Math Jams

Who Wants to Be a Mathematician, Qualifying Round

Go back to the Math Jam Archive

AoPS instructor David Patrick will discuss the problems on Round 1 of the 2017-2018 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 San Diego in January 2018.

Copyright © 2020 AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this file with others.

Facilitator: Dave Patrick

DPatrick 2017-09-27 19:30:37
Welcome to the 2017-18 Who Wants to Be a Mathematician Round 1 Math Jam!
DPatrick 2017-09-27 19:30:48
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 13 years, and I've written or co-written a few of our textbooks.
DPatrick 2017-09-27 19:31:01
In keeping with tonight's theme, I also once was a contestant on ABC's Who Wants to Be a Millionaire back before I started working at AoPS, way back when Regis Philbin was still the host. Here's a picture (I'm on the left, Regis is on the right):
DPatrick 2017-09-27 19:31:07
https://www.washington.edu/alumni/columns/march00/images/patrick.jpg
DPatrick 2017-09-27 19:31:24
Photo Credit: Maria Melin, copyright 1999 ABC Television.
DPatrick 2017-09-27 19:31:40
No, I didn't win the million bucks, but I did win enough to buy a new car.
DPatrick 2017-09-27 19:31:46
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 2017-09-27 19:32:01
The classroom is moderated, meaning that participants 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 2017-09-27 19:32:12
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick 2017-09-27 19:32:43
There are a lot of students here! As such, only a fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
DPatrick 2017-09-27 19:33:00
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 2017-09-27 19:33:17
We have an assistant here to help out tonight: Karthik Karnik (Pythagor25). Karthik is a sophomore at Caltech who enjoys solving puzzles, playing chess, programming UAVs, and robotics. He is majoring in math and computer science.
DPatrick 2017-09-27 19:33:24
He can try to help you if you have a question or are having some other difficulty. He may open a private window with you to chat if needed.
DPatrick 2017-09-27 19:33:52
Also joining us tonight are the co-creators of WWTBAM, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
mikebreen 2017-09-27 19:34:03
Hello, everyone. Thanks for joining in.
DPatrick 2017-09-27 19:34:06
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.
TPiR 2017-09-27 19:34:16
Hi everyone!
DPatrick 2017-09-27 19:34:20
Bill earned an undergraduate degree in mathematics from Santa Clara University and a Ph.D. from Northwestern University, and is currently an associate professor and associate chair of mathematics at DePaul University. He shares a life-long interest in game shows with colleague Mike Breen, with whom he works as the not-so-lovely assistant on the mathematics game show Who Wants to Be a Mathematician. In addition to authoring articles and presenting talks related to game-show mathematics, Bill served as mathematics consultant to the CBS television show The Price is Right from 1997 to 2009. (Hence, his username.)
DPatrick 2017-09-27 19:35:05
Thanks very much to Mike and Bill for joining us tonight. Mike will also be around to answer questions (as will I) after the Math Jam has concluded. As you can see, we have a lot of game show background here tonight!
DPatrick 2017-09-27 19:35:30
Who Wants to Be a Mathematician is a math contest for high school students 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.
Zachdog1 2017-09-27 19:35:46
What is the age requirement?
DPatrick 2017-09-27 19:36:06
There isn't one. Even though the contest is at the high school level, anyone high school or below can participate.
DPatrick 2017-09-27 19:36:26
Tonight we'll be talking about Round 1 of the national contest, which just concluded. It consisted of 10 questions, with a 15-minute time limit. So the problems are quick: you have an average of 90 seconds per question. (But as we'll see as we work through the problems, some of them shouldn't take you nearly that long.) No books, notes, calculators, or internet were permitted during the test.
yeskay 2017-09-27 19:36:49
You can have scratch paper right?
DPatrick 2017-09-27 19:36:51
Yes.
GeronimoStilton 2017-09-27 19:36:55
I took it online though.
DPatrick 2017-09-27 19:37:14
