod 7) basically just means the remainder when we divide by 7.)

DPatrick
2016-11-09 20:03:25

So the pattern of remainders of powers of 4 (modulo 7) is 4,2,1,4,2,1,..., where the 1's are powers that are multiples of 3. Hence $4^6 \equiv 1 \pmod{7}$.

So the pattern of remainders of powers of 4 (modulo 7) is 4,2,1,4,2,1,..., where the 1's are powers that are multiples of 3. Hence $4^6 \equiv 1 \pmod{7}$.

Jorvis
2016-11-09 20:03:30

Use similar logic for 3^6

Use similar logic for 3^6

DPatrick
2016-11-09 20:03:39

Similarly,

\begin{align*}

3^1 &\equiv 3 \pmod{7} \\

3^2 &\equiv 9 \equiv 2 \pmod{7} \\

3^3 &\equiv 6 \pmod{7} \\

3^4 &\equiv 18 \equiv 4 \pmod{7} \\

3^5 &\equiv 12 \equiv 5 \pmod{7} \\

3^6 &\equiv 15 \equiv 1 \pmod{7}

\end{align*}

Similarly,

\begin{align*}

3^1 &\equiv 3 \pmod{7} \\

3^2 &\equiv 9 \equiv 2 \pmod{7} \\

3^3 &\equiv 6 \pmod{7} \\

3^4 &\equiv 18 \equiv 4 \pmod{7} \\

3^5 &\equiv 12 \equiv 5 \pmod{7} \\

3^6 &\equiv 15 \equiv 1 \pmod{7}

\end{align*}

DPatrick
2016-11-09 20:03:50

And we conclude that $4^6 + 3^6 \equiv 1+1 = \boxed{2} \pmod{7}$.

And we conclude that $4^6 + 3^6 \equiv 1+1 = \boxed{2} \pmod{7}$.

DPatrick
2016-11-09 20:04:01

And if you know a bit more number theory...

And if you know a bit more number theory...

SkyFox163
2016-11-09 20:04:27

use fermat's little theorem

use fermat's little theorem

Kpatton
2016-11-09 20:04:27

Use Fermat's Little Theorem

Use Fermat's Little Theorem

fractal161
2016-11-09 20:04:27

Fermat's Little Theorem!

Fermat's Little Theorem!

EulerMacaroni
2016-11-09 20:04:27

fermat's little theorem

fermat's little theorem

primesandfractals
2016-11-09 20:04:27

use euler's theorem

use euler's theorem

Math1331Math
2016-11-09 20:04:27

also you can use a^{p-1}=1modp because a,p are relatively prime

also you can use a^{p-1}=1modp because a,p are relatively prime

DPatrick
2016-11-09 20:04:40

The most "slick" solution is to use

\[ a^{p-1} \equiv 1 \pmod{p}, \]

provided $p$ is prime and $a$ is not a multiple of $p$.

The most "slick" solution is to use

**Fermat's Little Theorem**. This theorem states that\[ a^{p-1} \equiv 1 \pmod{p}, \]

provided $p$ is prime and $a$ is not a multiple of $p$.

DPatrick
2016-11-09 20:04:59

(There's actually a more general version of this theorem, but I won't go into that right now.)

(There's actually a more general version of this theorem, but I won't go into that right now.)

DPatrick
2016-11-09 20:05:07

In our example $p=7$, and indeed the exponents are $p-1 = 6$, with the bases not being multiples of 7.

In our example $p=7$, and indeed the exponents are $p-1 = 6$, with the bases not being multiples of 7.

DPatrick
2016-11-09 20:05:21

So each term is $1 \pmod{7}$, and their sum is $1+1 \equiv \boxed{2} \pmod{7}$.

So each term is $1 \pmod{7}$, and their sum is $1+1 \equiv \boxed{2} \pmod{7}$.

mikebreen
2016-11-09 20:05:29

Great ideas on the solution. Were the 18 and 87 an accident?

Great ideas on the solution. Were the 18 and 87 an accident?

DPatrick
2016-11-09 20:05:38

I don't know...

I don't know...

*were*they an accident?
mikebreen
2016-11-09 20:05:54

A salute to Ramanujan, born in 1887

A salute to Ramanujan, born in 1887

DPatrick
2016-11-09 20:06:01

Ha! Clever.

Ha! Clever.

DPatrick
2016-11-09 20:06:16

Onwards:

Onwards:

DPatrick
2016-11-09 20:06:20

6. How many divisors (factors) of $10!$ (including $1$) are perfect squares?

6. How many divisors (factors) of $10!$ (including $1$) are perfect squares?

Wave-Particle
2016-11-09 20:06:39

Prime factorize

Prime factorize

thedoge
2016-11-09 20:06:39

you can use the prime factorization of 10!

you can use the prime factorization of 10!

themoocow
2016-11-09 20:06:39

prime factorize

prime factorize

Liopleurodon
2016-11-09 20:06:39

prime factorization of $10!$

prime factorization of $10!$

DPatrick
2016-11-09 20:06:45

To answer this, we probably need the prime factorization of $10!$.

To answer this, we probably need the prime factorization of $10!$.

DPatrick
2016-11-09 20:06:49

\begin{align*}

10! &= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \\

&= 2 \cdot 3 \cdot 2^2 \cdot 5 \cdot (2 \cdot 3) \cdot 7 \cdot 2^3 \cdot 3^2 \cdot (2 \cdot 5) \\

&= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7

\end{align*}

\begin{align*}

10! &= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \\

&= 2 \cdot 3 \cdot 2^2 \cdot 5 \cdot (2 \cdot 3) \cdot 7 \cdot 2^3 \cdot 3^2 \cdot (2 \cdot 5) \\

&= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7

\end{align*}

DPatrick
2016-11-09 20:07:10

So using this, what divisors of $10!$ are perfect squares?

