## 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 2019-20 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 Denver in January 2020.

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#### Facilitator: Dave Patrick

NerdyDude
2019-10-14 19:27:00

Weren't you on who wants to be a millionaire

Weren't you on who wants to be a millionaire

DPatrick
2019-10-14 19:27:07

*Photo Credit: Maria Melin, copyright 1999 ABC Television.*
DPatrick
2019-10-14 19:27:09

Yes.

Yes.

DPatrick
2019-10-14 19:27:24

I once was a contestant on ABC's

I 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):
RagingWar
2019-10-14 19:27:40

You won a car!

You won a car!

pow_h_2
2019-10-14 19:27:45

you won enough to buy a car, not full 1m

you won enough to buy a car, not full 1m

DPatrick
2019-10-14 19:27:55

Some of you know the story from past Math Jams...

Some of you know the story from past Math Jams...

asdf334
2019-10-14 19:28:33

it's that equator question

it's that equator question

DPatrick
2019-10-14 19:29:05

Yeah, you can search on the web to find all the gory details.

Yeah, you can search on the web to find all the gory details.

DPatrick
2019-10-14 19:30:05

**Welcome to the 2019-20***Who Wants to Be a Mathematician*Championship Round 2 Math Jam!
DPatrick
2019-10-14 19:30:16

I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens (probably hundreds!) of AoPS classes over the past 15 years, and I've written or co-written a few of our textbooks.

I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens (probably hundreds!) of AoPS classes over the past 15 years, and I've written or co-written a few of our textbooks.

DPatrick
2019-10-14 19:30:30

Before we get started I would like to take a moment to explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.

Before we get started I would like to take a moment to explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.

DPatrick
2019-10-14 19:30:37

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
2019-10-14 19:30:46

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

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

**well-written**comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick
2019-10-14 19:30:57

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
2019-10-14 19:31:18

Tonight we are going to discuss the problems and solutions from Round 2 of

Tonight we are going to discuss the problems and solutions from Round 2 of

*Who Wants to Be a Mathematician*, which concluded last week.
DPatrick
2019-10-14 19:31:31

*Who Wants to Be a Mathematician*(or WWTBAM for short) is conducted by the American Mathematical Society (AMS).
DPatrick
2019-10-14 19:31:50

The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society. The AMS is one of the two big professional mathematics organizations in the U.S. (The other is the Mathematical Association of America, or MAA, which as you may know runs the AMC/AIME/USA(J)MO contests.)

The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society. The AMS is one of the two big professional mathematics organizations in the U.S. (The other is the Mathematical Association of America, or MAA, which as you may know runs the AMC/AIME/USA(J)MO contests.)

DPatrick
2019-10-14 19:32:11

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

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

**mikebreen**) and Bill Butterworth (**TPiR**).
DPatrick
2019-10-14 19:32:22

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.
mikebreen
2019-10-14 19:32:38

Hello, everyone.

Hello, everyone.

DPatrick
2019-10-14 19:32:47

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.)
TPiR
2019-10-14 19:32:58

Hi everyone. Glad to be here.

Hi everyone. Glad to be here.

DPatrick
2019-10-14 19:33:16

And last but not least, we also have an assistant here to help out tonight: Zach Stein-Perlman (

And last but not least, 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.
ZachSteinPerlman
2019-10-14 19:33:20

Hello!

Hello!

DPatrick
2019-10-14 19:33:30

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
2019-10-14 19:33:55

But, as I said, 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!

But, as I said, we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!

DPatrick
2019-10-14 19:34:20

Round 1 of WWTBAM was held in September. Students who scored at least 7 out of 10 in Round 1 advanced to Round 2, which was held earlier this month.

Round 1 of WWTBAM was held in September. Students who scored at least 7 out of 10 in Round 1 advanced to Round 2, which was held earlier this month.

DPatrick
2019-10-14 19:34:34

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 shouldn't 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 shouldn't take you nearly that long.) No books, notes, calculators, or internet access was permitted during the contest.

DPatrick
2019-10-14 19:34:53

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
2019-10-14 19:35:20

Let's get started!

Let's get started!

DPatrick
2019-10-14 19:35:25

1. For how many primes $p < 300$ is $p^3 - 2p^2$ a positive perfect square (square of a positive integer)?

1. For how many primes $p < 300$ is $p^3 - 2p^2$ a positive perfect square (square of a positive integer)?

DPatrick
2019-10-14 19:35:44

What can we do with the given expression?

What can we do with the given expression?

RagingWar
2019-10-14 19:35:56

factor

factor

motorfinn
2019-10-14 19:35:56

Factoring p^3-2p^2 seems like a good idea

Factoring p^3-2p^2 seems like a good idea

ZhaoPow
2019-10-14 19:35:56

factor out p^2 first

factor out p^2 first

RagingWar
2019-10-14 19:35:56

factor $p^2$ out

factor $p^2$ out

AlexLikeMath
2019-10-14 19:35:56

factor p-2

factor p-2

t3sp
2019-10-14 19:35:56

factor?

factor?

amuthup
2019-10-14 19:35:56

factor

factor

asdf334
2019-10-14 19:35:56

factor p^2

factor p^2

SoulOfHeaven
2019-10-14 19:36:03

factor out p^2?

factor out p^2?

naman12
2019-10-14 19:36:03

Factor into $p^2(p-2)$

Factor into $p^2(p-2)$

aastha.sharma02
2019-10-14 19:36:03

for this one i factored into $p^2(p-2)$

for this one i factored into $p^2(p-2)$

DPatrick
2019-10-14 19:36:10

Right. We can factor out $p^2$. This makes the expression $p^2(p-2)$.

Right. We can factor out $p^2$. This makes the expression $p^2(p-2)$.

DPatrick
2019-10-14 19:36:15

When is this a perfect square?

When is this a perfect square?

will3145
2019-10-14 19:36:38

p-2 is a perfect squre

p-2 is a perfect squre

Qwerty71
2019-10-14 19:36:38

this means since p^2 is already a square, p-2 must be a perfect square

this means since p^2 is already a square, p-2 must be a perfect square

JustKeepRunning
2019-10-14 19:36:38

$p-2$ is perfect square

$p-2$ is perfect square

GammaZero
2019-10-14 19:36:38

when $p-2$ is a square

when $p-2$ is a square

mathstats
2019-10-14 19:36:38

then p - 2 must be a square

then p - 2 must be a square

Mathnerd1223334444
2019-10-14 19:36:38

When p-2 is a square

When p-2 is a square

Toinfinity
2019-10-14 19:36:38

When $p-2$ is a perfect square

When $p-2$ is a perfect square

DPatrick
2019-10-14 19:36:56

Yes. $p^2$ is already a perfect square, so the whole thing is a perfect square exactly when $p-2$ is a perfect square.

Yes. $p^2$ is already a perfect square, so the whole thing is a perfect square exactly when $p-2$ is a perfect square.

DPatrick
2019-10-14 19:37:04

Great, we've simplified the problem!

Great, we've simplified the problem!

DPatrick
2019-10-14 19:37:07

1. For how many primes $p < 300$ is $p-2$ a positive perfect square?

1. For how many primes $p < 300$ is $p-2$ a positive perfect square?

DPatrick
2019-10-14 19:37:17

We could list all the primes less than $300$ and check them, but what's simpler?

We could list all the primes less than $300$ and check them, but what's simpler?

mathstats
2019-10-14 19:37:38

pick all the squares from 1 - 300 and see which ones work

pick all the squares from 1 - 300 and see which ones work

Qwerty71
2019-10-14 19:37:38

there are 17 squares up until 300; now just add 2 and check if they are prime

there are 17 squares up until 300; now just add 2 and check if they are prime

asdf334
2019-10-14 19:37:38

add 2 to every square

add 2 to every square

aastha.sharma02
2019-10-14 19:37:38

so then i listed the squares below 300 because there are fewer squares then there are primes

so then i listed the squares below 300 because there are fewer squares then there are primes

MagnificentMathematician
2019-10-14 19:37:38

find all the squares

find all the squares

ZhaoPow
2019-10-14 19:37:38

list odd squares less than 300

list odd squares less than 300

melonder
2019-10-14 19:37:38

list perfect sqyuares and add 2

list perfect sqyuares and add 2

DPatrick
2019-10-14 19:37:47

Right. It's way simpler to do it in reverse: list all the positive perfect squares less than $300$ and check which are $2$ less than a prime.

Right. It's way simpler to do it in reverse: list all the positive perfect squares less than $300$ and check which are $2$ less than a prime.

DPatrick
2019-10-14 19:37:59

That is, we're answering the question:

That is, we're answering the question:

DPatrick
2019-10-14 19:38:02

1. For how many positive integers $n$ is $p = n^2 + 2$ a prime less than $300$?

1. For how many positive integers $n$ is $p = n^2 + 2$ a prime less than $300$?

DPatrick
2019-10-14 19:38:18

And as many of you have pointed out, we only have to check it for the odd $n$. (The prime $p=2$ comes from $n=0$ but that's not positive.)

And as many of you have pointed out, we only have to check it for the odd $n$. (The prime $p=2$ comes from $n=0$ but that's not positive.)

DPatrick
2019-10-14 19:38:28

So now we can make a chart:

So now we can make a chart:

DPatrick
2019-10-14 19:38:32

$$\begin{array}{r|r}

n & n^2 + 2 \\ \hline

1 & 3 \\

3 & 11 \\

5 & 27 \\

7 & 51 \\

9 & 83 \\

11 & 123 \\

13 & 171 \\

15 & 227 \\

17 & 291

\end{array}$$

$$\begin{array}{r|r}

n & n^2 + 2 \\ \hline

1 & 3 \\

3 & 11 \\

5 & 27 \\

7 & 51 \\

9 & 83 \\

11 & 123 \\

13 & 171 \\

15 & 227 \\

17 & 291

\end{array}$$

DPatrick
2019-10-14 19:38:45

$n=19$ gives $n^2 + 2 = 363$, which is too big. So this is all we have to check.

$n=19$ gives $n^2 + 2 = 363$, which is too big. So this is all we have to check.

DPatrick
2019-10-14 19:38:48

Which numbers in the right column are prime?

Which numbers in the right column are prime?

Mathnerd1223334444
2019-10-14 19:39:26

3, 11, 83, 227 I think

3, 11, 83, 227 I think

SoulOfHeaven
2019-10-14 19:39:26

3, 11, 83, 227

3, 11, 83, 227

melonder
2019-10-14 19:39:26

3,11.83.227

3,11.83.227

naman12
2019-10-14 19:39:26

$3,11,83,227$

$3,11,83,227$

RagingWar
2019-10-14 19:39:26

3,11,83,227

3,11,83,227

Student.
2019-10-14 19:39:26

3, 11, 83, 227

3, 11, 83, 227

Maskie
2019-10-14 19:39:26

3 11 83 227

3 11 83 227

kvedula2004
2019-10-14 19:39:26

all others divisible by 3

all others divisible by 3

DPatrick
2019-10-14 19:39:40

We can start with the easy ones: $3$ and $11$ are prime.

