## Who Wants to Be a Mathematician, Round 2

Go back to the Math Jam ArchiveAoPS instructor David Patrick will discuss the problems on Round 2 of the 2015-16 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 Seattle in January 2016.

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#### Facilitator: Dave Patrick

DPatrick
2015-11-03 19:35:32

**Welcome to the 2015-16***Who Wants to Be a Mathematician*Round 2 Math Jam!
DPatrick
2015-11-03 19:35:41

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

I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens of AoPS classes over the past 11 years, and I've written or co-written a few of our textbooks. I also once was a contestant on ABC's

*Who Wants to Be a Millionaire*, a few years before I started working at AoPS, when Regis was still the host. (No, I didn't win the million bucks.)
DPatrick
2015-11-03 19:35:58

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
2015-11-03 19:36:04

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
2015-11-03 19:36:28

This helps keep the session organized and on track.

This helps keep the session organized and on track.

DPatrick
2015-11-03 19:36:43

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
2015-11-03 19:36:58

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
2015-11-03 19:37:09

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
2015-11-03 19:37:15

Hello!

Hello!

DPatrick
2015-11-03 19:37:32

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
2015-11-03 19:37:37

Hello

Hello

DPatrick
2015-11-03 19:37:57

As you can see, between the three of us we have a lot of game show experience here tonight!

As you can see, between the three of us we have a lot of game show experience here tonight!

DPatrick
2015-11-03 19:38:13

*Who Wants to Be a Mathematician*is run by the American Mathematical Society (AMS). The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
DPatrick
2015-11-03 19:38:31

You can visit their website at www.ams.org.

You can visit their website at www.ams.org.

DPatrick
2015-11-03 19:38:54

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

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

DPatrick
2015-11-03 19:39:20

Of course, 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

Of course, 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
2015-11-03 19:39:34

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.

DPatrick
2015-11-03 19:39:37

Let's get started!

Let's get started!

DPatrick
2015-11-03 19:39:41

1. An integer between 2 and 7 inclusive is chosen at random. If a pair of fair dice is rolled, which sum (of the top numbers on the two dice) has the same probability of appearing as the randomly chosen integer?

1. An integer between 2 and 7 inclusive is chosen at random. If a pair of fair dice is rolled, which sum (of the top numbers on the two dice) has the same probability of appearing as the randomly chosen integer?

DPatrick
2015-11-03 19:40:06

(I'll always pop the current question to the top of the window. You can resize that top region by dragging on the horizontal bar above.)

(I'll always pop the current question to the top of the window. You can resize that top region by dragging on the horizontal bar above.)

AShane2000
2015-11-03 19:40:17

probablity of the integer is 1/6

probablity of the integer is 1/6

DeathLlama9
2015-11-03 19:40:17

Okay well the randomly chosen integer's probability is $\frac{1}{6}$

Okay well the randomly chosen integer's probability is $\frac{1}{6}$

DPatrick
2015-11-03 19:40:26

That's a good place to start. There are 6 possible integers, and each is equally likely. So each has a probability of $\dfrac16$ of being chosen.

That's a good place to start. There are 6 possible integers, and each is equally likely. So each has a probability of $\dfrac16$ of being chosen.

DPatrick
2015-11-03 19:40:38

Thus, we're looking for the sum between 2 and 7 inclusive that also has a probability of $\dfrac16$ of being rolled.

Thus, we're looking for the sum between 2 and 7 inclusive that also has a probability of $\dfrac16$ of being rolled.

DPatrick
2015-11-03 19:40:48

Which sum is that?

Which sum is that?

CantonMathGuy
2015-11-03 19:41:09

7

7

mathwhiz16
2015-11-03 19:41:09

From experience I know that the number is 7…

From experience I know that the number is 7…

nosaj
2015-11-03 19:41:09

7

7

mathsolver101
2015-11-03 19:41:09

We want to have six out of 36 successes in the dice part so the number 7 works

We want to have six out of 36 successes in the dice part so the number 7 works

gxah
2015-11-03 19:41:09

7

7

lax0000
2015-11-03 19:41:09

7=1+6=5+2=4+3=3+4=2+5=1+6

7=1+6=5+2=4+3=3+4=2+5=1+6

DPatrick
2015-11-03 19:41:25

Right. If you've seen enough of this sort of dice problem, you might already know that a 7 is the most likely sum to roll with two dice, and that its probability is $\dfrac16$.

Right. If you've seen enough of this sort of dice problem, you might already know that a 7 is the most likely sum to roll with two dice, and that its probability is $\dfrac16$.

DPatrick
2015-11-03 19:41:47

If you don't know the probabilities for two dice, how can we determine them?

If you don't know the probabilities for two dice, how can we determine them?

lax0000
2015-11-03 19:42:04

draw a table

draw a table

flyrain
2015-11-03 19:42:04

make a chart of all cases

make a chart of all cases

ninjataco
2015-11-03 19:42:04

list them out

list them out

DPatrick
2015-11-03 19:42:12

Right, we can make a table:

Right, we can make a table:

DPatrick
2015-11-03 19:42:15

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

&1&2&3&4&5&6 \\ \hline

1&2&3&4&5&6&7 \\

2&3&4&5&6&7&8 \\

3&4&5&6&7&8&9 \\

4&5&6&7&8&9&10 \\

5&6&7&8&9&10&11 \\

6&7&8&9&10&11&12

\end{array}\]

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

&1&2&3&4&5&6 \\ \hline

1&2&3&4&5&6&7 \\

2&3&4&5&6&7&8 \\

3&4&5&6&7&8&9 \\

4&5&6&7&8&9&10 \\

5&6&7&8&9&10&11 \\

6&7&8&9&10&11&12

\end{array}\]

DPatrick
2015-11-03 19:42:21

The roll of the first die is on top, and the roll of the second die is along the left side.

The roll of the first die is on top, and the roll of the second die is along the left side.

DPatrick
2015-11-03 19:42:32

We see that there are $6 \cdot 6 = 36$ equally-likely possible rolls.

We see that there are $6 \cdot 6 = 36$ equally-likely possible rolls.

DPatrick
2015-11-03 19:42:42

And 6 of those rolls give a sum of 7:

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

&1&2&3&4&5&6 \\ \hline

1&2&3&4&5&6&\mathbf{7} \\

2&3&4&5&6&\mathbf{7}&8 \\

3&4&5&6&\mathbf{7}&8&9 \\

4&5&6&\mathbf{7}&8&9&10 \\

5&6&\mathbf{7}&8&9&10&11 \\

6&\mathbf{7}&8&9&10&11&12

\end{array}\]

And 6 of those rolls give a sum of 7:

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

&1&2&3&4&5&6 \\ \hline

1&2&3&4&5&6&\mathbf{7} \\

2&3&4&5&6&\mathbf{7}&8 \\

3&4&5&6&\mathbf{7}&8&9 \\

4&5&6&\mathbf{7}&8&9&10 \\

5&6&\mathbf{7}&8&9&10&11 \\

6&\mathbf{7}&8&9&10&11&12

\end{array}\]

DPatrick
2015-11-03 19:42:59

So the probability of rolling a 7 is $\dfrac{6}{36} = \dfrac{1}{6}$.

So the probability of rolling a 7 is $\dfrac{6}{36} = \dfrac{1}{6}$.

DPatrick
2015-11-03 19:43:17

Thus $\boxed{7}$ is the answer: we have a $\dfrac16$ probability of selecting a 7 from 2 through 7 inclusive, which equals the $\dfrac16$ probability of rolling a sum of 7 on two dice.

Thus $\boxed{7}$ is the answer: we have a $\dfrac16$ probability of selecting a 7 from 2 through 7 inclusive, which equals the $\dfrac16$ probability of rolling a sum of 7 on two dice.

mathwhiz16
2015-11-03 19:43:25

The probabilities for each sum go 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36 and back down of rate number 1-12 respectively

The probabilities for each sum go 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36 and back down of rate number 1-12 respectively

DPatrick
2015-11-03 19:43:41

Indeed, that's right (except for the slight typo: the numbers in the sum go from 2 through 12).

Indeed, that's right (except for the slight typo: the numbers in the sum go from 2 through 12).

DPatrick
2015-11-03 19:43:48

Knowing the number of ways to roll any given sum on two dice is helpful in a lot of probability problems. (It's also helpful if you want to become an expert backgammon player.)

Knowing the number of ways to roll any given sum on two dice is helpful in a lot of probability problems. (It's also helpful if you want to become an expert backgammon player.)

DPatrick
2015-11-03 19:43:55

\[\begin{array}{r|l}

\text{sum} & \text{# of ways to roll} \\ \hline

2 & 1 \\

3 & 2 \\

4 & 3 \\

5 & 4 \\

6 & 5 \\

7 & 6 \\

8 & 5 \\

9 & 4 \\

10 & 3 \\

11 & 2 \\

12 & 1

\end{array}\]

\[\begin{array}{r|l}

\text{sum} & \text{# of ways to roll} \\ \hline

2 & 1 \\

3 & 2 \\

4 & 3 \\

5 & 4 \\

6 & 5 \\

7 & 6 \\

8 & 5 \\

9 & 4 \\

10 & 3 \\

11 & 2 \\

12 & 1

\end{array}\]

DeathLlama9
2015-11-03 19:44:23

DPatrick, are you an expert backgammon player?

DPatrick, are you an expert backgammon player?

DPatrick
2015-11-03 19:44:49

Sadly, no. But if you ever get the chance to go to the AMS's annual meeting in January, the mathematician Art Benjamin often runs a backgammon class!

Sadly, no. But if you ever get the chance to go to the AMS's annual meeting in January, the mathematician Art Benjamin often runs a backgammon class!

DPatrick
2015-11-03 19:45:05

On to #2:

On to #2:

DPatrick
2015-11-03 19:45:10

2. How many zeros are at the end (rightmost digits) of 2015!?

2. How many zeros are at the end (rightmost digits) of 2015!?

DPatrick
2015-11-03 19:45:27

How do we count the number of zeros at the end of a number?

How do we count the number of zeros at the end of a number?

DeathLlama9
2015-11-03 19:45:46

Find 10s

Find 10s

mathwhiz16
2015-11-03 19:45:46

Count the number of factors of 5.

Count the number of factors of 5.

ninjataco
2015-11-03 19:45:46

find the number of factors of 5

find the number of factors of 5

bearytasty
2015-11-03 19:45:46

its the same as counting the number of factors of 5 in the number

its the same as counting the number of factors of 5 in the number

Ancy
2015-11-03 19:45:46

find the number of factors with five

find the number of factors with five

bharatputra
2015-11-03 19:45:46

powers of 5

powers of 5

AShane2000
2015-11-03 19:45:46

Find the amount of times 10 goes into 2015?

Find the amount of times 10 goes into 2015?

bluephoenix
2015-11-03 19:45:46

find the highest power of 5 that divides it

find the highest power of 5 that divides it

gxah
2015-11-03 19:45:46

number of 5s and 2s but we mainly care about the 5s

number of 5s and 2s but we mainly care about the 5s

Liopleurodon
2015-11-03 19:45:46

count the number of times a multiple of 5 and 2 multiply together

count the number of times a multiple of 5 and 2 multiply together

nosaj
2015-11-03 19:45:46

Find the number of factors of 5 the number has.

Find the number of factors of 5 the number has.

mathsolver101
2015-11-03 19:45:51

Let's count how many groups of 5 and 2 there are in 2015! (though it's easier just to count the number of 5's)

Let's count how many groups of 5 and 2 there are in 2015! (though it's easier just to count the number of 5's)

DPatrick
2015-11-03 19:45:57

Right, we can count the number of times that 10 divides the number.

Right, we can count the number of times that 10 divides the number.

DPatrick
2015-11-03 19:46:12

But since 10 = 2 * 5, it's the smaller of the number of 2's that divides the number and the number of 5's that divides the number.

But since 10 = 2 * 5, it's the smaller of the number of 2's that divides the number and the number of 5's that divides the number.

Ancy
2015-11-03 19:46:29

The two is irrelevant because there are a lot more, so 5 is the limiting factor.

The two is irrelevant because there are a lot more, so 5 is the limiting factor.

yaredboy
2015-11-03 19:46:29

There are a lot of twos, so its best to count the number of fives

There are a lot of twos, so its best to count the number of fives

DPatrick
2015-11-03 19:46:36

Yes: there are a lot more 2's that divide 2015! than 5's: every other term of 2015! gives at least one factor of 2, but only every fifth term of 2015! gives at least one factor of 5.

Yes: there are a lot more 2's that divide 2015! than 5's: every other term of 2015! gives at least one factor of 2, but only every fifth term of 2015! gives at least one factor of 5.

DPatrick
2015-11-03 19:46:41

So the answer to our question is just the number of 5's that divides 2015!.

So the answer to our question is just the number of 5's that divides 2015!.

