Math Jams

Who Wants to Be a Mathematician, Round 2

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AoPS 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 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.
DPatrick 2015-11-03 19:36:04
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.
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!
DPatrick 2015-11-03 19:36:58
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 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!
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 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
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!
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.
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.)
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 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.
DPatrick 2015-11-03 19:39:37
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?
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.)
AShane2000 2015-11-03 19:40:17
probablity of the integer is 1/6
DeathLlama9 2015-11-03 19:40:17
Okay well the randomly chosen integer's probability is $\frac{1}{6}$
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.
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.
DPatrick 2015-11-03 19:40:48
Which sum is that?
CantonMathGuy 2015-11-03 19:41:09
7
mathwhiz16 2015-11-03 19:41:09
From experience I know that the number is 7…
nosaj 2015-11-03 19:41:09
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
gxah 2015-11-03 19:41:09
7
lax0000 2015-11-03 19:41:09
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 $\dfrac16$.
DPatrick 2015-11-03 19:41:47
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
flyrain 2015-11-03 19:42:04
make a chart of all cases
ninjataco 2015-11-03 19:42:04
list them out
DPatrick 2015-11-03 19:42:12
Right, we can make a table:
DPatrick 2015-11-03 19:42:15
\[\begin{array}{c|cccccc}
&1&2&3&4&5&6 \\ \hline
1&2&3&4&5&6&7 \\
2&3&4&5&6&7&8 \\
3&4&5&6&7&8&9 \\
4&5&6&7&8&9&10 \\
5&6&7&8&9&10&11 \\
6&7&8&9&10&11&12
\end{array}\]
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.
DPatrick 2015-11-03 19:42:32
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:
\[\begin{array}{c|cccccc}
&1&2&3&4&5&6 \\ \hline
1&2&3&4&5&6&\mathbf{7} \\
2&3&4&5&6&\mathbf{7}&8 \\
3&4&5&6&\mathbf{7}&8&9 \\
4&5&6&\mathbf{7}&8&9&10 \\
5&6&\mathbf{7}&8&9&10&11 \\
6&\mathbf{7}&8&9&10&11&12
\end{array}\]
DPatrick 2015-11-03 19:42:59
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.
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
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).
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.)
DPatrick 2015-11-03 19:43:55
\[\begin{array}{r|l}
\text{sum} & \text{# of ways to roll} \\ \hline
2 & 1 \\
3 & 2 \\
4 & 3 \\
5 & 4 \\
6 & 5 \\
7 & 6 \\
8 & 5 \\
9 & 4 \\
10 & 3 \\
11 & 2 \\
12 & 1
\end{array}\]
DeathLlama9 2015-11-03 19:44:23
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!
DPatrick 2015-11-03 19:45:05
On to #2:
DPatrick 2015-11-03 19:45:10
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?
DeathLlama9 2015-11-03 19:45:46
Find 10s
mathwhiz16 2015-11-03 19:45:46
Count the number of factors of 5.
ninjataco 2015-11-03 19:45:46
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
Ancy 2015-11-03 19:45:46
find the number of factors with five
bharatputra 2015-11-03 19:45:46
powers of 5
AShane2000 2015-11-03 19:45:46
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
gxah 2015-11-03 19:45:46
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
nosaj 2015-11-03 19:45:46
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)
DPatrick 2015-11-03 19:45:57
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.
Ancy 2015-11-03 19:46:29
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
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.
DPatrick 2015-11-03 19:46:41
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?
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
bearytasty 2015-11-03 19:47:23
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
yaredboy 2015-11-03 19:47:23
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
Mlux 2015-11-03 19:47:23
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...
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.
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.
flyrain 2015-11-03 19:47:54
what about the 25s?
nosaj 2015-11-03 19:47:54
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.