We can start with the easy ones: $3$ and $11$ are prime.

DPatrick
2019-10-14 19:40:02

And we can get rid of some easy ones: $27$, $51$, $123$, $171$, $291$ are all multiples of $3$ (their digits sum to a multiple of $3$), so are composite.

And we can get rid of some easy ones: $27$, $51$, $123$, $171$, $291$ are all multiples of $3$ (their digits sum to a multiple of $3$), so are composite.

DPatrick
2019-10-14 19:40:14

That just leaves $83$ and $227$ to check.

That just leaves $83$ and $227$ to check.

DPatrick
2019-10-14 19:40:38

...which you can do, and they indeed are both prime.

...which you can do, and they indeed are both prime.

bigwolfy
2019-10-14 19:40:50

4

4

muhandae17
2019-10-14 19:40:50

total of 4

total of 4

heidinet
2019-10-14 19:41:00

4

4

RagingWar
2019-10-14 19:41:00

4

4

Ausmumof3
2019-10-14 19:41:00

4

4

LovingPilot
2019-10-14 19:41:00

4

4

DPatrick
2019-10-14 19:41:01

That means that the only primes $p$ satisfying the condition of the problem are $3$, $11$, $83$, and $227$, for a total of $\boxed{4}$ primes.

That means that the only primes $p$ satisfying the condition of the problem are $3$, $11$, $83$, and $227$, for a total of $\boxed{4}$ primes.

DPatrick
2019-10-14 19:41:36

Surprisingly, this turned out to be one of the harder problems on the contest! Only about 28% of students taking the contest electronically got this right.

Surprisingly, this turned out to be one of the harder problems on the contest! Only about 28% of students taking the contest electronically got this right.

DPatrick
2019-10-14 19:41:54

Next up:

Next up:

DPatrick
2019-10-14 19:41:58

2. Compute $\cos\left(\tan^{-1}\left(\dfrac{-\sqrt{7}}{3}\right)\right)$ (where $\tan^{-1}$ denotes the inverse tangent function).

2. Compute $\cos\left(\tan^{-1}\left(\dfrac{-\sqrt{7}}{3}\right)\right)$ (where $\tan^{-1}$ denotes the inverse tangent function).

DPatrick
2019-10-14 19:42:36

Before we get too deep into this: what is the sign of our answer going to be? Will it be positive or negative?

Before we get too deep into this: what is the sign of our answer going to be? Will it be positive or negative?

yoyopianow
2019-10-14 19:43:05

Positive

Positive

kvedula2004
2019-10-14 19:43:05

positive, range of tan^-1 is -pi/2 to pi/2

positive, range of tan^-1 is -pi/2 to pi/2

DPatrick
2019-10-14 19:43:17

That's right. $\tan^{-1}$ always gives us a value between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ (or $-90^\circ$ and $90^\circ$, if you prefer).

That's right. $\tan^{-1}$ always gives us a value between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ (or $-90^\circ$ and $90^\circ$, if you prefer).

DPatrick
2019-10-14 19:43:32

But cosine is positive for any angle in that range.

But cosine is positive for any angle in that range.

DPatrick
2019-10-14 19:43:50

(You can think of the graph of cosine, or of the unit circle, to see this.)

(You can think of the graph of cosine, or of the unit circle, to see this.)

DPatrick
2019-10-14 19:44:06

So our answer must end up being positive.

So our answer must end up being positive.

DPatrick
2019-10-14 19:44:14

And now what?

And now what?

Mathnerd1223334444
2019-10-14 19:44:27

draw a triangle and stuff cancels

draw a triangle and stuff cancels

JustKeepRunning
2019-10-14 19:44:27

draw the triangle

draw the triangle

DPatrick
2019-10-14 19:44:43

Yes: I generally like to just draw a right triangle for problems like this.

Yes: I generally like to just draw a right triangle for problems like this.

DPatrick
2019-10-14 19:44:48

DPatrick
2019-10-14 19:45:08

That's the triangle with $\tan(\theta) = \dfrac{\sqrt7}{3}$.

That's the triangle with $\tan(\theta) = \dfrac{\sqrt7}{3}$.

DPatrick
2019-10-14 19:45:40

And since we know the answer we want is positive, we can use this triangle even though what we really want is $\tan(\theta) = -\dfrac{\sqrt7}{3}$ -- the cosine will be the same.

And since we know the answer we want is positive, we can use this triangle even though what we really want is $\tan(\theta) = -\dfrac{\sqrt7}{3}$ -- the cosine will be the same.

DPatrick
2019-10-14 19:45:52

So the quantity we want for our answer is $\cos(\theta)$.

So the quantity we want for our answer is $\cos(\theta)$.

RagingWar
2019-10-14 19:45:56

cosine is adjacent/hypotenuse

cosine is adjacent/hypotenuse

kvedula2004
2019-10-14 19:45:59

pythag thrm, hypotenuse = 4

pythag thrm, hypotenuse = 4

DPatrick
2019-10-14 19:46:05

Yep. The hypotenuse is $\sqrt{7 + 9} = \sqrt{16} = 4$.

Yep. The hypotenuse is $\sqrt{7 + 9} = \sqrt{16} = 4$.

Green4Applez
2019-10-14 19:46:15

so 3/4

so 3/4

t3sp
2019-10-14 19:46:15

so hypotenuse is 4 and cos(theta)=3/4

so hypotenuse is 4 and cos(theta)=3/4

Maskie
2019-10-14 19:46:15

so it would be 3/4

so it would be 3/4

DPatrick
2019-10-14 19:46:20

So $\cos\theta = \boxed{\dfrac34}$ is our answer.

So $\cos\theta = \boxed{\dfrac34}$ is our answer.

kvedula2004
2019-10-14 19:46:47

these are tough early questions...

these are tough early questions...

DPatrick
2019-10-14 19:46:59

This turned out to be the 3rd easiest...about 54% got it right.

This turned out to be the 3rd easiest...about 54% got it right.

DPatrick
2019-10-14 19:47:17

I think the contest as a whole was pretty tough this time around.

I think the contest as a whole was pretty tough this time around.

DPatrick
2019-10-14 19:47:25

On to #3:

On to #3:

DPatrick
2019-10-14 19:47:31

3 What proportion of the nine-digit numbers that can be formed by permuting the digits of $123456789$ are divisible by $36$?

3 What proportion of the nine-digit numbers that can be formed by permuting the digits of $123456789$ are divisible by $36$?

DPatrick
2019-10-14 19:47:49

How do we determine if a number is divisible by $36$?

How do we determine if a number is divisible by $36$?

asdf334
2019-10-14 19:48:01

9 and 4

9 and 4

SoulOfHeaven
2019-10-14 19:48:01

has to be a multiple of 4 and 9

has to be a multiple of 4 and 9

motorfinn
2019-10-14 19:48:01

Divisible by 9 and 4

Divisible by 9 and 4

aastha.sharma02
2019-10-14 19:48:01

div by 9 and 4

div by 9 and 4

ZhaoPow
2019-10-14 19:48:01

divisible by 9 and 4

divisible by 9 and 4

varunragu23
2019-10-14 19:48:01

divisible by 9 and 4

divisible by 9 and 4

AlexLikeMath
2019-10-14 19:48:01

divisible by 4 and 9

divisible by 4 and 9

Edwinyc
2019-10-14 19:48:05

by checking if it's divisible by 4 and 9

by checking if it's divisible by 4 and 9

DPatrick
2019-10-14 19:48:10

We can look at the prime factors of $36$ separately. That is, it has to be divisible by both $4$ and $9$.

We can look at the prime factors of $36$ separately. That is, it has to be divisible by both $4$ and $9$.

Orangecounty2
2019-10-14 19:48:40

all are divisible by 9

all are divisible by 9

ilikechocolate
2019-10-14 19:48:40

for 9 we add up digits, but no matter how we permutate, sum is always 45 so always divisible by 9

for 9 we add up digits, but no matter how we permutate, sum is always 45 so always divisible by 9

kvedula2004
2019-10-14 19:48:40

divisibility by 9 is gurenteed bc sum of digits in constant (45)

divisibility by 9 is gurenteed bc sum of digits in constant (45)

aastha.sharma02
2019-10-14 19:48:40

its automatically divisible by 9 because the sum is 45

its automatically divisible by 9 because the sum is 45

SoulOfHeaven
2019-10-14 19:48:40

9 is taken care of though because 1+2+3....+9 is a multilpe of 9

9 is taken care of though because 1+2+3....+9 is a multilpe of 9

DPatrick
2019-10-14 19:48:53

Aha...we know a number is divisible by $9$ if the sum of its digits is a multiple of $9$.

Aha...we know a number is divisible by $9$ if the sum of its digits is a multiple of $9$.

DPatrick
2019-10-14 19:48:59

But we're permuting the digits, so the sum is always the same! It's $1+2+\cdots+9 = 45$.

But we're permuting the digits, so the sum is always the same! It's $1+2+\cdots+9 = 45$.

DPatrick
2019-10-14 19:49:10

So every permutation is divisible by $9$. We only have to worry about divisibility by $4$.

So every permutation is divisible by $9$. We only have to worry about divisibility by $4$.

ilikechocolate
2019-10-14 19:49:32

for 4, we need to have the 2 final numbers be divisible by 4

for 4, we need to have the 2 final numbers be divisible by 4

NerdyDude
2019-10-14 19:49:32

last two digits as a number have to be divisible by 4

last two digits as a number have to be divisible by 4

mathstats
2019-10-14 19:49:32

for 4 last 2 digits div by 4

for 4 last 2 digits div by 4

GammaZero
2019-10-14 19:49:32

THe divisibility for $4$ is that the last 2 numbers of a quantity are divisible by 4

THe divisibility for $4$ is that the last 2 numbers of a quantity are divisible by 4

DPatrick
2019-10-14 19:49:50

And we know when is a number divisible by $4$: when its last two digits is a multiple of $4$.

And we know when is a number divisible by $4$: when its last two digits is a multiple of $4$.

DPatrick
2019-10-14 19:49:55

So we only have to look at the last two digits of our permutation.

So we only have to look at the last two digits of our permutation.

DPatrick
2019-10-14 19:50:05

How many possible last two digits are there?

How many possible last two digits are there?