DPatrick
2015-11-03 19:46:54

How do we compute that?

How do we compute that?

DeathLlama9
2015-11-03 19:47:23

There are 2015/5 = 403 powers of five, but we have to get the 25s and 125s and so on as well

There are 2015/5 = 403 powers of five, but we have to get the 25s and 125s and so on as well

bearytasty
2015-11-03 19:47:23

floor(2015/5) + floor(2015/5^2) + floor(2015/5^3) + ...

floor(2015/5) + floor(2015/5^2) + floor(2015/5^3) + ...

Liopleurodon
2015-11-03 19:47:23

2015/5+2015/25+2015/125+2015/625 rounded down

2015/5+2015/25+2015/125+2015/625 rounded down

yaredboy
2015-11-03 19:47:23

Remember 25 has two fives, and so does every other multiple of 25!

Remember 25 has two fives, and so does every other multiple of 25!

mathwhiz16
2015-11-03 19:47:23

divide 2015 by 5 repeatedly

divide 2015 by 5 repeatedly

Mlux
2015-11-03 19:47:23

factors of 5 between 1-2015

factors of 5 between 1-2015

rishis
2015-11-03 19:47:23

find number of factors of 2015 that are divisible of 5, 5^2, 5^3...

find number of factors of 2015 that are divisible of 5, 5^2, 5^3...

DPatrick
2015-11-03 19:47:32

Right. To start, every fifth term in $1 \cdot 2 \cdot \cdots \cdot 2014 \cdot 2015$ contributes a factor of 5.

Right. To start, every fifth term in $1 \cdot 2 \cdot \cdots \cdot 2014 \cdot 2015$ contributes a factor of 5.

DPatrick
2015-11-03 19:47:39

There are $\dfrac{2015}{5} = 403$ multiples of 5 in 2015!. So that's 403 factors of 5.

There are $\dfrac{2015}{5} = 403$ multiples of 5 in 2015!. So that's 403 factors of 5.

flyrain
2015-11-03 19:47:54

what about the 25s?

what about the 25s?

nosaj
2015-11-03 19:47:54

but there are some that give a second factor of 5

but there are some that give a second factor of 5

DPatrick
2015-11-03 19:48:02

Indeed, every fifth one of these is also a multiple of 25, so it contributes an extra factor of 5.

Indeed, every fifth one of these is also a multiple of 25, so it contributes an extra factor of 5.

DPatrick
2015-11-03 19:48:12

So that's $\dfrac{403}{5} = $ uh-oh, that's not an integer. What do we do?

So that's $\dfrac{403}{5} = $ uh-oh, that's not an integer. What do we do?

DeathLlama9
2015-11-03 19:48:24

round down

round down

Liopleurodon
2015-11-03 19:48:24

round down

round down

mathsolver101
2015-11-03 19:48:24

Floor function!!

Floor function!!

mathwhiz16
2015-11-03 19:48:24

round down

round down

Mlux
2015-11-03 19:48:24

least integer

least integer

CantonMathGuy
2015-11-03 19:48:24

round down to integer

round down to integer

bearytasty
2015-11-03 19:48:24

truncate it. the remainder doesnt matter

truncate it. the remainder doesnt matter

krishna10
2015-11-03 19:48:32

Keep it at 80, ignore the remainder

Keep it at 80, ignore the remainder

DPatrick
2015-11-03 19:48:35

We round down. That is, $\left\lfloor \dfrac{403}{5} \right\rfloor = \dfrac{400}{5} = 80$ terms in 2015! are multiples of 25, and contribute an extra factor of 5.

We round down. That is, $\left\lfloor \dfrac{403}{5} \right\rfloor = \dfrac{400}{5} = 80$ terms in 2015! are multiples of 25, and contribute an extra factor of 5.

DPatrick
2015-11-03 19:48:47

So that's 403+80 = 483 total factors of 5 so far.

So that's 403+80 = 483 total factors of 5 so far.

Mlux
2015-11-03 19:49:01

now 125

now 125

nosaj
2015-11-03 19:49:01

now for the multiples of 125!

now for the multiples of 125!

Ancy
2015-11-03 19:49:01

divide by 5 again

divide by 5 again

DPatrick
2015-11-03 19:49:10

Right, we go again! Every fifth one of these 80 is a multiple of 125, and gives an extra factor of 5.

Right, we go again! Every fifth one of these 80 is a multiple of 125, and gives an extra factor of 5.

DPatrick
2015-11-03 19:49:14

So that's $\dfrac{80}{5} = 16$ more, for a total so far of 483 + 16 = 499 factors of 5.

So that's $\dfrac{80}{5} = 16$ more, for a total so far of 483 + 16 = 499 factors of 5.

Ancy
2015-11-03 19:49:32

now divide by five to get three

now divide by five to get three

Mlux
2015-11-03 19:49:32

625's

625's

yaredboy
2015-11-03 19:49:32

Lol watch us do 625 XD

Lol watch us do 625 XD

flyrain
2015-11-03 19:49:32

there's still more

there's still more

DPatrick
2015-11-03 19:49:38

And again! Every fifth one of these 16 is a multiple of 625.

And again! Every fifth one of these 16 is a multiple of 625.

DPatrick
2015-11-03 19:49:43

So that's $\left\lfloor \dfrac{16}{5} \right\rfloor = \dfrac{15}{5} = 3$ more factors of 5, for a running total of 499 + 3 = 502.

So that's $\left\lfloor \dfrac{16}{5} \right\rfloor = \dfrac{15}{5} = 3$ more factors of 5, for a running total of 499 + 3 = 502.

Ancy
2015-11-03 19:49:55

we can stop at three because it is less than five, and will give you zero.

we can stop at three because it is less than five, and will give you zero.

Mlux
2015-11-03 19:49:55

that's it

that's it

DeathLlama9
2015-11-03 19:49:55

Nothing beyond that as $2015 < 3125$

Nothing beyond that as $2015 < 3125$

CantonMathGuy
2015-11-03 19:49:55

2015 < 3125

2015 < 3125

mathwhiz16
2015-11-03 19:49:55

sadly, no 3125 :(

sadly, no 3125 :(

DPatrick
2015-11-03 19:50:00

That's it. There are only three of these, so we can't get a multiple of 3125. (As expected, since 3125 > 2015.)

That's it. There are only three of these, so we can't get a multiple of 3125. (As expected, since 3125 > 2015.)

DPatrick
2015-11-03 19:50:09

So we have a total of 502 factors of 5 dividing 2015!, which means that 2015! ends in $\boxed{502}$ zeroes.

So we have a total of 502 factors of 5 dividing 2015!, which means that 2015! ends in $\boxed{502}$ zeroes.

DPatrick
2015-11-03 19:50:29

Fun fact: 2015! has 5786 digits total!

Fun fact: 2015! has 5786 digits total!

bearytasty
2015-11-03 19:50:44

Is there an easy way to compute that?

Is there an easy way to compute that?

DPatrick
2015-11-03 19:50:55

Yep -- you go to Wolfram Alpha and ask it to compute 2015!

Yep -- you go to Wolfram Alpha and ask it to compute 2015!

DPatrick
2015-11-03 19:51:01

3. In a geometric series $\displaystyle \sum_{n=1}^\infty a_n$, $a_2 = 54$ and $a_5 = 2$. What is the sum of the series?

3. In a geometric series $\displaystyle \sum_{n=1}^\infty a_n$, $a_2 = 54$ and $a_5 = 2$. What is the sum of the series?

DPatrick
2015-11-03 19:51:20

First of all, just so we're all on the same page: what is a geometric series?

First of all, just so we're all on the same page: what is a geometric series?

ninjataco
2015-11-03 19:51:49

constant ratio between consecutive terms

constant ratio between consecutive terms

flyrain
2015-11-03 19:51:49

a sequence with a common ratio

a sequence with a common ratio

bearytasty
2015-11-03 19:51:49

common ratio between terms

common ratio between terms

nosaj
2015-11-03 19:51:49

terms have a constant ratio

terms have a constant ratio

hwl0304
2015-11-03 19:51:49

a sum of terms in a geometric sequence

a sum of terms in a geometric sequence

mathwhiz16
2015-11-03 19:51:49

A series that goes a, ar, ar^2, ar^3,, …, Ar^n-1

A series that goes a, ar, ar^2, ar^3,, …, Ar^n-1

DPatrick
2015-11-03 19:51:58

Yes. It's a sum of terms in which the ratio between two consecutive terms is constant.

Yes. It's a sum of terms in which the ratio between two consecutive terms is constant.

DPatrick
2015-11-03 19:52:12

So what is a potential strategy when trying to work with this geometric series?

So what is a potential strategy when trying to work with this geometric series?

DeathLlama9
2015-11-03 19:52:27

Find the common ratio!

Find the common ratio!

ninjataco
2015-11-03 19:52:27

first find r

first find r

flyrain
2015-11-03 19:52:32

find r

find r

Liopleurodon
2015-11-03 19:52:35

find the ratio

find the ratio

DPatrick
2015-11-03 19:52:40

Sure. We'd like to find the common ratio between consecutive terms.

Sure. We'd like to find the common ratio between consecutive terms.

DPatrick
2015-11-03 19:52:47

But we're only given two terms that are three positions apart. What does that tell us?

But we're only given two terms that are three positions apart. What does that tell us?

nosaj
2015-11-03 19:53:15

Note that $\frac{a_5}{a_2}=r^3$.

Note that $\frac{a_5}{a_2}=r^3$.

ChrisY
2015-11-03 19:53:15

they are r^3 apart

they are r^3 apart

CantonMathGuy
2015-11-03 19:53:15

their ratio is the cube of the common ratio

their ratio is the cube of the common ratio

DeathLlama9
2015-11-03 19:53:15

Common ratio is cube root of their quotient

Common ratio is cube root of their quotient

hwl0304
2015-11-03 19:53:15

the ratio between them is the cube of the common ratio

the ratio between them is the cube of the common ratio

Liopleurodon
2015-11-03 19:53:15

r^3=2/54

r^3=2/54

bearytasty
2015-11-03 19:53:15

well between the 2nd term and 5th term u multiply by the ratio 3 times, so the ratio is 1/3

well between the 2nd term and 5th term u multiply by the ratio 3 times, so the ratio is 1/3

Ancy
2015-11-03 19:53:15

the fifth term divided by the second term gives us r^3

the fifth term divided by the second term gives us r^3

ChrisY
2015-11-03 19:53:15

the ratio is r^3

the ratio is r^3

AShane2000
2015-11-03 19:53:15

the difference between them is a magnitude of r^3

the difference between them is a magnitude of r^3

DPatrick
2015-11-03 19:53:24

Right. The ratio between those two given terms should be the

Right. The ratio between those two given terms should be the

*cube*of the common ratio!
DPatrick
2015-11-03 19:53:30

That is, if we let the common ratio be $r$, then we have (by definition) \[r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \frac{a_5}{a_4} = \cdots.\]

That is, if we let the common ratio be $r$, then we have (by definition) \[r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \frac{a_5}{a_4} = \cdots.\]

DPatrick
2015-11-03 19:53:40

But if we multiply the last three ratios I listed above, we get \[\frac{a_3}{a_2} \cdot \frac{a_4}{a_3} \cdot \frac{a_5}{a_4} = r \cdot r \cdot r = r^3.\]

But if we multiply the last three ratios I listed above, we get \[\frac{a_3}{a_2} \cdot \frac{a_4}{a_3} \cdot \frac{a_5}{a_4} = r \cdot r \cdot r = r^3.\]

DPatrick
2015-11-03 19:53:50

And we see that many of the terms of the left side cancel, leaving us with just $\dfrac{a_5}{a_2} = r^3$.

And we see that many of the terms of the left side cancel, leaving us with just $\dfrac{a_5}{a_2} = r^3$.

DPatrick
2015-11-03 19:54:15

And $a_5 = 2$ and $a_2 = 54$, so we have $\dfrac{2}{54} = r^3$.

And $a_5 = 2$ and $a_2 = 54$, so we have $\dfrac{2}{54} = r^3$.

Liopleurodon
2015-11-03 19:54:31

so r equals 1/3

so r equals 1/3

DeathLlama9
2015-11-03 19:54:31

$r = \frac{1}{3}$

$r = \frac{1}{3}$

flyrain
2015-11-03 19:54:31

r=1/3

r=1/3

Mlux
2015-11-03 19:54:31

r = 1/3

r = 1/3

Ancy
2015-11-03 19:54:31

so r = 1/3

so r = 1/3

ninjataco
2015-11-03 19:54:31

r=1/3

r=1/3

DPatrick
2015-11-03 19:54:35

This simplifies to $\dfrac{1}{27} = r^3$, so $r = \sqrt[3]{\dfrac{1}{27}} = \dfrac13$.