DPatrick 2015-11-03 19:48:12
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
Liopleurodon 2015-11-03 19:48:24
round down
mathsolver101 2015-11-03 19:48:24
Floor function!!
mathwhiz16 2015-11-03 19:48:24
round down
Mlux 2015-11-03 19:48:24
least integer
CantonMathGuy 2015-11-03 19:48:24
round down to integer
bearytasty 2015-11-03 19:48:24
truncate it. the remainder doesnt matter
krishna10 2015-11-03 19:48:32
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.
DPatrick 2015-11-03 19:48:47
So that's 403+80 = 483 total factors of 5 so far.
Mlux 2015-11-03 19:49:01
now 125
nosaj 2015-11-03 19:49:01
now for the multiples of 125!
Ancy 2015-11-03 19:49:01
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.
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.
Ancy 2015-11-03 19:49:32
now divide by five to get three
Mlux 2015-11-03 19:49:32
625's
yaredboy 2015-11-03 19:49:32
Lol watch us do 625 XD
flyrain 2015-11-03 19:49:32
there's still more
DPatrick 2015-11-03 19:49:38
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.
Ancy 2015-11-03 19:49:55
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
DeathLlama9 2015-11-03 19:49:55
Nothing beyond that as $2015 < 3125$
CantonMathGuy 2015-11-03 19:49:55
2015 < 3125
mathwhiz16 2015-11-03 19:49:55
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.)
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.
DPatrick 2015-11-03 19:50:29
Fun fact: 2015! has 5786 digits total!
bearytasty 2015-11-03 19:50:44
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!
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?
DPatrick 2015-11-03 19:51:20
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
flyrain 2015-11-03 19:51:49
a sequence with a common ratio
bearytasty 2015-11-03 19:51:49
common ratio between terms
nosaj 2015-11-03 19:51:49
terms have a constant ratio
hwl0304 2015-11-03 19:51:49
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
DPatrick 2015-11-03 19:51:58
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?
DeathLlama9 2015-11-03 19:52:27
Find the common ratio!
ninjataco 2015-11-03 19:52:27
first find r
flyrain 2015-11-03 19:52:32
find r
Liopleurodon 2015-11-03 19:52:35
find the ratio
DPatrick 2015-11-03 19:52:40
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?
nosaj 2015-11-03 19:53:15
Note that $\frac{a_5}{a_2}=r^3$.
ChrisY 2015-11-03 19:53:15
they are r^3 apart
CantonMathGuy 2015-11-03 19:53:15
their ratio is the cube of the common ratio
DeathLlama9 2015-11-03 19:53:15
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
Liopleurodon 2015-11-03 19:53:15
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
Ancy 2015-11-03 19:53:15
the fifth term divided by the second term gives us r^3
ChrisY 2015-11-03 19:53:15
the ratio is r^3
AShane2000 2015-11-03 19:53:15
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 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) \[r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \frac{a_5}{a_4} = \cdots.\]
DPatrick 2015-11-03 19:53:40
But if we multiply the last three ratios I listed above, we get \[\frac{a_3}{a_2} \cdot \frac{a_4}{a_3} \cdot \frac{a_5}{a_4} = r \cdot r \cdot r = r^3.\]
DPatrick 2015-11-03 19:53:50
And we see that many of the terms of the left side cancel, leaving us with just $\dfrac{a_5}{a_2} = r^3$.
DPatrick 2015-11-03 19:54:15
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
DeathLlama9 2015-11-03 19:54:31
$r = \frac{1}{3}$
flyrain 2015-11-03 19:54:31
r=1/3
Mlux 2015-11-03 19:54:31
r = 1/3
Ancy 2015-11-03 19:54:31
so r = 1/3
ninjataco 2015-11-03 19:54:31
r=1/3
DPatrick 2015-11-03 19:54:35
This simplifies to $\dfrac{1}{27} = r^3$, so $r = \sqrt[3]{\dfrac{1}{27}} = \dfrac13$.
bellehan 2015-11-03 19:54:45
now we can find the first term
DPatrick 2015-11-03 19:55:02
Indeed, what's the first term?