DPatrick
2019-10-14 19:50:28

(in total, not necessarily divisible by 4)

(in total, not necessarily divisible by 4)

varunragu23
2019-10-14 19:50:46

9*8 = 72

9*8 = 72

AlexLikeMath
2019-10-14 19:50:46

72

72

atmchallenge
2019-10-14 19:50:46

72

72

Green4Applez
2019-10-14 19:50:46

9*8=72

9*8=72

cj13609517288
2019-10-14 19:50:46

$8\cdot9=72$

$8\cdot9=72$

Giakki
2019-10-14 19:50:46

72

72

JustKeepRunning
2019-10-14 19:50:46

72

72

DPatrick
2019-10-14 19:51:05

There are $9 \cdot 8 = 72$ choices for the last two digits. (9 choices for the units then 8 choices for the tens.)

There are $9 \cdot 8 = 72$ choices for the last two digits. (9 choices for the units then 8 choices for the tens.)

DPatrick
2019-10-14 19:51:09

And how many of them are multiples of $4$?

And how many of them are multiples of $4$?

JustKeepRunning
2019-10-14 19:51:47

16

16

Puddles_Penguin
2019-10-14 19:51:47

16

16

RagingWar
2019-10-14 19:51:47

16

16

Giakki
2019-10-14 19:51:47

16

16

yoyopianow
2019-10-14 19:51:47

16

16

DPatrick
2019-10-14 19:51:58

We can list them: $12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96.$

We can list them: $12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96.$

DPatrick
2019-10-14 19:52:13

(These are all the two-digit multiples of $4$, except for those ending in $0$ like $20,40,60,80$ and those with a repeated digit like $44,88$.)

(These are all the two-digit multiples of $4$, except for those ending in $0$ like $20,40,60,80$ and those with a repeated digit like $44,88$.)

DPatrick
2019-10-14 19:52:34

There are $16$ numbers in our list.

There are $16$ numbers in our list.

DPatrick
2019-10-14 19:52:41

So what's the final answer?

So what's the final answer?

MegaminxHB2
2019-10-14 19:53:01

$\frac{2}{9}$

$\frac{2}{9}$

Maskie
2019-10-14 19:53:01

2/9

2/9

RWhite
2019-10-14 19:53:01

16/72 = 2/9

16/72 = 2/9

Alvin54968972
2019-10-14 19:53:01

So it’s just 16/72 = 2/9?

So it’s just 16/72 = 2/9?

naman12
2019-10-14 19:53:01

$\frac{2}{9}$

$\frac{2}{9}$

Student.
2019-10-14 19:53:01

2/9

2/9

DPatrick
2019-10-14 19:53:11

So the proportion of numbers that work is $\dfrac{16}{72}$.

So the proportion of numbers that work is $\dfrac{16}{72}$.

DPatrick
2019-10-14 19:53:16

In lowest terms, this is $\boxed{\dfrac{2}{9}}$.

In lowest terms, this is $\boxed{\dfrac{2}{9}}$.

DPatrick
2019-10-14 19:53:27

Note that we don't have to worry about the first seven digits: they don't affect the divisibility!

Note that we don't have to worry about the first seven digits: they don't affect the divisibility!

DPatrick
2019-10-14 19:53:42

Only the last two digits are relevant.

Only the last two digits are relevant.

Puddles_Penguin
2019-10-14 19:54:05

I assume a good majority got this. Correct?

I assume a good majority got this. Correct?

DPatrick
2019-10-14 19:54:20

Nope. Only 26%; it was the 2nd hardest (not counting #10 as we'll see later.)

Nope. Only 26%; it was the 2nd hardest (not counting #10 as we'll see later.)

DPatrick
2019-10-14 19:54:36

On to #4:

On to #4:

DPatrick
2019-10-14 19:54:45

4. Let $f(x) = (2x+3)^3$ and $g(x) = x^3+x^2-x-1$. What is the sum of the coefficients of the polynomial $f(g(x))$?

4. Let $f(x) = (2x+3)^3$ and $g(x) = x^3+x^2-x-1$. What is the sum of the coefficients of the polynomial $f(g(x))$?

DPatrick
2019-10-14 19:55:04

Note that this problem appeared differently on different versions of the contest, but the solution method is the same.

Note that this problem appeared differently on different versions of the contest, but the solution method is the same.

DPatrick
2019-10-14 19:55:20

We could try to write an algebraic expression for $f(g(x))$, but is there a simpler idea?

We could try to write an algebraic expression for $f(g(x))$, but is there a simpler idea?

kvedula2004
2019-10-14 19:55:41

plug in 1

plug in 1

asdf334
2019-10-14 19:55:41

plug in 1

plug in 1

serichaoo
2019-10-14 19:55:41

set x=1 to find coefficients

set x=1 to find coefficients

JustKeepRunning
2019-10-14 19:55:41

plug in $1$

plug in $1$

iamhungry
2019-10-14 19:55:41

Plug in $1$

Plug in $1$

jacoporizzo
2019-10-14 19:55:41

plug in 1 for x

plug in 1 for x

Ilikeapos
2019-10-14 19:55:41

Yes, just find $f(g(1))$

Yes, just find $f(g(1))$

AlexLikeMath
2019-10-14 19:55:41

Plug in 1

Plug in 1

DPatrick
2019-10-14 19:55:56

Right, that's the key. If $p(x)$ is any polynomial, then $p(1)$ is the sum of its coefficients.

Right, that's the key. If $p(x)$ is any polynomial, then $p(1)$ is the sum of its coefficients.

DPatrick
2019-10-14 19:56:07

So all we need to do is to compute $f(g(1))$.

So all we need to do is to compute $f(g(1))$.

Orangecounty2
2019-10-14 19:56:22

but g(1) = 0!!

but g(1) = 0!!

iamhungry
2019-10-14 19:56:22

$g(1)=0$

$g(1)=0$

aastha.sharma02
2019-10-14 19:56:22

g(1)=0

g(1)=0

DPatrick
2019-10-14 19:56:30

We have $g(1) = 1^3 + 1^2 - 1 - 1$.

We have $g(1) = 1^3 + 1^2 - 1 - 1$.

DPatrick
2019-10-14 19:56:36

This gives $g(1) = 0$.

This gives $g(1) = 0$.

DPatrick
2019-10-14 19:56:45

So our answer is $f(g(1)) = f(0)$.

So our answer is $f(g(1)) = f(0)$.

iamhungry
2019-10-14 19:57:05

$f(g(1))=f(0)=(3)^3=27$

$f(g(1))=f(0)=(3)^3=27$

SoulOfHeaven
2019-10-14 19:57:05

so 27

so 27

JustKeepRunning
2019-10-14 19:57:05

$f(0)=27$

$f(0)=27$

NerdyDude
2019-10-14 19:57:05

(0+3)^3=27

(0+3)^3=27

aastha.sharma02
2019-10-14 19:57:05

so $f(0)=27$

so $f(0)=27$

ilikechocolate
2019-10-14 19:57:05

f(0) = 27

f(0) = 27

DPatrick
2019-10-14 19:57:13

And this is just $f(0) = 3^3 = \boxed{27}$.

And this is just $f(0) = 3^3 = \boxed{27}$.

DPatrick
2019-10-14 19:57:40

Again, if you took the contest, you might have had a different $f(x)$ (but I believe the same $g(x)$), so you might have ended up with a different numeric answer.

Again, if you took the contest, you might have had a different $f(x)$ (but I believe the same $g(x)$), so you might have ended up with a different numeric answer.

RagingWar
2019-10-14 19:57:52

this one must have been easier

this one must have been easier

NerdyDude
2019-10-14 19:57:52

how many got this right

how many got this right

DPatrick
2019-10-14 19:57:59

This was about 34%.

This was about 34%.

DPatrick
2019-10-14 19:58:35

I should mention that these stats are the ones that the AMS sent me this morning, and only include electronic participants. If you took the contest on paper, those aren't included yet.

I should mention that these stats are the ones that the AMS sent me this morning, and only include electronic participants. If you took the contest on paper, those aren't included yet.

DPatrick
2019-10-14 19:58:45

On to #5:

On to #5:

DPatrick
2019-10-14 19:58:51

5. The

5. The

*incircle*(or*inscribed circle*) of a triangle is the circle that is tangent to all three sides of the triangle. What is the radius of the incircle in a triangle with side lengths $13$, $14$, and $15$?
DPatrick
2019-10-14 19:59:07

There is a very useful formula for the radius of the incircle of a triangle. Do you know it?

There is a very useful formula for the radius of the incircle of a triangle. Do you know it?

kvedula2004
2019-10-14 19:59:30

A=sr

A=sr

RagingWar
2019-10-14 19:59:30

rs=area

rs=area

NerdyDude
2019-10-14 19:59:30

A=rs

A=rs

xiaomage
2019-10-14 19:59:30

area=rs

area=rs

serichaoo
2019-10-14 19:59:30

A=rs

A=rs

Ilikeapos
2019-10-14 19:59:30

$A = sr$

$A = sr$

asdf334
2019-10-14 19:59:30

A=rs

A=rs

JustKeepRunning
2019-10-14 19:59:30

a=rs

a=rs

Orangecounty2
2019-10-14 19:59:30

[ABC] = rs

[ABC] = rs

Maskie
2019-10-14 19:59:30

Sr=A

Sr=A

aop2014
2019-10-14 19:59:30

r=a/s

r=a/s

Alvin54968972
2019-10-14 19:59:30

A = rs

A = rs

AlexLikeMath
2019-10-14 19:59:30

rs = A

rs = A

mathstats
2019-10-14 19:59:30

sr=A

sr=A

DPatrick
2019-10-14 19:59:36

The formula is $A = rs$, where $A$ is the area of the triangle, $r$ is the radius of the incircle, and $s$ is half of the perimeter of the triangle (this is called the

The formula is $A = rs$, where $A$ is the area of the triangle, $r$ is the radius of the incircle, and $s$ is half of the perimeter of the triangle (this is called the

*semiperimeter*).
SoulOfHeaven
2019-10-14 19:59:44

draw lines from center of circle to points of tangency

draw lines from center of circle to points of tangency

DPatrick
2019-10-14 19:59:56

Right -- even if you don't have this formula memorized, we can recover it.

Right -- even if you don't have this formula memorized, we can recover it.

DPatrick
2019-10-14 20:00:16

Let's draw a generic acute triangle and its incircle:

Let's draw a generic acute triangle and its incircle:

DPatrick
2019-10-14 20:00:19

DPatrick
2019-10-14 20:00:26

We can draw the radii from the center of the incircle (which is, not surprisingly, called the

We can draw the radii from the center of the incircle (which is, not surprisingly, called the

**incenter**) to the three sides:
DPatrick
2019-10-14 20:00:28

DPatrick
2019-10-14 20:00:42

And how does that help us prove the formula?