So using this, what divisors of $10!$ are perfect squares?

Wave-Particle
2016-11-09 20:07:17

perfect squares have even exponents (including 0)

perfect squares have even exponents (including 0)

Math1331Math
2016-11-09 20:07:17

notice perfect squares have even power of prime

notice perfect squares have even power of prime

DPatrick
2016-11-09 20:07:29

Right: to get a perfect square, we need each prime's exponent to be even.

Right: to get a perfect square, we need each prime's exponent to be even.

DPatrick
2016-11-09 20:07:42

(So we can get the square root by taking one-half of each exponent.)

(So we can get the square root by taking one-half of each exponent.)

DPatrick
2016-11-09 20:08:02

Any factor of $10!$ is of the form $2^a3^b5^c7^d$, where

$0 \le a \le 8$

$0 \le b \le 4$

$0 \le c \le 2$

$0 \le d \le 1$

Any factor of $10!$ is of the form $2^a3^b5^c7^d$, where

$0 \le a \le 8$

$0 \le b \le 4$

$0 \le c \le 2$

$0 \le d \le 1$

Wave-Particle
2016-11-09 20:08:13

For $2$, the exponent can be 0,2,4,6,8 which is 5 values. Using similar logic for 3,5,7 then multiply them

For $2$, the exponent can be 0,2,4,6,8 which is 5 values. Using similar logic for 3,5,7 then multiply them

EulerMacaroni
2016-11-09 20:08:13

5 choices for 2, 3 choices for 3, 2 choices for 5, 1 choice for 7

5 choices for 2, 3 choices for 3, 2 choices for 5, 1 choice for 7

themoocow
2016-11-09 20:08:13

for 2 we can pick $5$ possiblities, from 3 we can pick 3 possiblities, from 5 we can pick 2, and from 7 we can pick 1: 5*3*2*1=$30$

for 2 we can pick $5$ possiblities, from 3 we can pick 3 possiblities, from 5 we can pick 2, and from 7 we can pick 1: 5*3*2*1=$30$

thedoge
2016-11-09 20:08:13

the answer is 5*3*2=30

the answer is 5*3*2=30

DPatrick
2016-11-09 20:08:36

Right, since we must have the exponents be even:

$a$ must be 0, 2, 4, 6, or 8

$b$ must be 0, 2, or 4

$c$ must be 0 or 2

$d$ must be 0

Right, since we must have the exponents be even:

$a$ must be 0, 2, 4, 6, or 8

$b$ must be 0, 2, or 4

$c$ must be 0 or 2

$d$ must be 0

DPatrick
2016-11-09 20:08:50

This gives $5 \cdot 3 \cdot 2 = 30$ choices for exponents, and thus $\boxed{30}$ perfect square divisors.

This gives $5 \cdot 3 \cdot 2 = 30$ choices for exponents, and thus $\boxed{30}$ perfect square divisors.

DPatrick
2016-11-09 20:09:10

Next up:

Next up:

DPatrick
2016-11-09 20:09:20

7. Spheres of equal radius are stacked inside a cylinder of the same radius so that the spheres extend from the bottom of the cylinder to the top. The volume of the space inside the cylinder that is not taken up by the spheres is equal to the volume of three of the spheres. How many spheres are inside the cylinder?

7. Spheres of equal radius are stacked inside a cylinder of the same radius so that the spheres extend from the bottom of the cylinder to the top. The volume of the space inside the cylinder that is not taken up by the spheres is equal to the volume of three of the spheres. How many spheres are inside the cylinder?

thedoge
2016-11-09 20:09:52

set up a system of equations in terms of the radius of the spheres

set up a system of equations in terms of the radius of the spheres

Liopleurodon
2016-11-09 20:09:52

make variables

make variables

DPatrick
2016-11-09 20:10:01

What variable will be probably definitely need?

What variable will be probably definitely need?

Wave-Particle
2016-11-09 20:10:13

radius

radius

duruphi
2016-11-09 20:10:13

r

r

declanhiller
2016-11-09 20:10:13

r

r

Liopleurodon
2016-11-09 20:10:13

r for the radius

r for the radius

awin
2016-11-09 20:10:13

radius of spheres!

radius of spheres!

nvpthemvp
2016-11-09 20:10:13

r

r

mathpractice2011
2016-11-09 20:10:13

r

r

DPatrick
2016-11-09 20:10:16

Are you sure?

Are you sure?

DPatrick
2016-11-09 20:10:28

Does the radius of the spheres matter?

Does the radius of the spheres matter?

caro2be
2016-11-09 20:10:40

no

no

awin
2016-11-09 20:10:40

No!

No!

Liopleurodon
2016-11-09 20:10:40

not really

not really

Imaadrana
2016-11-09 20:10:40

no it does not matter

no it does not matter

fractal161
2016-11-09 20:10:50

Not really, because ratios are the same...

Not really, because ratios are the same...

primesandfractals
2016-11-09 20:10:52

let it be 1

let it be 1

DPatrick
2016-11-09 20:11:05

Indeed -- the whole problem just scales if we change the radius.

Indeed -- the whole problem just scales if we change the radius.

DPatrick
2016-11-09 20:11:16

So let's assume the radius of the spheres are 1. (This is a useful problem solving tactic: if an unknown value in a problem doesn't matter, you are free to assign it to whatever value is convenient. Don't introduce a variable you don't need!)