AShane2000 2015-11-03 19:55:21
(1/3)a=54
AShane2000 2015-11-03 19:55:21
a=162
bearytasty 2015-11-03 19:55:21
first term is 3*54 = 162
lkarhat 2015-11-03 19:55:21
162
DeathLlama9 2015-11-03 19:55:21
$54 \div \frac{1}{3} = 162$
sflorin 2015-11-03 19:55:21
162
thequantumguy 2015-11-03 19:55:25
54 * 3 = 162
krishna10 2015-11-03 19:55:25
54 times 3... 162
DPatrick 2015-11-03 19:55:30
We again have $\dfrac{a_2}{a_1} = r = \dfrac13$.
DPatrick 2015-11-03 19:55:42
Since $a_2 = 54$, this gives us $a_1 = 3a_2 = 3(54) = 162$.
DPatrick 2015-11-03 19:55:56
So our sum is the sum of a geometric series with first term 162 and common ratio $\frac13$.
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)
DeathLlama9 2015-11-03 19:56:10
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: $\frac{a}{1-r}$
bearytasty 2015-11-03 19:56:13
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 $r$.
DPatrick 2015-11-03 19:56:32
It's $\dfrac{a}{1-r}$...provided $|r| < 1$.
chopinwaltzes 2015-11-03 19:57:02
243!
mathwhiz16 2015-11-03 19:57:02
So we get $\frac{162}{\frac{2}{3}}$, which is 243.
rraj 2015-11-03 19:57:02
243
ChrisY 2015-11-03 19:57:02
The sum is 162/(2/3) = 243
Mlux 2015-11-03 19:57:02
it's 243
DPatrick 2015-11-03 19:57:06
Applying this formula, we see that our sum is $\dfrac{162}{1-\frac13}$.
DPatrick 2015-11-03 19:57:12
This evaluates as $\dfrac{162}{\frac23} = \frac32 \cdot 162 = 3 \cdot 81 = \boxed{243}$.
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
DPatrick 2015-11-03 19:57:38
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?
ninjataco 2015-11-03 19:58:16
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
nosaj 2015-11-03 19:58:27
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
DPatrick 2015-11-03 19:58:40
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 $r$.
DPatrick 2015-11-03 19:58:59
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 \[ rS = ar + ar^2 + ar^3 + ar^4 + \cdots.\]
DPatrick 2015-11-03 19:59:28
What happens when we subtract these two equations?
bearytasty 2015-11-03 19:59:45
BAM everything disappears!
mathsolver101 2015-11-03 19:59:45
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 $a$!
DPatrick 2015-11-03 20:00:27
So $S - rS = a$.
nosaj 2015-11-03 20:00:41
then we divide both sides by $1-r$
DeathLlama9 2015-11-03 20:00:41
$S(1 - r) = a$, so $\frac{a}{1-r} = S$
Liopleurodon 2015-11-03 20:00:48
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 $S = \dfrac{a}{1-r}$.
DPatrick 2015-11-03 20:01:09
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
bellehan 2015-11-03 20:01:36
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
mathwhiz16 2015-11-03 20:01:36
because a series with r>1 diverges
TheChampion 2015-11-03 20:01:36
Otherwise the sequence diverges.
Liopleurodon 2015-11-03 20:01:36
because if |r|>1, then the sum is infinite
Ancy 2015-11-03 20:01:36
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
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.)
DPatrick 2015-11-03 20:02:08
For example, it doesn't make sense to use the formula to conclude that \[1 + 2 + 4 + 8 + \cdots = \frac{1}{1-2} = -1\,???\]
DPatrick 2015-11-03 20:02:28
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 $r$.
bearytasty 2015-11-03 20:03:10
find the volumes of the two pieces
DPatrick 2015-11-03 20:03:17
OK. What's the volume of the seed?
bearytasty 2015-11-03 20:03:39
volume of seed is $4\pi / 3$
mathsolver101 2015-11-03 20:03:39
4pi/3
ninjataco 2015-11-03 20:03:39
$\frac{4}{3}\pi$
Liopleurodon 2015-11-03 20:03:39
4/3 pi
krishna10 2015-11-03 20:03:39
4/3pi
FTW 2015-11-03 20:03:39
$\frac{4}{3} \pi$
DPatrick 2015-11-03 20:03:42
The formula for the volume of a sphere of radius $r$ is $\frac43\pi r^3$.