And how does that help us prove the formula?

serichaoo
2019-10-14 20:01:01

Draw lengths from incentre to the vertices

Draw lengths from incentre to the vertices

RWhite
2019-10-14 20:01:01

we can draw lines from the vertices to the incenter

we can draw lines from the vertices to the incenter

cj13609517288
2019-10-14 20:01:09

spilt the whole are into pieces

spilt the whole are into pieces

DPatrick
2019-10-14 20:01:12

We can draw segments from the incenter to the three vertices, like so:

We can draw segments from the incenter to the three vertices, like so:

DPatrick
2019-10-14 20:01:14

Green4Applez
2019-10-14 20:01:23

then we have tree triangles and compute the sum of those areas

then we have tree triangles and compute the sum of those areas

kaiclc
2019-10-14 20:01:27

connect the incenter to the three vertices, three triangles combined area of sr

connect the incenter to the three vertices, three triangles combined area of sr

AlexLikeMath
2019-10-14 20:01:27

Add 3 triangles up

Add 3 triangles up

DPatrick
2019-10-14 20:01:30

And now we can compute the areas of the three blue triangles and sum them up!

And now we can compute the areas of the three blue triangles and sum them up!

DPatrick
2019-10-14 20:01:37

Each blue triangle has height $r$ and a base that's a side of the triangle.

Each blue triangle has height $r$ and a base that's a side of the triangle.

DPatrick
2019-10-14 20:01:56

So if the three sides of the big triangle are $a$, $b$, and $c$, the areas of the blue triangles are $\frac12ra$, $\frac12rb$, and $\frac12rc$.

So if the three sides of the big triangle are $a$, $b$, and $c$, the areas of the blue triangles are $\frac12ra$, $\frac12rb$, and $\frac12rc$.

DPatrick
2019-10-14 20:02:14

And when we sum them, we get the area of the whole triangle. So $A = \frac12ra + \frac12rb + \frac12rc$.

And when we sum them, we get the area of the whole triangle. So $A = \frac12ra + \frac12rb + \frac12rc$.

RWhite
2019-10-14 20:02:17

ra/2+rb/2+rc/2=r(a+b+c)/2

ra/2+rb/2+rc/2=r(a+b+c)/2

DPatrick
2019-10-14 20:02:25

But then this factors as $A = r\frac12(a+b+c) = rs$, just like we wanted!

But then this factors as $A = r\frac12(a+b+c) = rs$, just like we wanted!

DPatrick
2019-10-14 20:02:43

So back to our problem...the semiperimeter is $s = \frac12(13+14+15) = 21$.

So back to our problem...the semiperimeter is $s = \frac12(13+14+15) = 21$.

DPatrick
2019-10-14 20:02:48

Do we know the area?

Do we know the area?

NerdyDude
2019-10-14 20:03:15

herons formula!

herons formula!

cj13609517288
2019-10-14 20:03:15

we have heron! 84

we have heron! 84

SoulOfHeaven
2019-10-14 20:03:15

Use Heron's formula

Use Heron's formula

naman12
2019-10-14 20:03:15

Yes, $84$ by Heron's Formula

Yes, $84$ by Heron's Formula

mathstats
2019-10-14 20:03:15

herons

herons

Giakki
2019-10-14 20:03:15

heron formula

heron formula

RWhite
2019-10-14 20:03:15

heron's formula

heron's formula

yoyopianow
2019-10-14 20:03:15

Heron's?

Heron's?

lilyba
2019-10-14 20:03:15

use heron's formula

use heron's formula

DPatrick
2019-10-14 20:03:21

One method is to use

One method is to use

**Heron's Formula**, which states that the area of a triangle is $\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semiperimeter and $a$,$b$,$c$ are the three sides.
DPatrick
2019-10-14 20:03:31

If we plug our data in, we get $A = \sqrt{21(21-13)(21-14)(21-15)}$.

If we plug our data in, we get $A = \sqrt{21(21-13)(21-14)(21-15)}$.

DPatrick
2019-10-14 20:03:45

This simplifies to $A = \sqrt{21(8)(7)(6)}$, and that works out to $A = \sqrt{3 \cdot 7 \cdot 2^3 \cdot 7 \cdot 2 \cdot 3}$ in prime factors, so $A = 2^2 \cdot 3 \cdot 7 = 84$.

This simplifies to $A = \sqrt{21(8)(7)(6)}$, and that works out to $A = \sqrt{3 \cdot 7 \cdot 2^3 \cdot 7 \cdot 2 \cdot 3}$ in prime factors, so $A = 2^2 \cdot 3 \cdot 7 = 84$.

DPatrick
2019-10-14 20:03:53

But we don't need Heron's Formula. 13-14-15 triangles come up often in contest problems, because there's a neat little trick for them.

But we don't need Heron's Formula. 13-14-15 triangles come up often in contest problems, because there's a neat little trick for them.

JustKeepRunning
2019-10-14 20:04:01

Drop altitude to make nice pythag

Drop altitude to make nice pythag

Puddles_Penguin
2019-10-14 20:04:05

altitude to 14 is 12

altitude to 14 is 12

AlexLikeMath
2019-10-14 20:04:08

Yes, drop altitude to 14 side

Yes, drop altitude to 14 side

Maskie
2019-10-14 20:04:13

split the triangle into 9-12-15 and 5-12-13 triangles

split the triangle into 9-12-15 and 5-12-13 triangles

DPatrick
2019-10-14 20:04:23

What happens when we draw the altitude to the length $14$ side?

What happens when we draw the altitude to the length $14$ side?

DPatrick
2019-10-14 20:04:26

DPatrick
2019-10-14 20:04:36

If you know your basic Pythagorean triples, you might be able to guess the missing side lengths.

If you know your basic Pythagorean triples, you might be able to guess the missing side lengths.

xiaomage
2019-10-14 20:04:45

two right triangles

two right triangles

Alvin54968972
2019-10-14 20:04:45

Altitude is 12

Altitude is 12

aop2014
2019-10-14 20:04:45

the altitude is 12, and the base becomes 9 and 5

the altitude is 12, and the base becomes 9 and 5

DPatrick
2019-10-14 20:04:49

DPatrick
2019-10-14 20:04:58

That is: a 13-14-15 triangle is just a 9-12-15 right triangle and a 5-12-13 right triangle glued together along their common length 12 sides!

That is: a 13-14-15 triangle is just a 9-12-15 right triangle and a 5-12-13 right triangle glued together along their common length 12 sides!

DPatrick
2019-10-14 20:05:16

And we see that the area is $\frac12(12)(14) = 84$.

And we see that the area is $\frac12(12)(14) = 84$.

SoulOfHeaven
2019-10-14 20:05:35

84/21 = 4

84/21 = 4

xiaomage
2019-10-14 20:05:35

radius=4

radius=4

RWhite
2019-10-14 20:05:35

so the answer would be 84=r(21) r = 4

so the answer would be 84=r(21) r = 4

Edwinyc
2019-10-14 20:05:35

so r=4

so r=4

DPatrick
2019-10-14 20:05:42

And finally, using $A = rs$ with $A = 84$ and $s = 21$, we get $r = \boxed{4}$.

And finally, using $A = rs$ with $A = 84$ and $s = 21$, we get $r = \boxed{4}$.

DPatrick
2019-10-14 20:06:11

This one was 41% correct.

This one was 41% correct.

DPatrick
2019-10-14 20:06:19

On to #6:

On to #6:

DPatrick
2019-10-14 20:06:26

6. The integers $3$, $4$, $5$, $6$, $12$, and $13$ are arranged, without repetition, in a horizontal row so that the sum of any two numbers in adjacent positions is a perfect square (square of a positive integer). What is the sum of the first and last numbers in the arrangement?

6. The integers $3$, $4$, $5$, $6$, $12$, and $13$ are arranged, without repetition, in a horizontal row so that the sum of any two numbers in adjacent positions is a perfect square (square of a positive integer). What is the sum of the first and last numbers in the arrangement?

aastha.sharma02
2019-10-14 20:06:34

which one was easiest?

which one was easiest?

DPatrick
2019-10-14 20:06:38

This one.

This one.

DPatrick
2019-10-14 20:07:01

I don't know any clever way to solve this except to just get our handy dirty and try to construct this arrangement.

I don't know any clever way to solve this except to just get our handy dirty and try to construct this arrangement.

NerdyDude
2019-10-14 20:07:13

pair the numbers to find the two that can only be paired with one othe rnumber

pair the numbers to find the two that can only be paired with one othe rnumber

skyguy88
2019-10-14 20:07:18

If you check which numbers can form only one perfect square with the other integers...

If you check which numbers can form only one perfect square with the other integers...

DPatrick
2019-10-14 20:07:25

Yes. I'd solve it like a jigsaw puzzle: look for the end pieces first.

Yes. I'd solve it like a jigsaw puzzle: look for the end pieces first.

DPatrick
2019-10-14 20:07:39

That is, are there any numbers who only have one other number to pair with to get a perfect square?

That is, are there any numbers who only have one other number to pair with to get a perfect square?

Green4Applez
2019-10-14 20:08:08

to start, six should be first (or last) since it can only add with one other number to be a perfect square

to start, six should be first (or last) since it can only add with one other number to be a perfect square

asdf334
2019-10-14 20:08:08

6and5

6and5

varunragu23
2019-10-14 20:08:08

5 and 6

5 and 6

RagingWar
2019-10-14 20:08:08

6 has to be a last one, can only be paired with 3

6 has to be a last one, can only be paired with 3

kaiclc
2019-10-14 20:08:08

5 and 6

5 and 6

NerdyDude
2019-10-14 20:08:08

6 can only pair with 3

6 can only pair with 3

mtqchen
2019-10-14 20:08:08

5 and 6

5 and 6

chessking1
2019-10-14 20:08:08

6 and 5

6 and 5

DPatrick
2019-10-14 20:08:14

$5$ only pairs with $4$.

$5$ only pairs with $4$.

DPatrick
2019-10-14 20:08:18

And $6$ only pairs with $3$.

And $6$ only pairs with $3$.

yoyopianow
2019-10-14 20:08:29

So we're done?

So we're done?

DPatrick
2019-10-14 20:08:35

I think so. We could actually stop here!

I think so. We could actually stop here!

DPatrick
2019-10-14 20:08:41

If we have a valid arrangement, we know that numbers with just one pair must go on the ends.

If we have a valid arrangement, we know that numbers with just one pair must go on the ends.

DPatrick
2019-10-14 20:08:55

So since we've just determined that the ends must be $5$ and $6$, our answer is $5 + 6 = \boxed{11}$.