So let's assume the radius of the spheres are 1. (This is a useful problem solving tactic: if an unknown value in a problem doesn't matter, you are free to assign it to whatever value is convenient. Don't introduce a variable you don't need!)

thedoge
2016-11-09 20:11:23

and the number of spheres

and the number of spheres

DPatrick
2016-11-09 20:11:40

We do almost certainly need a variable for the number of spheres -- after all, that's what we're trying to find! Let's call it $n$.

We do almost certainly need a variable for the number of spheres -- after all, that's what we're trying to find! Let's call it $n$.

NewtonFTW
2016-11-09 20:12:06

If we let r = 1, then the volume of each sphere is 4/3 * pi

If we let r = 1, then the volume of each sphere is 4/3 * pi

DPatrick
2016-11-09 20:12:13

Indeed, the volume of a sphere of radius 1 is $\dfrac43\pi$.

Indeed, the volume of a sphere of radius 1 is $\dfrac43\pi$.

Liopleurodon
2016-11-09 20:12:22

the height is $2n$

the height is $2n$

DPatrick
2016-11-09 20:12:37

Good -- this is where you could easily mess up. The height of the cylinder is $2n$ (NOT $n$!!)

Good -- this is where you could easily mess up. The height of the cylinder is $2n$ (NOT $n$!!)

DPatrick
2016-11-09 20:12:47

So what's the volume of the cylinder?

So what's the volume of the cylinder?

Liopleurodon
2016-11-09 20:13:17

$2n\pi$

$2n\pi$

thedoge
2016-11-09 20:13:17

$1^2*\pi*2n=2n\pi$

$1^2*\pi*2n=2n\pi$

SkyFox163
2016-11-09 20:13:21

pi 2 n if the radius is 1

pi 2 n if the radius is 1

nvpthemvp
2016-11-09 20:13:21

2n times pi

2n times pi

Irshad
2016-11-09 20:13:21

2npi

2npi

DPatrick
2016-11-09 20:13:31

Yes -- the cylinder has a base of area $\pi$ and a height of $2n$ (since each sphere has diameter 2). So the cylinder's volume is $2n\pi$.

Yes -- the cylinder has a base of area $\pi$ and a height of $2n$ (since each sphere has diameter 2). So the cylinder's volume is $2n\pi$.

DPatrick
2016-11-09 20:13:45

How do we set up an equation now that we have all this data?

How do we set up an equation now that we have all this data?

DPatrick
2016-11-09 20:14:27

"volume of cylinder not taken up by spheres" = "volume of 3 spheres"

"volume of cylinder not taken up by spheres" = "volume of 3 spheres"

declanhiller
2016-11-09 20:14:45

2npi-n*4/3pi=3*4/3pi?

2npi-n*4/3pi=3*4/3pi?

Liopleurodon
2016-11-09 20:14:45

$2n\pi-\frac{4}{3}n\pi=4\pi$

$2n\pi-\frac{4}{3}n\pi=4\pi$

vonnan
2016-11-09 20:14:45

2n\pi-4/3\pi*n=3*4/3\pi

2n\pi-4/3\pi*n=3*4/3\pi

gshabanets
2016-11-09 20:14:45

2n*pi - n*4/3*pi = 3*4/3*pi

2n*pi - n*4/3*pi = 3*4/3*pi

DPatrick
2016-11-09 20:14:57

Exactly.

Exactly.

DPatrick
2016-11-09 20:15:05

The $n$ spheres together take up $\dfrac43n\pi$ of the volume.

The $n$ spheres together take up $\dfrac43n\pi$ of the volume.

DPatrick
2016-11-09 20:15:13

This means that the volume of the cylinder

This means that the volume of the cylinder

*not*taken up by the spheres is $2n\pi - \dfrac43n\pi = \dfrac23n\pi$.
DPatrick
2016-11-09 20:15:22

This is equal to 3 spheres' volume, which is $3 \cdot \dfrac43\pi = 4\pi$.

This is equal to 3 spheres' volume, which is $3 \cdot \dfrac43\pi = 4\pi$.

DPatrick
2016-11-09 20:15:32

Thus, $\dfrac23n\pi = 4\pi$.

Thus, $\dfrac23n\pi = 4\pi$.

Hanting_Z
2016-11-09 20:15:38

6

6

mamis511
2016-11-09 20:15:38

6

6

SkyFox163
2016-11-09 20:15:38

So $n=6$

So $n=6$

Liopleurodon
2016-11-09 20:15:44

$n=6$

$n=6$

Wave-Particle
2016-11-09 20:15:44

so n=6

so n=6

awin
2016-11-09 20:15:44

Yep, and then n = 6

Yep, and then n = 6

duruphi
2016-11-09 20:15:44

n=6

n=6

Superman2333
2016-11-09 20:15:44

6

6

DPatrick
2016-11-09 20:15:47

Solving gives us $n = \boxed{6}$, so there are 6 spheres in the cylinder.

Solving gives us $n = \boxed{6}$, so there are 6 spheres in the cylinder.

DPatrick
2016-11-09 20:16:08

Two more to go, then killer #10!

Two more to go, then killer #10!

DPatrick
2016-11-09 20:16:14

8. Suppose that $a$, $b$, $c$, and $d$ are positive integers such that $a^3 = b^2$, $c^3 = d^2$, and $c-a = 9$. Find $a+b+c+d$.

8. Suppose that $a$, $b$, $c$, and $d$ are positive integers such that $a^3 = b^2$, $c^3 = d^2$, and $c-a = 9$. Find $a+b+c+d$.

DPatrick
2016-11-09 20:16:30

How can we approach this?