DPatrick 2015-11-03 20:03:47
So the volume of the seed is $\frac43\pi$.
DPatrick 2015-11-03 20:03:53
What's the volume of the entire fruit?
ninjataco 2015-11-03 20:04:25
$\frac{32}{3}\pi$
DeathLlama9 2015-11-03 20:04:25
28/3 + 4/3 = 32pi/3
AShane2000 2015-11-03 20:04:25
32pi/3
Liopleurodon 2015-11-03 20:04:25
4/3 pi r^3
FTW 2015-11-03 20:04:25
$\frac{4}{3} \pi r^3$
sflorin 2015-11-03 20:04:25
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 $\frac43\pi \cdot 7$.
DPatrick 2015-11-03 20:04:37
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.
DPatrick 2015-11-03 20:05:04
Hence, $\frac43 \pi \cdot 8 = \frac43\pi r^3$.
xiaodragon 2015-11-03 20:05:22
r=2
Mlux 2015-11-03 20:05:22
r = 2
eveningstarandlion 2015-11-03 20:05:22
r=2
thequantumguy 2015-11-03 20:05:22
r is 2
mathwhiz16 2015-11-03 20:05:22
so r=2!
bearytasty 2015-11-03 20:05:22
r=2
CantonMathGuy 2015-11-03 20:05:22
$r^3=8\implies r=2$
DPatrick 2015-11-03 20:05:27
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 $8x$
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 $2:1$.
Generic_Username 2015-11-03 20:06:08
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 cube of the ratio of lengths.
DPatrick 2015-11-03 20:06:41
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
AShane2000 2015-11-03 20:07:17
a
lkarhat 2015-11-03 20:07:17
a
thequantumguy 2015-11-03 20:07:17
a
xiaodragon 2015-11-03 20:07:17
(a
eveningstarandlion 2015-11-03 20:07:17
John Von Neumann
trumpeter 2015-11-03 20:07:17
von neumann
Ancy 2015-11-03 20:07:17
A?
ninjataco 2015-11-03 20:07:17
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:
DPatrick 2015-11-03 20:07:27

//cdn.artofproblemsolving.com/images/e/3/c/e3c8f8620ae218da3f0c3ee7bb54e3ae3d903edf.jpg
DPatrick 2015-11-03 20:07:44
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.
DPatrick 2015-11-03 20:08:23
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.
DPatrick 2015-11-03 20:08:52

//cdn.artofproblemsolving.com/images/f/a/8/fa84827fd271621725d649dc5402858728380eea.jpg
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 .
mathwhiz16 2015-11-03 20:09:17
Kurt godel had his famous incompleteness theorem…
DPatrick 2015-11-03 20:09:22
That's right.
DPatrick 2015-11-03 20:09:32

//cdn.artofproblemsolving.com/images/5/7/c/57c4e2e75d4883370d678f95e2140a7648ceef12.jpg
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 How to Solve It about problem solving.
DPatrick 2015-11-03 20:10:06

//cdn.artofproblemsolving.com/images/1/b/4/1b4c4be06ef939292da7f3534827d01438e9aa5d.jpg
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 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.
DPatrick 2015-11-03 20:11:07
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
DPatrick 2015-11-03 20:11:32
You are right. We'll get to that in a bit...
DPatrick 2015-11-03 20:11:40
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?
DeathLlama9 2015-11-03 20:12:13
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.
DPatrick 2015-11-03 20:12:36
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
mathsolver101 2015-11-03 20:12:52
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.
DPatrick 2015-11-03 20:13:29
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).