This simplifies to $\dfrac{1}{27} = r^3$, so $r = \sqrt[3]{\dfrac{1}{27}} = \dfrac13$.

bellehan
2015-11-03 19:54:45

now we can find the first term

now we can find the first term

DPatrick
2015-11-03 19:55:02

Indeed, what's the first term?

Indeed, what's the first term?

AShane2000
2015-11-03 19:55:21

(1/3)a=54

(1/3)a=54

AShane2000
2015-11-03 19:55:21

a=162

a=162

bearytasty
2015-11-03 19:55:21

first term is 3*54 = 162

first term is 3*54 = 162

lkarhat
2015-11-03 19:55:21

162

162

DeathLlama9
2015-11-03 19:55:21

$54 \div \frac{1}{3} = 162$

$54 \div \frac{1}{3} = 162$

sflorin
2015-11-03 19:55:21

162

162

thequantumguy
2015-11-03 19:55:25

54 * 3 = 162

54 * 3 = 162

krishna10
2015-11-03 19:55:25

54 times 3... 162

54 times 3... 162

DPatrick
2015-11-03 19:55:30

We again have $\dfrac{a_2}{a_1} = r = \dfrac13$.

We again have $\dfrac{a_2}{a_1} = r = \dfrac13$.

DPatrick
2015-11-03 19:55:42

Since $a_2 = 54$, this gives us $a_1 = 3a_2 = 3(54) = 162$.

Since $a_2 = 54$, this gives us $a_1 = 3a_2 = 3(54) = 162$.

DPatrick
2015-11-03 19:55:56

So our sum is the sum of a geometric series with first term 162 and common ratio $\frac13$.

So our sum is the sum of a geometric series with first term 162 and common ratio $\frac13$.

Ancy
2015-11-03 19:56:10

Now we can use the formula to find the sum of a geometric series, (first term)/(1-r)

Now we can use the formula to find the sum of a geometric series, (first term)/(1-r)

DeathLlama9
2015-11-03 19:56:10

Now we can use the equation $\frac{a}{1-r}$!

Now we can use the equation $\frac{a}{1-r}$!

mathwhiz16
2015-11-03 19:56:10

And now we use the formula for a infinite geometric series: $\frac{a}{1-r}$

And now we use the formula for a infinite geometric series: $\frac{a}{1-r}$

bearytasty
2015-11-03 19:56:13

use the formula a/(1-r) for sum of infinite geo series

use the formula a/(1-r) for sum of infinite geo series

DPatrick
2015-11-03 19:56:22

Right. You might know the formula for a geometric series with first term $a$ and common ratio $r$.

Right. You might know the formula for a geometric series with first term $a$ and common ratio $r$.

DPatrick
2015-11-03 19:56:32

It's $\dfrac{a}{1-r}$...

It's $\dfrac{a}{1-r}$...

**provided**$|r| < 1$.
chopinwaltzes
2015-11-03 19:57:02

243!

243!

mathwhiz16
2015-11-03 19:57:02

So we get $\frac{162}{\frac{2}{3}}$, which is 243.

So we get $\frac{162}{\frac{2}{3}}$, which is 243.

rraj
2015-11-03 19:57:02

243

243

ChrisY
2015-11-03 19:57:02

The sum is 162/(2/3) = 243

The sum is 162/(2/3) = 243

Mlux
2015-11-03 19:57:02

it's 243

it's 243

DPatrick
2015-11-03 19:57:06

Applying this formula, we see that our sum is $\dfrac{162}{1-\frac13}$.

Applying this formula, we see that our sum is $\dfrac{162}{1-\frac13}$.

DPatrick
2015-11-03 19:57:12

This evaluates as $\dfrac{162}{\frac23} = \frac32 \cdot 162 = 3 \cdot 81 = \boxed{243}$.

This evaluates as $\dfrac{162}{\frac23} = \frac32 \cdot 162 = 3 \cdot 81 = \boxed{243}$.

mikebreen
2015-11-03 19:57:27

We don't ask calc questions but figured people who haven't taken calculus can still do geometric series

We don't ask calc questions but figured people who haven't taken calculus can still do geometric series

DPatrick
2015-11-03 19:57:38

Indeed...even if you didn't know this formula, how can you sum the series?

Indeed...even if you didn't know this formula, how can you sum the series?

DPatrick
2015-11-03 19:57:55

Or to ask another way: how do we prove that formula?

Or to ask another way: how do we prove that formula?

ninjataco
2015-11-03 19:58:16

let S be the sum of the series, multiply by 1/3, do S - 1/3 S

let S be the sum of the series, multiply by 1/3, do S - 1/3 S

krishna10
2015-11-03 19:58:16

Basically be writing a second equation, 1\3 of the first and cancel terms

Basically be writing a second equation, 1\3 of the first and cancel terms

nosaj
2015-11-03 19:58:27

Multiply by the sum by $r$ and then subtract the original and lost of things cancel! :D

Multiply by the sum by $r$ and then subtract the original and lost of things cancel! :D

ChrisY
2015-11-03 19:58:27

you have a geometric series s, and another that is sr, then you subtract

you have a geometric series s, and another that is sr, then you subtract

DPatrick
2015-11-03 19:58:40

Right. Let me quickly show you how it works for a general geometric series.

Right. Let me quickly show you how it works for a general geometric series.

DPatrick
2015-11-03 19:58:45

Let's set $S$ to be the sum of the geometric series with first term $a$ and common ratio $r$.

Let's set $S$ to be the sum of the geometric series with first term $a$ and common ratio $r$.

DPatrick
2015-11-03 19:58:59

Then we can write \[ S = a + ar + ar^2 + ar^3 + \cdots.\]

Then we can write \[ S = a + ar + ar^2 + ar^3 + \cdots.\]

DPatrick
2015-11-03 19:59:15

If we multiply by $r$, we have \[ rS = ar + ar^2 + ar^3 + ar^4 + \cdots.\]

If we multiply by $r$, we have \[ rS = ar + ar^2 + ar^3 + ar^4 + \cdots.\]

DPatrick
2015-11-03 19:59:28

What happens when we subtract these two equations?

What happens when we subtract these two equations?

bearytasty
2015-11-03 19:59:45

BAM everything disappears!

BAM everything disappears!

mathsolver101
2015-11-03 19:59:45

A lot of things cancel and everyone is happy

A lot of things cancel and everyone is happy

DPatrick
2015-11-03 20:00:20

Right. On the left side, we get $S - rS$, and on the right side, all of the terms cancel out except for the initial $a$!

Right. On the left side, we get $S - rS$, and on the right side, all of the terms cancel out except for the initial $a$!

DPatrick
2015-11-03 20:00:27

So $S - rS = a$.

So $S - rS = a$.

nosaj
2015-11-03 20:00:41

then we divide both sides by $1-r$

then we divide both sides by $1-r$

DeathLlama9
2015-11-03 20:00:41

$S(1 - r) = a$, so $\frac{a}{1-r} = S$

$S(1 - r) = a$, so $\frac{a}{1-r} = S$

Liopleurodon
2015-11-03 20:00:48

Factor out the S and divide by 1-r

Factor out the S and divide by 1-r

DPatrick
2015-11-03 20:00:50

The left side can factor out an $S$ to give us $S(1-r) = a$, and then dividing by $1-r$ gives us our formula $S = \dfrac{a}{1-r}$.

The left side can factor out an $S$ to give us $S(1-r) = a$, and then dividing by $1-r$ gives us our formula $S = \dfrac{a}{1-r}$.

DPatrick
2015-11-03 20:01:09

And lastly before we move on: why does this only work for $|r| < 1$?

And lastly before we move on: why does this only work for $|r| < 1$?

AShane2000
2015-11-03 20:01:36

if r>1 the sum tends to infinity

if r>1 the sum tends to infinity

bellehan
2015-11-03 20:01:36

because of r>1 the sum is iniffinity

because of r>1 the sum is iniffinity

bearytasty
2015-11-03 20:01:36

because if $|r|>1$ then the sum is some sort of infinity

because if $|r|>1$ then the sum is some sort of infinity

mathwhiz16
2015-11-03 20:01:36

because a series with r>1 diverges

because a series with r>1 diverges

TheChampion
2015-11-03 20:01:36

Otherwise the sequence diverges.

Otherwise the sequence diverges.

Liopleurodon
2015-11-03 20:01:36

because if |r|>1, then the sum is infinite

because if |r|>1, then the sum is infinite

Ancy
2015-11-03 20:01:36

Then the series diverges and is +- infinity

Then the series diverges and is +- infinity

mathsolver101
2015-11-03 20:01:36

Because otherwise the inifiite series will just get bigger and bigger and bigger

Because otherwise the inifiite series will just get bigger and bigger and bigger

DPatrick
2015-11-03 20:02:03

Right. If $|r| \ge 1$, then the sum would be infinite: it would grow without bound. (Well, if $r = -1$ the issue is a little different.)

Right. If $|r| \ge 1$, then the sum would be infinite: it would grow without bound. (Well, if $r = -1$ the issue is a little different.)

DPatrick
2015-11-03 20:02:08

For example, it doesn't make sense to use the formula to conclude that \[1 + 2 + 4 + 8 + \cdots = \frac{1}{1-2} = -1\,???\]

For example, it doesn't make sense to use the formula to conclude that \[1 + 2 + 4 + 8 + \cdots = \frac{1}{1-2} = -1\,???\]

DPatrick
2015-11-03 20:02:28

OK, on to #4!

OK, on to #4!

DPatrick
2015-11-03 20:02:44

4. A piece of fruit is a perfect sphere of radius $r$ and has a hard spherical seed at its center of radius 1. If the seed is removed, the volume of the remaining fruit is 7 times the volume of the seed. Find $r$.

4. A piece of fruit is a perfect sphere of radius $r$ and has a hard spherical seed at its center of radius 1. If the seed is removed, the volume of the remaining fruit is 7 times the volume of the seed. Find $r$.

bearytasty
2015-11-03 20:03:10

find the volumes of the two pieces

find the volumes of the two pieces

DPatrick
2015-11-03 20:03:17

OK. What's the volume of the seed?

OK. What's the volume of the seed?

bearytasty
2015-11-03 20:03:39

volume of seed is $4\pi / 3$

volume of seed is $4\pi / 3$

mathsolver101
2015-11-03 20:03:39

4pi/3

4pi/3

ninjataco
2015-11-03 20:03:39

$\frac{4}{3}\pi$

$\frac{4}{3}\pi$

Liopleurodon
2015-11-03 20:03:39

4/3 pi

4/3 pi

krishna10
2015-11-03 20:03:39

4/3pi

4/3pi

FTW
2015-11-03 20:03:39

$\frac{4}{3} \pi$

$\frac{4}{3} \pi$

DPatrick
2015-11-03 20:03:42

The formula for the volume of a sphere of radius $r$ is $\frac43\pi r^3$.

The formula for the volume of a sphere of radius $r$ is $\frac43\pi r^3$.

DPatrick
2015-11-03 20:03:47

So the volume of the seed is $\frac43\pi$.

So the volume of the seed is $\frac43\pi$.

DPatrick
2015-11-03 20:03:53

What's the volume of the entire fruit?

What's the volume of the entire fruit?

ninjataco
2015-11-03 20:04:25

$\frac{32}{3}\pi$

$\frac{32}{3}\pi$

DeathLlama9
2015-11-03 20:04:25

28/3 + 4/3 = 32pi/3

28/3 + 4/3 = 32pi/3

AShane2000
2015-11-03 20:04:25

32pi/3

32pi/3

Liopleurodon
2015-11-03 20:04:25

4/3 pi r^3

4/3 pi r^3

FTW
2015-11-03 20:04:25

$\frac{4}{3} \pi r^3$

$\frac{4}{3} \pi r^3$

sflorin
2015-11-03 20:04:25

4/3pir^3

4/3pir^3

DPatrick
2015-11-03 20:04:32

On the one hand, the seed is $\frac43\pi$, and the rest of the fruit is 7 times the seed, or $\frac43\pi \cdot 7$.

On the one hand, the seed is $\frac43\pi$, and the rest of the fruit is 7 times the seed, or $\frac43\pi \cdot 7$.

DPatrick
2015-11-03 20:04:37

So the entire fruit is $\frac43\pi + \frac43\pi \cdot 7 = \frac43\pi \cdot 8$.

So the entire fruit is $\frac43\pi + \frac43\pi \cdot 7 = \frac43\pi \cdot 8$.

DPatrick
2015-11-03 20:04:59

On the other hand, it's just $\frac43 \pi r^3$, where $r$ is the radius we want.

On the other hand, it's just $\frac43 \pi r^3$, where $r$ is the radius we want.

DPatrick
2015-11-03 20:05:04

Hence, $\frac43 \pi \cdot 8 = \frac43\pi r^3$.