How can we approach this?

thedoge
2016-11-09 20:16:45

we know that a is a perfect square

we know that a is a perfect square

atmchallenge
2016-11-09 20:16:45

We must have $a,c$ are perfect squares and $b,d$ are perfect cubes,

We must have $a,c$ are perfect squares and $b,d$ are perfect cubes,

DPatrick
2016-11-09 20:17:06

Aha. If a perfect cube $a^3$ is also a perfect square, then $a$ itself must also be a perfect square.

Aha. If a perfect cube $a^3$ is also a perfect square, then $a$ itself must also be a perfect square.

DPatrick
2016-11-09 20:17:11

Think about this in terms of the prime factorization of $a^3$. Since $a^3$ is a perfect square, all of the powers of its primes must be even. When we take its cube root to get back to $a$, we divide all the exponents by 3. But dividing an even number by 3 still leaves an even number. So all of the exponents in the prime factorization of $a$ are even too, which means that $a$ is itself a perfect square.

Think about this in terms of the prime factorization of $a^3$. Since $a^3$ is a perfect square, all of the powers of its primes must be even. When we take its cube root to get back to $a$, we divide all the exponents by 3. But dividing an even number by 3 still leaves an even number. So all of the exponents in the prime factorization of $a$ are even too, which means that $a$ is itself a perfect square.

skiboy32
2016-11-09 20:17:57

now, we find 2 positive squares that have a difference of 9

now, we find 2 positive squares that have a difference of 9

DPatrick
2016-11-09 20:18:10

Right, now the $c-a = 9$ equation becomes a difference of squares!

Right, now the $c-a = 9$ equation becomes a difference of squares!

DPatrick
2016-11-09 20:18:17

We have $y^2 - x^2 = 9$. What can we do with this?

We have $y^2 - x^2 = 9$. What can we do with this?

DPatrick
2016-11-09 20:18:45

Oops, I accidentally deleted my definitions: $a = x^2$ and $c = y^2$ for some positive integers $x$ and $y$.

Oops, I accidentally deleted my definitions: $a = x^2$ and $c = y^2$ for some positive integers $x$ and $y$.

Liopleurodon
2016-11-09 20:18:56

$(y-x)(y+x)=9$

$(y-x)(y+x)=9$

fractal161
2016-11-09 20:18:56

factor $y^2-x^2=(y+x)(y-x)$

factor $y^2-x^2=(y+x)(y-x)$

yuvi18
2016-11-09 20:18:56

(y+x)(y-x) = 9

(y+x)(y-x) = 9

DPatrick
2016-11-09 20:19:02

Factor! $(y-x)(y+x) = 9$.

Factor! $(y-x)(y+x) = 9$.

DPatrick
2016-11-09 20:19:10

We can't have both factors be 3, because then $x=0$, and we're told all the quantities are positive.

We can't have both factors be 3, because then $x=0$, and we're told all the quantities are positive.

DPatrick
2016-11-09 20:19:23

Therefore, $y-x = 1$ and $y+x = 9$ is the only possibility.

Therefore, $y-x = 1$ and $y+x = 9$ is the only possibility.

yuvi18
2016-11-09 20:19:38

y = 5 x = 4

y = 5 x = 4

Liopleurodon
2016-11-09 20:19:38

$y=5,x=4$

$y=5,x=4$

manick
2016-11-09 20:19:38

y=5 and x=4

y=5 and x=4

DPatrick
2016-11-09 20:19:41

This gives $y=5$ and $x=4$.

This gives $y=5$ and $x=4$.

Puzzled417
2016-11-09 20:20:02

$y^2 = 25$ and $x^2 = 16$

$y^2 = 25$ and $x^2 = 16$

yuvi18
2016-11-09 20:20:02

c = 25 a = 16

c = 25 a = 16

DPatrick
2016-11-09 20:20:08

We have $a = x^2 = 16$ and $c = y^2 = 25$.

We have $a = x^2 = 16$ and $c = y^2 = 25$.

DPatrick
2016-11-09 20:20:26

Then $b = a^{\frac32} = 64$ and $d = c^{\frac32} = 125$.

Then $b = a^{\frac32} = 64$ and $d = c^{\frac32} = 125$.

bogstop320
2016-11-09 20:20:33

230

230

mamis511
2016-11-09 20:20:33

230

230

DPatrick
2016-11-09 20:20:36

Finally $a+b+c+d = 16 + 64 + 25 + 125 = \boxed{230}$.

Finally $a+b+c+d = 16 + 64 + 25 + 125 = \boxed{230}$.

DPatrick
2016-11-09 20:20:59

On to my favorite, #9:

On to my favorite, #9:

DPatrick
2016-11-09 20:21:04

9. How many distinct real solutions are there to

\[

(x+1)^7 + (x+1)^6(x-1) + (x+1)^5(x-1)^2 + \cdots + (x+1)(x-1)^6 + (x-1)^7 = 0 ?

\]

9. How many distinct real solutions are there to

\[

(x+1)^7 + (x+1)^6(x-1) + (x+1)^5(x-1)^2 + \cdots + (x+1)(x-1)^6 + (x-1)^7 = 0 ?

\]

fractal161
2016-11-09 20:21:48

Write $a=x+1$ and $b=x-1$?

Write $a=x+1$ and $b=x-1$?

DPatrick
2016-11-09 20:22:01

That is an excellent idea. Let's set $a = x+1$ and $b = x-1$ to make this equation a little easier to work with:

\[

a^7 + a^6b + a^5b^2 + \cdots + ab^6 + b^7 = 0.

\]

What can we do with this?

That is an excellent idea. Let's set $a = x+1$ and $b = x-1$ to make this equation a little easier to work with:

\[

a^7 + a^6b + a^5b^2 + \cdots + ab^6 + b^7 = 0.