DPatrick 2015-11-03 20:13:51
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
bearytasty 2015-11-03 20:14:24
5+2*4 = 13
nosaj 2015-11-03 20:14:27
13
DPatrick 2015-11-03 20:14:31
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!)
DeathLlama9 2015-11-03 20:15:01
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.
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.
mathwhiz16 2015-11-03 20:15:38
so 13*24!
mathwhiz16 2015-11-03 20:15:38
not 24 factorial, though
nosaj 2015-11-03 20:15:45
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.
DPatrick 2015-11-03 20:16:05
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) $m * n = \max\{m,n\}$
DPatrick 2015-11-03 20:16:28
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)
DeathLlama9 2015-11-03 20:17:05
Hint: Associative Property a * (b * c) = (a * b) * c
thequantumguy 2015-11-03 20:17:05
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 $a,b,c$.
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.
DPatrick 2015-11-03 20:17:35
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.
DPatrick 2015-11-03 20:18:21
Is (a) associative?
DaVinci 2015-11-03 20:18:36
no
geniusofart 2015-11-03 20:18:36
No.
Mlux 2015-11-03 20:18:36
no
flyrain 2015-11-03 20:18:45
even a=b=c=1 fails in (a)
mathwhiz16 2015-11-03 20:18:45
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
DPatrick 2015-11-03 20:18:49
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.
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*}
DPatrick 2015-11-03 20:19:16
How about (b)?
Benq 2015-11-03 20:19:40
no
Parni10 2015-11-03 20:19:40
No.
Generic_Username 2015-11-03 20:19:40
No
nosaj 2015-11-03 20:19:40
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*}
flyrain 2015-11-03 20:20:00
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*}
DPatrick 2015-11-03 20:20:19
How about (c)?
bearytasty 2015-11-03 20:20:40
nope
geniusofart 2015-11-03 20:20:40
No.
AShane2000 2015-11-03 20:20:40
no
CantonMathGuy 2015-11-03 20:20:40
no
DPatrick 2015-11-03 20:20:45
For (c), it's not clear that it's an operation at all!
DPatrick 2015-11-03 20:20:50
In particular, what's $-1 * \dfrac12$?
DeathLlama9 2015-11-03 20:21:11
i?
Mlux 2015-11-03 20:21:11
sqrt of -1
nosaj 2015-11-03 20:21:11
$i$
FTW 2015-11-03 20:21:11
$i$
Mlux 2015-11-03 20:21:11
$i$
Stephenpiano 2015-11-03 20:21:11
or Imaginary
Ancy 2015-11-03 20:21:11
oh wow .-. $i$
geniusofart 2015-11-03 20:21:11
$i$
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.
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*}
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 $8^4 = 2^{12}$.)
Mlux 2015-11-03 20:21:51
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.
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
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
DPatrick 2015-11-03 20:22:34
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 $c$.
DPatrick 2015-11-03 20:22:58
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
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 $c$.
Ancy 2015-11-03 20:23:27
so its just $max{a,b,c}$
DPatrick 2015-11-03 20:23:33
Right. Usually we'd just write it as $\max\{a,b,c\}$.
DPatrick 2015-11-03 20:23:36
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.
DPatrick 2015-11-03 20:24:00
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?
bearytasty 2015-11-03 20:24:34
reduce modulo 3?
Benq 2015-11-03 20:24:34
Take the equation mod 3.
DPatrick 2015-11-03 20:25:08
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.
DPatrick 2015-11-03 20:25:31
So $a_0 \equiv 1 \pmod{3}$.
CantonMathGuy 2015-11-03 20:25:36
reduce the recurrence mod $3$, so $a_n\equiv a_{n-1}+(n+1)$
mathwhiz16 2015-11-03 20:25:36
so you get $a_n=a_{n-1}+n+1$
DeathLlama9 2015-11-03 20:25:36
$a_n = a_{n-1} + n + 1$
DPatrick 2015-11-03 20:25:42
Right, since $100 \equiv 10 \equiv 1 \pmod{3}$, we have \[a_n \equiv a_{n-1} + n + 1 \pmod{3}.\]
DeathLlama9 2015-11-03 20:25:54
Calculate a few values, find a pattern
nosaj 2015-11-03 20:25:54
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?
