## Who Wants to Be a Mathematician, Qualifying Round

Go back to the Math Jam ArchiveAoPS instructor David Patrick will discuss the problems on Round 1 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
2016-10-05 19:31:03

**Welcome to the 2016-17***Who Wants to Be a Mathematician*Round 1 Math Jam!
DPatrick
2016-10-05 19:31:15

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'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.

DPatrick
2016-10-05 19:31:28

I also once was a contestant on ABC's

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
2016-10-05 19:32:05

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
2016-10-05 19:32:18

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
2016-10-05 19:32:33

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 shared into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick
2016-10-05 19:32:57

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
2016-10-05 19:33:22

Joining us tonight is the co-creator of WWTBAM, Mike Breen (

Joining us tonight is the co-creator of WWTBAM, Mike Breen (

**mikebreen**).
mikebreen
2016-10-05 19:33:38

Very glad to be here!

Very glad to be here!

DPatrick
2016-10-05 19:33:42

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

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

*Who Wants to Be a Mathematician*for the American Mathematical Society in 2001. 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.
caro2be
2016-10-05 19:33:57

Hi Mike!

Hi Mike!

yiqun
2016-10-05 19:33:57

HI MIKE!!!

HI MIKE!!!

piphi
2016-10-05 19:33:57

Hi mikebreen

Hi mikebreen

DPatrick
2016-10-05 19:34:11

*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.
mikebreen
2016-10-05 19:34:19

Hello, everyone. Love all the interest and enthusiasm.

Hello, everyone. Love all the interest and enthusiasm.

DPatrick
2016-10-05 19:34:28

Round 1 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 1 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
2016-10-05 19:34:51

We'll take a bit longer than 15 minutes tonight, probably more like 45-60 minutes, 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, probably more like 45-60 minutes, 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.
Python54
2016-10-05 19:35:10

What is the difficulty of the problems?

What is the difficulty of the problems?

DPatrick
2016-10-05 19:35:46

You'll probably get a good idea as we go. Some are more-or-less basic algebra, but some are quite tricky. (And I've seen the Round 2 problems, and they're even trickier still.)

You'll probably get a good idea as we go. Some are more-or-less basic algebra, but some are quite tricky. (And I've seen the Round 2 problems, and they're even trickier still.)

DPatrick
2016-10-05 19:35:56

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.

mikebreen
2016-10-05 19:36:10

Easy to difficult. Round two much harder.

Easy to difficult. Round two much harder.

DPatrick
2016-10-05 19:36:38

Please hold your questions about the contest administration until the end, when Mike and I will try to answer them. Right now, let's get on with the math!

Please hold your questions about the contest administration until the end, when Mike and I will try to answer them. Right now, let's get on with the math!

DPatrick
2016-10-05 19:36:49

1. What is the only positive solution to $3x^2 + 17x = 28$?

1. What is the only positive solution to $3x^2 + 17x = 28$?

DPatrick
2016-10-05 19:37:17

Like I said, the early questions on the contest, like this one, are often more-or-less just basic algebra.

Like I said, the early questions on the contest, like this one, are often more-or-less just basic algebra.

hwl0304
2016-10-05 19:37:25

quadratic formula!

quadratic formula!

Christina26
2016-10-05 19:37:25

quadratic formula

quadratic formula

AOPS12142015
2016-10-05 19:37:25

Quadratic fornula

Quadratic fornula

Generic_Username
2016-10-05 19:37:25

QF

QF

yiqun
2016-10-05 19:37:25

quadratic formula

quadratic formula

DPatrick
2016-10-05 19:37:41

Certainly we can write it in a more standard form as \[ 3x^2 + 17x - 28 = 0 \] and use the quadratic formula.

Certainly we can write it in a more standard form as \[ 3x^2 + 17x - 28 = 0 \] and use the quadratic formula.

pkandarpa
2016-10-05 19:37:50

Factor the question.

Factor the question.

supermathkid
2016-10-05 19:37:50

(3x-4)*(x+7)

(3x-4)*(x+7)

DPatrick
2016-10-05 19:37:54

Or, if we guess that the solutions will be rational numbers, we can try to factor it.

Or, if we guess that the solutions will be rational numbers, we can try to factor it.

DPatrick
2016-10-05 19:37:59

Indeed, it factors as $(3x - 4)(x + 7) = 0$.

Indeed, it factors as $(3x - 4)(x + 7) = 0$.

DPatrick
2016-10-05 19:38:08

So what are the solutions, and in particular what is the positive solution?

So what are the solutions, and in particular what is the positive solution?

Patrickzcy
2016-10-05 19:38:30

4/3

4/3

Fridaychimp
2016-10-05 19:38:30

4/3

4/3

Liopleurodon
2016-10-05 19:38:30

4/3

4/3

EpicRuler101
2016-10-05 19:38:30

4/3

4/3

Intellectuality123
2016-10-05 19:38:30

4/3

4/3

rafa2be
2016-10-05 19:38:30

Positive one is $4/3$.

Positive one is $4/3$.

Shaktiala
2016-10-05 19:38:30

Is the answer 4/3?

Is the answer 4/3?

pramodmitikiri
2016-10-05 19:38:30

the only positive number would be 4/3, although both answers are 4/3 and -7

the only positive number would be 4/3, although both answers are 4/3 and -7

Python54
2016-10-05 19:38:30

$-7,4/3 \implies 4/3$ is answer

$-7,4/3 \implies 4/3$ is answer

DPatrick
2016-10-05 19:38:40

Right. The solutions are the solutions to $3x - 4 = 0$ and $x + 7 = 0$.

Right. The solutions are the solutions to $3x - 4 = 0$ and $x + 7 = 0$.

DPatrick
2016-10-05 19:38:52

The second factor gives us $x = -7$, which is negative.

The second factor gives us $x = -7$, which is negative.

DPatrick
2016-10-05 19:39:02

But the first factor gives us $x = \boxed{\dfrac43}$, which is positive, so this is our final answer.

But the first factor gives us $x = \boxed{\dfrac43}$, which is positive, so this is our final answer.

DPatrick
2016-10-05 19:39:21

By the way, how can we tell quickly, without much computation, that there must be a positive root?

By the way, how can we tell quickly, without much computation, that there must be a positive root?

Patrickzcy
2016-10-05 19:40:13

(b^2)-4ac

(b^2)-4ac

themoocow
2016-10-05 19:40:13

discrinimant

discrinimant

cynjinli
2016-10-05 19:40:13

b is positive but c is negative

b is positive but c is negative

Mega-bot
2016-10-05 19:40:19

-a/b in the equation is negative, meaning there must be a negative root and a positive one

-a/b in the equation is negative, meaning there must be a negative root and a positive one

Mudkipswims42
2016-10-05 19:40:19

The product of the roots is negative as the constant is negative, so one must be positive

The product of the roots is negative as the constant is negative, so one must be positive

DPatrick
2016-10-05 19:40:30

Right. Because the constant term is negative and the other two terms are positive, we can quickly see that the discriminant $17^2 - 4(3)(-28)$ is positive.

Right. Because the constant term is negative and the other two terms are positive, we can quickly see that the discriminant $17^2 - 4(3)(-28)$ is positive.

DPatrick
2016-10-05 19:40:41

That tells us that there must be two distinct real roots.

That tells us that there must be two distinct real roots.

DPatrick
2016-10-05 19:40:54

And, one of Vieta's Formulas tells us that the product of the roots is $-\dfrac{28}{3}$, which is negative. So one root must be positive and one root must be negative.

And, one of Vieta's Formulas tells us that the product of the roots is $-\dfrac{28}{3}$, which is negative. So one root must be positive and one root must be negative.

DPatrick
2016-10-05 19:41:06

In general, Vieta's Formulas tell us that, given the quadratic $ax^2 + bx + c = 0$, the sum of the roots is $-\dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$. (See if you can prove this yourself!) Vieta's Formulas also generalize to higher-degree polynomials, but that's a topic for another day.

In general, Vieta's Formulas tell us that, given the quadratic $ax^2 + bx + c = 0$, the sum of the roots is $-\dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$. (See if you can prove this yourself!) Vieta's Formulas also generalize to higher-degree polynomials, but that's a topic for another day.

DPatrick
2016-10-05 19:41:35

2. What is the ones digit of $2017^{2015}$?

2. What is the ones digit of $2017^{2015}$?

Christina26
2016-10-05 19:42:13

find pattern for 7's

find pattern for 7's

letsgomath
2016-10-05 19:42:13

ones digit of 7^2015

ones digit of 7^2015

popcorn1
2016-10-05 19:42:13

Ones digit of $7^2015$

Ones digit of $7^2015$

guluguluga
2016-10-05 19:42:13

We can look at just the 7, because the other parts doesn't affect the ones digi

We can look at just the 7, because the other parts doesn't affect the ones digi

yiqun
2016-10-05 19:42:13

we only need the units digit. So, 2017 turns into 7!

we only need the units digit. So, 2017 turns into 7!

DPatrick
2016-10-05 19:42:20

Great. The units digits of any product only depends on the units digits of the numbers that we're multiplying.

Great. The units digits of any product only depends on the units digits of the numbers that we're multiplying.

DPatrick
2016-10-05 19:42:29

So the units digit of $2017^{2015}$ is the same as the units digit of $7^{2015}$.

So the units digit of $2017^{2015}$ is the same as the units digit of $7^{2015}$.

DPatrick
2016-10-05 19:42:35

That helps, but still I don't think we can compute $7^{2015}$.

That helps, but still I don't think we can compute $7^{2015}$.

ptxpotterhead
2016-10-05 19:42:56

try the first few powers and find a pattern

try the first few powers and find a pattern

csasomachar
2016-10-05 19:42:56

patterns!

patterns!

NewtonFTW
2016-10-05 19:43:13

There are 4 cycles of the units digit when a number that ends in 7 is raised to a power: 7, 9, 3, and 1.

There are 4 cycles of the units digit when a number that ends in 7 is raised to a power: 7, 9, 3, and 1.

BobaFett101
2016-10-05 19:43:13

the units digit cycles with period 4: 7, 9, 3, 1

the units digit cycles with period 4: 7, 9, 3, 1

winnertakeover
2016-10-05 19:43:13

The units digit follows this pattern 7,9,3,1

The units digit follows this pattern 7,9,3,1

letsgomath
2016-10-05 19:43:13

pattern for powers of 7: 7,9,3,1,7,9,3,1,.....

pattern for powers of 7: 7,9,3,1,7,9,3,1,.....

geniusofart
2016-10-05 19:43:13

Note the cycle of 7, 9, 3, 1 of the units digits.

Note the cycle of 7, 9, 3, 1 of the units digits.

caro2be
2016-10-05 19:43:13

the possible units digits are 7, 9, 3, and 1

the possible units digits are 7, 9, 3, and 1

DPatrick
2016-10-05 19:43:20

Right. We can find the units digits of small powers of 7, and see if we can find a pattern.

Right. We can find the units digits of small powers of 7, and see if we can find a pattern.

DPatrick
2016-10-05 19:43:32

\begin{align*}

7^1 &= 7, \\

7^2 &= 49, \\

7^3 &= 243, \\

\vdots

\end{align*}

\begin{align*}

7^1 &= 7, \\

7^2 &= 49, \\

7^3 &= 243, \\

\vdots

\end{align*}

DPatrick
2016-10-05 19:43:38

But remember, we only care about the units digits!