\]

What can we do with this?

mathgandalfjr
2016-11-09 20:22:20

binomail theorem again??

binomail theorem again??

thedoge
2016-11-09 20:22:20

looks like binomial theorem

looks like binomial theorem

skiboy32
2016-11-09 20:22:20

binomial theorem.

binomial theorem.

Puzzled417
2016-11-09 20:22:20

That is a binomial expansion.

That is a binomial expansion.

DPatrick
2016-11-09 20:22:26

...except we don't have the binomial coefficients.

...except we don't have the binomial coefficients.

DPatrick
2016-11-09 20:22:36

But in fact, this is even easier to work with.

But in fact, this is even easier to work with.

Hanting_Z
2016-11-09 20:22:45

mult by a - b?

mult by a - b?

zetaintegrator
2016-11-09 20:22:45

*(a-b)

*(a-b)

DPatrick
2016-11-09 20:22:55

We can multiply both sides by $(a-b)$, then the whole thing telescopes!

We can multiply both sides by $(a-b)$, then the whole thing telescopes!

thedoge
2016-11-09 20:23:05

we can also use (a^8-b^8)=(a-b)(a^7+a^6*b+a^5*b^2...+b^7)

we can also use (a^8-b^8)=(a-b)(a^7+a^6*b+a^5*b^2...+b^7)

DPatrick
2016-11-09 20:23:41

Exactly, if we multiply by $a-b$ (which is just $2$ if you remember what $a$ and $b$ are), we end up with the very simple $a^8 - b^8 = 0$.

Exactly, if we multiply by $a-b$ (which is just $2$ if you remember what $a$ and $b$ are), we end up with the very simple $a^8 - b^8 = 0$.

Liopleurodon
2016-11-09 20:23:46

$a^8=b^8$

$a^8=b^8$

DPatrick
2016-11-09 20:24:00

Indeed: when do two numbers have the same 8th powers?

Indeed: when do two numbers have the same 8th powers?

DPatrick
2016-11-09 20:24:07

Two

Two

*real*numbers, remember!
Liopleurodon
2016-11-09 20:24:18

$a=\pm b$

$a=\pm b$

NewtonFTW
2016-11-09 20:24:18

when they are equal or opposites

when they are equal or opposites

Irshad
2016-11-09 20:24:18

a=b or a=-b

a=b or a=-b

DPatrick
2016-11-09 20:24:49

Right, if $a$ and $b$ are real, then for $a^8 = b^8$ we need $a = \pm b$. (If they're complex number we have a lot more options, but we don't care about complex roots in this problem.)

Right, if $a$ and $b$ are real, then for $a^8 = b^8$ we need $a = \pm b$. (If they're complex number we have a lot more options, but we don't care about complex roots in this problem.)

Wave-Particle
2016-11-09 20:24:59

clearly $x-1\neq x+1$ tho

clearly $x-1\neq x+1$ tho

DPatrick
2016-11-09 20:25:15

Right: we can't have $a=b$, because that's $x+1 = x-1$, which is no solution for $x$.

Right: we can't have $a=b$, because that's $x+1 = x-1$, which is no solution for $x$.

DPatrick
2016-11-09 20:25:26

But we can have $x+1 = -(x-1)$.

But we can have $x+1 = -(x-1)$.

thedoge
2016-11-09 20:25:42

There's only one solution--x=0

There's only one solution--x=0

Liopleurodon
2016-11-09 20:25:42

x=0

x=0

vonnan
2016-11-09 20:25:42

x=0, one real solution

x=0, one real solution

skiboy32
2016-11-09 20:25:42

so x=0 only. So the answer is $\boxed{1}$

so x=0 only. So the answer is $\boxed{1}$

Wave-Particle
2016-11-09 20:25:42

x=0

x=0

DPatrick
2016-11-09 20:25:54

Right: that simplifies to $x+1 = -x+1$, so $x=0$ is the solution.

Right: that simplifies to $x+1 = -x+1$, so $x=0$ is the solution.

DPatrick
2016-11-09 20:26:00

So there is only $\boxed{1}$ real solution: namely, $x=0$.

So there is only $\boxed{1}$ real solution: namely, $x=0$.

DPatrick
2016-11-09 20:26:33

And finally, on to #10:

And finally, on to #10:

DPatrick
2016-11-09 20:26:37

10. What is the smallest (minimum) value of $n$ such that $(n+1)!$ has four more digits than $n!$?

10. What is the smallest (minimum) value of $n$ such that $(n+1)!$ has four more digits than $n!$?

*Note: this is the tie-breaking question. In the event of a tie for high scores in a region, the winner is the person with the closest to the correct answer without going over.*
zetaintegrator
2016-11-09 20:26:56

the price is right :L

the price is right :L

DPatrick
2016-11-09 20:27:17

(yes, this uses "The Price is Right" bidding rules: closest without going over wins the showcase)

(yes, this uses "The Price is Right" bidding rules: closest without going over wins the showcase)

TPiR
2016-11-09 20:27:25

LOL

LOL

DPatrick
2016-11-09 20:27:34

This is pretty tricky. Any ideas?

This is pretty tricky. Any ideas?

skiboy32
2016-11-09 20:27:50

well, we know that the answer is less than 10001...

well, we know that the answer is less than 10001...

DPatrick
2016-11-09 20:27:58

This is a good starting point to think about.

This is a good starting point to think about.

DPatrick
2016-11-09 20:28:04

To get from $n!$ to $(n+1)!$, we multiply by $n+1$.

To get from $n!$ to $(n+1)!$, we multiply by $n+1$.

DPatrick
2016-11-09 20:28:36

If $n=9999$, then multiplying by 10000 will add 4 zeros to the number. So $n=9999$ definitely satisfies the property.