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.
DPatrick 2015-11-03 20:26:25
For example, by the formula, $a_1 \equiv a_0 + 1 + 1 \equiv 1 + 2 \equiv 0 \pmod{3}$.
DeathLlama9 2015-11-03 20:26:43
$a_2 = 0 + 2 + 1$ which also works
Ancy 2015-11-03 20:26:43
the next one is also 0
DPatrick 2015-11-03 20:26:47
Again by the formula, $a_2 \equiv a_1 + 2 + 1 \equiv 0 + 3 \equiv 0 \pmod{3}$.
bearytasty 2015-11-03 20:27:09
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
Snowie 2015-11-03 20:27:22
I got remainder 1,0,0,1,0,0
DPatrick 2015-11-03 20:27:26
Again by the formula, $a_3 \equiv a_2 + 3 + 1 \equiv 0 + 1 \equiv 1 \pmod{3}$.
DPatrick 2015-11-03 20:27:48
Do we have enough terms to stop now and get the general pattern?
AShane2000 2015-11-03 20:28:16
yes
nosaj 2015-11-03 20:28:16
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.
DPatrick 2015-11-03 20:28:38
So (starting at $n=0$) the sequence of residues (mod 3) are: \[1,0,0,1,0,0,1,0,0,1,\ldots.\]
DPatrick 2015-11-03 20:29:03
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?
nosaj 2015-11-03 20:29:31
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 $0 \le n \le 100$.
DPatrick 2015-11-03 20:29:41
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
bearytasty 2015-11-03 20:29:59
we can use complementary counting. 101 - 34 = 67?
mathwhiz16 2015-11-03 20:29:59
101-34
geniusofart 2015-11-03 20:29:59
101-34
DeathLlama9 2015-11-03 20:30:06
101 - (multiples of three), which are 0 to 99 or 34, so 101 - 34 = $\boxed{67}$
DPatrick 2015-11-03 20:30:10
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.
DPatrick 2015-11-03 20:30:27
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.
DPatrick 2015-11-03 20:31:00
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: 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.
DPatrick 2015-11-03 20:31:57
9. Suppose $\sqrt{9+4\sqrt5} = a + b\sqrt5$ where $a$ and $b$ are integers. Find $a+b$.
bearytasty 2015-11-03 20:32:25
square it
Generic_Username 2015-11-03 20:32:25
Square both sides
CantonMathGuy 2015-11-03 20:32:25
square
FTW 2015-11-03 20:32:25
square it
DeathLlama9 2015-11-03 20:32:25
Square both sides of the equation!
mathsolver101 2015-11-03 20:32:25
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.
DPatrick 2015-11-03 20:32:44
What happens if we square the equation $\sqrt{9+4\sqrt5} = a + b\sqrt5$?
DeathLlama9 2015-11-03 20:33:15
You get $9 + 4\sqrt{5} = a^2 + 5b^2 + 2ab\sqrt{5}$
Eugenis 2015-11-03 20:33:15
you get $9+4\sqrt{5}=a^2+2ab\sqrt{5}+5b^2$
mihirb 2015-11-03 20:33:15
$9+4\sqrt{5} = a^2+2ab\sqrt{5}+5b^2$
FTW 2015-11-03 20:33:15
$9+4\sqrt{5}=a^2+5b^2+2\sqrt{5}ab$
thepiercingarrow 2015-11-03 20:33:15
We get $9+4\sqrt5=a^2+5b^2+2ab\sqrt5$
bearytasty 2015-11-03 20:33:15
9 + 4sqrt(5) = (a^2 + 5b^2) + 2absqrt(5)
Liopleurodon 2015-11-03 20:33:15
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.)