But remember, we only care about the units digits!

NewtonFTW
2016-10-05 19:43:53

7, 9, 3, 1 and repeat

7, 9, 3, 1 and repeat

popcorn1
2016-10-05 19:43:53

7, 9, 3, 1

7, 9, 3, 1

DPatrick
2016-10-05 19:43:58

\[\begin{array}{c|c}

\text{Power of 7} & \text{Units digit} \\ \hline

7^1 & 7 \\

7^2 & 9 \\

7^3 & 3 \\

7^4 & 1

\end{array}\]

\[\begin{array}{c|c}

\text{Power of 7} & \text{Units digit} \\ \hline

7^1 & 7 \\

7^2 & 9 \\

7^3 & 3 \\

7^4 & 1

\end{array}\]

DPatrick
2016-10-05 19:44:10

And then we get a repeat: the units digit of $7^5$ is the same as the units digit of $7 \cdot 7^4$, which is $7 \cdot 1 = 7$.

And then we get a repeat: the units digit of $7^5$ is the same as the units digit of $7 \cdot 7^4$, which is $7 \cdot 1 = 7$.

DPatrick
2016-10-05 19:44:21

\[\begin{array}{c|c}

\text{Power of 7} & \text{Units digit} \\ \hline

7^1 & 7 \\

7^2 & 9 \\

7^3 & 3 \\

7^4 & 1 \\

7^5 & 7

\end{array}\]

\[\begin{array}{c|c}

\text{Power of 7} & \text{Units digit} \\ \hline

7^1 & 7 \\

7^2 & 9 \\

7^3 & 3 \\

7^4 & 1 \\

7^5 & 7

\end{array}\]

DPatrick
2016-10-05 19:44:32

And so on. The pattern of units digits of powers of 7 will repeat in a cycle of 4:

And so on. The pattern of units digits of powers of 7 will repeat in a cycle of 4:

DPatrick
2016-10-05 19:44:35

\[\begin{array}{c|c}

\text{Power of 7} & \text{Units digit} \\ \hline

7^1,7^5,7^9,7^{13},\ldots & 7 \\

7^2,7^6,7^{10},7^{14},\ldots & 9 \\

7^3,7^7,7^{11},7^{15},\ldots & 3 \\

7^4,7^8,7^{12},7^{16},\ldots & 1

\end{array}\]

\[\begin{array}{c|c}

\text{Power of 7} & \text{Units digit} \\ \hline

7^1,7^5,7^9,7^{13},\ldots & 7 \\

7^2,7^6,7^{10},7^{14},\ldots & 9 \\

7^3,7^7,7^{11},7^{15},\ldots & 3 \\

7^4,7^8,7^{12},7^{16},\ldots & 1

\end{array}\]

DPatrick
2016-10-05 19:44:42

We need to figure out which row of the chart contains $7^{2015}$. How do we do that?

We need to figure out which row of the chart contains $7^{2015}$. How do we do that?

supermessi
2016-10-05 19:45:08

So we do 2015 (mod 4)

So we do 2015 (mod 4)

divijleisha
2016-10-05 19:45:08

so we do 2015mod4

so we do 2015mod4

supermathkid
2016-10-05 19:45:08

since that works then it must repeat every 4 so 2015/4 has remainder of 3 so 7^2015 must end in the same thing as 7^3 which is 3 so the answer: 3

since that works then it must repeat every 4 so 2015/4 has remainder of 3 so 7^2015 must end in the same thing as 7^3 which is 3 so the answer: 3

elmaso
2016-10-05 19:45:08

i mean, 2015 is 3 mod 4, and the pattern goes: 7,9,3,1

i mean, 2015 is 3 mod 4, and the pattern goes: 7,9,3,1

NewtonFTW
2016-10-05 19:45:08

We can divide 2015 by 4 to find out how many cycles will be completed, and that simplifies to 503 and 3/4

We can divide 2015 by 4 to find out how many cycles will be completed, and that simplifies to 503 and 3/4

Liopleurodon
2016-10-05 19:45:08

we find what 2015 mod 4 is

we find what 2015 mod 4 is

csasomachar
2016-10-05 19:45:08

find the reaminder when 2105 is divided by 4

find the reaminder when 2105 is divided by 4

seanwang2001
2016-10-05 19:45:08

2015 mod 4

2015 mod 4

DPatrick
2016-10-05 19:45:29

Right. Or, to state it more simply, you perhaps notice that the bottom row of the chart is all the powers of 7 where the exponent is a multiple of 4.

Right. Or, to state it more simply, you perhaps notice that the bottom row of the chart is all the powers of 7 where the exponent is a multiple of 4.

DPatrick
2016-10-05 19:45:39

Similarly:

The top row is all exponents that are 1 more than a multiple of 7.

The 2nd row is all exponents that are 2 more than a multiple of 7.

The 3rd row is all exponents that are 3 more than a multiple of 7.

Similarly:

The top row is all exponents that are 1 more than a multiple of 7.

The 2nd row is all exponents that are 2 more than a multiple of 7.

The 3rd row is all exponents that are 3 more than a multiple of 7.

divijleisha
2016-10-05 19:45:53

2012 is multiple

2012 is multiple

DPatrick
2016-10-05 19:45:55

2012 is a multiple of 4, so 2015 is 3 more than a multiple of 4.

2012 is a multiple of 4, so 2015 is 3 more than a multiple of 4.

DPatrick
2016-10-05 19:46:04

So $7^{2015}$ lies in the third row, and thus its units digit is $\boxed{3}$.

So $7^{2015}$ lies in the third row, and thus its units digit is $\boxed{3}$.

DPatrick
2016-10-05 19:46:27

If you've studying some more number theory, you might know of

If you've studying some more number theory, you might know of

**Euler's Theorem**, which states that, if $a$ and $n$ are relatively prime, then \[ a^{\phi(n)} \equiv 1 \pmod{n}, \] where $\phi(n)$ is the number of positive integers less than $n$ that are relatively prime to $n$.
DPatrick
2016-10-05 19:46:47

In this problem, we can use $n=10$, noting $\phi(10) = 4$ (the numbers less than 10 that are relatively prime to 10 are 1, 3, 7, and 9), and $a=2017$ to get

\[ 2017^4 \equiv 1 \pmod{10}.\]

In this problem, we can use $n=10$, noting $\phi(10) = 4$ (the numbers less than 10 that are relatively prime to 10 are 1, 3, 7, and 9), and $a=2017$ to get

\[ 2017^4 \equiv 1 \pmod{10}.\]

DPatrick
2016-10-05 19:47:13

From there, we get \[2017^{2012} = (2017^4)^{503} \equiv 1^{503} \equiv 1 \pmod{10},\] so the units digit of $2017^{2015}$ is the same as the units digit of $2017^3$, which is the same as the units digit of $7^3 = 243$, so the answer is 3.

From there, we get \[2017^{2012} = (2017^4)^{503} \equiv 1^{503} \equiv 1 \pmod{10},\] so the units digit of $2017^{2015}$ is the same as the units digit of $2017^3$, which is the same as the units digit of $7^3 = 243$, so the answer is 3.

DPatrick
2016-10-05 19:47:35

It's basically the same work we just did, but with fancier language and notation.

It's basically the same work we just did, but with fancier language and notation.

DPatrick
2016-10-05 19:47:48

Onwards:

Onwards:

DPatrick
2016-10-05 19:47:52

3. $(15+i)(15-i) = $

3. $(15+i)(15-i) = $

csasomachar
2016-10-05 19:48:13

difference of squares!

difference of squares!

arowaaron
2016-10-05 19:48:13

difference of squares

difference of squares

pramodmitikiri
2016-10-05 19:48:13

difference of sqaure

difference of sqaure

cinamon
2016-10-05 19:48:13

it's the difference of squares

it's the difference of squares

DPatrick
2016-10-05 19:48:23

Yes, it's very helpful if you know the difference-of-squares formula $(a+b)(a-b) = a^2 - b^2$. This formula comes up over and over and over again in math contests.

Yes, it's very helpful if you know the difference-of-squares formula $(a+b)(a-b) = a^2 - b^2$. This formula comes up over and over and over again in math contests.

DPatrick
2016-10-05 19:48:30

So, $(15+i)(15-i) = 15^2 - i^2$.

So, $(15+i)(15-i) = 15^2 - i^2$.

DPatrick
2016-10-05 19:48:47

What the heck is that $i$ thingy?

What the heck is that $i$ thingy?

pkandarpa
2016-10-05 19:49:08

imaginary number

imaginary number

EpicRuler101
2016-10-05 19:49:08

i = sqrt(-1) so i^2 = -1

i = sqrt(-1) so i^2 = -1

Intellectuality123
2016-10-05 19:49:08

$\sqrt{-1}$

$\sqrt{-1}$

Liopleurodon
2016-10-05 19:49:08

$\sqrt{-1}$

$\sqrt{-1}$

fruitarian
2016-10-05 19:49:08

the square root of negative 1

the square root of negative 1

mathmaniacjulia
2016-10-05 19:49:08

square root of -1

square root of -1

Apurple
2016-10-05 19:49:08

square root of -1

square root of -1

EpicRuler101
2016-10-05 19:49:08

i is the root of -1

i is the root of -1

pkandarpa
2016-10-05 19:49:08

i=sqrt-1

i=sqrt-1

rafa2be
2016-10-05 19:49:08

$\sqrt{-1}$

$\sqrt{-1}$

DPatrick
2016-10-05 19:49:14

$i$ is a special number, called an

$i$ is a special number, called an

*imaginary number*, with the property that $i^2 = -1$. (As you probably know, no__real__number has a negative square, so we have to invent a new number to fill this role.)
Patrickzcy
2016-10-05 19:49:44

(15+i)(15-i)=(15^2)-(i^2)=225-(-1)=226

(15+i)(15-i)=(15^2)-(i^2)=225-(-1)=226

mathgandalfjr
2016-10-05 19:49:44

so the answer to this problem is 225 -(-1) which is 226

so the answer to this problem is 225 -(-1) which is 226

DPatrick
2016-10-05 19:49:47

So now we can finish: $15^2 = 225$ and $i^2 = -1$, so this is just $225 - (-1) = \boxed{226}$.

So now we can finish: $15^2 = 225$ and $i^2 = -1$, so this is just $225 - (-1) = \boxed{226}$.

DPatrick
2016-10-05 19:50:12

That was pretty straightforward, as long as you're familiar with what $i$ means.

That was pretty straightforward, as long as you're familiar with what $i$ means.

DPatrick
2016-10-05 19:50:23

4. A cone of radius $r$ and height $h$ has a volume equal to that of a right circular cylinder having the same height. What is the radius of the right circular cylinder?

4. A cone of radius $r$ and height $h$ has a volume equal to that of a right circular cylinder having the same height. What is the radius of the right circular cylinder?

letsgomath
2016-10-05 19:50:52

make an equation

make an equation

DPatrick
2016-10-05 19:51:19

Right. It may seem obvious, but when the word "equal" is in a word problem, that's a pretty good clue that we'll have two quantities that we can equate using an equation.

Right. It may seem obvious, but when the word "equal" is in a word problem, that's a pretty good clue that we'll have two quantities that we can equate using an equation.