If $n=9999$, then multiplying by 10000 will add 4 zeros to the number. So $n=9999$ definitely satisfies the property.

DPatrick
2016-11-09 20:29:08

Similarly, if $n=999$, then multiplying by 1000 only adds 3 digits. So $n=999$ definitely does not satisfy the property.

Similarly, if $n=999$, then multiplying by 1000 only adds 3 digits. So $n=999$ definitely does not satisfy the property.

vonnan
2016-11-09 20:29:14

1000-9999

1000-9999

DPatrick
2016-11-09 20:29:23

The answer is definitely in the range $1000 \le n \le 9999$.

The answer is definitely in the range $1000 \le n \le 9999$.

bearytasty
2016-11-09 20:29:34

some intuition is that if your number is big enough (leading digit of 9), you can multiply by something near 1000 and it will "skip" over another digit and increase by 4 digits, not 3

some intuition is that if your number is big enough (leading digit of 9), you can multiply by something near 1000 and it will "skip" over another digit and increase by 4 digits, not 3

DPatrick
2016-11-09 20:29:47

Yes! Multiplying by 1001

Yes! Multiplying by 1001

*might*add four digits, if the number we started with was already pretty close to needing an extra digit.
DPatrick
2016-11-09 20:29:52

For example, $9999 \cdot 1001 = 10008999$, gaining 4 digits in the process.

For example, $9999 \cdot 1001 = 10008999$, gaining 4 digits in the process.

DPatrick
2016-11-09 20:30:13

But how can we quantify this in a way that lets us make a reasonable guess?

But how can we quantify this in a way that lets us make a reasonable guess?

thedoge
2016-11-09 20:30:22

we could use logarithms

we could use logarithms

DPatrick
2016-11-09 20:30:39

There's a thought. Using logs, what's the formula for the number of digits of a number $x$?

There's a thought. Using logs, what's the formula for the number of digits of a number $x$?

DPatrick
2016-11-09 20:30:58

For example, if $2 \le \log_{10} x < 3$, what can we conclude?

For example, if $2 \le \log_{10} x < 3$, what can we conclude?

thedoge
2016-11-09 20:31:10

x has 3 digits

x has 3 digits

Wave-Particle
2016-11-09 20:31:10

3 digits

3 digits

Liopleurodon
2016-11-09 20:31:19

it has 3 digits

it has 3 digits

Irshad
2016-11-09 20:31:19

3 digits

3 digits

manick
2016-11-09 20:31:19

3 digits

3 digits

DPatrick
2016-11-09 20:31:22

Right. Removing the logs (by taking 10 to the power of each term) gives $100 \le x < 1000$, which exactly means that $x$ has 3 digits.

Right. Removing the logs (by taking 10 to the power of each term) gives $100 \le x < 1000$, which exactly means that $x$ has 3 digits.

DPatrick
2016-11-09 20:31:52

So the formula for the number of digits of $x$ is $$\lfloor \log_{10} x \rfloor + 1,$$ where $\lfloor \; \rfloor$ means to round down to the nearest integer.

So the formula for the number of digits of $x$ is $$\lfloor \log_{10} x \rfloor + 1,$$ where $\lfloor \; \rfloor$ means to round down to the nearest integer.

NewtonFTW
2016-11-09 20:32:06

floor function

floor function

DPatrick
2016-11-09 20:32:15

Yes, that's usually called the

Yes, that's usually called the

**floor**function.
DPatrick
2016-11-09 20:32:21

Great. How are $\log_{10} n!$ and $\log_{10} (n+1)!$ related?

Great. How are $\log_{10} n!$ and $\log_{10} (n+1)!$ related?

thedoge
2016-11-09 20:32:54

log((n+1)!)=log(n!)+log(n+1)

log((n+1)!)=log(n!)+log(n+1)

Jorvis
2016-11-09 20:32:59

log n! + log (n+1) = log (n+1)!

log n! + log (n+1) = log (n+1)!

DPatrick
2016-11-09 20:33:03

Since $(n+1)! = (n+1) \cdot n!$, we have $\log_{10} (n+1)! = \log_{10} n! + \log_{10} (n+1)$.

Since $(n+1)! = (n+1) \cdot n!$, we have $\log_{10} (n+1)! = \log_{10} n! + \log_{10} (n+1)$.

DPatrick
2016-11-09 20:33:23

So if we think of $\log_{10} x$ as one less than the digit count (where we through away "fractional" digits), we're adding $\log_{10} (n+1)$ digits every time we go from $n!$ to $(n+1)!$.

So if we think of $\log_{10} x$ as one less than the digit count (where we through away "fractional" digits), we're adding $\log_{10} (n+1)$ digits every time we go from $n!$ to $(n+1)!$.

DPatrick
2016-11-09 20:33:44

Let's recap: $\log_{10} 1000 = 3$, which is consistent with "multiplying by 1000 adds 3 digits always".

Let's recap: $\log_{10} 1000 = 3$, which is consistent with "multiplying by 1000 adds 3 digits always".

DPatrick
2016-11-09 20:33:55

$\log_{10} 1001 = 3 + \text{ a tiny bit extra}$, so multiplying by 1000 will almost always add just 3 digits, unless the "tiny bit extra" rolls the fractional part of $\log_{10}(1000!)$ over to a new integer, in which case we get a 4th digit.

$\log_{10} 1001 = 3 + \text{ a tiny bit extra}$, so multiplying by 1000 will almost always add just 3 digits, unless the "tiny bit extra" rolls the fractional part of $\log_{10}(1000!)$ over to a new integer, in which case we get a 4th digit.