So since we've just determined that the ends must be $5$ and $6$, our answer is $5 + 6 = \boxed{11}$.

Puddles_Penguin
2019-10-14 20:09:19

5,4,12,13,3,6 so 11

5,4,12,13,3,6 so 11

asdf334
2019-10-14 20:09:19

it's 5,4,12,13,3,6

it's 5,4,12,13,3,6

DPatrick
2019-10-14 20:09:29

To be safe, and to check, we could construct the entire arrangement.

To be safe, and to check, we could construct the entire arrangement.

DPatrick
2019-10-14 20:09:34

The arrangement is $5,4,12,13,3,6$. (Or it could be reversed as $6,3,13,12,4,5$.) These are the only possibilities.

The arrangement is $5,4,12,13,3,6$. (Or it could be reversed as $6,3,13,12,4,5$.) These are the only possibilities.

RWhite
2019-10-14 20:09:46

what percentage got this right?

what percentage got this right?

DPatrick
2019-10-14 20:09:50

About 80%.

About 80%.

DPatrick
2019-10-14 20:09:59

This was the easiest by a wide margin.

This was the easiest by a wide margin.

DPatrick
2019-10-14 20:10:15

And the next problem was the second-easiest:

And the next problem was the second-easiest:

DPatrick
2019-10-14 20:10:19

7. Which of the following numbers is the product of three consecutive prime numbers?

$\phantom{hi}$

a. 1223 b. 1309 c. 1989 d. 2431 e. 2717

7. Which of the following numbers is the product of three consecutive prime numbers?

$\phantom{hi}$

a. 1223 b. 1309 c. 1989 d. 2431 e. 2717

NerdyDude
2019-10-14 20:10:48

estimate with perfect cubes??

estimate with perfect cubes??

asdf334
2019-10-14 20:10:48

find lower and upper bounds

find lower and upper bounds

ilikechocolate
2019-10-14 20:10:54

estimate

estimate

Bole
2019-10-14 20:10:54

look at which cubes are closest

look at which cubes are closest

DPatrick
2019-10-14 20:11:02

Indeed. All the numbers are slightly larger than $1000 = 10^3$.

Indeed. All the numbers are slightly larger than $1000 = 10^3$.

DPatrick
2019-10-14 20:11:14

$11^3 = 1331$ is in the ballpark of the choices. So is $13^3 = 2197$.

$11^3 = 1331$ is in the ballpark of the choices. So is $13^3 = 2197$.

t3sp
2019-10-14 20:11:22

can you use a calculator on this test?

can you use a calculator on this test?

DPatrick
2019-10-14 20:11:26

Nope.

Nope.

mathstats
2019-10-14 20:11:34

do 7 ,11 ,13

do 7 ,11 ,13

AlexLikeMath
2019-10-14 20:11:37

7*11*13 = 1001, check 11*13*17

7*11*13 = 1001, check 11*13*17

DPatrick
2019-10-14 20:11:58

Sure, we can check that $7 \cdot 11 \cdot 13 = 1001$. (This is a neat fact that comes up every once in a while in problems.)

Sure, we can check that $7 \cdot 11 \cdot 13 = 1001$. (This is a neat fact that comes up every once in a while in problems.)

DPatrick
2019-10-14 20:12:24

That's too small, so the next one to check is $11 \cdot 13 \cdot 17$.

That's too small, so the next one to check is $11 \cdot 13 \cdot 17$.

skyguy88
2019-10-14 20:12:50

11 times 13 times 17 is 2431

11 times 13 times 17 is 2431

SoulOfHeaven
2019-10-14 20:12:50

11*13*17 = 2431

11*13*17 = 2431

Alvin54968972
2019-10-14 20:12:50

11, 13, 17 gets 2431 (D)

11, 13, 17 gets 2431 (D)

SoulOfHeaven
2019-10-14 20:12:50

aha it's D

aha it's D

RWhite
2019-10-14 20:12:50

2431=11*13*17

2431=11*13*17

xiaomage
2019-10-14 20:12:50

It's 2431

It's 2431

Giakki
2019-10-14 20:12:59

11x13x17 = 2431

11x13x17 = 2431

DPatrick
2019-10-14 20:13:12

$13 \cdot 17 = 221$, so $11 \cdot 13 \cdot 17 = 11 \cdot 221 = 2210 + 221 = 2431$.

$13 \cdot 17 = 221$, so $11 \cdot 13 \cdot 17 = 11 \cdot 221 = 2210 + 221 = 2431$.

DPatrick
2019-10-14 20:13:19

(That's how I did it without a calculator.)

(That's how I did it without a calculator.)

DPatrick
2019-10-14 20:13:26

Aha! That's choice $\boxed{\text{d}}$.

Aha! That's choice $\boxed{\text{d}}$.

DPatrick
2019-10-14 20:13:41

This was about 57% correct.

This was about 57% correct.

yoyopianow
2019-10-14 20:13:56

Were some of the problems multiple choice and some short answer?

Were some of the problems multiple choice and some short answer?

DPatrick
2019-10-14 20:14:02

That was the only problem that was multiple choice.

That was the only problem that was multiple choice.

DPatrick
2019-10-14 20:14:12

Next up is #8:

Next up is #8:

DPatrick
2019-10-14 20:14:16

8. What is the sum of all real solutions to $\sqrt{x+15-8\sqrt{x-1}} = 2$?

8. What is the sum of all real solutions to $\sqrt{x+15-8\sqrt{x-1}} = 2$?

motorfinn
2019-10-14 20:14:53

Square it

Square it

RagingWar
2019-10-14 20:14:53

square it!!!

square it!!!

NerdyDude
2019-10-14 20:14:53

square both sides

square both sides

Bole
2019-10-14 20:14:53

get rid of the square roots

get rid of the square roots

Green4Applez
2019-10-14 20:14:53

we can start off by squaring both sides

we can start off by squaring both sides

alphaone001
2019-10-14 20:14:53

square and then isolate the radical?

square and then isolate the radical?

montana_mathlete
2019-10-14 20:14:53

Square it.

Square it.

aop2014
2019-10-14 20:14:53

first square both sides

first square both sides

DPatrick
2019-10-14 20:15:02

Good idea -- we can square both sides to get rid of the outside radical.

Good idea -- we can square both sides to get rid of the outside radical.

DPatrick
2019-10-14 20:15:09

That gives us $x + 15 - 8\sqrt{x-1} = 4$.

That gives us $x + 15 - 8\sqrt{x-1} = 4$.

DPatrick
2019-10-14 20:15:13

Now what?

Now what?

asdf334
2019-10-14 20:15:37

subtract x+15 from both

subtract x+15 from both

Creator
2019-10-14 20:15:37

take 8sqrtx-1 to the other side and square both sides

take 8sqrtx-1 to the other side and square both sides

mwizard
2019-10-14 20:15:37

move root to one side, and square again

move root to one side, and square again

Orangecounty2
2019-10-14 20:15:37

isolate square root and squarwe again

isolate square root and squarwe again

DPatrick
2019-10-14 20:15:47

Right: we want to isolate the radical so that we can square again.

Right: we want to isolate the radical so that we can square again.

DPatrick
2019-10-14 20:15:52

That is, we rearrange to get $x + 11 = 8\sqrt{x-1}$.

That is, we rearrange to get $x + 11 = 8\sqrt{x-1}$.

DPatrick
2019-10-14 20:16:04

And now we can square again.

And now we can square again.

DPatrick
2019-10-14 20:16:11

We get $(x+11)^2 = 64(x-1)$.

We get $(x+11)^2 = 64(x-1)$.

motorfinn
2019-10-14 20:16:25

x^2+22x+121=64x-64

x^2+22x+121=64x-64

JayEmSea
2019-10-14 20:16:25

Move x terms to one side

Move x terms to one side

DPatrick
2019-10-14 20:16:27

This gives us $x^2 + 22x + 121 = 64x - 64$.

This gives us $x^2 + 22x + 121 = 64x - 64$.

DPatrick
2019-10-14 20:16:31

And this simplifies to $x^2 - 42x + 185 = 0$.

And this simplifies to $x^2 - 42x + 185 = 0$.

DPatrick
2019-10-14 20:17:32

You might think "We don't need to solve this: we know that the sum of the solutions to this is $42$ by Vieta's formulas, so that's the answer."

You might think "We don't need to solve this: we know that the sum of the solutions to this is $42$ by Vieta's formulas, so that's the answer."

DPatrick
2019-10-14 20:17:36

What's a potential problem with this?

What's a potential problem with this?

naman12
2019-10-14 20:18:08

Extranneous roots

Extranneous roots

GammaZero
2019-10-14 20:18:08

extrataneous solutions

extrataneous solutions

AlexLikeMath
2019-10-14 20:18:08

erraneous solutions

erraneous solutions

mathstats
2019-10-14 20:18:08

extraneous solutions

extraneous solutions

varunragu23
2019-10-14 20:18:08

oh, square root might have negative numbers inside of it and we can't have that happen

oh, square root might have negative numbers inside of it and we can't have that happen

DPatrick
2019-10-14 20:18:14

We might have introduced extraneous solutions when we squared.

We might have introduced extraneous solutions when we squared.

DPatrick
2019-10-14 20:18:31

For example, consider the much simpler equation $\sqrt{x} = x - 1$.

For example, consider the much simpler equation $\sqrt{x} = x - 1$.

DPatrick
2019-10-14 20:18:42

If we square this, we get $x = x^2 - 2x + 1$, so $0 = x^2 - 3x + 1$.

If we square this, we get $x = x^2 - 2x + 1$, so $0 = x^2 - 3x + 1$.

DPatrick
2019-10-14 20:18:52

And by the quadratic formula, the solutions are $x = \dfrac{3 \pm \sqrt5}{2}$.

And by the quadratic formula, the solutions are $x = \dfrac{3 \pm \sqrt5}{2}$.

DPatrick
2019-10-14 20:19:02

However, the solution $x = \dfrac{3 - \sqrt5}{2}$ is

However, the solution $x = \dfrac{3 - \sqrt5}{2}$ is

**extraneous**: if we go back to our original equation and plug this in, we get $$\sqrt{\dfrac{3-\sqrt5}{2}} = \dfrac{3-\sqrt5}{2} - 1 = \dfrac{1 - \sqrt5}{2},$$ which cannot be true because the left side is positive (by definition) but the right side is negative.
DPatrick
2019-10-14 20:19:38

So, back to our problem: to be sure the solutions to our quadratic actually work in the original equation, we should find the actual solutions and check that they actually work.

So, back to our problem: to be sure the solutions to our quadratic actually work in the original equation, we should find the actual solutions and check that they actually work.