EpicRuler101
2016-10-05 19:51:25

volume of cone = volume of cylinder

volume of cone = volume of cylinder

elmaso
2016-10-05 19:51:25

equal both equations, since it's the same volume

equal both equations, since it's the same volume

DPatrick
2016-10-05 19:51:43

Let's call the radius of the cylinder $x$, since that's what we're trying to find.

Let's call the radius of the cylinder $x$, since that's what we're trying to find.

DPatrick
2016-10-05 19:51:49

What's the volume of the cylinder?

What's the volume of the cylinder?

acegikmoqsuwy2000
2016-10-05 19:52:25

$\pi x^2h$

$\pi x^2h$

cynjinli
2016-10-05 19:52:25

x^2*h*pi

x^2*h*pi

Mathisfun04
2016-10-05 19:52:25

x squared * pi * h

x squared * pi * h

letsgomath
2016-10-05 19:52:25

x^2*h*pi

x^2*h*pi

Liopleurodon
2016-10-05 19:52:25

$x^2h\pi$

$x^2h\pi$

cinamon
2016-10-05 19:52:25

pi x^2 * h

pi x^2 * h

Christina26
2016-10-05 19:52:25

so pix^2h

so pix^2h

DPatrick
2016-10-05 19:52:44

It's $\pi x^2 h$. (The volume of a cylinder is the area of the base times the height.)

It's $\pi x^2 h$. (The volume of a cylinder is the area of the base times the height.)

DPatrick
2016-10-05 19:52:49

And what's the volume of the cone?

And what's the volume of the cone?

acegikmoqsuwy2000
2016-10-05 19:53:27

$\dfrac 13\pi r^2h$

$\dfrac 13\pi r^2h$

AnaT129
2016-10-05 19:53:27

1/3hr^2

1/3hr^2

slackroadia
2016-10-05 19:53:27

1/3 * r^2 * pi * h

1/3 * r^2 * pi * h

NewtonFTW
2016-10-05 19:53:27

1/3pir^2h

1/3pir^2h

DPatrick
2016-10-05 19:53:42

It's $\frac13 \pi r^2 h$. (The volume of a cone is 1/3 times the area of the base times the height.)

It's $\frac13 \pi r^2 h$. (The volume of a cone is 1/3 times the area of the base times the height.)

DPatrick
2016-10-05 19:53:49

The volumes are equal, so we must have $\pi x^2 h = \frac13 \pi r^2 h$.

The volumes are equal, so we must have $\pi x^2 h = \frac13 \pi r^2 h$.

slackroadia
2016-10-05 19:54:04

divide by pi and h

divide by pi and h

NewtonFTW
2016-10-05 19:54:04

We can cancel out pi and h

We can cancel out pi and h

Mathisfun04
2016-10-05 19:54:07

you would cancel out from here

you would cancel out from here

DPatrick
2016-10-05 19:54:11

We can divide by $\pi h$ and we get $x^2 = \frac13 r^2$.

We can divide by $\pi h$ and we get $x^2 = \frac13 r^2$.

winnertakeover
2016-10-05 19:54:21

Take the root

Take the root

DPatrick
2016-10-05 19:54:27

So, taking the square root, we see that $x = \boxed{\frac{1}{\sqrt3}r}$.

So, taking the square root, we see that $x = \boxed{\frac{1}{\sqrt3}r}$.

acegikmoqsuwy2000
2016-10-05 19:54:44

so $x=\dfrac{r\sqrt 3} {3}$

so $x=\dfrac{r\sqrt 3} {3}$

acegikmoqsuwy2000
2016-10-05 19:54:44

if we rationalized, does the answer count as correct?

if we rationalized, does the answer count as correct?

DPatrick
2016-10-05 19:54:48

Absolutely.

Absolutely.

NewtonFTW
2016-10-05 19:55:10

In the contest, will we have to rationalize the denominator

In the contest, will we have to rationalize the denominator

EpicRuler101
2016-10-05 19:55:10

we dont need to rationalize the denominator?

we dont need to rationalize the denominator?

Christina26
2016-10-05 19:55:10

if you dont rationalize, is the answer wrong?

if you dont rationalize, is the answer wrong?

DPatrick
2016-10-05 19:55:25

It should also be correct. (Though I'm not 100% sure of the rules for answer format on the contest.)

It should also be correct. (Though I'm not 100% sure of the rules for answer format on the contest.)

mikebreen
2016-10-05 19:55:31

No, Dave's answer OK.

No, Dave's answer OK.

DPatrick
2016-10-05 19:55:42

There you go -- leaving it the way I did is correct too.

There you go -- leaving it the way I did is correct too.

DPatrick
2016-10-05 19:56:02

Let's continue:

Let's continue:

DPatrick
2016-10-05 19:56:06

5. A palindromic number is one whose digits read the same backward and forward, for example 484 or 909. Which of the following prime numbers is a factor of every four-digit palindromic number?

a. 3 b. 7 c. 11 d. 13 e. There is no such prime number

5. A palindromic number is one whose digits read the same backward and forward, for example 484 or 909. Which of the following prime numbers is a factor of every four-digit palindromic number?

a. 3 b. 7 c. 11 d. 13 e. There is no such prime number

popcorn1
2016-10-05 19:56:40

Let the number be $abba$

Let the number be $abba$

slackroadia
2016-10-05 19:56:40

we have to write out the base 10 form of a palindromic number

we have to write out the base 10 form of a palindromic number

DPatrick
2016-10-05 19:57:02

Right. Our number has to look like ABBA for some digits A and B. (I like using capital letters for digits, to remind me that they're not just random variables.)

Right. Our number has to look like ABBA for some digits A and B. (I like using capital letters for digits, to remind me that they're not just random variables.)

DPatrick
2016-10-05 19:57:13

Plus maybe I like 70s-era Swedish pop bands.

Plus maybe I like 70s-era Swedish pop bands.

elmaso
2016-10-05 19:57:38

divisibility properties

divisibility properties

popcorn1
2016-10-05 19:57:38

Divisibility rule for 3, 7, 11, 13

Divisibility rule for 3, 7, 11, 13

Mathisfun04
2016-10-05 19:57:38

from here, we could generalize and use the divisibility tricks

from here, we could generalize and use the divisibility tricks

DPatrick
2016-10-05 19:58:00

Right. There are easy tests for divisibility by 3 or 11.

Right. There are easy tests for divisibility by 3 or 11.

themoocow
2016-10-05 19:58:29

Obviously a-b+b-a=0 divisible by 11

Obviously a-b+b-a=0 divisible by 11

csasomachar
2016-10-05 19:58:29

it should be 11, by divisibilityy rules

it should be 11, by divisibilityy rules

slackroadia
2016-10-05 19:58:29

It should be divisible by 11 because the rule is A-B+B-A = 0, which is divisible by 11

It should be divisible by 11 because the rule is A-B+B-A = 0, which is divisible by 11

Liopleurodon
2016-10-05 19:58:29

11 works!

11 works!

DPatrick
2016-10-05 19:58:42

Yes! To test for divisibility by 11, we take the "alternating sum" of the digits: the first digit - the second digit + the third digit - the fourth digit. (And if it were a longer number, we'd keep going, alternately adding and subtracting digits.)

Yes! To test for divisibility by 11, we take the "alternating sum" of the digits: the first digit - the second digit + the third digit - the fourth digit. (And if it were a longer number, we'd keep going, alternately adding and subtracting digits.)

DPatrick
2016-10-05 19:58:49

If the alternating sum is 11, then the number is a multiple of 11.

If the alternating sum is 11, then the number is a multiple of 11.

DPatrick
2016-10-05 19:59:00

So for ABBA, the alternating sum is A - B + B - A = 0.

So for ABBA, the alternating sum is A - B + B - A = 0.

DPatrick
2016-10-05 19:59:11

This is a multiple of 11!

This is a multiple of 11!

DPatrick
2016-10-05 19:59:30

So every number of the form ABBA is a multiple of 11. The answer is $\boxed{\text{c. 11}}$.

So every number of the form ABBA is a multiple of 11. The answer is $\boxed{\text{c. 11}}$.

elmaso
2016-10-05 19:59:45

for ABBA to be divisible by 3: 2(A+B) must be divisible by 3

for ABBA to be divisible by 3: 2(A+B) must be divisible by 3

Z_Math404
2016-10-05 19:59:45

2(a+b) isn't necessarily divisible by 3

2(a+b) isn't necessarily divisible by 3

Yanlord
2016-10-05 19:59:45

3 doesn't work e.g. 4774

3 doesn't work e.g. 4774

DPatrick
2016-10-05 19:59:56

Right, it's pretty clear that 3 doesn't work.

Right, it's pretty clear that 3 doesn't work.

DPatrick
2016-10-05 20:00:07

The sum of the digits is 2A + 2B. There's no reason this needs to be a multiple of 3. (It's often not. For example, 2002 is not a multiple of 3.)

The sum of the digits is 2A + 2B. There's no reason this needs to be a multiple of 3. (It's often not. For example, 2002 is not a multiple of 3.)

DPatrick
2016-10-05 20:00:54

There's no easy test for when a number is divisible by 7 or 13, but there are lots of examples of ABBA's that aren't divisible by 7 or 13. (2112 is such an example I believe.)

There's no easy test for when a number is divisible by 7 or 13, but there are lots of examples of ABBA's that aren't divisible by 7 or 13. (2112 is such an example I believe.)

winnertakeover
2016-10-05 20:01:10

$1000a+100b+10b+a=1001a+110b=11(91a+10b)$

$1000a+100b+10b+a=1001a+110b=11(91a+10b)$

DPatrick
2016-10-05 20:01:22

That's a nice algebraic proof that it works!

That's a nice algebraic proof that it works!

DPatrick
2016-10-05 20:01:58

And this gives insight into

$10^0 = 1$ is one

$10^1 = 10$ is one

$10^2 = 100$ is one

$10^3 = 1000$ is one

And so on.

And this gives insight into

*why*the divisibility test for 11 works: it's because the powers of 10 alternate between being 1 more and 1 less than a multiple of 11:$10^0 = 1$ is one

**more**than a multiple of 11 (namely, 0).$10^1 = 10$ is one

**less**than a multiple of 11 (namely, 11).$10^2 = 100$ is one

**more**than a multiple of 11 (namely, 99).$10^3 = 1000$ is one

**less**than a multiple of 11 (namely, 1001).And so on.

DPatrick
2016-10-05 20:02:33

OK, on to what was the second-hardest problem on the contest:

OK, on to what was the second-hardest problem on the contest:

DPatrick
2016-10-05 20:02:37

6. How many solutions are there to the equation $\cos 2x - \sin x = 1$, for $0 \le x < 2\pi$ ($x$ in radians)?

6. How many solutions are there to the equation $\cos 2x - \sin x = 1$, for $0 \le x < 2\pi$ ($x$ in radians)?

DeathLlama9
2016-10-05 20:03:13

Double angle!

Double angle!

Christina26
2016-10-05 20:03:13

cos and sin identities

cos and sin identities

hliu70
2016-10-05 20:03:13

start with double angle?

start with double angle?

Catoc17
2016-10-05 20:03:13

cos(2x) = 1-2*sin^2(x)

cos(2x) = 1-2*sin^2(x)

DPatrick
2016-10-05 20:03:33

Sounds promising. A goal is to relate the cosine and sine terms somehow.

Sounds promising. A goal is to relate the cosine and sine terms somehow.