Hence, $\frac43 \pi \cdot 8 = \frac43\pi r^3$.

xiaodragon
2015-11-03 20:05:22

r=2

r=2

Mlux
2015-11-03 20:05:22

r = 2

r = 2

eveningstarandlion
2015-11-03 20:05:22

r=2

r=2

thequantumguy
2015-11-03 20:05:22

r is 2

r is 2

mathwhiz16
2015-11-03 20:05:22

so r=2!

so r=2!

bearytasty
2015-11-03 20:05:22

r=2

r=2

CantonMathGuy
2015-11-03 20:05:22

$r^3=8\implies r=2$

$r^3=8\implies r=2$

DPatrick
2015-11-03 20:05:27

So $8 = r^3$, and thus $r = \boxed{2}$.

So $8 = r^3$, and thus $r = \boxed{2}$.

eveningstarandlion
2015-11-03 20:05:38

Wait if the seed is removed, then we get $7x$ so that means the see is $x$, so the whole sphere is $8x$

Wait if the seed is removed, then we get $7x$ so that means the see is $x$, so the whole sphere is $8x$

DPatrick
2015-11-03 20:05:48

Right: we really didn't need to carry that $\frac43\pi$ term around. We could just reason out that the fruit is 8 times the volume of the seed. And since the ratio of volumes of similar objects is the cube of the ratio of their corresponding lengths, the ratio of the radii of the fruit to the side must be the cube root of $8:1$, or $2:1$.

Right: we really didn't need to carry that $\frac43\pi$ term around. We could just reason out that the fruit is 8 times the volume of the seed. And since the ratio of volumes of similar objects is the cube of the ratio of their corresponding lengths, the ratio of the radii of the fruit to the side must be the cube root of $8:1$, or $2:1$.

Generic_Username
2015-11-03 20:06:08

We can also do similarity ratio stuff for 3-D objects

We can also do similarity ratio stuff for 3-D objects

DPatrick
2015-11-03 20:06:28

Exactly: it's the same as it is for 2-D objects, except the ratio of volumes is the

Exactly: it's the same as it is for 2-D objects, except the ratio of volumes is the

**cube**of the ratio of lengths.
DPatrick
2015-11-03 20:06:41

Math history time!

Math history time!

DPatrick
2015-11-03 20:06:45

5. Which of the following mathematicians was one of the inventors of game theory?

(a) John von Neumann (b) Kurt Gödel (c) George Pólya (d) Paul Erdős

5. Which of the following mathematicians was one of the inventors of game theory?

(a) John von Neumann (b) Kurt Gödel (c) George Pólya (d) Paul Erdős

AShane2000
2015-11-03 20:07:17

a

a

lkarhat
2015-11-03 20:07:17

a

a

thequantumguy
2015-11-03 20:07:17

a

a

xiaodragon
2015-11-03 20:07:17

(a

(a

eveningstarandlion
2015-11-03 20:07:17

John Von Neumann

John Von Neumann

trumpeter
2015-11-03 20:07:17

von neumann

von neumann

Ancy
2015-11-03 20:07:17

A?

A?

ninjataco
2015-11-03 20:07:17

von Neumann!

von Neumann!

DPatrick
2015-11-03 20:07:22

The answer is $\boxed{\text{(a)}}$, John von Neumann. Here is a picture of von Neumann on a postage stamp:

The answer is $\boxed{\text{(a)}}$, John von Neumann. Here is a picture of von Neumann on a postage stamp:

DPatrick
2015-11-03 20:07:27

DPatrick
2015-11-03 20:07:44

If you look up "Game theory" on Wikipedia, the first sentence of the second paragraph states:

If you look up "Game theory" on Wikipedia, the first sentence of the second paragraph states:

*Modern game theory began with the idea regarding the existence of mixed-strategy equilibria in two-person zero-sum games and its proof by John von Neumann.*
DPatrick
2015-11-03 20:08:01

Von Neumann grew up in Hungary, but moved to the United States in his late 20s and lived here the rest of his life (which, sadly, was all too short: he died of cancer at age 53). He was one of the most prolific American mathematicians of the mid 20th-century, and did important work in many different branches of mathematics and physics.

Von Neumann grew up in Hungary, but moved to the United States in his late 20s and lived here the rest of his life (which, sadly, was all too short: he died of cancer at age 53). He was one of the most prolific American mathematicians of the mid 20th-century, and did important work in many different branches of mathematics and physics.

DPatrick
2015-11-03 20:08:23

He was also a

He was also a

*bon vivant*: he liked to dress sharp and throw lavish parties. One of his biographers said "The parties at the von Neumann's house were frequent, and famous, and long." He also reportedly rode to the bottom of the Grand Canyon on a mule, while wearing a three-piece suit. (Von Neumann was wearing the suit, not the mule.)
DPatrick
2015-11-03 20:08:47

Quickly, let's see who the other three people are.

Quickly, let's see who the other three people are.

DPatrick
2015-11-03 20:08:52

DPatrick
2015-11-03 20:09:09

Kurt Godel (1906-1978) grew up in Austria and moved to the U.S. in his 30s. He was one of the leading mathematicians in the field of mathematical logic, and is one of the three title luminaries featured in Douglas Hofstadter's book .

Kurt Godel (1906-1978) grew up in Austria and moved to the U.S. in his 30s. He was one of the leading mathematicians in the field of mathematical logic, and is one of the three title luminaries featured in Douglas Hofstadter's book .

mathwhiz16
2015-11-03 20:09:17

Kurt godel had his famous incompleteness theorem…

Kurt godel had his famous incompleteness theorem…

DPatrick
2015-11-03 20:09:22

That's right.

That's right.

DPatrick
2015-11-03 20:09:32

DPatrick
2015-11-03 20:09:42

George Polya (1887-1985) grew up in Hungary and spent his professional life around Europe and the U.S. He did research in many different fields of mathematics but is today perhaps best known for his 1945 book

George Polya (1887-1985) grew up in Hungary and spent his professional life around Europe and the U.S. He did research in many different fields of mathematics but is today perhaps best known for his 1945 book

*How to Solve It*about problem solving.
DPatrick
2015-11-03 20:10:06

DPatrick
2015-11-03 20:10:17

Paul Erdos (1913-1996) also grew up in Hungary, but basically lived all over the world during his adult life. He was perhaps the most prolific mathematician of the 20th century, and is believed to have authored or co-authored over 1,500 articles, primarily in number theory but in many other branches of mathematics as well. He's also the source of the

Paul Erdos (1913-1996) also grew up in Hungary, but basically lived all over the world during his adult life. He was perhaps the most prolific mathematician of the 20th century, and is believed to have authored or co-authored over 1,500 articles, primarily in number theory but in many other branches of mathematics as well. He's also the source of the

*Erdos Number*: if you wrote a paper with Erdos, your number is 1 (it is believed that there are 511 such people); if you wrote a paper with someone whose number is 1, your number is 2; and so on. (My Erdos Number is 3.)
mathwhiz16
2015-11-03 20:10:52

Is the math history question the most missed question on the competition? Seems like it.

Is the math history question the most missed question on the competition? Seems like it.

DPatrick
2015-11-03 20:11:07

No, it was actually the 3rd most missed. 50.3% got this question right.

No, it was actually the 3rd most missed. 50.3% got this question right.

nosaj
2015-11-03 20:11:24

i'm guessing 10 was the most missed

i'm guessing 10 was the most missed

DPatrick
2015-11-03 20:11:32

You are right. We'll get to that in a bit...

You are right. We'll get to that in a bit...

DPatrick
2015-11-03 20:11:40

Let's keep going...

Let's keep going...

DPatrick
2015-11-03 20:11:44

6. How many even six-digit numbers use every one of the six digits 0,1,2,3,4,5?

6. How many even six-digit numbers use every one of the six digits 0,1,2,3,4,5?

DeathLlama9
2015-11-03 20:12:13

Obviously 0 can't be first and 0, 2, 4 needs to be last

Obviously 0 can't be first and 0, 2, 4 needs to be last

DPatrick
2015-11-03 20:12:23

Right, those are our two restrictions: the first digit cannot be 0, and the last digit must be 0, 2, or 4.

Right, those are our two restrictions: the first digit cannot be 0, and the last digit must be 0, 2, or 4.

DPatrick
2015-11-03 20:12:36

In counting, we usually try to deal with the most restrictive condition first.

In counting, we usually try to deal with the most restrictive condition first.

bearytasty
2015-11-03 20:12:52

casework on the last digit

casework on the last digit

mathsolver101
2015-11-03 20:12:52

Cases when 0 is last and when 0 is not last

Cases when 0 is last and when 0 is not last

DPatrick
2015-11-03 20:13:09

Right: the condition that the last digit must be even is the most restrictive, so that's where we can focus our casework.

Right: the condition that the last digit must be even is the most restrictive, so that's where we can focus our casework.

DPatrick
2015-11-03 20:13:29

And then after that the first digit is the most restricted digit -- it can't be 0.

And then after that the first digit is the most restricted digit -- it can't be 0.

DPatrick
2015-11-03 20:13:42

If the last digit is 0, we have 5 choices for the first digit (any of 1-5).

But if the last digit is 2 or 4, we only have 4 choices for the first digit (any of 1-5 except for the 2 or 4).

If the last digit is 0, we have 5 choices for the first digit (any of 1-5).

But if the last digit is 2 or 4, we only have 4 choices for the first digit (any of 1-5 except for the 2 or 4).

DPatrick
2015-11-03 20:13:51

So how many choices do we have for the first and last digit combined?

So how many choices do we have for the first and last digit combined?

DeathLlama9
2015-11-03 20:14:24

5 + 2*4 = 13

5 + 2*4 = 13

bearytasty
2015-11-03 20:14:24

5+2*4 = 13

5+2*4 = 13

nosaj
2015-11-03 20:14:27

13

13

DPatrick
2015-11-03 20:14:31

We have $1 \cdot 5 + 2 \cdot 4 = 13$ possibilities.

We have $1 \cdot 5 + 2 \cdot 4 = 13$ possibilities.

DPatrick
2015-11-03 20:14:39

(Note we could also have counted this by considering that if we choose 1,3,5 for the first digit, then we have 3 choices for the last digit, but if we choose 2 or 4 for the first digit, then we have 2 choices for the last digit. That gives $3 \cdot 3 + 2 \cdot 2 = 13$ possibilities. It's also a nice check of our work so far!)

(Note we could also have counted this by considering that if we choose 1,3,5 for the first digit, then we have 3 choices for the last digit, but if we choose 2 or 4 for the first digit, then we have 2 choices for the last digit. That gives $3 \cdot 3 + 2 \cdot 2 = 13$ possibilities. It's also a nice check of our work so far!)

DeathLlama9
2015-11-03 20:15:01

Other than that it's just 4!

Other than that it's just 4!

Generic_Username
2015-11-03 20:15:08

We can note that there are 4! ways to order the middle four numbers regardless of what they are.

We can note that there are 4! ways to order the middle four numbers regardless of what they are.

DPatrick
2015-11-03 20:15:24

Indeed, there are no more restrictions (other than all the remaining digits each gets used once), and the remaining 4 digits can be placed in $4! = 24$ ways.

Indeed, there are no more restrictions (other than all the remaining digits each gets used once), and the remaining 4 digits can be placed in $4! = 24$ ways.

mathwhiz16
2015-11-03 20:15:38

so 13*24!

so 13*24!

mathwhiz16
2015-11-03 20:15:38

not 24 factorial, though

not 24 factorial, though

nosaj
2015-11-03 20:15:45

so the answer is $13\times 4!=312$.

so the answer is $13\times 4!=312$.

DPatrick
2015-11-03 20:15:48

So there are $13 \cdot 24 = \boxed{312}$ possible numbers.

So there are $13 \cdot 24 = \boxed{312}$ possible numbers.

DPatrick
2015-11-03 20:16:05

Next is another multiple-choice:

Next is another multiple-choice:

DPatrick
2015-11-03 20:16:09

7. Which of the following definitions of the binary operation $*$ on the nonzero rational numbers defines an associative operation? ("max" below denotes the maximum, if $m=n$, choose $m$)

(a) $m * n = m - n$

(b) $m * n = 2m + 4n$

(c) $m * n = m^n$

(d) $m * n = \max\{m,n\}$

7. Which of the following definitions of the binary operation $*$ on the nonzero rational numbers defines an associative operation? ("max" below denotes the maximum, if $m=n$, choose $m$)

(a) $m * n = m - n$

(b) $m * n = 2m + 4n$

(c) $m * n = m^n$

(d) $m * n = \max\{m,n\}$

DPatrick
2015-11-03 20:16:28

Of course, first we should be sure that we know what

Of course, first we should be sure that we know what

*associative*means.
mathwhiz16
2015-11-03 20:17:05

It means that (m*n)*p=m*(n*p)

It means that (m*n)*p=m*(n*p)

DeathLlama9
2015-11-03 20:17:05

Hint: Associative Property a * (b * c) = (a * b) * c

Hint: Associative Property a * (b * c) = (a * b) * c

thequantumguy
2015-11-03 20:17:05

such as ( a × b ) × c = a × ( b × c )

such as ( a × b ) × c = a × ( b × c )

DPatrick
2015-11-03 20:17:10

Associative means that $(a * b) * c = a * (b * c)$ for all $a,b,c$.