Indeed, most students participated online. 4100 students total did Round 1, with 3600 of them taking the test online.
DPatrick 2017-09-27 19:37:51
The mean score was 3.78 out of 10. A score of 8 out of 10 was sufficient to advance to Round 2, which will be in October.
kenisaka 2017-09-27 19:38:03
may i still take the test or is it over?
DPatrick 2017-09-27 19:38:18
Round 1 has already concluded. We'll be going over the problems tonight.
DPatrick 2017-09-27 19:38:40
You can visit the contest website at ams.org/wwtbam to learn more about the contest for next year.
Fortified 2017-09-27 19:38:53
How many people qualified for the second round???
DPatrick 2017-09-27 19:38:55
About 500.
DPatrick 2017-09-27 19:39:07
We'll answer more questions at the end -- for now, let's move on to the problems!
DPatrick 2017-09-27 19:39:18
As I said, Round 1 consisted of 10 questions, with a 15-minute time limit.
DPatrick 2017-09-27 19:39:26
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 2017-09-27 19:39:49
Some of the questions may have interesting sidetracks, so we'll also stop and view some of the scenery along the way.
DPatrick 2017-09-27 19:40:29
Let's start at the beginning:
DPatrick 2017-09-27 19:40:33
1. Find the $y$-intercept ($y$-coordinate only) of the line whose equation is $2x + 5y = 7$.
GeronimoStilton 2017-09-27 19:41:12
Plug in $x = 0$.
Designerd 2017-09-27 19:41:12
just test $x=0$
yeskay 2017-09-27 19:41:12
the y intercept is when x is zero
sgadekar 2017-09-27 19:41:12
plug in x=0
fishy15 2017-09-27 19:41:16
this means x = 0, so sub in
DPatrick 2017-09-27 19:41:25
Right. All the points on the $y$-axis have in common that $x=0$.
DPatrick 2017-09-27 19:41:31
So we can just plug in $x=0$ into our equation and solve for $y$.
seaskyline 2017-09-27 19:41:53
You can plug in 0 for x to get the equation 5y = 7 and divide by 5 to get y=7/5
Liping 2017-09-27 19:41:53
THe y-intercept is the coordinate where x is 0. Then its 5y =7 and y=7/5
pokemonduel 2017-09-27 19:41:59
and you get 7/5
DPatrick 2017-09-27 19:42:08
Exactly. This gives $5y = 7$, or $y = \boxed{\dfrac75}$.
pianoman24 2017-09-27 19:42:14
That's like chapter Mathcounts Countdown round level
DPatrick 2017-09-27 19:42:27
Indeed, this was the pretty straightforward warm-up problem.
DPatrick 2017-09-27 19:42:37
Nearly 85% of participants got this correct.
DPatrick 2017-09-27 19:43:08
It gets harder almost right away. All of #2-#9 were solved by somewhere between 24% and 42% of students.
DPatrick 2017-09-27 19:43:15
2. $2 + 4 + 6 + 8 + \cdots + 2018 = \;?$
mxzhang 2017-09-27 19:43:57
Arithmetic sequence!
mxzhang 2017-09-27 19:43:57
There are $1009$ terms.
rick101 2017-09-27 19:43:57
Pair up numbers: 2 and 2018, 4 and 2016, etc.
DPatrick 2017-09-27 19:44:05
Indeed, this is an arithmetic series: the difference between two consecutive summands is constant (in this case, 2).
DPatrick 2017-09-27 19:44:37
Its sum is equal to (the number of terms in the series) * (the average number in the series).
DPatrick 2017-09-27 19:45:08
Indeed, you can see this by pairing up terms: 2 and 2018, 4 and 2016, 6 and 2014, and so on. This gives pairs of terms each of whose average is the average of the entire series.
yali306 2017-09-27 19:45:27
1009*1010
pianoman24 2017-09-27 19:45:27
Average number = first+last/2
DPatrick 2017-09-27 19:45:41
There are 1009 terms (the series is the sum of the first $2018/2 = 1009$ even positive integers.)
DPatrick 2017-09-27 19:45:58
And yes, the average term of an arithmetic sequence is just the average of the first and last term. So the average term is $(2+2018)/2 = 2020/2 = 1010$.
prajna1225 2017-09-27 19:46:09
1009 . 1010
iks92 2017-09-27 19:46:09
1009*1010=1019090
Cidkip 2017-09-27 19:46:09
So the answer is $1009\cdot 1010 = 1009000 + 10090 = 1019090$
DPatrick 2017-09-27 19:46:15
Therefore, our sum is the product $(1009)(1010)$.
DPatrick 2017-09-27 19:46:22
And to finish, $(1009)(1010) = 1010000 + 9090 = \boxed{1019090}$.
sde7 2017-09-27 19:46:30
factor out 2 and then sum of first 1009 numbers formula
neelbhalla 2017-09-27 19:46:36
divide the sequence by 2 giving you the sum of all numbers from 1...1009 which can be generalized by the formula n(n+1)/2
DPatrick 2017-09-27 19:46:53
Definitely, as an alternative you could have gotten this by rewriting the given sum as $2(1+2+3+4+\cdots+1009)$, and then using the formula that the sum of the first $n$ positive integers is $\dfrac{n(n+1)}{2}$.