DPatrick
2019-10-14 20:19:41

What are the solutions to $x^2 - 42x + 185 = 0$?

What are the solutions to $x^2 - 42x + 185 = 0$?

TheTallGuy2.0
2019-10-14 20:20:05

(x-5)(x-37)

(x-5)(x-37)

RWhite
2019-10-14 20:20:05

x = 5 or 37

x = 5 or 37

GammaZero
2019-10-14 20:20:05

5 and 37

5 and 37

xiaomage
2019-10-14 20:20:05

(x-37)(x-5), so x=5 or 37

(x-37)(x-5), so x=5 or 37

AlexLikeMath
2019-10-14 20:20:05

5 & 37

5 & 37

SoulOfHeaven
2019-10-14 20:20:11

37 and 5

37 and 5

Edwinyc
2019-10-14 20:20:11

x=37, x=5

x=37, x=5

DPatrick
2019-10-14 20:20:21

We could certainly use the quadratic formula here...

We could certainly use the quadratic formula here...

DPatrick
2019-10-14 20:20:30

...or notice that this factors as $(x-5)(x-37) = 0$, so the solutions are $x=5$ and $x=37$.

...or notice that this factors as $(x-5)(x-37) = 0$, so the solutions are $x=5$ and $x=37$.

DPatrick
2019-10-14 20:20:44

And so let's check them.

And so let's check them.

DPatrick
2019-10-14 20:20:52

Using $x=5$ we get $\sqrt{5+15-8\sqrt{5-1}} = \sqrt{5+15-8(2)} = \sqrt{4} = 2$. OK!

Using $x=5$ we get $\sqrt{5+15-8\sqrt{5-1}} = \sqrt{5+15-8(2)} = \sqrt{4} = 2$. OK!

DPatrick
2019-10-14 20:21:06

Using $x=37$ we get $\sqrt{37+15-8\sqrt{37-1}} = \sqrt{37+15-48} = \sqrt{4} = 2$. OK!

Using $x=37$ we get $\sqrt{37+15-8\sqrt{37-1}} = \sqrt{37+15-48} = \sqrt{4} = 2$. OK!

Maskie
2019-10-14 20:21:22

42!

42!

kvedula2004
2019-10-14 20:21:22

now our answer is 42

now our answer is 42

NerdyDude
2019-10-14 20:21:22

so it is 42

so it is 42

DPatrick
2019-10-14 20:21:25

So both solutions are valid, and their sum is $\boxed{42}$.

So both solutions are valid, and their sum is $\boxed{42}$.

DPatrick
2019-10-14 20:21:38

This was medium difficult: 40% got it right.

This was medium difficult: 40% got it right.

DPatrick
2019-10-14 20:22:02

Next is #9, which was the hardest of 1-9 (#10 was special as we'll soon see):

Next is #9, which was the hardest of 1-9 (#10 was special as we'll soon see):

DPatrick
2019-10-14 20:22:06

9. A jar contains ten balls, numbered $1$ to $10$. Three balls are randomly drawn from the jar without replacement. What is the probability that no two of the three balls are labeled with consecutive integers?

9. A jar contains ten balls, numbered $1$ to $10$. Three balls are randomly drawn from the jar without replacement. What is the probability that no two of the three balls are labeled with consecutive integers?

DPatrick
2019-10-14 20:22:44

This is a tricky problem. There are a number of ways to approach it.

This is a tricky problem. There are a number of ways to approach it.

kvedula2004
2019-10-14 20:23:02

complmentary counting...

complmentary counting...

NerdyDude
2019-10-14 20:23:02

1-P(balls have consecutive integers)

1-P(balls have consecutive integers)

motorfinn
2019-10-14 20:23:02

Complementary Counting

Complementary Counting

alphaone001
2019-10-14 20:23:02

complementary counting?

complementary counting?

DPatrick
2019-10-14 20:23:12

It might be easiest to use complementary counting here: let's determine the probability that we

It might be easiest to use complementary counting here: let's determine the probability that we

**do**draw balls where two of them are congruent.
DPatrick
2019-10-14 20:23:20

How could we draw a set of balls that fail?

How could we draw a set of balls that fail?

alphaone001
2019-10-14 20:23:50

all three are consecutive, or 2 are consecutive and one is not

all three are consecutive, or 2 are consecutive and one is not

asdf334
2019-10-14 20:23:50

all consecutive or 2 consecutive and one is away from those 2

all consecutive or 2 consecutive and one is away from those 2

DPatrick
2019-10-14 20:24:04

Right. There are two cases:

1. All three balls are consecutive (like 6, 7, 8)

2. Two of the balls are consecutive but the third isn't (like 3, 4, 9)

Right. There are two cases:

1. All three balls are consecutive (like 6, 7, 8)

2. Two of the balls are consecutive but the third isn't (like 3, 4, 9)

DPatrick
2019-10-14 20:24:28

How many sets of balls in case 1? (Let's count them without regard to order, so 6,7,8 is the same as 7,6,8.)

How many sets of balls in case 1? (Let's count them without regard to order, so 6,7,8 is the same as 7,6,8.)

Giakki
2019-10-14 20:25:01

8

8

montana_mathlete
2019-10-14 20:25:01

$8$

$8$

naman12
2019-10-14 20:25:01

$8$

$8$

Student.
2019-10-14 20:25:01

8

8

Yagato
2019-10-14 20:25:01

8 cases*

8 cases*

DPatrick
2019-10-14 20:25:08

There are exactly $8$: the middle ball can be any of 2 through 9 (inclusive).

There are exactly $8$: the middle ball can be any of 2 through 9 (inclusive).

DPatrick
2019-10-14 20:25:37

How many sets of balls in case 2, with two balls consecutive and the third not?

How many sets of balls in case 2, with two balls consecutive and the third not?

DPatrick
2019-10-14 20:26:13

Well...first, how many ways to choose two consecutive balls?

Well...first, how many ways to choose two consecutive balls?

varunragu23
2019-10-14 20:26:32

9 ways

9 ways

Yagato
2019-10-14 20:26:32

9 ways

9 ways

xiaomage
2019-10-14 20:26:32

9

9

SoulOfHeaven
2019-10-14 20:26:32

9

9

bob321326
2019-10-14 20:26:32

9 ways

9 ways

AlexLikeMath
2019-10-14 20:26:32

9

9

NerdyDude
2019-10-14 20:26:32

9 same method as before

9 same method as before

aop2014
2019-10-14 20:26:32

9

9

DPatrick
2019-10-14 20:26:47

Right: we can have any pair from 1-2 up to 9-10, so there are $9$ possibilities.

Right: we can have any pair from 1-2 up to 9-10, so there are $9$ possibilities.

DPatrick
2019-10-14 20:26:54

Then, how many ways to choose the third ball?

Then, how many ways to choose the third ball?

Maskie
2019-10-14 20:27:23

depends on which two

depends on which two

asdf334
2019-10-14 20:27:23

depends

depends

JustKeepRunning
2019-10-14 20:27:23

Is determined by first choice

Is determined by first choice

Student.
2019-10-14 20:27:23

6 or 7 if the first two are 1 and 2 or 9 and 10

6 or 7 if the first two are 1 and 2 or 9 and 10

DPatrick
2019-10-14 20:27:28

Right: it depends!

Right: it depends!

DPatrick
2019-10-14 20:27:49

If the pair is either 1-2 or 9-10, there are $7$ choices for the single ball. (For example, if the pair is 1-2, then any of 4 through 10 could be the single ball).

If the pair is either 1-2 or 9-10, there are $7$ choices for the single ball. (For example, if the pair is 1-2, then any of 4 through 10 could be the single ball).

DPatrick
2019-10-14 20:28:09

For any other pair (and there are $7$ others), there are $6$ choices for the single ball (we must exclude the two balls in the pair and the two balls on either side).

For any other pair (and there are $7$ others), there are $6$ choices for the single ball (we must exclude the two balls in the pair and the two balls on either side).

Student.
2019-10-14 20:28:20

7 * 6 + 2 * 7 = 56

7 * 6 + 2 * 7 = 56

Maskie
2019-10-14 20:28:26

7*2+7*6=56

7*2+7*6=56

DPatrick
2019-10-14 20:28:44

Right: adding these up, there are $2 \cdot 7 + 7 \cdot 6 = 56$ ways to choose a pair and then a third ball.

Right: adding these up, there are $2 \cdot 7 + 7 \cdot 6 = 56$ ways to choose a pair and then a third ball.

asdf334
2019-10-14 20:28:52

case 1+case 2=64

case 1+case 2=64

RagingWar
2019-10-14 20:28:52

+ the other 8 combinations

+ the other 8 combinations

NerdyDude
2019-10-14 20:28:52

56+8 =64 total unfavorable cases

56+8 =64 total unfavorable cases

DPatrick
2019-10-14 20:29:07

And then adding up the cases, there are $8 + 56 = 64$ sets of three balls that fail.

And then adding up the cases, there are $8 + 56 = 64$ sets of three balls that fail.

DPatrick
2019-10-14 20:29:15

How do we finish from here?

How do we finish from here?

knchen
2019-10-14 20:29:30

find total amount of cases

find total amount of cases

mtqchen
2019-10-14 20:29:30

Subtract from the total possibilities

Subtract from the total possibilities

abcb1030
2019-10-14 20:29:36

Find the total number of ways the 3 balls can be drawn

Find the total number of ways the 3 balls can be drawn

aop2014
2019-10-14 20:29:40

subtract from $\dbinom{10}{3}$

subtract from $\dbinom{10}{3}$

Yagato
2019-10-14 20:29:47

a total of 10 choose 3 cases = 120

a total of 10 choose 3 cases = 120

DPatrick
2019-10-14 20:29:52

Right. There are $\dbinom{10}{3} = \dfrac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120$ ways to choose any three balls.

Right. There are $\dbinom{10}{3} = \dfrac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120$ ways to choose any three balls.

DPatrick
2019-10-14 20:29:59

And $64$ of them fail.

And $64$ of them fail.

DPatrick
2019-10-14 20:30:02

So $120 - 64 = 56$ of them succeed.

So $120 - 64 = 56$ of them succeed.

DPatrick
2019-10-14 20:30:21

And thus what's the probability of success?

And thus what's the probability of success?

asdf334
2019-10-14 20:30:45

56/120=7/15!