DPatrick 2015-11-03 20:33:31
That simplifies to $9 + 4\sqrt5 = a^2 + 2ab\sqrt5 + 5b^2$.
DPatrick 2015-11-03 20:33:38
What can we conclude?
geniusofart 2015-11-03 20:33:52
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
mihirb 2015-11-03 20:33:52
$a^2+5b^2 = 9$, $2ab = 4$
mathwhiz16 2015-11-03 20:33:52
So $a^2+5b^2=9$ and $2ab=4$.
Eugenis 2015-11-03 20:33:52
compare coefficients
FTW 2015-11-03 20:33:56
$a^2+5b^2=9$, $ab=2$
DPatrick 2015-11-03 20:34:00
Exactly. The integer parts must be equal, and the $\sqrt5$ parts must be equal.
DPatrick 2015-11-03 20:34:06
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.
Dragon2kz 2015-11-03 20:34:28
a=2 b=1
va2010 2015-11-03 20:34:28
a = 2 b = 1
FTW 2015-11-03 20:34:28
$a=2, b=1$
CantonMathGuy 2015-11-03 20:34:28
$2$ is small so we use trial and error to get $(a,b)=(2,1)$
DPatrick 2015-11-03 20:34:36
We can quickly see that $a=2$ and $b=1$ is the solution to both equations.
DPatrick 2015-11-03 20:34:44
So $\sqrt{9 + 4\sqrt5} = 2 + \sqrt5$.
DPatrick 2015-11-03 20:34:58
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 $-2-\sqrt{5}<0$
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 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...
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?
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.
DPatrick 2015-11-03 20:36:17
Are any of them easy?
DaVinci 2015-11-03 20:36:37
11
gxah 2015-11-03 20:36:37
11 is prime
ninjataco 2015-11-03 20:36:37
11
nosaj 2015-11-03 20:36:37
yes 11 is obviously prime
minimario 2015-11-03 20:36:37
11 easy
mathsolver101 2015-11-03 20:36:37
11
DPatrick 2015-11-03 20:36:41
11 is prime!
Snowie 2015-11-03 20:37:00
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
thepiercingarrow 2015-11-03 20:37:37
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
Mlux 2015-11-03 20:37:37
multiples of $3$
thepiercingarrow 2015-11-03 20:37:37
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
DeathLlama9 2015-11-03 20:37:37
$111$ 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.
DPatrick 2015-11-03 20:37:50
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
DPatrick 2015-11-03 20:38:05
Right, there's that too.
mathwhiz16 2015-11-03 20:38:11
that leaves us with 11, 11111, and 1111111
nosaj 2015-11-03 20:38:11
so just 11111 and 1111111 to check now
DPatrick 2015-11-03 20:38:15
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.
eveningstarandlion 2015-11-03 20:38:36
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.
DPatrick 2015-11-03 20:39:03
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.
DPatrick 2015-11-03 20:39:26
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
DPatrick 2015-11-03 20:39:48
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.
DPatrick 2015-11-03 20:40:08
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
Mlux 2015-11-03 20:40:28
roots of unity
nosaj 2015-11-03 20:40:28
roots of unity?
Generic_Username 2015-11-03 20:40:28
5th roots of unity
Benq 2015-11-03 20:40:30
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 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 $x^5 = 1$.
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!
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 $\cos(72^\circ) = \dfrac{\sqrt5 - 1}{4}$.
Generic_Username 2015-11-03 20:41:45
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 $b$.
DPatrick 2015-11-03 20:42:27
If we expand the right side out, we get $$f(x) = x^4 + (a+b)x^3 + (2+ab)x^2 + (a+b)x + 1.$$
mathwhiz16 2015-11-03 20:42:51
a+b=1, and 2+ab=1
vatatmaja 2015-11-03 20:42:59
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 $ab = -1$.
DPatrick 2015-11-03 20:43:15
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
DPatrick 2015-11-03 20:43:57
Right! By Vieta's Formulas, $a$ and $b$ are the roots of the quadratic $(t-a)(t-b) = t^2 - (a+b)t + ab = t^2 -t - 1 = 0$.