Associative means that $(a * b) * c = a * (b * c)$ for all $a,b,c$.

DPatrick
2015-11-03 20:17:17

In other words, we can write expressions like $a * b * c$ without worrying about which $*$ we have to do first.

In other words, we can write expressions like $a * b * c$ without worrying about which $*$ we have to do first.

DPatrick
2015-11-03 20:17:35

In order to show an operation is

In order to show an operation is

*not*associative, we just have to find one example that doesn't work.
DPatrick
2015-11-03 20:18:07

So unless we see the answer that we're sure is associative, we can just work through the answer choices one at a time and try to find an example that's not associative.

So unless we see the answer that we're sure is associative, we can just work through the answer choices one at a time and try to find an example that's not associative.

DPatrick
2015-11-03 20:18:21

Is (a) associative?

Is (a) associative?

DaVinci
2015-11-03 20:18:36

no

no

geniusofart
2015-11-03 20:18:36

No.

No.

Mlux
2015-11-03 20:18:36

no

no

flyrain
2015-11-03 20:18:45

even a=b=c=1 fails in (a)

even a=b=c=1 fails in (a)

mathwhiz16
2015-11-03 20:18:45

no, pick m=1, n=2, p=3

no, pick m=1, n=2, p=3

nosaj
2015-11-03 20:18:45

1-(1-1) is not equal to (1-1)-1

1-(1-1) is not equal to (1-1)-1

DPatrick
2015-11-03 20:18:49

No. It is not true in general that $(a-b)-c = a-(b-c)$.

No. It is not true in general that $(a-b)-c = a-(b-c)$.

DPatrick
2015-11-03 20:18:54

We would usually simplify this to $a-b-c = a-b+c$, and we can see that this is true only if $c$ is zero.

We would usually simplify this to $a-b-c = a-b+c$, and we can see that this is true only if $c$ is zero.

DPatrick
2015-11-03 20:19:05

Or, you could just test some numbers:

\begin{align*}

(1 * 2) * 3 &= (1-2) - 3 = -1 - 3 = -4, \\

1 * (2 * 3) &= 1 - (2-3) = 1 - (-1) = 2.

\end{align*}

Or, you could just test some numbers:

\begin{align*}

(1 * 2) * 3 &= (1-2) - 3 = -1 - 3 = -4, \\

1 * (2 * 3) &= 1 - (2-3) = 1 - (-1) = 2.

\end{align*}

DPatrick
2015-11-03 20:19:16

How about (b)?

How about (b)?

Benq
2015-11-03 20:19:40

no

no

Parni10
2015-11-03 20:19:40

No.

No.

Generic_Username
2015-11-03 20:19:40

No

No

nosaj
2015-11-03 20:19:40

nope

nope

DPatrick
2015-11-03 20:19:48

We can write the expressions:

\begin{align*}

(a * b) * c &= (2a + 4b) * c = 2(2a+4b) + 4c = 4a + 8b + 4c, \\

a * (b*c) &= a * (2b + 4c) = 2a + 4(2b+4c) = 2a + 8b + 16c.

\end{align*}

We can write the expressions:

\begin{align*}

(a * b) * c &= (2a + 4b) * c = 2(2a+4b) + 4c = 4a + 8b + 4c, \\

a * (b*c) &= a * (2b + 4c) = 2a + 4(2b+4c) = 2a + 8b + 16c.

\end{align*}

flyrain
2015-11-03 20:20:00

again a=b=c=1 failz

again a=b=c=1 failz

DPatrick
2015-11-03 20:20:06

Indeed, we can also plug in numbers:

\begin{align*}

(1 * 2) * 3 &= 10 * 3 = 32, \\

1 * (2 * 3) &= 1 * 16 = 66.

\end{align*}

Indeed, we can also plug in numbers:

\begin{align*}

(1 * 2) * 3 &= 10 * 3 = 32, \\

1 * (2 * 3) &= 1 * 16 = 66.

\end{align*}

DPatrick
2015-11-03 20:20:19

How about (c)?

How about (c)?

bearytasty
2015-11-03 20:20:40

nope

nope

geniusofart
2015-11-03 20:20:40

No.

No.

AShane2000
2015-11-03 20:20:40

no

no

CantonMathGuy
2015-11-03 20:20:40

no

no

DPatrick
2015-11-03 20:20:45

For (c), it's not clear that it's an operation at all!

For (c), it's not clear that it's an operation at all!

DPatrick
2015-11-03 20:20:50

In particular, what's $-1 * \dfrac12$?

In particular, what's $-1 * \dfrac12$?

DeathLlama9
2015-11-03 20:21:11

i?

i?

Mlux
2015-11-03 20:21:11

sqrt of -1

sqrt of -1

nosaj
2015-11-03 20:21:11

$i$

$i$

FTW
2015-11-03 20:21:11

$i$

$i$

Mlux
2015-11-03 20:21:11

$i$

$i$

Stephenpiano
2015-11-03 20:21:11

or Imaginary

or Imaginary

Ancy
2015-11-03 20:21:11

oh wow .-. $i$

oh wow .-. $i$

geniusofart
2015-11-03 20:21:11

$i$

$i$

DPatrick
2015-11-03 20:21:18

That's the complex number $i$, which is not a "nonzero rational number" as stated in the problem.

That's the complex number $i$, which is not a "nonzero rational number" as stated in the problem.

DPatrick
2015-11-03 20:21:26

But even if we restrict (c) to positive integers, it's not associative:

\begin{align*}

(2 * 3) * 4 &= (2^3) * 4 = 8 * 4 = 8^4 = 4096, \\

2 * (3*4) &= 2 * (3^4) = 2 * 81 = 2^{81} = 2417851639229258349412352.

\end{align*}

But even if we restrict (c) to positive integers, it's not associative:

\begin{align*}

(2 * 3) * 4 &= (2^3) * 4 = 8 * 4 = 8^4 = 4096, \\

2 * (3*4) &= 2 * (3^4) = 2 * 81 = 2^{81} = 2417851639229258349412352.

\end{align*}

DPatrick
2015-11-03 20:21:35

(I may have used a computer to calculate that last number. But even without a computer, it's clear that $2^{81}$ cannot equal $8^4 = 2^{12}$.)

(I may have used a computer to calculate that last number. But even without a computer, it's clear that $2^{81}$ cannot equal $8^4 = 2^{12}$.)

Mlux
2015-11-03 20:21:51

so it must be d

so it must be d

DPatrick
2015-11-03 20:21:57

By process of elimination, $\boxed{\text{(d)}}$ must be associative. Let's check and make sure.

By process of elimination, $\boxed{\text{(d)}}$ must be associative. Let's check and make sure.

bearytasty
2015-11-03 20:22:09

(d) works because regardless of what order you perform them in, it will always give u the maximum of the set. yay the answer is D

(d) works because regardless of what order you perform them in, it will always give u the maximum of the set. yay the answer is D

mathwhiz16
2015-11-03 20:22:19

It works because the maximum of three numbers stays the same no matter what order you check in

It works because the maximum of three numbers stays the same no matter what order you check in

DPatrick
2015-11-03 20:22:34

Right. By definition we have \[(a*b) * c = \max\{a,b\} * c = \max\{\max\{a,b\},c\}.\]

Right. By definition we have \[(a*b) * c = \max\{a,b\} * c = \max\{\max\{a,b\},c\}.\]

DPatrick
2015-11-03 20:22:42

So that means take the larger of $a$ and $b$, and then take the larger of that and $c$.

So that means take the larger of $a$ and $b$, and then take the larger of that and $c$.

DPatrick
2015-11-03 20:22:58

But we're just taking the largest of the three of $a$, $b$, and $c$.

But we're just taking the largest of the three of $a$, $b$, and $c$.

Matt17
2015-11-03 20:23:04

and if there's a tie, it will always go to the leftmost max

and if there's a tie, it will always go to the leftmost max

DPatrick
2015-11-03 20:23:18

Right, but ties don't matter either. In the same way, \[a * (b * c) = a * \max\{b,c\} = \max\{a,\max\{b,c\}\}, \] is also just the largest of $a$, $b$, and $c$.

Right, but ties don't matter either. In the same way, \[a * (b * c) = a * \max\{b,c\} = \max\{a,\max\{b,c\}\}, \] is also just the largest of $a$, $b$, and $c$.

Ancy
2015-11-03 20:23:27

so its just $max{a,b,c}$

so its just $max{a,b,c}$

DPatrick
2015-11-03 20:23:33

Right. Usually we'd just write it as $\max\{a,b,c\}$.

Right. Usually we'd just write it as $\max\{a,b,c\}$.

DPatrick
2015-11-03 20:23:36

So this operation is associative.

So this operation is associative.

mikebreen
2015-11-03 20:23:39

Nice work by everyone. You could almost guess this one based on the extra details we had to provide for the correct choice.

Nice work by everyone. You could almost guess this one based on the extra details we had to provide for the correct choice.

DPatrick
2015-11-03 20:24:00

On to #8:

On to #8:

DPatrick
2015-11-03 20:24:04

8. Let $a_0 = 10$ and for each positive integer $n$, let $a_n = 100a_{n-1} + (n+10)$. For how many $n$, $0 \le n \le 100$, is it true that $a_n$ is a multiple of 3?

8. Let $a_0 = 10$ and for each positive integer $n$, let $a_n = 100a_{n-1} + (n+10)$. For how many $n$, $0 \le n \le 100$, is it true that $a_n$ is a multiple of 3?

bearytasty
2015-11-03 20:24:34

reduce modulo 3?

reduce modulo 3?

Benq
2015-11-03 20:24:34

Take the equation mod 3.

Take the equation mod 3.

DPatrick
2015-11-03 20:25:08

This looks like a job for modular arithmetic, specifically arithmetic modulo 3.

This looks like a job for modular arithmetic, specifically arithmetic modulo 3.

DPatrick
2015-11-03 20:25:23

"Modulo 3" means we're just going to keep track of the remainders upon division by 3.

"Modulo 3" means we're just going to keep track of the remainders upon division by 3.

DPatrick
2015-11-03 20:25:31

So $a_0 \equiv 1 \pmod{3}$.

So $a_0 \equiv 1 \pmod{3}$.

CantonMathGuy
2015-11-03 20:25:36

reduce the recurrence mod $3$, so $a_n\equiv a_{n-1}+(n+1)$

reduce the recurrence mod $3$, so $a_n\equiv a_{n-1}+(n+1)$

mathwhiz16
2015-11-03 20:25:36

so you get $a_n=a_{n-1}+n+1$

so you get $a_n=a_{n-1}+n+1$

DeathLlama9
2015-11-03 20:25:36

$a_n = a_{n-1} + n + 1$

$a_n = a_{n-1} + n + 1$

DPatrick
2015-11-03 20:25:42

Right, since $100 \equiv 10 \equiv 1 \pmod{3}$, we have \[a_n \equiv a_{n-1} + n + 1 \pmod{3}.\]

Right, since $100 \equiv 10 \equiv 1 \pmod{3}$, we have \[a_n \equiv a_{n-1} + n + 1 \pmod{3}.\]

DeathLlama9
2015-11-03 20:25:54

Calculate a few values, find a pattern

Calculate a few values, find a pattern

nosaj
2015-11-03 20:25:54

Analyze the terms of the sequence modulo 3.

Analyze the terms of the sequence modulo 3.

geniusofart
2015-11-03 20:25:54

Can we see what happens by just listing out the first couple terms?

Can we see what happens by just listing out the first couple terms?

DPatrick
2015-11-03 20:25:59

Right: we can compute a few terms using this recursion and see if we find a pattern that repeats.

Right: we can compute a few terms using this recursion and see if we find a pattern that repeats.

DPatrick
2015-11-03 20:26:25

For example, by the formula, $a_1 \equiv a_0 + 1 + 1 \equiv 1 + 2 \equiv 0 \pmod{3}$.

For example, by the formula, $a_1 \equiv a_0 + 1 + 1 \equiv 1 + 2 \equiv 0 \pmod{3}$.

DeathLlama9
2015-11-03 20:26:43

$a_2 = 0 + 2 + 1$ which also works

$a_2 = 0 + 2 + 1$ which also works

Ancy
2015-11-03 20:26:43

the next one is also 0

the next one is also 0

DPatrick
2015-11-03 20:26:47

Again by the formula, $a_2 \equiv a_1 + 2 + 1 \equiv 0 + 3 \equiv 0 \pmod{3}$.