56/120=7/15!

kvedula2004
2019-10-14 20:30:45

56/120 = 7/15

56/120 = 7/15

RWhite
2019-10-14 20:30:45

is the answer 7/15

is the answer 7/15

Orangecounty2
2019-10-14 20:30:45

7/15

7/15

mathstats
2019-10-14 20:30:45

7/15

7/15

motorfinn
2019-10-14 20:30:45

56/120=7/15

56/120=7/15

gmannnnnnn
2019-10-14 20:30:45

$7/15$

$7/15$

GammaZero
2019-10-14 20:30:45

7/15

7/15

Puddles_Penguin
2019-10-14 20:30:45

7/15

7/15

GammaZero
2019-10-14 20:30:45

$\frac{7}{15}$

$\frac{7}{15}$

Tenthgrader
2019-10-14 20:30:45

7/15

7/15

DPatrick
2019-10-14 20:30:49

It's $\dfrac{56}{120}$, which simplifies to $\boxed{\dfrac{7}{15}}$.

It's $\dfrac{56}{120}$, which simplifies to $\boxed{\dfrac{7}{15}}$.

DPatrick
2019-10-14 20:31:13

There is a fairly slick solution that lets us get the whole thing at once.

There is a fairly slick solution that lets us get the whole thing at once.

AlexLikeMath
2019-10-14 20:31:40

Stars and bars

Stars and bars

Alvin54968972
2019-10-14 20:31:40

Could you use stars and bars?

Could you use stars and bars?

DPatrick
2019-10-14 20:31:52

Right. The way to construct a set of three balls that work is as follows: imagine that the 7 balls that we

X X X X X X X

Right. The way to construct a set of three balls that work is as follows: imagine that the 7 balls that we

*don't*pick are in a row, like so:X X X X X X X

DPatrick
2019-10-14 20:32:18

Then picking 3 balls that are nonconsecutive is in 1-1 correspondence to picking 3 slots from among the 8 available slots between the X's:

_ X _ X _ X _ X _ X _ X _ X _

Then picking 3 balls that are nonconsecutive is in 1-1 correspondence to picking 3 slots from among the 8 available slots between the X's:

_ X _ X _ X _ X _ X _ X _ X _

DPatrick
2019-10-14 20:32:31

For example, if we pick the three slots shown below:

* X _ X _ X * X * X _ X _ X _

then reading left-to right, the three balls in our set are 1, 5, and 7:

1 2 _ 3 _ 4 5 6 7 8 _ 9 _ 10 _

For example, if we pick the three slots shown below:

* X _ X _ X * X * X _ X _ X _

then reading left-to right, the three balls in our set are 1, 5, and 7:

1 2 _ 3 _ 4 5 6 7 8 _ 9 _ 10 _

DPatrick
2019-10-14 20:32:58

Picking them this way ensure that there's always at least one ball between the balls we pick.

Picking them this way ensure that there's always at least one ball between the balls we pick.

DPatrick
2019-10-14 20:33:10

There are 8 slots and we need to pick three of them, so that's $\dbinom83$ ways.

There are 8 slots and we need to pick three of them, so that's $\dbinom83$ ways.

DPatrick
2019-10-14 20:33:18

And we know that there are $\dbinom{10}{3}$ total possible sets of 3 balls. So the probability is $\dfrac{\binom83}{\binom{10}{3}}$.

And we know that there are $\dbinom{10}{3}$ total possible sets of 3 balls. So the probability is $\dfrac{\binom83}{\binom{10}{3}}$.

DPatrick
2019-10-14 20:33:25

This simplifies to $\dfrac{8 \cdot 7 \cdot 6}{10 \cdot 9 \cdot 8} = \dfrac{42}{90} = \boxed{\dfrac{7}{15}}$.

This simplifies to $\dfrac{8 \cdot 7 \cdot 6}{10 \cdot 9 \cdot 8} = \dfrac{42}{90} = \boxed{\dfrac{7}{15}}$.

SoulOfHeaven
2019-10-14 20:33:43

slick

slick

montana_mathlete
2019-10-14 20:33:43

Wow that's much faster

Wow that's much faster

DPatrick
2019-10-14 20:33:56

If you've had a

If you've had a

**lot**of experience with counting problems, you might have seen this sort of solution before.
RagingWar
2019-10-14 20:34:00

how many people got this right?

how many people got this right?

DPatrick
2019-10-14 20:34:10

About 13%. It was by far the hardest of the first nine.

About 13%. It was by far the hardest of the first nine.

DPatrick
2019-10-14 20:34:20

And at last, on to #10:

And at last, on to #10:

DPatrick
2019-10-14 20:34:24

10. What positive integer is closest to $(e + 2\pi)^{9/2}$?

10. What positive integer is closest to $(e + 2\pi)^{9/2}$?

DPatrick
2019-10-14 20:34:33

This is the tie-breaker question: it's unlikely you'll come up with the exact answer, but you want to make the best estimate that you can.

This is the tie-breaker question: it's unlikely you'll come up with the exact answer, but you want to make the best estimate that you can.

DPatrick
2019-10-14 20:34:39

**And no calculators!**
kvedula2004
2019-10-14 20:35:07

e+2pi is scarily close to 9

e+2pi is scarily close to 9

asdf334
2019-10-14 20:35:07

2.71+6.28=8.99 about 9

2.71+6.28=8.99 about 9

NerdyDude
2019-10-14 20:35:17

approx 2.71 for e and 3.14 for pi the sum is really close to 9

approx 2.71 for e and 3.14 for pi the sum is really close to 9

Alvin54968972
2019-10-14 20:35:23

e is approximately 2.718 and 2π is about 6.283

e is approximately 2.718 and 2π is about 6.283

DPatrick
2019-10-14 20:35:52

Indeed, if you know a few decimal places for $e$ and $\pi$, or even just one or two decimal places, you'll see that $e + 2\pi$ is really close to $9$.

Indeed, if you know a few decimal places for $e$ and $\pi$, or even just one or two decimal places, you'll see that $e + 2\pi$ is really close to $9$.

DPatrick
2019-10-14 20:36:17

So $9^{9/2}$ is already a pretty good guess.

So $9^{9/2}$ is already a pretty good guess.

motorfinn
2019-10-14 20:36:34

sqrt(9^9)

sqrt(9^9)

skyguy88
2019-10-14 20:36:34

3^9 = 19683

3^9 = 19683

aop2014
2019-10-14 20:36:34

so it is about sqrt(9^9), which is 3^9 which is 19683

so it is about sqrt(9^9), which is 3^9 which is 19683

kvedula2004
2019-10-14 20:36:34

19683

19683

vanilla.froyo
2019-10-14 20:36:38

its about 3^9

its about 3^9

Creator
2019-10-14 20:36:38

(sqrt9)^9 which is 3^9

(sqrt9)^9 which is 3^9

DPatrick
2019-10-14 20:36:55

And yes, we have $9^{9/2} = 3^9 = 19683$.

And yes, we have $9^{9/2} = 3^9 = 19683$.

DPatrick
2019-10-14 20:37:02

And this is a really good guess.

And this is a really good guess.

DPatrick
2019-10-14 20:37:18

But let me show you what I did to get an even better guess.

But let me show you what I did to get an even better guess.

Alvin54968972
2019-10-14 20:37:25

But e + 2π is a little bit more than 9

But e + 2π is a little bit more than 9

DPatrick
2019-10-14 20:37:32

Right.

Right.

DPatrick
2019-10-14 20:37:41

It turns out that I know both $e$ and $\pi$ to 5 decimal places.

\begin{align*}

e &= 2.71828\ldots \\

\pi &= 3.14159\ldots

\end{align*}

It turns out that I know both $e$ and $\pi$ to 5 decimal places.

\begin{align*}

e &= 2.71828\ldots \\

\pi &= 3.14159\ldots

\end{align*}

DPatrick
2019-10-14 20:37:47

(I actually know $\pi$ to a couple more and $e$ to a few more, but let's stick with this.)

(I actually know $\pi$ to a couple more and $e$ to a few more, but let's stick with this.)

DPatrick
2019-10-14 20:37:59

That gives us $e + 2\pi = 9.00146\ldots$.

That gives us $e + 2\pi = 9.00146\ldots$.

DPatrick
2019-10-14 20:38:14

We might have the wrong rounded in the rightmost digit because of the "...", so lets use $e + 2\pi = 9.0015$ going forward.

We might have the wrong rounded in the rightmost digit because of the "...", so lets use $e + 2\pi = 9.0015$ going forward.

DPatrick
2019-10-14 20:38:33

So how can we estimate $(9.0015)^{9/2}$?

So how can we estimate $(9.0015)^{9/2}$?

DPatrick
2019-10-14 20:38:45

I'll write what we're trying to compute as $(9 + \epsilon)^{9/2}$, where $\epsilon = 0.0015$. (The Greek letter $\epsilon$ is commonly used for very small numbers.)

I'll write what we're trying to compute as $(9 + \epsilon)^{9/2}$, where $\epsilon = 0.0015$. (The Greek letter $\epsilon$ is commonly used for very small numbers.)

AlexLikeMath
2019-10-14 20:38:49

binomial theorem

binomial theorem

ForeverAPenguin47
2019-10-14 20:38:53

binom thereom

binom thereom

DPatrick
2019-10-14 20:39:00

Yes! The "trick" is that we'll go ahead treat this like a binomial expansion, and use the Binomial Theorem on this.

Yes! The "trick" is that we'll go ahead treat this like a binomial expansion, and use the Binomial Theorem on this.

DPatrick
2019-10-14 20:39:12

If $n$ were an integer, how would we expand $(x+y)^n$?

If $n$ were an integer, how would we expand $(x+y)^n$?

DPatrick
2019-10-14 20:39:40

I don't need all the terms, just the first two or three...

I don't need all the terms, just the first two or three...

Puddles_Penguin
2019-10-14 20:40:05

x^n+nx^(n-1)y...

x^n+nx^(n-1)y...

JustKeepRunning
2019-10-14 20:40:13

$\binom n 0 x^n+\binom n 1x^{n-1}y+\binom n 2x^{n-2}y^2\cdots$

$\binom n 0 x^n+\binom n 1x^{n-1}y+\binom n 2x^{n-2}y^2\cdots$

amuthup
2019-10-14 20:40:16

$x^n + nx^{n-1}y + (n(n-1)/2)x^{n-2}y^2$

$x^n + nx^{n-1}y + (n(n-1)/2)x^{n-2}y^2$

DPatrick
2019-10-14 20:40:23

Right. It starts $x^n + nx^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \cdots$.

Right. It starts $x^n + nx^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \cdots$.

DPatrick
2019-10-14 20:40:34

But if $y$ is our $\epsilon$, which is a really small number, then the terms with $y^2$, $y^3$, etc. are really really small numbers.

But if $y$ is our $\epsilon$, which is a really small number, then the terms with $y^2$, $y^3$, etc. are really really small numbers.

DPatrick
2019-10-14 20:40:47

So we can forget about those terms when we approximate.

So we can forget about those terms when we approximate.