DPatrick
2016-10-05 20:03:52

We have the identity $\cos 2x = \cos^2 x - \sin^2 x$. (This is the only way that I remember it.)

We have the identity $\cos 2x = \cos^2 x - \sin^2 x$. (This is the only way that I remember it.)

DPatrick
2016-10-05 20:04:08

But we also know that $\cos^2 x = 1 - \sin^2 x$. So $\cos 2x = 1 - 2 \sin^2 x$. How does that help?

But we also know that $\cos^2 x = 1 - \sin^2 x$. So $\cos 2x = 1 - 2 \sin^2 x$. How does that help?

ptes77
2016-10-05 20:04:35

We can solve a quadratic in sinx

We can solve a quadratic in sinx

winnertakeover
2016-10-05 20:04:35

we can solve for $\sin(x)$!

we can solve for $\sin(x)$!

slackroadia
2016-10-05 20:04:35

its easier to solve a trigonometric equation with the same functions

its easier to solve a trigonometric equation with the same functions

cynjinli
2016-10-05 20:04:35

set sin x as a new variable and solve quadratic

set sin x as a new variable and solve quadratic

BobaFett101
2016-10-05 20:04:35

rewrite as a quadratic in sin x

rewrite as a quadratic in sin x

pkandarpa
2016-10-05 20:04:42

get rid of the cosines and only leave the sines

get rid of the cosines and only leave the sines

divijleisha
2016-10-05 20:04:42

we could substitute for cos2x

we could substitute for cos2x

acegikmoqsuwy2000
2016-10-05 20:04:42

substitute in, we get a quadratic in $\sin x$

substitute in, we get a quadratic in $\sin x$

DPatrick
2016-10-05 20:04:53

Good idea! It might be useful to make a new variable $u = \sin x$.

Good idea! It might be useful to make a new variable $u = \sin x$.

DPatrick
2016-10-05 20:05:04

Then, we just found out that $\cos 2x = 1 - 2u^2$, so our original equation becomes

\[

1 - 2u^2 - u = 1.

\]

Then, we just found out that $\cos 2x = 1 - 2u^2$, so our original equation becomes

\[

1 - 2u^2 - u = 1.

\]

acegikmoqsuwy2000
2016-10-05 20:05:27

rewrite as $u(2u+1)=0$

rewrite as $u(2u+1)=0$

NewtonFTW
2016-10-05 20:05:30

-2u^2-u = 0

-2u^2-u = 0

supermathkid
2016-10-05 20:05:30

2u^2 + u= 0

2u^2 + u= 0

DPatrick
2016-10-05 20:05:34

This simplifies to $2u^2 + u = 0$, or $u(2u+1) = 0$.

This simplifies to $2u^2 + u = 0$, or $u(2u+1) = 0$.

slackroadia
2016-10-05 20:05:52

therefore u = 0 or 2u + 1 = 0

therefore u = 0 or 2u + 1 = 0

Liopleurodon
2016-10-05 20:05:52

u=0,u=-1/2

u=0,u=-1/2

DPatrick
2016-10-05 20:05:55

Right. So either $u=0$ or $u = -\frac12$.

Right. So either $u=0$ or $u = -\frac12$.

DPatrick
2016-10-05 20:06:08

And so back to the original question: how many solutions does this give us with $0 \le x < 2\pi$?

And so back to the original question: how many solutions does this give us with $0 \le x < 2\pi$?

yamyamx2
2016-10-05 20:06:17

u=sinx

u=sinx

DPatrick
2016-10-05 20:06:22

Right. We need $\sin x = 0$ or $\sin x = -\frac12$.

Right. We need $\sin x = 0$ or $\sin x = -\frac12$.

ImpossibleCube
2016-10-05 20:06:41

Each gives $2$ solutions, so $\boxed{4}$

Each gives $2$ solutions, so $\boxed{4}$

acegikmoqsuwy2000
2016-10-05 20:06:41

4 solutions, but just to be safe we should check they all work

4 solutions, but just to be safe we should check they all work

DPatrick
2016-10-05 20:06:55

That's right. There are two $x$ for for each: $\sin x = 0$ for $x=0$ or $x=\pi$, and $\sin x = -\frac12$ for $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$.

That's right. There are two $x$ for for each: $\sin x = 0$ for $x=0$ or $x=\pi$, and $\sin x = -\frac12$ for $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$.

DPatrick
2016-10-05 20:07:17

The actual values of $x$ don't matter, of course -- all that matters is that there are 2 values of $u$ that work, and then 2 values of $x$ for each value of $u$. But I also agree that it's a good check of our work.

The actual values of $x$ don't matter, of course -- all that matters is that there are 2 values of $u$ that work, and then 2 values of $x$ for each value of $u$. But I also agree that it's a good check of our work.

DPatrick
2016-10-05 20:07:25

So in total there are $\boxed{4}$ solutions.

So in total there are $\boxed{4}$ solutions.

DPatrick
2016-10-05 20:07:36

Did anybody solve this problem in a different way? (I did!)

Did anybody solve this problem in a different way? (I did!)

Generic_Username
2016-10-05 20:07:42

you can also draw a picture

you can also draw a picture

DPatrick
2016-10-05 20:08:01

Yes!

Yes!

DPatrick
2016-10-05 20:08:06

I wrote the equation as $\cos 2x = \sin x + 1$, and sketched the graphs:

I wrote the equation as $\cos 2x = \sin x + 1$, and sketched the graphs:

DPatrick
2016-10-05 20:08:10

DPatrick
2016-10-05 20:08:30

Even though I drew this graph with software for tonight's Math Jam, you don't need it to be so accurate to see how many times they intersect along $0 \le x < 2\pi$.

Even though I drew this graph with software for tonight's Math Jam, you don't need it to be so accurate to see how many times they intersect along $0 \le x < 2\pi$.

DPatrick
2016-10-05 20:08:38

Note that the graph of $\cos 2x$ (in red) is just the graph of $\cos x$, but compressed horizontally by a factor of 2.

Note that the graph of $\cos 2x$ (in red) is just the graph of $\cos x$, but compressed horizontally by a factor of 2.

DPatrick
2016-10-05 20:08:50

And the graph of $\sin x + 1$ is the graph of $\sin x$ shifted up 1 unit.

And the graph of $\sin x + 1$ is the graph of $\sin x$ shifted up 1 unit.

DPatrick
2016-10-05 20:09:07

We can clearly see 4 intersection points: one at $x=0$, one at $x=\pi$, and then two with $\pi < x < 2\pi$.

We can clearly see 4 intersection points: one at $x=0$, one at $x=\pi$, and then two with $\pi < x < 2\pi$.

DPatrick
2016-10-05 20:09:29

7. Which of the following is closest to $1 + \dfrac{1}{1+\frac{1}{1+\frac{1}{1+1}}}$?

a. $\dfrac{1+\sqrt3}{2}$ b. $\sqrt2$ c. $\dfrac{1+\sqrt5}{2}$ d. $\sqrt{\pi}$ e. $\dfrac{2\pi}{3}$

7. Which of the following is closest to $1 + \dfrac{1}{1+\frac{1}{1+\frac{1}{1+1}}}$?

a. $\dfrac{1+\sqrt3}{2}$ b. $\sqrt2$ c. $\dfrac{1+\sqrt5}{2}$ d. $\sqrt{\pi}$ e. $\dfrac{2\pi}{3}$

caro2be
2016-10-05 20:10:03

Start from the bottom and work up!

Start from the bottom and work up!

DPatrick
2016-10-05 20:10:17

I suppose we could try to simplify it.

I suppose we could try to simplify it.

supermathkid
2016-10-05 20:10:33

8/5 = the big fraction

8/5 = the big fraction

EpicRuler101
2016-10-05 20:10:33

8/5

8/5

BobaFett101
2016-10-05 20:10:33

8/5 is the final result

8/5 is the final result

NewtonFTW
2016-10-05 20:10:33

8/5

8/5

DPatrick
2016-10-05 20:10:35

To start, $\dfrac{1}{1+1}$ is just $\dfrac12$:

\[

1 + \frac{1}{1+\frac{1}{1+\frac12}}.

\]

To start, $\dfrac{1}{1+1}$ is just $\dfrac12$:

\[

1 + \frac{1}{1+\frac{1}{1+\frac12}}.

\]

DPatrick
2016-10-05 20:10:46

Then $1 + \frac12 = \frac32$, so $\frac{1}{1+\frac12} = \frac23$:

\[

1 + \frac{1}{1+\frac23}.

\]

Then $1 + \frac12 = \frac32$, so $\frac{1}{1+\frac12} = \frac23$:

\[

1 + \frac{1}{1+\frac23}.

\]

DPatrick
2016-10-05 20:10:59

Finally, $1+ \frac23 = \frac53$, so $\frac{1}{1+\frac23} = \frac35$, and the whole thing is equal to $\frac85$.

Finally, $1+ \frac23 = \frac53$, so $\frac{1}{1+\frac23} = \frac35$, and the whole thing is equal to $\frac85$.

DPatrick
2016-10-05 20:11:08

Which of those answer choices is closest to $\frac85$ (or $1.6$ if you prefer -- decimals are often easier to work with when approximating)?

Which of those answer choices is closest to $\frac85$ (or $1.6$ if you prefer -- decimals are often easier to work with when approximating)?

hliu70
2016-10-05 20:11:58

C=1.61

C=1.61

ImpossibleCube
2016-10-05 20:11:58

c, it's the golden ratio, which is about 1.618

c, it's the golden ratio, which is about 1.618

mathguy5041
2016-10-05 20:11:58

The choices are 1.36, 1.41, 1.62, 1.78, >2 so obviously c

The choices are 1.36, 1.41, 1.62, 1.78, >2 so obviously c

DPatrick
2016-10-05 20:12:15

Right, we can eyeball these quantities, even without a calculator (which isn't permitted).

Right, we can eyeball these quantities, even without a calculator (which isn't permitted).

DPatrick
2016-10-05 20:12:27

$\sqrt3$ is less than 2, so (a) is less than the average of 1 and 2, which is 1.5.

$\sqrt3$ is less than 2, so (a) is less than the average of 1 and 2, which is 1.5.

DPatrick
2016-10-05 20:12:40

Similarly (b) is a bit less than 1.5.

Similarly (b) is a bit less than 1.5.

DPatrick
2016-10-05 20:12:53

$\sqrt5$ is a little larger than 2, so (c) is a little larger than the average of 1 and 2, which is 1.5. That's promising.

$\sqrt5$ is a little larger than 2, so (c) is a little larger than the average of 1 and 2, which is 1.5. That's promising.

DPatrick
2016-10-05 20:13:00

Indeed, to be a little more precise, $\sqrt5$ is between 2.2 and 2.3, so (c) is between 1.6 and 1.65. That's certainly closer than (a) and (b), which are each less than 1.5.

Indeed, to be a little more precise, $\sqrt5$ is between 2.2 and 2.3, so (c) is between 1.6 and 1.65. That's certainly closer than (a) and (b), which are each less than 1.5.

DPatrick
2016-10-05 20:13:28

(And many of you know more exactly what (c) is, because it's a famous number. I'll say more about that in just a minute.)

(And many of you know more exactly what (c) is, because it's a famous number. I'll say more about that in just a minute.)