Again by the formula, $a_2 \equiv a_1 + 2 + 1 \equiv 0 + 3 \equiv 0 \pmod{3}$.

bearytasty
2015-11-03 20:27:09

1, 0, 0, 1, 0, 0 ...?

1, 0, 0, 1, 0, 0 ...?

Ancy
2015-11-03 20:27:14

teh next one is 1, then 0, 0, 1,0,0,1,0,0

teh next one is 1, then 0, 0, 1,0,0,1,0,0

Snowie
2015-11-03 20:27:22

I got remainder 1,0,0,1,0,0

I got remainder 1,0,0,1,0,0

DPatrick
2015-11-03 20:27:26

Again by the formula, $a_3 \equiv a_2 + 3 + 1 \equiv 0 + 1 \equiv 1 \pmod{3}$.

Again by the formula, $a_3 \equiv a_2 + 3 + 1 \equiv 0 + 1 \equiv 1 \pmod{3}$.

DPatrick
2015-11-03 20:27:48

Do we have enough terms to stop now and get the general pattern?

Do we have enough terms to stop now and get the general pattern?

AShane2000
2015-11-03 20:28:16

yes

yes

nosaj
2015-11-03 20:28:16

yep

yep

DPatrick
2015-11-03 20:28:28

Yes! Our cycle will now repeat, since each subsequent term depends only on the mod-3 residue of $a_{n-1}$ and $n$. Since these are equal for $n=1$ and $n=4$, the sequence will repeat from this point.

Yes! Our cycle will now repeat, since each subsequent term depends only on the mod-3 residue of $a_{n-1}$ and $n$. Since these are equal for $n=1$ and $n=4$, the sequence will repeat from this point.

DPatrick
2015-11-03 20:28:38

So (starting at $n=0$) the sequence of residues (mod 3) are: \[1,0,0,1,0,0,1,0,0,1,\ldots.\]

So (starting at $n=0$) the sequence of residues (mod 3) are: \[1,0,0,1,0,0,1,0,0,1,\ldots.\]

DPatrick
2015-11-03 20:29:03

That is, $a_n$ is a multiple of 3 if and only if $n$ is

That is, $a_n$ is a multiple of 3 if and only if $n$ is

*not*a multiple of 3.
DPatrick
2015-11-03 20:29:12

How do we finish the problem from here?

How do we finish the problem from here?

nosaj
2015-11-03 20:29:31

count the number of numbers between 0 and 100 that aren't multiples of 3

count the number of numbers between 0 and 100 that aren't multiples of 3

DPatrick
2015-11-03 20:29:35

We need to count the non-multiples of 3 among $0 \le n \le 100$.

We need to count the non-multiples of 3 among $0 \le n \le 100$.

DPatrick
2015-11-03 20:29:41

You have to be really careful at this final step to not be off by one!

You have to be really careful at this final step to not be off by one!

Ancy
2015-11-03 20:29:59

Count the multiples of three, subtract from 101

Count the multiples of three, subtract from 101

bearytasty
2015-11-03 20:29:59

we can use complementary counting. 101 - 34 = 67?

we can use complementary counting. 101 - 34 = 67?

mathwhiz16
2015-11-03 20:29:59

101-34

101-34

geniusofart
2015-11-03 20:29:59

101-34

101-34

DeathLlama9
2015-11-03 20:30:06

101 - (multiples of three), which are 0 to 99 or 34, so 101 - 34 = $\boxed{67}$

101 - (multiples of three), which are 0 to 99 or 34, so 101 - 34 = $\boxed{67}$

DPatrick
2015-11-03 20:30:10

Good! There are 101 total numbers in the range $0 \le n \le 100$.

Good! There are 101 total numbers in the range $0 \le n \le 100$.

DPatrick
2015-11-03 20:30:20

Since $n$ can range from $0 \cdot 3$ to $33 \cdot 3$, there are 34 multiples of 3 in this list.

Since $n$ can range from $0 \cdot 3$ to $33 \cdot 3$, there are 34 multiples of 3 in this list.

DPatrick
2015-11-03 20:30:27

So there are $101 - 34 = \boxed{67}$ non-multiples of 3 in our list.

So there are $101 - 34 = \boxed{67}$ non-multiples of 3 in our list.

mikebreen
2015-11-03 20:30:45

Funny that most people that missed, missed by getting 34 or 33, not 66.

Funny that most people that missed, missed by getting 34 or 33, not 66.

DPatrick
2015-11-03 20:31:00

Indeed, answering the wrong question is also a common way to end up with a wrong answer.

Indeed, answering the wrong question is also a common way to end up with a wrong answer.

DPatrick
2015-11-03 20:31:14

That's why my First Rule of Problem Solving is:

That's why my First Rule of Problem Solving is:

**Read The Problem Carefully!**
DPatrick
2015-11-03 20:31:45

And that's also probably why this was the 2nd-most missed problem on the contest.

And that's also probably why this was the 2nd-most missed problem on the contest.

DPatrick
2015-11-03 20:31:57

9. Suppose $\sqrt{9+4\sqrt5} = a + b\sqrt5$ where $a$ and $b$ are integers. Find $a+b$.

9. Suppose $\sqrt{9+4\sqrt5} = a + b\sqrt5$ where $a$ and $b$ are integers. Find $a+b$.

bearytasty
2015-11-03 20:32:25

square it

square it

Generic_Username
2015-11-03 20:32:25

Square both sides

Square both sides

CantonMathGuy
2015-11-03 20:32:25

square

square

FTW
2015-11-03 20:32:25

square it

square it

DeathLlama9
2015-11-03 20:32:25

Square both sides of the equation!

Square both sides of the equation!

mathsolver101
2015-11-03 20:32:25

Square both sides!

Square both sides!

DPatrick
2015-11-03 20:32:41

Sure -- the easiest way to get rid of a square root is to square both sides.

Sure -- the easiest way to get rid of a square root is to square both sides.

DPatrick
2015-11-03 20:32:44

What happens if we square the equation $\sqrt{9+4\sqrt5} = a + b\sqrt5$?

What happens if we square the equation $\sqrt{9+4\sqrt5} = a + b\sqrt5$?

DeathLlama9
2015-11-03 20:33:15

You get $9 + 4\sqrt{5} = a^2 + 5b^2 + 2ab\sqrt{5}$

You get $9 + 4\sqrt{5} = a^2 + 5b^2 + 2ab\sqrt{5}$

Eugenis
2015-11-03 20:33:15

you get $9+4\sqrt{5}=a^2+2ab\sqrt{5}+5b^2$

you get $9+4\sqrt{5}=a^2+2ab\sqrt{5}+5b^2$

mihirb
2015-11-03 20:33:15

$9+4\sqrt{5} = a^2+2ab\sqrt{5}+5b^2$

$9+4\sqrt{5} = a^2+2ab\sqrt{5}+5b^2$

FTW
2015-11-03 20:33:15

$9+4\sqrt{5}=a^2+5b^2+2\sqrt{5}ab$

$9+4\sqrt{5}=a^2+5b^2+2\sqrt{5}ab$

thepiercingarrow
2015-11-03 20:33:15

We get $9+4\sqrt5=a^2+5b^2+2ab\sqrt5$

We get $9+4\sqrt5=a^2+5b^2+2ab\sqrt5$

bearytasty
2015-11-03 20:33:15

9 + 4sqrt(5) = (a^2 + 5b^2) + 2absqrt(5)

9 + 4sqrt(5) = (a^2 + 5b^2) + 2absqrt(5)

Liopleurodon
2015-11-03 20:33:15

9+4sqrt5=a^2+5b^2+2absqrt5

9+4sqrt5=a^2+5b^2+2absqrt5

DPatrick
2015-11-03 20:33:21

We get \[9 + 4\sqrt5 = \left(a + b\sqrt5\right)^2 = a^2 + 2a(b\sqrt5) + (b\sqrt5)^2.\] (Notice that we're using the expansion $(x+y)^2 = x^2 + 2xy + y^2$ on the right.)

We get \[9 + 4\sqrt5 = \left(a + b\sqrt5\right)^2 = a^2 + 2a(b\sqrt5) + (b\sqrt5)^2.\] (Notice that we're using the expansion $(x+y)^2 = x^2 + 2xy + y^2$ on the right.)

DPatrick
2015-11-03 20:33:31

That simplifies to $9 + 4\sqrt5 = a^2 + 2ab\sqrt5 + 5b^2$.

That simplifies to $9 + 4\sqrt5 = a^2 + 2ab\sqrt5 + 5b^2$.

DPatrick
2015-11-03 20:33:38

What can we conclude?

What can we conclude?

geniusofart
2015-11-03 20:33:52

Clearly the $2ab\sqrt5$ term must be $4\sqrt5$.

Clearly the $2ab\sqrt5$ term must be $4\sqrt5$.

mathsolver101
2015-11-03 20:33:52

We can match the integer parts on the RHS with the integer parts on the LHS and the with the radical parts

We can match the integer parts on the RHS with the integer parts on the LHS and the with the radical parts

mihirb
2015-11-03 20:33:52

$a^2+5b^2 = 9$, $2ab = 4$

$a^2+5b^2 = 9$, $2ab = 4$

mathwhiz16
2015-11-03 20:33:52

So $a^2+5b^2=9$ and $2ab=4$.

So $a^2+5b^2=9$ and $2ab=4$.

Eugenis
2015-11-03 20:33:52

compare coefficients

compare coefficients

FTW
2015-11-03 20:33:56

$a^2+5b^2=9$, $ab=2$

$a^2+5b^2=9$, $ab=2$

DPatrick
2015-11-03 20:34:00

Exactly. The integer parts must be equal, and the $\sqrt5$ parts must be equal.

Exactly. The integer parts must be equal, and the $\sqrt5$ parts must be equal.

DPatrick
2015-11-03 20:34:06

That is, $a^2 + 5b^2 = 9$ and $2ab = 4$.

That is, $a^2 + 5b^2 = 9$ and $2ab = 4$.

DPatrick
2015-11-03 20:34:12

The second equation is just $ab=2$, and that doesn't have a whole lot of solutions in integers.

The second equation is just $ab=2$, and that doesn't have a whole lot of solutions in integers.

Dragon2kz
2015-11-03 20:34:28

a=2 b=1

a=2 b=1

va2010
2015-11-03 20:34:28

a = 2 b = 1

a = 2 b = 1

FTW
2015-11-03 20:34:28

$a=2, b=1$

$a=2, b=1$

CantonMathGuy
2015-11-03 20:34:28

$2$ is small so we use trial and error to get $(a,b)=(2,1)$

$2$ is small so we use trial and error to get $(a,b)=(2,1)$

DPatrick
2015-11-03 20:34:36

We can quickly see that $a=2$ and $b=1$ is the solution to both equations.

We can quickly see that $a=2$ and $b=1$ is the solution to both equations.

DPatrick
2015-11-03 20:34:44

So $\sqrt{9 + 4\sqrt5} = 2 + \sqrt5$.

So $\sqrt{9 + 4\sqrt5} = 2 + \sqrt5$.

DPatrick
2015-11-03 20:34:58

And that gives $a+b = 2+1 = \boxed{3}$.

And that gives $a+b = 2+1 = \boxed{3}$.

CantonMathGuy
2015-11-03 20:35:08

also obviously $(a,b)$ is not $(-2,-1)$ since $-2-\sqrt{5}<0$

also obviously $(a,b)$ is not $(-2,-1)$ since $-2-\sqrt{5}<0$

DPatrick
2015-11-03 20:35:24

Right. Although that works in our equation, we must remember that the $\sqrt{\phantom{2}}$ symbol, by definition, means the

Right. Although that works in our equation, we must remember that the $\sqrt{\phantom{2}}$ symbol, by definition, means the

**positive**square root of a number. So we must have $a + b\sqrt5 > 0$.
DPatrick
2015-11-03 20:35:45

In a not-obvious way, problems 9 and 10 are related, as I'll show you in a minute...

In a not-obvious way, problems 9 and 10 are related, as I'll show you in a minute...

DPatrick
2015-11-03 20:35:50

10. How many elements of the set $\{11,111,1111,\ldots,1111111111\}$ (base 10; starting with eleven, then one hundred eleven, etc. 10 1's in the last number) are prime?

10. How many elements of the set $\{11,111,1111,\ldots,1111111111\}$ (base 10; starting with eleven, then one hundred eleven, etc. 10 1's in the last number) are prime?

DPatrick
2015-11-03 20:36:11

We have the numbers from "two 1's" to "ten 1's" in our set, so we have 9 numbers.

We have the numbers from "two 1's" to "ten 1's" in our set, so we have 9 numbers.

DPatrick
2015-11-03 20:36:17

Are any of them easy?

Are any of them easy?