DPatrick
2016-11-09 20:34:05

Then continuing:

\begin{align*}

\log_{10} 1002 &= 3 + \text{ a slightly larger but still tiny bit extra} \\

\log_{10} 1003 &= 3 + \text{ an even larger but still pretty tiny bit extra }

\end{align*}

Then continuing:

\begin{align*}

\log_{10} 1002 &= 3 + \text{ a slightly larger but still tiny bit extra} \\

\log_{10} 1003 &= 3 + \text{ an even larger but still pretty tiny bit extra }

\end{align*}

DPatrick
2016-11-09 20:34:18

The question is: at what point does adding up all those "extras" result in an extra digit?

The question is: at what point does adding up all those "extras" result in an extra digit?

DPatrick
2016-11-09 20:34:36

To me, the biggest problem in this computation is that we have no good way of telling what the fractional part of $\log_{10}(1000!)$ is to begin with.

To me, the biggest problem in this computation is that we have no good way of telling what the fractional part of $\log_{10}(1000!)$ is to begin with.

DPatrick
2016-11-09 20:35:06

Certainly if we want to be absolutely sure that we don't go over (under these Price is Right rules), we could guess 1000. (This is the equivalent of the "ONE DOLLAR!" bid on TPiR.)

Certainly if we want to be absolutely sure that we don't go over (under these Price is Right rules), we could guess 1000. (This is the equivalent of the "ONE DOLLAR!" bid on TPiR.)

DPatrick
2016-11-09 20:35:38

But if we want to try something a bit more clever, here is maybe what I'd do.

But if we want to try something a bit more clever, here is maybe what I'd do.

DPatrick
2016-11-09 20:36:12

$\log_{10}(1001)$ is about $3.0005$. (It turns out that if $\epsilon$ is very small, then $10^\epsilon \approx 1 + 2\epsilon$ is a pretty good approximation.)

$\log_{10}(1001)$ is about $3.0005$. (It turns out that if $\epsilon$ is very small, then $10^\epsilon \approx 1 + 2\epsilon$ is a pretty good approximation.)

DPatrick
2016-11-09 20:36:30

Actually, if you know a little calculus, you may know that $10^\epsilon \approx 1 + \ln(10)\epsilon$ is an even better approximation, and $\ln(10) \approx 2.3$.

Actually, if you know a little calculus, you may know that $10^\epsilon \approx 1 + \ln(10)\epsilon$ is an even better approximation, and $\ln(10) \approx 2.3$.

DPatrick
2016-11-09 20:36:57

But let's stick with $\log_{10}(1001) \approx 3.0005$. And more generally, $\log_{10}(1000+k) = 3 + .0005k$.

But let's stick with $\log_{10}(1001) \approx 3.0005$. And more generally, $\log_{10}(1000+k) = 3 + .0005k$.

DPatrick
2016-11-09 20:37:17

Admittedly, you kind of have to know this from experience in working with these sorts of calculations.

Admittedly, you kind of have to know this from experience in working with these sorts of calculations.

DPatrick
2016-11-09 20:37:34

So the fractional amount of digits we get by multiplying by 1001, 1002, etc. up through $1000+(k+1)$ is approximately

\[

\dfrac{1+2+\cdots+(k+1)}{2000} = \dfrac{(k+1)(k+2)}{4000}.

\]

So the fractional amount of digits we get by multiplying by 1001, 1002, etc. up through $1000+(k+1)$ is approximately

\[

\dfrac{1+2+\cdots+(k+1)}{2000} = \dfrac{(k+1)(k+2)}{4000}.

\]

DPatrick
2016-11-09 20:37:52

And since $\sqrt{4000} \approx 63$, this suggests $n \le 62$ will make the extra fractional digits add up to about 1.

And since $\sqrt{4000} \approx 63$, this suggests $n \le 62$ will make the extra fractional digits add up to about 1.

DPatrick
2016-11-09 20:38:22

So now I've got a really crude estimate that the answer can't possibly be too much bigger than about 1062.

So now I've got a really crude estimate that the answer can't possibly be too much bigger than about 1062.

DPatrick
2016-11-09 20:38:32

But we also have to keep in mind that $\log_{10}(1000!)$ probably started with some fractional digits to begin with.

But we also have to keep in mind that $\log_{10}(1000!)$ probably started with some fractional digits to begin with.

DPatrick
2016-11-09 20:38:52

So we might hedge our bets and guess 1031 or even 1020. And we want to be absolutely sure that we don't go over, we could still always guess 1000.

So we might hedge our bets and guess 1031 or even 1020. And we want to be absolutely sure that we don't go over, we could still always guess 1000.

DPatrick
2016-11-09 20:39:02

Anybody want to guess at the actual exact answer?

Anybody want to guess at the actual exact answer?

zetaintegrator
2016-11-09 20:39:14

1042

1042

1-1 is 3
2016-11-09 20:39:21

1042?

1042?

adamz
2016-11-09 20:39:21

1042 xd

1042 xd

DPatrick
2016-11-09 20:39:28

Indeed, it's 1042.

Indeed, it's 1042.

zetaintegrator
2016-11-09 20:39:36

42 is the number of the gods

42 is the number of the gods

DPatrick
2016-11-09 20:39:42

Yep, it's a good guess.

Yep, it's a good guess.

DPatrick
2016-11-09 20:39:48

That's it for Round 2!

That's it for Round 2!

DPatrick
2016-11-09 20:39:55

I do have some stats on the contest that Mike Breen kindly provided.

I do have some stats on the contest that Mike Breen kindly provided.

mikebreen
2016-11-09 20:39:59

Successful guesses were about what you said: 1020 and 1030, but we did get 1042.