DPatrick
2016-10-05 20:13:41

In (d), $\sqrt{\pi} > \sqrt3 > 1.7$, so that's further than (c).

In (d), $\sqrt{\pi} > \sqrt3 > 1.7$, so that's further than (c).

NewtonFTW
2016-10-05 20:13:49

But 2pi/3 is certainly too large

But 2pi/3 is certainly too large

DPatrick
2016-10-05 20:14:02

Right, (e) is easiest: it's more than 2 (since $\pi > 3$), so it's a lot farther than (c).

Right, (e) is easiest: it's more than 2 (since $\pi > 3$), so it's a lot farther than (c).

DPatrick
2016-10-05 20:14:12

So the answer is $\boxed{\text{c}}$.

So the answer is $\boxed{\text{c}}$.

DPatrick
2016-10-05 20:14:38

That was the ugly way to plow through the answer choices. But as many of you noted, there's a more "big-picture" way to think about this problem.

That was the ugly way to plow through the answer choices. But as many of you noted, there's a more "big-picture" way to think about this problem.

yamyamx2
2016-10-05 20:14:57

I just assumed infinitly going on

I just assumed infinitly going on

winnertakeover
2016-10-05 20:14:57

You could instead evaluate it as an infinite series, and use that...

You could instead evaluate it as an infinite series, and use that...

hinna
2016-10-05 20:14:57

What happens when we continue that fraction infinitely?

What happens when we continue that fraction infinitely?

DPatrick
2016-10-05 20:15:11

Suppose that, instead of being a finite fraction, the fraction continued on with this pattern repeated infinitely often. Let's name this new number $z$:

\[

z = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}.

\]

Suppose that, instead of being a finite fraction, the fraction continued on with this pattern repeated infinitely often. Let's name this new number $z$:

\[

z = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}.

\]

DPatrick
2016-10-05 20:15:25

This should be very close to the given number, because the stuff we've added is buried under several layers of denominator. (Each "layer" makes the overall denominator a little bit bigger, but the deeper we go, the less and less effect it has on the overall quantity.)

This should be very close to the given number, because the stuff we've added is buried under several layers of denominator. (Each "layer" makes the overall denominator a little bit bigger, but the deeper we go, the less and less effect it has on the overall quantity.)

bobjoe123
2016-10-05 20:15:50

$z = 1 + \frac{1}{z}$

$z = 1 + \frac{1}{z}$

csasomachar
2016-10-05 20:15:50

z=1+(1/z)

z=1+(1/z)

ImpossibleCube
2016-10-05 20:15:50

$z = 1 + \frac{1}{z}$

$z = 1 + \frac{1}{z}$

AnaT129
2016-10-05 20:15:50

since it is infinite, you can substitute z in, getting 1+(1/z)

since it is infinite, you can substitute z in, getting 1+(1/z)

winnertakeover
2016-10-05 20:15:50

$z=1+\frac{1}{z}$

$z=1+\frac{1}{z}$

mathgandalfjr
2016-10-05 20:15:50

z=1+(1/z)

z=1+(1/z)

DPatrick
2016-10-05 20:15:57

Aha! Notice that $z$ is "self-similar" -- everything other than the first two 1's (the initial "1+" and the 1 in the numerator) is just another copy of $z$!

Aha! Notice that $z$ is "self-similar" -- everything other than the first two 1's (the initial "1+" and the 1 in the numerator) is just another copy of $z$!

DPatrick
2016-10-05 20:16:09

That is, $z = 1+\dfrac{1}{z}$.

That is, $z = 1+\dfrac{1}{z}$.

DPatrick
2016-10-05 20:16:16

This is an equation that we can solve!

This is an equation that we can solve!

popcorn1
2016-10-05 20:16:44

z^2 = z+1

z^2 = z+1

csasomachar
2016-10-05 20:16:44

z= (1+sqrt5)/2

z= (1+sqrt5)/2

supermathkid
2016-10-05 20:16:44

z^2 - z -1 =0

z^2 - z -1 =0

Shaktiala
2016-10-05 20:16:44

Z^2 = z + 1

Z^2 = z + 1

DPatrick
2016-10-05 20:16:50

Multiplying by $z$ and simplifying gives us $z^2 - z - 1 = 0$.

Multiplying by $z$ and simplifying gives us $z^2 - z - 1 = 0$.

DPatrick
2016-10-05 20:16:59

This is a quadratic, so there are two solutions: $z = \dfrac{1 \pm \sqrt{5}}{2}$.

This is a quadratic, so there are two solutions: $z = \dfrac{1 \pm \sqrt{5}}{2}$.

DPatrick
2016-10-05 20:17:09

But we know $z$ is positive, so we exclude the negative solution and we conclude $z = \dfrac{1+\sqrt5}{2}$.

But we know $z$ is positive, so we exclude the negative solution and we conclude $z = \dfrac{1+\sqrt5}{2}$.

Liopleurodon
2016-10-05 20:17:15

That's the golden ratio! C.

That's the golden ratio! C.

DPatrick
2016-10-05 20:17:41

That's answer choice (c)! While this certainly doesn't

That's answer choice (c)! While this certainly doesn't

*prove*that (c) is the closest, because our $z$ is not exactly the quantity we started with, it's pretty strong evidence.
DPatrick
2016-10-05 20:17:57

As you may know, this number $\dfrac{1+\sqrt5}{2}$ is a pretty famous number in mathematics.

It's called the

As you may know, this number $\dfrac{1+\sqrt5}{2}$ is a pretty famous number in mathematics.

It's called the

**golden ratio**and is often denoted by the Greek letter $\varphi$. Indeed, this ratio was known to the ancient Greeks. A rectangle whose sides are in the golden ratio was widely considered to be aesthetically pleasing:
DPatrick
2016-10-05 20:18:07

DPatrick
2016-10-05 20:18:23

The geometric property here is that if we cut off a square from one side of this rectangle, what remains is a smaller rectangle whose sides also satisfy the golden ratio:

The geometric property here is that if we cut off a square from one side of this rectangle, what remains is a smaller rectangle whose sides also satisfy the golden ratio:

DPatrick
2016-10-05 20:18:29

DPatrick
2016-10-05 20:18:48

And this pattern can be repeated indefinitely, to form a pleasing spiral:

And this pattern can be repeated indefinitely, to form a pleasing spiral:

DPatrick
2016-10-05 20:18:52

DPatrick
2016-10-05 20:19:11

We have a bunch of white squares, and the small red rectangle is still in the golden ratio, so we could continue the process forever.

We have a bunch of white squares, and the small red rectangle is still in the golden ratio, so we could continue the process forever.

EpicRuler101
2016-10-05 20:19:19

Fibonnaci Sequence!

Fibonnaci Sequence!

BCharwhis50
2016-10-05 20:19:19

this relates to the fibonacci sequence

this relates to the fibonacci sequence

DPatrick
2016-10-05 20:19:35

Indeed, the golden ratio is very closely related to the Fibonacci numbers!

Indeed, the golden ratio is very closely related to the Fibonacci numbers!

DPatrick
2016-10-05 20:20:10

That is a really interesting topic to explore deeply -- I encourage you to investigate further on your own!

That is a really interesting topic to explore deeply -- I encourage you to investigate further on your own!

DPatrick
2016-10-05 20:20:16

Let's move on:

Let's move on:

DPatrick
2016-10-05 20:20:20

8. A right triangle has legs $a$ and $b$, hypotenuse $c$ and perimeter $2d$. Find $\sqrt{d(d-a)(d-b)(d-c)}$.

8. A right triangle has legs $a$ and $b$, hypotenuse $c$ and perimeter $2d$. Find $\sqrt{d(d-a)(d-b)(d-c)}$.

supermathkid
2016-10-05 20:20:49

heron's formula

heron's formula

ballislife21
2016-10-05 20:20:49

This looks like herons formula

This looks like herons formula

slackroadia
2016-10-05 20:20:49

this is herons formula

this is herons formula

hliu70
2016-10-05 20:20:49

heron's formula!

heron's formula!

mathgandalfjr
2016-10-05 20:20:49

isn't that the Herons formula or something

isn't that the Herons formula or something

Catoc17
2016-10-05 20:20:49

Heron's Formula

Heron's Formula

Generic_Username
2016-10-05 20:20:49

heron

heron

ImpossibleCube
2016-10-05 20:20:49

It's heron's formula

It's heron's formula

letsgomath
2016-10-05 20:20:49

heron's formula!

heron's formula!

NewtonFTW
2016-10-05 20:20:49

That is Heron's Formula

That is Heron's Formula

math129
2016-10-05 20:20:49

Looks like heron's

Looks like heron's

hwl0304
2016-10-05 20:20:49

heron's

heron's

sas4
2016-10-05 20:20:49

Heron's Formula!

Heron's Formula!

DPatrick
2016-10-05 20:20:58

Yes. This is actually a famous formula involving triangles, called

Yes. This is actually a famous formula involving triangles, called

**Heron's Formula**.
DPatrick
2016-10-05 20:21:19

Heron's Formula states that this quantity equals the area of the triangle. (You often see Heron's Formula with an $s$, for

Heron's Formula states that this quantity equals the area of the triangle. (You often see Heron's Formula with an $s$, for

*semiperimeter*.)
divijleisha
2016-10-05 20:21:33

area of a triangle

area of a triangle

NewtonFTW
2016-10-05 20:21:33

Therefore, the formula equals to the area of the triangle

Therefore, the formula equals to the area of the triangle

DPatrick
2016-10-05 20:21:39

Right: our answer is just the area of the triangle.

Right: our answer is just the area of the triangle.

NewtonFTW
2016-10-05 20:21:58

ab/2

ab/2

bobjoe123
2016-10-05 20:21:58

ab/2

ab/2

winnertakeover
2016-10-05 20:21:58

it's ab/2

it's ab/2

cinamon
2016-10-05 20:21:58

so a*b/2

so a*b/2

Intellectuality123
2016-10-05 20:21:58

it's essentially asking for the area of the triangle which is also equal to ab/2

it's essentially asking for the area of the triangle which is also equal to ab/2

acegikmoqsuwy2000
2016-10-05 20:21:58

$\dfrac{ab}{2}$

$\dfrac{ab}{2}$

Catoc17
2016-10-05 20:21:58

which is 1/2 * a * b

which is 1/2 * a * b

hinna
2016-10-05 20:21:58

Or $ab/2$

Or $ab/2$

supermathkid
2016-10-05 20:21:58

a * b /2

a * b /2

Shaktiala
2016-10-05 20:21:58

The area is (a*b)/2

The area is (a*b)/2

DPatrick
2016-10-05 20:22:02

So, since this is a right triangle with legs $a$ and $b$, the area (and thus our answer) is $\boxed{\dfrac12ab}$.

So, since this is a right triangle with legs $a$ and $b$, the area (and thus our answer) is $\boxed{\dfrac12ab}$.

DPatrick
2016-10-05 20:22:27

Heron's Formula is not that difficult to prove: it's mostly just using the Pythagorean Theorem and then plowing through a bunch of algebra. (I say "not that difficult" in the present day -- it's remarkable that the ancient Greeks knew how to prove this over 2000 years ago.)

Heron's Formula is not that difficult to prove: it's mostly just using the Pythagorean Theorem and then plowing through a bunch of algebra. (I say "not that difficult" in the present day -- it's remarkable that the ancient Greeks knew how to prove this over 2000 years ago.)