DaVinci
2015-11-03 20:36:37

11

11

gxah
2015-11-03 20:36:37

11 is prime

11 is prime

ninjataco
2015-11-03 20:36:37

11

11

nosaj
2015-11-03 20:36:37

yes 11 is obviously prime

yes 11 is obviously prime

minimario
2015-11-03 20:36:37

11 easy

11 easy

mathsolver101
2015-11-03 20:36:37

11

11

DPatrick
2015-11-03 20:36:41

11 is prime!

11 is prime!

Snowie
2015-11-03 20:37:00

Obviously you can only have an odd number of digits for that to work

Obviously you can only have an odd number of digits for that to work

DPatrick
2015-11-03 20:37:12

Right: all the other numbers with an even number of 1's are multiples of 11 (and thus not prime):

1111 = 11 * 101

111111 = 11 * 10101

11111111 = 11 * 1010101

1111111111 = 11 * 101010101

Right: all the other numbers with an even number of 1's are multiples of 11 (and thus not prime):

1111 = 11 * 101

111111 = 11 * 10101

11111111 = 11 * 1010101

1111111111 = 11 * 101010101

thepiercingarrow
2015-11-03 20:37:37

those with multiple of 3 digits are divisible by 3, 11 is prime...

those with multiple of 3 digits are divisible by 3, 11 is prime...

mathwhiz16
2015-11-03 20:37:37

we can see right off the bat 111, 111111, and 111111111 aren’t prime-divisible by 3

we can see right off the bat 111, 111111, and 111111111 aren’t prime-divisible by 3

Mlux
2015-11-03 20:37:37

multiples of $3$

multiples of $3$

thepiercingarrow
2015-11-03 20:37:37

3 ones, 6 ones, 9 ones....... are NOT prime

3 ones, 6 ones, 9 ones....... are NOT prime

va2010
2015-11-03 20:37:37

111, 111111, and 111111111 all divide 3, so their out

111, 111111, and 111111111 all divide 3, so their out

DeathLlama9
2015-11-03 20:37:37

$111$ and $111111111$ are both div. by three

$111$ and $111111111$ are both div. by three

DPatrick
2015-11-03 20:37:44

If the number of 1's is a multiple of 3, then the digit sum of the number is a multiple of 3, so the number itself is a multiple of 3.

If the number of 1's is a multiple of 3, then the digit sum of the number is a multiple of 3, so the number itself is a multiple of 3.

DPatrick
2015-11-03 20:37:50

Thus, 111 and 111111111 (nine 1's) are each multiples of 3, and not prime.

Thus, 111 and 111111111 (nine 1's) are each multiples of 3, and not prime.

CantonMathGuy
2015-11-03 20:37:57

111111111 = 111 * 1001001 also

111111111 = 111 * 1001001 also

DPatrick
2015-11-03 20:38:05

Right, there's that too.

Right, there's that too.

mathwhiz16
2015-11-03 20:38:11

that leaves us with 11, 11111, and 1111111

that leaves us with 11, 11111, and 1111111

nosaj
2015-11-03 20:38:11

so just 11111 and 1111111 to check now

so just 11111 and 1111111 to check now

DPatrick
2015-11-03 20:38:15

So that leaves just 11111 and 1111111 to check.

So that leaves just 11111 and 1111111 to check.

DPatrick
2015-11-03 20:38:29

Unfortunately, I don't know of an easy way of determining whether these are prime or composite.

Unfortunately, I don't know of an easy way of determining whether these are prime or composite.

eveningstarandlion
2015-11-03 20:38:36

11111 is divisible by 41 -_-

11111 is divisible by 41 -_-

DPatrick
2015-11-03 20:38:45

That's right. It turns out that 11111 = 41 * 271, so five 1's is composite.

That's right. It turns out that 11111 = 41 * 271, so five 1's is composite.

DPatrick
2015-11-03 20:39:03

Let me show you one possible systematic approach by using some algebra.

Let me show you one possible systematic approach by using some algebra.

DPatrick
2015-11-03 20:39:16

Specifically, if $f(x) = x^4 + x^3 + x^2 + x + 1$, then $f(10) = 11111$ is the number we're considering.

Specifically, if $f(x) = x^4 + x^3 + x^2 + x + 1$, then $f(10) = 11111$ is the number we're considering.

DPatrick
2015-11-03 20:39:26

So we could try to factor $f(x) = x^4 + x^3 + x^2 + x + 1$ algebraically.

So we could try to factor $f(x) = x^4 + x^3 + x^2 + x + 1$ algebraically.

CantonMathGuy
2015-11-03 20:39:41

except you can't over the integers

except you can't over the integers

DPatrick
2015-11-03 20:39:48

True.

True.

DPatrick
2015-11-03 20:39:54

Since all the roots are complex numbers, this won't factor into linear factors over the reals.

Since all the roots are complex numbers, this won't factor into linear factors over the reals.

DPatrick
2015-11-03 20:40:08

By the way, before we go on, what are the roots of $f(x)$?

By the way, before we go on, what are the roots of $f(x)$?

mihirb
2015-11-03 20:40:28

5th roots of unity

5th roots of unity

Mlux
2015-11-03 20:40:28

roots of unity

roots of unity

nosaj
2015-11-03 20:40:28

roots of unity?

roots of unity?

Generic_Username
2015-11-03 20:40:28

5th roots of unity

5th roots of unity

Benq
2015-11-03 20:40:30

5th roots of unity besides 1

5th roots of unity besides 1

DPatrick
2015-11-03 20:40:35

They're the primitive 5th roots of unity. That is, its roots are the complex numbers that are the 5th roots of 1. (We say that $f(x)$ is the 5th

They're the primitive 5th roots of unity. That is, its roots are the complex numbers that are the 5th roots of 1. (We say that $f(x)$ is the 5th

*cyclotomic polynomial*.) That's because we could write $f(x) = \dfrac{x^5-1}{x-1}$ for all $x \not= 1$.
DPatrick
2015-11-03 20:41:01

So any $x \not= 1$ that's a root must satisfy $x^5 - 1 = 0$, or $x^5 = 1$.

So any $x \not= 1$ that's a root must satisfy $x^5 - 1 = 0$, or $x^5 = 1$.

DPatrick
2015-11-03 20:41:23

So, since all the roots are complex numbers, this won't factor into linear factors over the reals. But we should be able to factor it as the product of two quadratic factors over the reals!

So, since all the roots are complex numbers, this won't factor into linear factors over the reals. But we should be able to factor it as the product of two quadratic factors over the reals!

DPatrick
2015-11-03 20:41:38

One way to do that is to use the fact (if you know it, or can derive it) that $\cos(72^\circ) = \dfrac{\sqrt5 - 1}{4}$.

One way to do that is to use the fact (if you know it, or can derive it) that $\cos(72^\circ) = \dfrac{\sqrt5 - 1}{4}$.

Generic_Username
2015-11-03 20:41:45

quadratic factors with undetermined coefficients

quadratic factors with undetermined coefficients

DPatrick
2015-11-03 20:41:55

Another, perhaps more straightforward way, is to use undetermined coefficients: suppose \[f(x) = (x^2 + ax + 1)(x^2 + bx + 1)\] for some unknown $a$ and $b$.

Another, perhaps more straightforward way, is to use undetermined coefficients: suppose \[f(x) = (x^2 + ax + 1)(x^2 + bx + 1)\] for some unknown $a$ and $b$.

DPatrick
2015-11-03 20:42:27

If we expand the right side out, we get $$f(x) = x^4 + (a+b)x^3 + (2+ab)x^2 + (a+b)x + 1.$$

If we expand the right side out, we get $$f(x) = x^4 + (a+b)x^3 + (2+ab)x^2 + (a+b)x + 1.$$

mathwhiz16
2015-11-03 20:42:51

a+b=1, and 2+ab=1

a+b=1, and 2+ab=1

vatatmaja
2015-11-03 20:42:59

ab=-1, a+b=1

ab=-1, a+b=1

DPatrick
2015-11-03 20:43:01

Right: matching this up to the coefficients of $f(x)$, we get $a+b = 1$ and $2+ab = 1$, hence $ab = -1$.

Right: matching this up to the coefficients of $f(x)$, we get $a+b = 1$ and $2+ab = 1$, hence $ab = -1$.

DPatrick
2015-11-03 20:43:15

But what does that tell us about $a$ and $b$?

But what does that tell us about $a$ and $b$?

Ancy
2015-11-03 20:43:48

plug it in to a quadratic equation to get the roots using vietas

plug it in to a quadratic equation to get the roots using vietas

DPatrick
2015-11-03 20:43:57

Right! By Vieta's Formulas, $a$ and $b$ are the roots of the quadratic $(t-a)(t-b) = t^2 - (a+b)t + ab = t^2 -t - 1 = 0$.

Right! By Vieta's Formulas, $a$ and $b$ are the roots of the quadratic $(t-a)(t-b) = t^2 - (a+b)t + ab = t^2 -t - 1 = 0$.

DPatrick
2015-11-03 20:44:24

And we can use the quadratic formula to find those roots: \[t = \dfrac{1 \pm \sqrt5}{2}.\]

And we can use the quadratic formula to find those roots: \[t = \dfrac{1 \pm \sqrt5}{2}.\]

DPatrick
2015-11-03 20:44:37

Therefore, \[f(x) = \left(x^2 + \left(\dfrac{1+\sqrt5}{2}\right)x + 1\right)\left(x^2 + \left(\dfrac{1-\sqrt5}{2}\right)x + 1\right).\]

Therefore, \[f(x) = \left(x^2 + \left(\dfrac{1+\sqrt5}{2}\right)x + 1\right)\left(x^2 + \left(\dfrac{1-\sqrt5}{2}\right)x + 1\right).\]

DPatrick
2015-11-03 20:45:13

And now we go back to our original plan, and plug in $x=10$:

And now we go back to our original plan, and plug in $x=10$:

DPatrick
2015-11-03 20:45:21

We get:

\begin{align*}

11111 = f(10) &= \left(100 + 5(1+\sqrt5)+1\right)\left(100+5(1-\sqrt5)+1\right) \\&= (106+5\sqrt5)(106-5\sqrt5).

\end{align*}

We get:

\begin{align*}

11111 = f(10) &= \left(100 + 5(1+\sqrt5)+1\right)\left(100+5(1-\sqrt5)+1\right) \\&= (106+5\sqrt5)(106-5\sqrt5).

\end{align*}

DPatrick
2015-11-03 20:45:44

Big deal, perhaps: that's not an integer factorization.

Big deal, perhaps: that's not an integer factorization.

DeathLlama9
2015-11-03 20:45:49

Diff of squares = 106^2 - 125 = 11236 - 125 = 11111

Diff of squares = 106^2 - 125 = 11236 - 125 = 11111

lkarhat
2015-11-03 20:45:56

106^2 - 125

106^2 - 125

DPatrick
2015-11-03 20:46:00

Right: that verifies that what we did worked so far.

Right: that verifies that what we did worked so far.

DPatrick
2015-11-03 20:46:17

Now if you're really clever, and make good guesses, you might notice that

\begin{align*}

106 + 5\sqrt5 &= (11+4\sqrt5)(26-9\sqrt5), \\

106 - 5\sqrt5 &= (11-4\sqrt5)(26+9\sqrt5). \\

\end{align*}

Now if you're really clever, and make good guesses, you might notice that

\begin{align*}

106 + 5\sqrt5 &= (11+4\sqrt5)(26-9\sqrt5), \\

106 - 5\sqrt5 &= (11-4\sqrt5)(26+9\sqrt5). \\

\end{align*}

CantonMathGuy
2015-11-03 20:46:47

now multiply the $11\pm4\sqrt{5}$ together and same for the other two

now multiply the $11\pm4\sqrt{5}$ together and same for the other two

DPatrick
2015-11-03 20:46:53

And now for the punchline: if we recombine the terms, we get that \[(11+4\sqrt5)(11-4\sqrt5) = 11^2 - (4\sqrt5)^2 = 121 - 80 = 41\] is a factor of 11111, and so is \[(26+9\sqrt5)(26-9\sqrt5) = 26^2 - (9\sqrt5)^2 = 676 - 405 = 271.\]

And now for the punchline: if we recombine the terms, we get that \[(11+4\sqrt5)(11-4\sqrt5) = 11^2 - (4\sqrt5)^2 = 121 - 80 = 41\] is a factor of 11111, and so is \[(26+9\sqrt5)(26-9\sqrt5) = 26^2 - (9\sqrt5)^2 = 676 - 405 = 271.\]

DPatrick
2015-11-03 20:47:06

Thus 11111 = 41 * 271, so 11111 is composite.

Thus 11111 = 41 * 271, so 11111 is composite.

DaVinci
2015-11-03 20:47:25

is this really the most efficient way to solve the problem?

is this really the most efficient way to solve the problem?

DPatrick
2015-11-03 20:47:35

Probably not. But it's cool!

Probably not. But it's cool!