DPatrick 2015-11-03 20:44:24
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, \[f(x) = \left(x^2 + \left(\dfrac{1+\sqrt5}{2}\right)x + 1\right)\left(x^2 + \left(\dfrac{1-\sqrt5}{2}\right)x + 1\right).\]
DPatrick 2015-11-03 20:45:13
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*}
DPatrick 2015-11-03 20:45:44
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
lkarhat 2015-11-03 20:45:56
106^2 - 125
DPatrick 2015-11-03 20:46:00
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*}
CantonMathGuy 2015-11-03 20:46:47
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 \[(26+9\sqrt5)(26-9\sqrt5) = 26^2 - (9\sqrt5)^2 = 676 - 405 = 271.\]
DPatrick 2015-11-03 20:47:06
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?
DPatrick 2015-11-03 20:47:35
Probably not. But it's cool!
DPatrick 2015-11-03 20:47:41
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?
DPatrick 2015-11-03 20:48:16
It's much like any factorization: on some level it becomes educated guesswork.
DPatrick 2015-11-03 20:48:28
Anyway, how about 1111111?
eveningstarandlion 2015-11-03 20:48:44
divisible by 239 I think
Generic_Username 2015-11-03 20:48:47
same thing?
Liopleurodon 2015-11-03 20:48:47
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.
DPatrick 2015-11-03 20:49:11
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:
\[ 1111111 = (1044 + 55\sqrt{-7})(1044 - 55\sqrt{-7}).\]
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*}
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 \[(43+20\sqrt{-7})(43-20\sqrt{-7}) = 43^2 - (20\sqrt{-7})^2 = 1849 + 7 \cdot 400 = 4649.\]
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
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.
DPatrick 2015-11-03 20:50:53
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/
DPatrick 2015-11-03 20:51:25
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.)
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.)
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.
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.
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?
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?
va2010 2015-11-03 20:53:30
mersenne
trumpeter 2015-11-03 20:53:30
mersenne numbers
mathwhiz16 2015-11-03 20:53:30
2^n-1
DPatrick 2015-11-03 20:53:35
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 $\dfrac{2^n-1}{2-1} = 2^n-1$.
DPatrick 2015-11-03 20:53:48
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.
DPatrick 2015-11-03 20:54:11
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!
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!
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.
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.
nosaj 2015-11-03 20:55:50
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:
DPatrick 2015-11-03 20:56:13
#1 82.9%
DPatrick 2015-11-03 20:56:17
#2 67.1%
DPatrick 2015-11-03 20:56:20
#3 78.6%
DPatrick 2015-11-03 20:56:25
#4 91.0%
DPatrick 2015-11-03 20:56:27
#5 50.3%
DPatrick 2015-11-03 20:56:30
#6 56.9%
DPatrick 2015-11-03 20:56:33
#7 66.2%
DPatrick 2015-11-03 20:56:36
#8 38.2%
DPatrick 2015-11-03 20:56:39
#9 71.4%
DPatrick 2015-11-03 20:56:42
#10 17.1%
mikebreen 2015-11-03 20:56:53
#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)?
DPatrick 2015-11-03 20:57:34
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?
DPatrick 2015-11-03 20:57:52
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?
DPatrick 2015-11-03 20:58:07
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
DeathLlama9 2015-11-03 20:59:10
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!
nosaj 2015-11-03 20:59:48
How many people participated in Round 2?
DPatrick 2015-11-03 20:59:50
346
eveningstarandlion 2015-11-03 21:00:18
What's the number in round 1?
gxah 2015-11-03 21:00:18
what about round 1?
DPatrick 2015-11-03 21:00:24
I think it was about 2300.
mikebreen 2015-11-03 21:00:32
Yes
pi37 2015-11-03 21:00:47
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.
DPatrick 2015-11-03 21:01:44
OK, I'll guess we'll wrap up here. Thanks for coming tonight!

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