DPatrick
2016-10-05 20:22:44

Next up is what turned out to be the hardest problem on the contest:

Next up is what turned out to be the hardest problem on the contest:

DPatrick
2016-10-05 20:22:50

9. A perfect number is a number greater than 1 that is equal to the sum of its proper factors/divisors (including 1, but not including itself). Example: $6 = 1 + 2 + 3$. How many perfect numbers are less than $10{,}000$?

9. A perfect number is a number greater than 1 that is equal to the sum of its proper factors/divisors (including 1, but not including itself). Example: $6 = 1 + 2 + 3$. How many perfect numbers are less than $10{,}000$?

DPatrick
2016-10-05 20:23:09

6 is the smallest perfect number: as in the given example, its proper divisors are 1, 2, 3, and $1+2+3=6$.

6 is the smallest perfect number: as in the given example, its proper divisors are 1, 2, 3, and $1+2+3=6$.

ballislife21
2016-10-05 20:23:14

We could find a pattern

We could find a pattern

DPatrick
2016-10-05 20:23:28

Indeed. Does anyone know the next perfect number?

Indeed. Does anyone know the next perfect number?

popcorn1
2016-10-05 20:23:42

The next is 28

The next is 28

j2005
2016-10-05 20:23:42

As is 28

As is 28

letsgomath
2016-10-05 20:23:42

28

28

ImpossibleCube
2016-10-05 20:23:42

28

28

mathgandalfjr
2016-10-05 20:23:42

28

28

acegikmoqsuwy2000
2016-10-05 20:23:42

28

28

hwl0304
2016-10-05 20:23:42

28

28

Catoc17
2016-10-05 20:23:42

28

28

popcorn1
2016-10-05 20:23:42

$28$

$28$

j2005
2016-10-05 20:23:42

28

28

ptes77
2016-10-05 20:23:42

28

28

Generic_Username
2016-10-05 20:23:42

28

28

DPatrick
2016-10-05 20:23:48

The next one is 28. Its prime factorization is $2^2 \cdot 7$, so its proper divisors are 1, 2, 4, 7, and 14, and indeed $1+2+4+7+14 = 28$.

The next one is 28. Its prime factorization is $2^2 \cdot 7$, so its proper divisors are 1, 2, 4, 7, and 14, and indeed $1+2+4+7+14 = 28$.

hwl0304
2016-10-05 20:24:10

there is no reasonable pattern, it's just you know it or you don't

there is no reasonable pattern, it's just you know it or you don't

DPatrick
2016-10-05 20:24:21

Actually, there

Actually, there

*is*a pattern to the perfect numbers!
DPatrick
2016-10-05 20:24:38

It's not easy, though.

It's not easy, though.

DPatrick
2016-10-05 20:25:04

We have $6 = 2 \cdot 3$ and $28 = 2^2 \cdot 7$. Do you see any pattern in those prime factorizations?

We have $6 = 2 \cdot 3$ and $28 = 2^2 \cdot 7$. Do you see any pattern in those prime factorizations?

hinna
2016-10-05 20:25:20

Both have 2

Both have 2

sathvi_k
2016-10-05 20:25:20

powers of 2

powers of 2

scienceFTW
2016-10-05 20:25:33

the 2s are increasing by a power of 1

the 2s are increasing by a power of 1

hinna
2016-10-05 20:25:33

Also 2^2-1=3

Also 2^2-1=3

mathgandalfjr
2016-10-05 20:25:43

yes, 3 or 7 plus 1 is a power of 2

yes, 3 or 7 plus 1 is a power of 2

DPatrick
2016-10-05 20:25:58

Indeed. It's a power of 2, times a prime that is one less than the next power of 2.

Indeed. It's a power of 2, times a prime that is one less than the next power of 2.

ptes77
2016-10-05 20:26:09

Catoc17
2016-10-05 20:26:18

(2^(n-1))(2^n -1)

(2^(n-1))(2^n -1)

ninjataco
2016-10-05 20:26:25

of the form $2^k (2^{k+1} - 1)$

of the form $2^k (2^{k+1} - 1)$

DPatrick
2016-10-05 20:26:38

Right: the examples we've found are of the form $2^n(2^{n+1}-1)$ for some $n$. Can we say any more?

Right: the examples we've found are of the form $2^n(2^{n+1}-1)$ for some $n$. Can we say any more?

EpicRuler101
2016-10-05 20:26:51

so the next would be 2^3 x 15

so the next would be 2^3 x 15

DPatrick
2016-10-05 20:27:07

Right, we can try $2^3 \cdot 15 = 120$.

Right, we can try $2^3 \cdot 15 = 120$.

winnertakeover
2016-10-05 20:27:10

$2^{n+1}-1$ must be prime.

$2^{n+1}-1$ must be prime.

DPatrick
2016-10-05 20:27:23

Right: this doesn't work because 15 isn't prime. This has too many factors.

Right: this doesn't work because 15 isn't prime. This has too many factors.

popcorn1
2016-10-05 20:27:42

n = 4, then 496

n = 4, then 496

divijleisha
2016-10-05 20:27:42

and then would be 2^4*31

and then would be 2^4*31

EpicDragonSlayr
2016-10-05 20:27:42

then 2^4*31

then 2^4*31

ImpossibleCube
2016-10-05 20:27:42

$2^4 *31$ works

$2^4 *31$ works

agbdmrbirdyface
2016-10-05 20:27:42

the next is actually 496

the next is actually 496

DPatrick
2016-10-05 20:27:52

Next is 496. Its prime factorization is $2^4 \cdot 31$, so its proper divisors are 1, 2, 4, 8, 16, 31, 62, 124, and 248. Indeed $1+2+4+8+16+31+62+124+248 = 496$.

Next is 496. Its prime factorization is $2^4 \cdot 31$, so its proper divisors are 1, 2, 4, 8, 16, 31, 62, 124, and 248. Indeed $1+2+4+8+16+31+62+124+248 = 496$.

DPatrick
2016-10-05 20:28:35

Before we go on...why does this work? Why is $2^{n-1}(2^n-1)$ perfect if $2^n-1$ is prime?

Before we go on...why does this work? Why is $2^{n-1}(2^n-1)$ perfect if $2^n-1$ is prime?

DPatrick
2016-10-05 20:29:24

I'm going to write it as $2^{p-1}(2^p-1)$ where $p$ is prime and $2^p-1$ is prime, because $2^p-1$ can only be prime if $p$ is prime (it's a neat exercise to try to prove this yourself).

I'm going to write it as $2^{p-1}(2^p-1)$ where $p$ is prime and $2^p-1$ is prime, because $2^p-1$ can only be prime if $p$ is prime (it's a neat exercise to try to prove this yourself).

Derive_Foiler
2016-10-05 20:29:33

add all factors

add all factors

Derive_Foiler
2016-10-05 20:29:33

try adding all factors to get the same number again

try adding all factors to get the same number again

DPatrick
2016-10-05 20:29:46

Right, we can just list the factors for this number!

Right, we can just list the factors for this number!

DPatrick
2016-10-05 20:30:08

If we set $q = 2^p -1$, then the proper factors are just $1, 2, 4, \ldots 2^{p-1}$ and $q, 2q, 4q, \ldots 2^{p-2}q$.

If we set $q = 2^p -1$, then the proper factors are just $1, 2, 4, \ldots 2^{p-1}$ and $q, 2q, 4q, \ldots 2^{p-2}q$.

NewtonFTW
2016-10-05 20:30:23

1 + 2 + 4...+ 2^n-1 = 2^n-1

1 + 2 + 4...+ 2^n-1 = 2^n-1

DPatrick
2016-10-05 20:30:44

Right. Moreover, we've got

\begin{align*}

1 + 2 + \cdots + 2^{p-1} &= 2^p - 1, \\

q + 2q + \cdots + 2^{p-2}q &= q(2^{p-1}-1).

\end{align*}

Right. Moreover, we've got

*two*geometric series worth of factors:\begin{align*}

1 + 2 + \cdots + 2^{p-1} &= 2^p - 1, \\

q + 2q + \cdots + 2^{p-2}q &= q(2^{p-1}-1).

\end{align*}

popcorn1
2016-10-05 20:30:58

they add up!

they add up!

DPatrick
2016-10-05 20:31:06

That first sum is just $q$ itself, so the divisors sum to $q + q(2^{p-1}-1) = 2^{p-1}q$, which is the number we started with!

That first sum is just $q$ itself, so the divisors sum to $q + q(2^{p-1}-1) = 2^{p-1}q$, which is the number we started with!

DPatrick
2016-10-05 20:31:19

So we've proved a theorem:

If $2^p-1$ is prime, then $2^{p-1}(2^p-1)$ is a perfect number.

So we've proved a theorem:

If $2^p-1$ is prime, then $2^{p-1}(2^p-1)$ is a perfect number.

Derive_Foiler
2016-10-05 20:31:31

and we're done! but how do we know that these are the $only$ perfect numbers?

and we're done! but how do we know that these are the $only$ perfect numbers?

DPatrick
2016-10-05 20:32:06

Well, we don't based just on what we've done above -- that's a good point. But it is known that these are the only possibilities for even perfect numbers.

Well, we don't based just on what we've done above -- that's a good point. But it is known that these are the only possibilities for even perfect numbers.

DPatrick
2016-10-05 20:32:12

In fact, the theorem goes the other way too: all even perfect numbers are of this form. This is called the

In fact, the theorem goes the other way too: all even perfect numbers are of this form. This is called the

**Euclid-Euler Theorem**. (Euclid proved the direction that we just proved, over 2000 years ago! Euler proved the converse in the 18th century.)
DPatrick
2016-10-05 20:32:40

So let's see if we can generate any other perfect numbers under 10000 using this theorem. Recall we've already got:

\begin{align*}

6 &= 2 \cdot 3 = 2^1(2^2-1) \\

28 &= 2^2 \cdot 7 = 2^2(2^3-1) \\

496 &= 2^4 \cdot 31 = 2^4(2^5-1)

\end{align*}

So let's see if we can generate any other perfect numbers under 10000 using this theorem. Recall we've already got:

\begin{align*}

6 &= 2 \cdot 3 = 2^1(2^2-1) \\

28 &= 2^2 \cdot 7 = 2^2(2^3-1) \\

496 &= 2^4 \cdot 31 = 2^4(2^5-1)

\end{align*}

DPatrick
2016-10-05 20:32:50

What's next to check?

What's next to check?

supermathkid
2016-10-05 20:32:58

8128

8128

mathgandalfjr
2016-10-05 20:32:58

and 8128

and 8128

mathguy5041
2016-10-05 20:32:58

$127*64=8128$

$127*64=8128$

NewtonFTW
2016-10-05 20:33:01

2^6 * 127

2^6 * 127

Mathisfun04
2016-10-05 20:33:03

8128?

8128?

DPatrick
2016-10-05 20:33:08

We'll try $p=7$, giving us the number $2^6(2^7-1)$.

We'll try $p=7$, giving us the number $2^6(2^7-1)$.

DPatrick
2016-10-05 20:33:13

This is $64 \cdot 127 = 8128$. (You don't actually need to compute it -- it's enough for our purposes to see that it's less than 10000).

This is $64 \cdot 127 = 8128$. (You don't actually need to compute it -- it's enough for our purposes to see that it's less than 10000).