DPatrick
2015-11-03 20:47:41

This is all an example of

This is all an example of

**algebraic number theory**, and is a really deep and interesting field of mathematics.
vatatmaja
2015-11-03 20:48:00

How would we get that factorization?

How would we get that factorization?

DPatrick
2015-11-03 20:48:16

It's much like any factorization: on some level it becomes educated guesswork.

It's much like any factorization: on some level it becomes educated guesswork.

DPatrick
2015-11-03 20:48:28

Anyway, how about 1111111?

Anyway, how about 1111111?

eveningstarandlion
2015-11-03 20:48:44

divisible by 239 I think

divisible by 239 I think

Generic_Username
2015-11-03 20:48:47

same thing?

same thing?

Liopleurodon
2015-11-03 20:48:47

same thing but longer????????????????????????????

same thing but longer????????????????????????????

DPatrick
2015-11-03 20:48:52

It turns out that 1111111 = 239 * 4649, and in fact that is its prime factorization. So 1111111 is composite too.

It turns out that 1111111 = 239 * 4649, and in fact that is its prime factorization. So 1111111 is composite too.

DPatrick
2015-11-03 20:49:11

We can do the same sort of algebra to factor 1111111 too.

We can do the same sort of algebra to factor 1111111 too.

DPatrick
2015-11-03 20:49:26

We can factor this using difference of squares, so long as we introduce the complex number $\sqrt{-7}$ as a factor:

\[ 1111111 = (1044 + 55\sqrt{-7})(1044 - 55\sqrt{-7}).\]

We can factor this using difference of squares, so long as we introduce the complex number $\sqrt{-7}$ as a factor:

\[ 1111111 = (1044 + 55\sqrt{-7})(1044 - 55\sqrt{-7}).\]

DPatrick
2015-11-03 20:49:41

And now you have to get especially creative to use:

\begin{align*}

1044 + 55\sqrt{-7} &= (8 + 5\sqrt{-7})(43 - 20\sqrt{-7}), \\

1044 - 55\sqrt{-7} &= (8 - 5\sqrt{-7})(43 + 20\sqrt{-7}).

\end{align*}

And now you have to get especially creative to use:

\begin{align*}

1044 + 55\sqrt{-7} &= (8 + 5\sqrt{-7})(43 - 20\sqrt{-7}), \\

1044 - 55\sqrt{-7} &= (8 - 5\sqrt{-7})(43 + 20\sqrt{-7}).

\end{align*}

DPatrick
2015-11-03 20:49:59

So if we recombine the terms, we get that \[(8+5\sqrt{-7})(8-5\sqrt{-7}) = 8^2 - (5\sqrt{-7})^2 = 64 + 5 \cdot 49 = 239\] is a factor of 1111111, and so is \[(43+20\sqrt{-7})(43-20\sqrt{-7}) = 43^2 - (20\sqrt{-7})^2 = 1849 + 7 \cdot 400 = 4649.\]

So if we recombine the terms, we get that \[(8+5\sqrt{-7})(8-5\sqrt{-7}) = 8^2 - (5\sqrt{-7})^2 = 64 + 5 \cdot 49 = 239\] is a factor of 1111111, and so is \[(43+20\sqrt{-7})(43-20\sqrt{-7}) = 43^2 - (20\sqrt{-7})^2 = 1849 + 7 \cdot 400 = 4649.\]

Eugenis
2015-11-03 20:50:10

it turns out many identities are a consequence of trying to factor something as a difference of squares

it turns out many identities are a consequence of trying to factor something as a difference of squares

DPatrick
2015-11-03 20:50:47

Indeed, a lot of this particular area of algebraic number theory comes down to factoring $a^2 + pb^2$ where $p$ is an odd prime.

Indeed, a lot of this particular area of algebraic number theory comes down to factoring $a^2 + pb^2$ where $p$ is an odd prime.

DPatrick
2015-11-03 20:50:53

Or $a^2 - pb^2$.

Or $a^2 - pb^2$.

nosaj
2015-11-03 20:51:14

Here's an interesting blog post relating to this problem: https://dyinglovegrape.wordpress.com/2011/07/01/when-is-111111-prime/

Here's an interesting blog post relating to this problem: https://dyinglovegrape.wordpress.com/2011/07/01/when-is-111111-prime/

DPatrick
2015-11-03 20:51:25

A number that is composed of all 1's as its digits is called a

A number that is composed of all 1's as its digits is called a

**repunit**.
DPatrick
2015-11-03 20:51:39

The first repunit after 11 that's prime is 1,111,111,111,111,111,111. (That's 19 1's.)

The first repunit after 11 that's prime is 1,111,111,111,111,111,111. (That's 19 1's.)

DPatrick
2015-11-03 20:51:47

(17 1's fools a lot of people, but the prime factorization is 11,111,111,111,111,111 = 2,071,723 * 5,363,222,357.)

(17 1's fools a lot of people, but the prime factorization is 11,111,111,111,111,111 = 2,071,723 * 5,363,222,357.)

DPatrick
2015-11-03 20:52:02

It is an open question as to whether there are infinitely many repunit primes or not. The largest known repunit prime is 1031 1's.

It is an open question as to whether there are infinitely many repunit primes or not. The largest known repunit prime is 1031 1's.

DPatrick
2015-11-03 20:52:30

At any rate, only 11 in our original list is prime. The answer is that only $\boxed{1}$ number in the list is prime.

At any rate, only 11 in our original list is prime. The answer is that only $\boxed{1}$ number in the list is prime.

DPatrick
2015-11-03 20:52:48

By the way: if we consider this problem in base 2, rather than base 10, what do we get?

By the way: if we consider this problem in base 2, rather than base 10, what do we get?

DPatrick
2015-11-03 20:53:17

That is, if we look at $111\ldots1_2$ and there are $n$ 1's, what is this number?

That is, if we look at $111\ldots1_2$ and there are $n$ 1's, what is this number?

va2010
2015-11-03 20:53:30

mersenne

mersenne

trumpeter
2015-11-03 20:53:30

mersenne numbers

mersenne numbers

mathwhiz16
2015-11-03 20:53:30

2^n-1

2^n-1

DPatrick
2015-11-03 20:53:35

Right. It's $2^{n-1} + 2^{n-2} + \cdots + 2 + 1$.

Right. It's $2^{n-1} + 2^{n-2} + \cdots + 2 + 1$.

DPatrick
2015-11-03 20:53:41

But we can write sum this finite geometric series using the same technique that we saw in problem #3! This sums to $\dfrac{2^n-1}{2-1} = 2^n-1$.

But we can write sum this finite geometric series using the same technique that we saw in problem #3! This sums to $\dfrac{2^n-1}{2-1} = 2^n-1$.

DPatrick
2015-11-03 20:53:48

You might know that primes of the form $2^n-1$ are called

You might know that primes of the form $2^n-1$ are called

**Mersenne primes**(after the French mathematician Marin Mersenne).
DPatrick
2015-11-03 20:53:58

In fact, the largest known prime number is the Mersenne prime $2^{57885161} - 1$, which has 17425170 digits.

In fact, the largest known prime number is the Mersenne prime $2^{57885161} - 1$, which has 17425170 digits.

DPatrick
2015-11-03 20:54:11

This was the hardest problem on the contest. Only 17% of the participants got it correct.

This was the hardest problem on the contest. Only 17% of the participants got it correct.

DPatrick
2015-11-03 20:54:21

That's it for Round 2!

That's it for Round 2!

DPatrick
2015-11-03 20:54:29

The top scorer on Round 2 in each 9 regions, plus the top scorer in the Seattle area, will advance to the National semifinals, to be held live at the 2016 Joint Mathematics Meetings in Seattle on January 7. The National semifinals and finals will also be live-streamed on the web. The AMS should be announcing the National semifinalists next week. Perhaps I'll see you there!

The top scorer on Round 2 in each 9 regions, plus the top scorer in the Seattle area, will advance to the National semifinals, to be held live at the 2016 Joint Mathematics Meetings in Seattle on January 7. The National semifinals and finals will also be live-streamed on the web. The AMS should be announcing the National semifinalists next week. Perhaps I'll see you there!

DPatrick
2015-11-03 20:54:43

There are also going to be regional versions of the game this spring in Rhode Island, Connecticut, and Washington, DC. Local teachers will be contacted about these regional games.

There are also going to be regional versions of the game this spring in Rhode Island, Connecticut, and Washington, DC. Local teachers will be contacted about these regional games.

DPatrick
2015-11-03 20:54:53

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

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

nosaj
2015-11-03 20:55:50

Can you give us the statistics (percent correct) for each problem?

Can you give us the statistics (percent correct) for each problem?

DPatrick
2015-11-03 20:56:09

Yep, Mike sent me these statistics this morning:

Yep, Mike sent me these statistics this morning:

DPatrick
2015-11-03 20:56:13

#1 82.9%

#1 82.9%

DPatrick
2015-11-03 20:56:17

#2 67.1%

#2 67.1%

DPatrick
2015-11-03 20:56:20

#3 78.6%

#3 78.6%

DPatrick
2015-11-03 20:56:25

#4 91.0%

#4 91.0%

DPatrick
2015-11-03 20:56:27

#5 50.3%

#5 50.3%

DPatrick
2015-11-03 20:56:30

#6 56.9%

#6 56.9%

DPatrick
2015-11-03 20:56:33

#7 66.2%

#7 66.2%

DPatrick
2015-11-03 20:56:36

#8 38.2%

#8 38.2%

DPatrick
2015-11-03 20:56:39

#9 71.4%

#9 71.4%

DPatrick
2015-11-03 20:56:42

#10 17.1%

#10 17.1%

mikebreen
2015-11-03 20:56:53

#1 and #10 were perfect complements

#1 and #10 were perfect complements

ninjataco
2015-11-03 20:57:25

do you happen to have score breakdowns (ie what percentage of people got each score)?

do you happen to have score breakdowns (ie what percentage of people got each score)?

DPatrick
2015-11-03 20:57:34

I don't have that info -- I just have the question data.

I don't have that info -- I just have the question data.

pi37
2015-11-03 20:57:45

do you have the top scores by region?

do you have the top scores by region?

DPatrick
2015-11-03 20:57:52

They are still working those out.

They are still working those out.

AShane2000
2015-11-03 20:57:58

i was wondering how a hypothetical tiebreaker between two seniors with perfect scores would work out?

i was wondering how a hypothetical tiebreaker between two seniors with perfect scores would work out?

DPatrick
2015-11-03 20:58:07

This is on the AMS website ... let me go grab it...

This is on the AMS website ... let me go grab it...

DPatrick
2015-11-03 20:58:31

Tie-breakers. Because there are so few semifinal slots available for the national game and so many good students across the country, we often have many ties for first in a region, even many with perfect scores. In addition to the scores, further considerations for contest selection are gender (an effort is made to include an equal number of males and females); grade (a high school senior receives preference over younger students who will have another chance before they graduate); the correct answer on a particular qualifying test question (varies from year to year, usually question 10); and the free-form answers to the non-math questions. - See more at: http://www.ams.org/programs/students/wwtbam/qualify#ties

Tie-breakers. Because there are so few semifinal slots available for the national game and so many good students across the country, we often have many ties for first in a region, even many with perfect scores. In addition to the scores, further considerations for contest selection are gender (an effort is made to include an equal number of males and females); grade (a high school senior receives preference over younger students who will have another chance before they graduate); the correct answer on a particular qualifying test question (varies from year to year, usually question 10); and the free-form answers to the non-math questions. - See more at: http://www.ams.org/programs/students/wwtbam/qualify#ties

DeathLlama9
2015-11-03 20:59:10

Can you do this event alone if your school doesn't support it?

Can you do this event alone if your school doesn't support it?

DPatrick
2015-11-03 20:59:37

No -- it has to be arranged through a school. But the contest is free and easy to administer, so there shouldn't be any barrier to getting your school to sign up!

No -- it has to be arranged through a school. But the contest is free and easy to administer, so there shouldn't be any barrier to getting your school to sign up!

nosaj
2015-11-03 20:59:48

How many people participated in Round 2?

How many people participated in Round 2?

DPatrick
2015-11-03 20:59:50

346

346

eveningstarandlion
2015-11-03 21:00:18

What's the number in round 1?

What's the number in round 1?

gxah
2015-11-03 21:00:18

what about round 1?

what about round 1?

DPatrick
2015-11-03 21:00:24

I think it was about 2300.

I think it was about 2300.

mikebreen
2015-11-03 21:00:32

Yes

Yes

pi37
2015-11-03 21:00:47

do you know how many participated by region? (or around how many?)

do you know how many participated by region? (or around how many?)

DPatrick
2015-11-03 21:00:56

Sorry, I don't have that information.

Sorry, I don't have that information.

DPatrick
2015-11-03 21:01:44

OK, I'll guess we'll wrap up here. Thanks for coming tonight!

OK, I'll guess we'll wrap up here. Thanks for coming tonight!