Successful guesses were about what you said: 1020 and 1030, but we did get 1042.

DPatrick
2016-11-09 20:40:07

377 students qualified for Round 2. The median score was 6 (out of 10) and the mean was 5.9.

377 students qualified for Round 2. The median score was 6 (out of 10) and the mean was 5.9.

DPatrick
2016-11-09 20:40:31

As I mentioned earlier, Problem 3 (the history question about Ramanujan and Hardy) was the hardest of the first nine, with about 33% getting it right. Problem 4 (the machine making defective parts) was the next hardest with about 36% getting it right.

As I mentioned earlier, Problem 3 (the history question about Ramanujan and Hardy) was the hardest of the first nine, with about 33% getting it right. Problem 4 (the machine making defective parts) was the next hardest with about 36% getting it right.

DPatrick
2016-11-09 20:40:46

All of the remaining problems (except #10) had success rates of over 50%.

All of the remaining problems (except #10) had success rates of over 50%.

Wave-Particle
2016-11-09 20:40:56

how many people got 10

how many people got 10

DPatrick
2016-11-09 20:41:11

I believe just a single person had all of the first nine right and also was correct on #10.

I believe just a single person had all of the first nine right and also was correct on #10.

mikebreen
2016-11-09 20:41:24

Students do much better than I would have in high school. Yes, one person got a 10.

Students do much better than I would have in high school. Yes, one person got a 10.

DPatrick
2016-11-09 20:41:27

The top scorer on Round 2 in each of 9 regions, plus the top scorer in the Atlanta area, will advance to the National semifinals, to be held live at the 2017 Joint Mathematics Meetings in Atlanta in January.

The top scorer on Round 2 in each of 9 regions, plus the top scorer in the Atlanta area, will advance to the National semifinals, to be held live at the 2017 Joint Mathematics Meetings in Atlanta in January.

DPatrick
2016-11-09 20:41:31

zetaintegrator
2016-11-09 20:41:47

how many 9's moo..

how many 9's moo..

DPatrick
2016-11-09 20:42:07

Mike sent me that data, but I lost where I put it. Perhaps he knows.

Mike sent me that data, but I lost where I put it. Perhaps he knows.

mikebreen
2016-11-09 20:42:25

About 9% of the 377 scored 9.

About 9% of the 377 scored 9.

bogstop320
2016-11-09 20:42:34

When will the winners be announced?

When will the winners be announced?

DPatrick
2016-11-09 20:42:56

I believe they're in the processing of contacting winners to be sure they can travel to Atlanta in January.

I believe they're in the processing of contacting winners to be sure they can travel to Atlanta in January.

mikebreen
2016-11-09 20:43:13

We're waiting for the contestants to confirm that they can participate. We hope at the end of this week or beginning of next.

We're waiting for the contestants to confirm that they can participate. We hope at the end of this week or beginning of next.

DPatrick
2016-11-09 20:43:42

And Mike, Bill, and I will all be in Atlanta in person at the finals as well. (Mike had better be -- he's the host!)

And Mike, Bill, and I will all be in Atlanta in person at the finals as well. (Mike had better be -- he's the host!)

DPatrick
2016-11-09 20:43:53

Visit http://www.ams.org/programs/students/wwtbam/wwtbam to learn more about the game and to visit the archive of past years' contests.

Visit http://www.ams.org/programs/students/wwtbam/wwtbam to learn more about the game and to visit the archive of past years' contests.

mikebreen
2016-11-09 20:44:21

Yes, we'll be there. Watch us live at http://www.livestream.com/wwtbam2017.

Yes, we'll be there. Watch us live at http://www.livestream.com/wwtbam2017.

DPatrick
2016-11-09 20:44:28

That's it for our Math Jam tonight -- thanks for attending!

That's it for our Math Jam tonight -- thanks for attending!

DPatrick
2016-11-09 20:44:46

If you missed any part of our discussion, a transcript will be up on the AoPS website soon.

If you missed any part of our discussion, a transcript will be up on the AoPS website soon.

mikebreen
2016-11-09 20:44:51

Thanks, everyone! Great work by you and Dave.

Thanks, everyone! Great work by you and Dave.

TPiR
2016-11-09 20:44:51

Thanks, Dave. Great job, as always.

Thanks, Dave. Great job, as always.

mathsm
2016-11-09 20:48:05

How do you access the transcript?

How do you access the transcript?

DPatrick
2016-11-09 20:48:26

Once it's up (probably in about 15-30 minutes), go to "Math Jams" under the Online School menu of the website.

Once it's up (probably in about 15-30 minutes), go to "Math Jams" under the Online School menu of the website.

DPatrick
2016-11-09 20:48:32

Then click on the "Previous Math Jams" button.

Then click on the "Previous Math Jams" button.

zetaintegrator
2016-11-09 20:49:21

can you release who was the perfect scorer? xD

can you release who was the perfect scorer? xD

DPatrick
2016-11-09 20:49:37

I suspect they will soon. Maybe not though.

I suspect they will soon. Maybe not though.

DPatrick
2016-11-09 20:49:53

(I don't know who it is.)

(I don't know who it is.)

mikebreen
2016-11-09 20:50:35

Not now. He is one of the 10, though. We may post it on the page profiling the 10 contestants and talk about it at the game in Atlanta.

Not now. He is one of the 10, though. We may post it on the page profiling the 10 contestants and talk about it at the game in Atlanta.

DPatrick
2016-11-09 20:52:35

It's getting late in the east and I suspect many of you were up late last night. So let's end it here. Have a great evening!

It's getting late in the east and I suspect many of you were up late last night. So let's end it here. Have a great evening!