DPatrick
2016-10-05 20:33:23

(We do need to check that 127 is prime, and indeed it is.)

(We do need to check that 127 is prime, and indeed it is.)

csasomachar
2016-10-05 20:33:34

these are the only 4

these are the only 4

divijleisha
2016-10-05 20:33:34

so the answer is 4

so the answer is 4

DPatrick
2016-10-05 20:33:41

The next possibility on our list would be $2^{10}(2^{11}-1) = 1024 \cdot 2047$, but that's already way too big. And it's not perfect, since $2047 = 23 \cdot 89$ isn't prime.

The next possibility on our list would be $2^{10}(2^{11}-1) = 1024 \cdot 2047$, but that's already way too big. And it's not perfect, since $2047 = 23 \cdot 89$ isn't prime.

DPatrick
2016-10-05 20:33:51

So, the only perfect numbers less than 10000 are 6, 28, 496, and 8128. There are $\boxed{4}$ of them.

So, the only perfect numbers less than 10000 are 6, 28, 496, and 8128. There are $\boxed{4}$ of them.

Hero_of_Hyrule
2016-10-05 20:34:01

Are there any odd perfect numbers?

Are there any odd perfect numbers?

DPatrick
2016-10-05 20:34:17

This is one of the great unsolved problems in mathematics. Nobody knows if there any odd perfect numbers. Our theorem only works for even perfect numbers. No one has found an odd perfect number yet, but no one has been able to prove that they don't exist.

This is one of the great unsolved problems in mathematics. Nobody knows if there any odd perfect numbers. Our theorem only works for even perfect numbers. No one has found an odd perfect number yet, but no one has been able to prove that they don't exist.

DPatrick
2016-10-05 20:34:45

To add one more point: you may have heard of

To add one more point: you may have heard of

**Mersenne primes**: these are primes of the form $2^p - 1$ where $p$ is itself prime. So the Euclid-Euler Theorem tells us that all even perfect numbers are Mersenne primes times a power of 2.
hliu70
2016-10-05 20:35:05

do we know there aren't any under 10,000?

do we know there aren't any under 10,000?

DPatrick
2016-10-05 20:35:27

Yes -- we know in fact that there are no odd perfect numbers less than $10^{1500}$.

Yes -- we know in fact that there are no odd perfect numbers less than $10^{1500}$.

DPatrick
2016-10-05 20:35:39

But there might be a bigger one out there somewhere!

But there might be a bigger one out there somewhere!

DPatrick
2016-10-05 20:36:00

It is also unknown if there are infinitely many perfect numbers.

It is also unknown if there are infinitely many perfect numbers.

DPatrick
2016-10-05 20:36:05

Only 49 of them are known so far.

Only 49 of them are known so far.

DPatrick
2016-10-05 20:36:15

The largest known Mersenne prime is $2^{74207281}-1$, which means that the largest known perfect number is $2^{74207280}(2^{74207281}-1)$. This number has over 44 million digits!

The largest known Mersenne prime is $2^{74207281}-1$, which means that the largest known perfect number is $2^{74207280}(2^{74207281}-1)$. This number has over 44 million digits!

acegikmoqsuwy2000
2016-10-05 20:36:29

GIMPS!

GIMPS!

DPatrick
2016-10-05 20:36:39

Yes: the Great Internet Mersenne Prime Search

Yes: the Great Internet Mersenne Prime Search

slackroadia
2016-10-05 20:36:46

shouldnt there be infinite mersenne primes if there are infinite primes?

shouldnt there be infinite mersenne primes if there are infinite primes?

DPatrick
2016-10-05 20:36:53

That's exactly right: these two questions are equivalent.

That's exactly right: these two questions are equivalent.

DPatrick
2016-10-05 20:37:01

Anyway, it's getting late, so on to #10:

Anyway, it's getting late, so on to #10:

DPatrick
2016-10-05 20:37:05

10. Which of the following is largest?

a. $2016^{2016}$ b. $2016!$ c. $20^{(16^{20})}$ d. $16^{(20^{20})}$ e. $20^{(20^{16})}$

10. Which of the following is largest?

a. $2016^{2016}$ b. $2016!$ c. $20^{(16^{20})}$ d. $16^{(20^{20})}$ e. $20^{(20^{16})}$

NewtonFTW
2016-10-05 20:37:32

Eliminate b easily

Eliminate b easily

cinamon
2016-10-05 20:37:32

I can see already that a>b

I can see already that a>b

scienceFTW
2016-10-05 20:37:32

We can immediately rule out B as A is larger

We can immediately rule out B as A is larger

somepersonoverhere
2016-10-05 20:37:32

rule out b, and look at logarithms of the answer choices

rule out b, and look at logarithms of the answer choices

DPatrick
2016-10-05 20:37:45

Let start with the easiest comparison. We can easily compare (a) and (b).

Let start with the easiest comparison. We can easily compare (a) and (b).

DPatrick
2016-10-05 20:37:52

Both are the product of 2016 numbers. But the numbers in (a)'s product are all 2016's, whereas almost all of the numbers in (b)'s product are less than 2016.

Both are the product of 2016 numbers. But the numbers in (a)'s product are all 2016's, whereas almost all of the numbers in (b)'s product are less than 2016.

DPatrick
2016-10-05 20:38:00

So (a) is greater than (b). Thus (b) cannot be the answer. Let's delete it from our choices:

So (a) is greater than (b). Thus (b) cannot be the answer. Let's delete it from our choices:

DPatrick
2016-10-05 20:38:04

10. Which of the following is largest?

a. $2016^{2016}$ c. $20^{(16^{20})}$ d. $16^{(20^{20})}$ e. $20^{(20^{16})}$

10. Which of the following is largest?

a. $2016^{2016}$ c. $20^{(16^{20})}$ d. $16^{(20^{20})}$ e. $20^{(20^{16})}$

DPatrick
2016-10-05 20:38:18

(a) looks a lot different than the others, so let's see if we can rule it in or out.

(a) looks a lot different than the others, so let's see if we can rule it in or out.

DPatrick
2016-10-05 20:38:30

About how big is (a)? Can we estimate its number of digits?

About how big is (a)? Can we estimate its number of digits?

Derive_Foiler
2016-10-05 20:39:08

2016 ~ 2*1000, so 3000 digits? 600?

2016 ~ 2*1000, so 3000 digits? 600?

Derive_Foiler
2016-10-05 20:39:08

6000?

6000?

ptes77
2016-10-05 20:39:08

6666

6666

to_chicken
2016-10-05 20:39:08

6400 ish

6400 ish

jasonhu4
2016-10-05 20:39:08

less than 8000

less than 8000

DPatrick
2016-10-05 20:39:13

Pretty good estimates.

Pretty good estimates.

DPatrick
2016-10-05 20:39:35

An often-useful estimating tool is $2^{10} \approx 10^3$. So we get that $$2016^{10} \approx (2 \cdot 10)^{10} \approx 10^{33}.$$

An often-useful estimating tool is $2^{10} \approx 10^3$. So we get that $$2016^{10} \approx (2 \cdot 10)^{10} \approx 10^{33}.$$

DPatrick
2016-10-05 20:40:14

So, $2016^{2016} \approx (10^{33})^{201.6} \approx 10^{6653}$. Therefore, (a) has approximately 6654 digits.

So, $2016^{2016} \approx (10^{33})^{201.6} \approx 10^{6653}$. Therefore, (a) has approximately 6654 digits.

DPatrick
2016-10-05 20:40:30

(This is a pretty good approximation. I put it into my computer and it actually has 6662 digits.)

(This is a pretty good approximation. I put it into my computer and it actually has 6662 digits.)

DPatrick
2016-10-05 20:40:43

How does this compare to the number of digits for any of (c) through (e)?

How does this compare to the number of digits for any of (c) through (e)?

acegikmoqsuwy2000
2016-10-05 20:40:54

6654 is way less than the n umber of digits of c,d,e

6654 is way less than the n umber of digits of c,d,e

ptes77
2016-10-05 20:40:54

A whole lot smaller

A whole lot smaller

somepersonoverhere
2016-10-05 20:40:54

It pales in comparison

It pales in comparison

DPatrick
2016-10-05 20:41:05

Not even close. (c),(d),(e) all have a base larger than 10, so the number of digits is larger than the value of the exponent.

Not even close. (c),(d),(e) all have a base larger than 10, so the number of digits is larger than the value of the exponent.

DPatrick
2016-10-05 20:41:18

But those exponents are huge -- they're all at least 20-digit numbers themselves, so each of (c),(d),(e) have a number of digits that itself is at least a 20-digit number. (Way more than the 4-digit number of digits given by 6654.)

But those exponents are huge -- they're all at least 20-digit numbers themselves, so each of (c),(d),(e) have a number of digits that itself is at least a 20-digit number. (Way more than the 4-digit number of digits given by 6654.)

DPatrick
2016-10-05 20:41:34

So the answer must be one of (c), (d), or (e).

So the answer must be one of (c), (d), or (e).

DPatrick
2016-10-05 20:41:38

10. Which of the following is largest?

c. $20^{(16^{20})}$ d. $16^{(20^{20})}$ e. $20^{(20^{16})}$

10. Which of the following is largest?

c. $20^{(16^{20})}$ d. $16^{(20^{20})}$ e. $20^{(20^{16})}$

DPatrick
2016-10-05 20:42:22

Before we try to compute anything, do you have intuition? These all use the same three numbers (a 16 and two 20's). Do we want bigger numbers in the base or in the exponent?

Before we try to compute anything, do you have intuition? These all use the same three numbers (a 16 and two 20's). Do we want bigger numbers in the base or in the exponent?

ninjataco
2016-10-05 20:42:46

exponent

exponent

Liopleurodon
2016-10-05 20:42:46

exponent

exponent

popcorn1
2016-10-05 20:42:46

exponent

exponent

Z_Math404
2016-10-05 20:42:46

exponent

exponent

mathgandalfjr
2016-10-05 20:42:46

the bigger exponent will do so I think d is the answer

the bigger exponent will do so I think d is the answer

supermathkid
2016-10-05 20:42:46

exponent

exponent

divijleisha
2016-10-05 20:42:46

exponent

exponent

Intellectuality123
2016-10-05 20:42:46

exponent

exponent

UrInvalid
2016-10-05 20:42:46

exponent

exponent

DPatrick
2016-10-05 20:42:56

The intuitive idea is that, if the numbers are big enough, we're generally better off making the exponent as big as possible.

The intuitive idea is that, if the numbers are big enough, we're generally better off making the exponent as big as possible.

DPatrick
2016-10-05 20:43:01

To look at some simple examples, $3^5 = 243$ is larger than $5^3 = 125$, and $2^{10} = 1024$ is larger than $10^2 = 100$.

To look at some simple examples, $3^5 = 243$ is larger than $5^3 = 125$, and $2^{10} = 1024$ is larger than $10^2 = 100$.

winnertakeover
2016-10-05 20:43:11

d) could be the largest.

d) could be the largest.

Apurple
2016-10-05 20:43:11

the answer is d

the answer is d

DPatrick
2016-10-05 20:43:32

This intuition should lead us to answer choice (d) -- it puts the smaller number 16 in the base, and the bigger numbers in both exponents.