DPatrick
2019-10-14 20:40:59

That is, $(x+\epsilon)^n \approx x^n + nx^{n-1}\epsilon$.

That is, $(x+\epsilon)^n \approx x^n + nx^{n-1}\epsilon$.

DPatrick
2019-10-14 20:41:11

And the beauty of this is that this works even if $n$ isn't an integer! (Although to rigorously prove this, you probably need a little calculus.)

And the beauty of this is that this works even if $n$ isn't an integer! (Although to rigorously prove this, you probably need a little calculus.)

DPatrick
2019-10-14 20:41:29

So we will use the approximation $$(9+\epsilon)^{9/2} \approx 9^{9/2} + \frac92 \cdot 9^{7/2} \cdot \epsilon.$$

So we will use the approximation $$(9+\epsilon)^{9/2} \approx 9^{9/2} + \frac92 \cdot 9^{7/2} \cdot \epsilon.$$

DPatrick
2019-10-14 20:41:55

And plugging in $\epsilon$, this is just $3^9 + \frac92 \cdot 3^7 \cdot 0.0015$.

And plugging in $\epsilon$, this is just $3^9 + \frac92 \cdot 3^7 \cdot 0.0015$.

DPatrick
2019-10-14 20:42:16

Conveniently, $9 \cdot 3^7$ is the same as $3^9$, so it's $3^9 \left(1 + \dfrac12 \cdot 0.0015\right)$.

Conveniently, $9 \cdot 3^7$ is the same as $3^9$, so it's $3^9 \left(1 + \dfrac12 \cdot 0.0015\right)$.

DPatrick
2019-10-14 20:42:27

So this is $19683(1.00075)$.

So this is $19683(1.00075)$.

DPatrick
2019-10-14 20:43:04

That tells us how much we should increase our original estimate of $3^9 = 19683$ by: about $0.075\%$ of $19683$.

That tells us how much we should increase our original estimate of $3^9 = 19683$ by: about $0.075\%$ of $19683$.

asdf334
2019-10-14 20:43:19

0.00075 is 3/4000

0.00075 is 3/4000

asdf334
2019-10-14 20:43:26

and 19683 is abuot 20000

and 19683 is abuot 20000

DPatrick
2019-10-14 20:43:34

Right, so without a calculator I'd use $0.00075 \cdot 19683 \approx 15$ (it's about $75\%$ of 19.683).

Right, so without a calculator I'd use $0.00075 \cdot 19683 \approx 15$ (it's about $75\%$ of 19.683).

mtqchen
2019-10-14 20:43:46

Which is about 19698

Which is about 19698

DPatrick
2019-10-14 20:43:59

Exactly. My final guess was $19683 + 15 = 19698$.

Exactly. My final guess was $19683 + 15 = 19698$.

kvedula2004
2019-10-14 20:44:11

really close to exact answer of 19697

really close to exact answer of 19697

asdf334
2019-10-14 20:44:11

with calculator i got 19697

with calculator i got 19697

vanilla.froyo
2019-10-14 20:44:11

That's really good

That's really good

DPatrick
2019-10-14 20:44:19

Turns out I'm only off by one. The actual answer is $\boxed{19697}$. (Which you can check by calculator.)

Turns out I'm only off by one. The actual answer is $\boxed{19697}$. (Which you can check by calculator.)

DPatrick
2019-10-14 20:44:31

I was so close because the mathematics I used was legit! In particular $(x+\epsilon)^n \approx x^n + nx^{n-1}\epsilon$ is a really really good approximation provided $\epsilon$ is very small in comparison to $x$.

I was so close because the mathematics I used was legit! In particular $(x+\epsilon)^n \approx x^n + nx^{n-1}\epsilon$ is a really really good approximation provided $\epsilon$ is very small in comparison to $x$.

DPatrick
2019-10-14 20:44:59

I'm told that there were a handful of people who got the exact answer.

I'm told that there were a handful of people who got the exact answer.

purnimak
2019-10-14 20:45:04

so how would you get the exact answer without a calculator?

so how would you get the exact answer without a calculator?

DPatrick
2019-10-14 20:45:19

If you ran this estimate out to a couple more decimal places, you'd probably get it exactly.

If you ran this estimate out to a couple more decimal places, you'd probably get it exactly.

DPatrick
2019-10-14 20:45:33

So that's it for Round 2!

So that's it for Round 2!

DPatrick
2019-10-14 20:45:42

After the scoring for Round 2 is complete, 12 students will be invited to compete in the Championship Finals, live in Denver at the 2020 Joint Mathematics Meetings in January. Travel costs to and from Denver will be covered by the AMS.

After the scoring for Round 2 is complete, 12 students will be invited to compete in the Championship Finals, live in Denver at the 2020 Joint Mathematics Meetings in January. Travel costs to and from Denver will be covered by the AMS.

DPatrick
2019-10-14 20:45:52

Here's how the 12 finalists will be determined: 10 of the 12 will be the top scorer from Round 2 in each of the following regions:

Here's how the 12 finalists will be determined: 10 of the 12 will be the top scorer from Round 2 in each of the following regions:

DPatrick
2019-10-14 20:45:57

DPatrick
2019-10-14 20:46:09

The other two contestants will be the top scorer in the United Kingdom and the top scorer in the Denver metro area (or, I suppose, the second-highest scorer, if the highest scorer in Region 8 happens to be from Denver).

The other two contestants will be the top scorer in the United Kingdom and the top scorer in the Denver metro area (or, I suppose, the second-highest scorer, if the highest scorer in Region 8 happens to be from Denver).

DPatrick
2019-10-14 20:46:19

Question #10 will be used to break ties for high scores within a region.

Question #10 will be used to break ties for high scores within a region.

DPatrick
2019-10-14 20:46:30

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

The Championship Finals are held live in front of an audience at the Joint Mathematics Meetings, and are also live streamed on the web. (You can watch the archives of past years' finals on the WWTBAM website.) Contestants will compete directly against each other in semi-final rounds, with the semi-final winners advancing to a

*Jeopardy!*-style buzz-in final round to determine a champion.
DPatrick
2019-10-14 20:47:01

I believe the exact date and time have been set, but I don't have that info in front of me.

I believe the exact date and time have been set, but I don't have that info in front of me.

mikebreen
2019-10-14 20:47:24

Sat, Jan. 18 at noon Eastern (US)

Sat, Jan. 18 at noon Eastern (US)

DPatrick
2019-10-14 20:47:31

Great, thanks Mike.

Great, thanks Mike.

DPatrick
2019-10-14 20:48:02

Though I'm not sure that's right...I thought it was afternoon Denver time, which would be late afternoon ET -- am I wrong?

Though I'm not sure that's right...I thought it was afternoon Denver time, which would be late afternoon ET -- am I wrong?

DPatrick
2019-10-14 20:48:07

(Wouldn't be the first time if I were.)

(Wouldn't be the first time if I were.)

DPatrick
2019-10-14 20:49:00

I'll try to look it up while we continue chatting.

I'll try to look it up while we continue chatting.

DPatrick
2019-10-14 20:49:14

...though I don't have much more to say. I don't have any scoring info other than the stats I've given you.

...though I don't have much more to say. I don't have any scoring info other than the stats I've given you.

mikebreen
2019-10-14 20:49:18

We had to switch from our normal time. It will be at 10 Denver time.

We had to switch from our normal time. It will be at 10 Denver time.

DPatrick
2019-10-14 20:49:49

Aha -- that change hasn't made it into the program for the Joint Math Meetings yet (which still shows 1 pm MT).

Aha -- that change hasn't made it into the program for the Joint Math Meetings yet (which still shows 1 pm MT).

DPatrick
2019-10-14 20:50:43

That's all for tonight. Thanks for attending!

That's all for tonight. Thanks for attending!

asdf334
2019-10-14 20:50:46

thanks so much for doing this math jam!

thanks so much for doing this math jam!

DPatrick
2019-10-14 20:50:49

You're welcome!

You're welcome!

mikebreen
2019-10-14 20:50:57

Thanks to all the participants in tonight's Jam--wonderful solutions--and thanks to Dave and AoPS for hosting.

Thanks to all the participants in tonight's Jam--wonderful solutions--and thanks to Dave and AoPS for hosting.

TPiR
2019-10-14 20:51:08

Thanks, everyone.

Thanks, everyone.

Puddles_Penguin
2019-10-14 20:52:23

Thanks

Thanks

ForeverAPenguin47
2019-10-14 20:52:23

thank you!

thank you!

mtqchen
2019-10-14 20:52:23

Until next year

Until next year

NerdyDude
2019-10-14 20:52:23

thanks

thanks

Nmath101
2019-10-14 20:52:23

thank you

thank you

NT23
2019-10-14 20:52:23

Wow this was hard but fun

Wow this was hard but fun

aastha.sharma02
2019-10-14 20:52:23

thank you so much!

thank you so much!

mathstats
2019-10-14 20:52:23

thank you !!

thank you !!

asdf334
2019-10-14 20:53:16

when and what is the next math jam?

when and what is the next math jam?

DPatrick
2019-10-14 20:53:37

The next Math Jam is this Thursday the 17th.

The next Math Jam is this Thursday the 17th.

DPatrick
2019-10-14 20:53:49

We will be discussing the problems from Round 1 of the USA Math Talent Search.

We will be discussing the problems from Round 1 of the USA Math Talent Search.

DPatrick
2019-10-14 20:53:58

The full schedule is at https://artofproblemsolving.com/school/mathjams

The full schedule is at https://artofproblemsolving.com/school/mathjams

DPatrick
2019-10-14 20:55:19

I think we're ready to shut it down

I think we're ready to shut it down

devenware
2019-10-14 20:55:29

I think you're right!

I think you're right!

DPatrick
2019-10-14 20:55:30

Everyone have a good night!

Everyone have a good night!

ZachSteinPerlman
2019-10-14 20:55:53

Thanks everyone!

Thanks everyone!

eminentabyss
2019-10-14 20:56:02

One last huge thanks to Dave, Mike, Bill, Zach and all of you who could make it out tonight!

One last huge thanks to Dave, Mike, Bill, Zach and all of you who could make it out tonight!

eminentabyss
2019-10-14 20:56:17

This classroom will close shortly.

This classroom will close shortly.

asdf334
2019-10-14 20:56:26

goodbye!!!

goodbye!!!

asdf334
2019-10-14 20:56:26

goodbye!

goodbye!

piphi
2019-10-14 20:56:26

thx guys!

thx guys!

aastha.sharma02
2019-10-14 20:56:26

yep, thank you!

yep, thank you!

aastha.sharma02
2019-10-14 20:56:26

byeeeee

byeeeee

motorfinn
2019-10-14 20:56:57

Bye <3

Bye <3