This intuition should lead us to answer choice (d) -- it puts the smaller number 16 in the base, and the bigger numbers in both exponents.

jasonhu4
2016-10-05 20:43:45

this is hardly rigorous

this is hardly rigorous

DPatrick
2016-10-05 20:43:52

I agree. Let's see if we can approximate it better.

I agree. Let's see if we can approximate it better.

DPatrick
2016-10-05 20:44:09

A lot of you earlier suggested using logarithms. That's a good idea.

A lot of you earlier suggested using logarithms. That's a good idea.

DPatrick
2016-10-05 20:44:17

We have the helpful rule that $\log(a^b) = b\log(a)$.

We have the helpful rule that $\log(a^b) = b\log(a)$.

DPatrick
2016-10-05 20:44:23

So we can replace our question with:

So we can replace our question with:

DPatrick
2016-10-05 20:44:27

10. Which of the following is largest?

c. $16^{20} \log 20$ d. $20^{20} \log 16$ e. $20^{16} \log 20$

10. Which of the following is largest?

c. $16^{20} \log 20$ d. $20^{20} \log 16$ e. $20^{16} \log 20$

DPatrick
2016-10-05 20:44:51

Any why not use that rule again?

Any why not use that rule again?

DPatrick
2016-10-05 20:46:28

$\log(16^{20} \log 20) = \log(16^{20}) + \log(\log 20) = 20 \log 16 + \log(\log 20)$

$\log(16^{20} \log 20) = \log(16^{20}) + \log(\log 20) = 20 \log 16 + \log(\log 20)$

DPatrick
2016-10-05 20:47:18

10. Which of the following is largest?

c. $20 \log(16) + \log(\log 20)$ d. $20 \log(20) + \log(\log 16)$ e. $16 \log(20) + \log(\log 20)$

10. Which of the following is largest?

c. $20 \log(16) + \log(\log 20)$ d. $20 \log(20) + \log(\log 16)$ e. $16 \log(20) + \log(\log 20)$

acegikmoqsuwy2000
2016-10-05 20:47:31

now $\log(\log(x))$ is very close to 1 for such small numbers like 16 and 20

now $\log(\log(x))$ is very close to 1 for such small numbers like 16 and 20

DPatrick
2016-10-05 20:47:42

Right. If, say, we use 10 as the base of our logs, roughly what are $\log_{10}(16)$ and $\log_{10}(20)$?

Right. If, say, we use 10 as the base of our logs, roughly what are $\log_{10}(16)$ and $\log_{10}(20)$?

yamyamx2
2016-10-05 20:48:03

1.2

1.2

jasonhu4
2016-10-05 20:48:03

1.2 or so

1.2 or so

DPatrick
2016-10-05 20:48:08

Estimating using logs is a bit of an art form. Here's how you can estimate $\log_{10}(20)$: write it as \[\log_{10}(20) = \log_{10}(10 \cdot 2) = 1 + \log_{10}2 = 1 + \frac{1}{\log_2{10}}.\] Then $2^3 = 8$ and $2^4 = 16$, so $\log_2{10}$ is about $3\frac13$, so $\frac{1}{\log_2{10}}$ is about $0.3$, and hence $\log_{10}(20) \approx 1.3$ should be a pretty good estimate.

Estimating using logs is a bit of an art form. Here's how you can estimate $\log_{10}(20)$: write it as \[\log_{10}(20) = \log_{10}(10 \cdot 2) = 1 + \log_{10}2 = 1 + \frac{1}{\log_2{10}}.\] Then $2^3 = 8$ and $2^4 = 16$, so $\log_2{10}$ is about $3\frac13$, so $\frac{1}{\log_2{10}}$ is about $0.3$, and hence $\log_{10}(20) \approx 1.3$ should be a pretty good estimate.

DPatrick
2016-10-05 20:48:19

A similar exercise gives the approximation $\log_{10}(16) \approx 1.2$.

A similar exercise gives the approximation $\log_{10}(16) \approx 1.2$.

DPatrick
2016-10-05 20:48:59

And then $\log(\log 16)$ and $\log(\log 20)$ are each about $\log(1.2)$ and $\log(1.3)$, so these are really close to 0.

And then $\log(\log 16)$ and $\log(\log 20)$ are each about $\log(1.2)$ and $\log(1.3)$, so these are really close to 0.

DPatrick
2016-10-05 20:49:33

So (c) is about 32, (d) is about 36, and (e) is about 21.

So (c) is about 32, (d) is about 36, and (e) is about 21.

Derive_Foiler
2016-10-05 20:49:40

So, we want the 20log 20, and we get d

So, we want the 20log 20, and we get d

Liopleurodon
2016-10-05 20:49:40

so the answer is d, since the multiplies the largest numbers

so the answer is d, since the multiplies the largest numbers

DPatrick
2016-10-05 20:49:48

Right. $\boxed{(\text{d})}$ is the largest quantity.

Right. $\boxed{(\text{d})}$ is the largest quantity.

DPatrick
2016-10-05 20:50:03

That's it for Round 1!

That's it for Round 1!

DPatrick
2016-10-05 20:50:15

If you officially participated in the contest, and you got at least 8 of these 10 problems correct, then you'll move on to Round 2 later this month.

If you officially participated in the contest, and you got at least 8 of these 10 problems correct, then you'll move on to Round 2 later this month.

DPatrick
2016-10-05 20:50:26

Official notification of this should go out to your teacher early next week.

Official notification of this should go out to your teacher early next week.

DPatrick
2016-10-05 20:50:34

We will have a Round 2 Math Jam on Wednesday, November 2 at 7:30 PM Eastern / 4:30 PM Pacific to discuss the Round 2 problems and solutions.

We will have a Round 2 Math Jam on Wednesday, November 2 at 7:30 PM Eastern / 4:30 PM Pacific to discuss the Round 2 problems and solutions.

DPatrick
2016-10-05 20:50:45

Then, 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.

Then, 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
2016-10-05 20:50:54

If you're curious, here are the 9 regions:

If you're curious, here are the 9 regions:

DPatrick
2016-10-05 20:50:57

DPatrick
2016-10-05 20:51:11

Perhaps I'll see you in Atlanta in January!

Perhaps I'll see you in Atlanta in January!

DPatrick
2016-10-05 20:51:19

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

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

DPatrick
2016-10-05 20:51:37

That's all for tonight's Math Jam -- have a good evening!

That's all for tonight's Math Jam -- have a good evening!

mikebreen
2016-10-05 20:51:43

Good luck in Round Two!

Good luck in Round Two!

adik7
2016-10-05 20:52:10

Is your round 1 score taken into account when deciding what top scorer (in case of ties) will advance to semifinals?

Is your round 1 score taken into account when deciding what top scorer (in case of ties) will advance to semifinals?

DPatrick
2016-10-05 20:52:27

I don't believe so. The tiebreaker criteria are on the AMS's website that I linked to above.

I don't believe so. The tiebreaker criteria are on the AMS's website that I linked to above.

mikebreen
2016-10-05 20:52:52

Right. Round One doesn't count for final selection.

Right. Round One doesn't count for final selection.

DPatrick
2016-10-05 20:53:42

For those who missed the intro at the beginning, Mike Breen is the AMS Public Awareness Officer, and the co-creator and host of

For those who missed the intro at the beginning, Mike Breen is the AMS Public Awareness Officer, and the co-creator and host of

*Who Wants to Be a Mathematician*. So his answers are definitive.
Shaktiala
2016-10-05 20:53:57

Are there any sample questions to prepare?

Are there any sample questions to prepare?

DPatrick
2016-10-05 20:54:06

There is an extensive archive of past contests on the AMS's website.

There is an extensive archive of past contests on the AMS's website.

mikebreen
2016-10-05 20:54:36

Tests linked to from http://www.ams.org/programs/students/wwtbam/tests

Tests linked to from http://www.ams.org/programs/students/wwtbam/tests

bobjoe123
2016-10-05 20:54:41

WHEN IS IT NEXT YEAR

WHEN IS IT NEXT YEAR

DPatrick
2016-10-05 20:54:59

The national contest qualifying round always runs in late September through early October.

The national contest qualifying round always runs in late September through early October.

DPatrick
2016-10-05 20:55:18

There are also various regional contests throughout the year. Check the AMS website for a schedule (as they become announced).

There are also various regional contests throughout the year. Check the AMS website for a schedule (as they become announced).

DPatrick
2016-10-05 20:55:43

I see that there's one in late October in Fort Worth, Texas.

I see that there's one in late October in Fort Worth, Texas.

seanwang2001
2016-10-05 20:55:51

will there be a transcript for this

will there be a transcript for this

DPatrick
2016-10-05 20:56:09

Yes, the full transcript will be posted as soon as we're done answering questions (which we'll do for a few more minutes).

Yes, the full transcript will be posted as soon as we're done answering questions (which we'll do for a few more minutes).

Hero_of_Hyrule
2016-10-05 20:56:14

Will we review the semifinal problems later here?

Will we review the semifinal problems later here?

DPatrick
2016-10-05 20:56:34

Yes, there's a Math Jam scheduled for November 2 to go over the Round 2 problems.

Yes, there's a Math Jam scheduled for November 2 to go over the Round 2 problems.

DPatrick
2016-10-05 20:56:44

And the Finals from Atlanta in January will be webcast live.

And the Finals from Atlanta in January will be webcast live.

mikebreen
2016-10-05 20:56:56

RI on Pi Day, MI in March (?), DC in late April (National Math Festival)

RI on Pi Day, MI in March (?), DC in late April (National Math Festival)

DPatrick
2016-10-05 20:57:08

These are some upcoming regional contests coming this winter/spring.

These are some upcoming regional contests coming this winter/spring.

cinamon
2016-10-05 20:57:14

Did Mr. Breen's hand actually bleed on the wheel of fortune?

Did Mr. Breen's hand actually bleed on the wheel of fortune?

DPatrick
2016-10-05 20:57:23

I don't think he made it up.

I don't think he made it up.

mikebreen
2016-10-05 20:57:46

Yes. I didn't realize it until I sat back down in the audience and saw the stage hands cleaning up

Yes. I didn't realize it until I sat back down in the audience and saw the stage hands cleaning up

DPatrick
2016-10-05 20:58:04

Mike tells me that the national Final will be on Saturday, January 7 at 1 PM Eastern.

Mike tells me that the national Final will be on Saturday, January 7 at 1 PM Eastern.

DPatrick
2016-10-05 20:59:03

I think we'll end it there for tonight. If you have further questions, please check out the website at http://www.ams.org/programs/students/wwtbam/wwtbam.

I think we'll end it there for tonight. If you have further questions, please check out the website at http://www.ams.org/programs/students/wwtbam/wwtbam.

DPatrick
2016-10-05 20:59:14

Thanks for coming, and hopefully I'll see you again in 4 weeks for Round 2!

Thanks for coming, and hopefully I'll see you again in 4 weeks for Round 2!

mikebreen
2016-10-05 20:59:17

Thanks, everyone.

Thanks, everyone.

bobjoe123
2016-10-05 20:59:57

THANK YOU!!!!!!!!!!!!!!!!!

THANK YOU!!!!!!!!!!!!!!!!!

Liopleurodon
2016-10-05 21:01:57

thank you for the Math Jam!

thank you for the Math Jam!

yiqun
2016-10-05 21:01:57

thank you!! @DPatrick and Naysh and mikebreen!!!!

thank you!! @DPatrick and Naysh and mikebreen!!!!