DPatrick 2017-09-27 19:47:37
I suspect that most of the wrong answers either were off by a factor of 2, or computed $1009 \cdot 1008$ instead.
DPatrick 2017-09-27 19:47:44
On to #3:
DPatrick 2017-09-27 19:47:47
3. $\left(\cos\dfrac{\pi}{12}+\sin\dfrac{\pi}{12}\right)\left(\cos\dfrac{\pi}{12}-\sin\dfrac{\pi}{12}\right) = \;?$
DPatrick 2017-09-27 19:47:51
(We are instructed that no trig functions may appear in our final answer.)
GeronimoStilton 2017-09-27 19:48:21
Use difference of squares first.
reddragon644 2017-09-27 19:48:21
Difference of squares?
pokemonduel 2017-09-27 19:48:21
its a difference of squares
DPatrick 2017-09-27 19:48:43
Do you know what $\cos\dfrac{\pi}{12}$ or $\sin\dfrac{\pi}{12}$ are? I don't off the top of my head. But we can simplify a bit first.
DPatrick 2017-09-27 19:49:00
It may be clearer to see what's going on if we set $c = \cos\dfrac{\pi}{12}$ and $s = \sin\dfrac{\pi}{12}$, so that our expression becomes $(c+s)(c-s)$. (This is often a good tactic: replace a complicated chunk in an expression with a variable, especially if it appears in multiple places.)
Unlimited 2017-09-27 19:49:23
$c^2-s^2$!
ConanC 2017-09-27 19:49:23
c^2-s^2
DPatrick 2017-09-27 19:49:37
And as many of you pointed out, this is the difference-of-squares factorization: $(c+s)(c-s) = c^2 - s^2$.
DPatrick 2017-09-27 19:49:52
So our expression is equal to $\cos^2\dfrac{\pi}{12} - \sin^2\dfrac{\pi}{12}$.
iks92 2017-09-27 19:50:14
double angle formula for cos
Cidkip 2017-09-27 19:50:14
$\cos^2 \pi/12 - \sin^2 \pi/12,$ which looks a lot like the double angle identity...
mxzhang 2017-09-27 19:50:14
The left hand side is $\cos^2\frac\pi{12}-\sin^2\frac\pi{12}=\cos^22\cdot\frac\pi{12}$.
vatatmaja 2017-09-27 19:50:14
this is just double angle for cosine
stronto 2017-09-27 19:50:14
$ = \cos \frac{\pi}{6}$ by double-angle identities
mxzhang 2017-09-27 19:50:14
Cosine double angle formula!
sketchcomedyrules 2017-09-27 19:50:14
This is the cosine double angle formula!
DPatrick 2017-09-27 19:50:22
Right! Now we can use the double-angle formula $\cos 2x = \cos^2x - \sin^2x$.
DPatrick 2017-09-27 19:50:31
So our expression is just $\cos\dfrac{\pi}{6}$.
awesomemaths 2017-09-27 19:50:54
sqrt3/2
rapturt9 2017-09-27 19:50:59
$\sqrt{3}/2$
DPatrick 2017-09-27 19:51:02
And that's $\boxed{\dfrac{\sqrt3}{2}}$.
DPatrick 2017-09-27 19:51:30
This was actually the hardest out of the batch #2-#9: only 24% got this correct.
DPatrick 2017-09-27 19:51:40
(#10 was a lot harder, though, as we'll see in a bit.)
mikebreen 2017-09-27 19:51:44
That surprised us.
DPatrick 2017-09-27 19:52:15
On to #4, which is one of the unique features of WWTBAM -- math history or trivia.
DPatrick 2017-09-27 19:52:19
4. What ancient Greek mathematician has a "sieve," used for finding prime numbers, named after him?
a. Archimedes b. Eratosthenes c. Euclid d. Pythagoras
OmicronGamma 2017-09-27 19:52:51
ERATOSTHENES' SIEVE
GeronimoStilton 2017-09-27 19:52:51
Sieve of Eratosthenes
albertlu00 2017-09-27 19:52:51
sieve of eratosthenes
pianoman24 2017-09-27 19:52:51
Sieve of Eratosthenes I coded that
DPatrick 2017-09-27 19:52:56
It's $\boxed{\text{b. Eratosthenes}}$
DPatrick 2017-09-27 19:53:03
//cdn.artofproblemsolving.com/images/c/b/8/cb82c8eea68bbfb8ec705c5bc081acdadb647e71.jpg
DPatrick 2017-09-27 19:53:20
Eratosthenes lived 276-194 BC and was the chief librarian at the Library of Alexandria (in present-day Egypt, then part of the Greek empire).
DPatrick 2017-09-27 19:53:30
The Sieve of Eratosthenes is a method for creating a list of prime numbers. A typical example is to find all the primes less than 100. We start with a chart from 2 to 100 (we don't include 1 because 1 is special -- it's neither prime nor composite):
DPatrick 2017-09-27 19:53:39
DPatrick 2017-09-27 19:53:58
Those of you who have seen it before: how does it work?
neelbhalla 2017-09-27 19:54:32
take away all numbers divisble by 2, then take away all numbers divisible by 3 until you reach the max number(100)
pianoman24 2017-09-27 19:54:32
2 is the first prime, so now all the multiples of 2 can be erased since they are not primes
Cidkip 2017-09-27 19:54:32
Start with 2. Circle it, then cross off all multiples of 2
DPatrick 2017-09-27 19:54:38
Right. The smallest number in the chart -- 2 -- is prime. And any multiple of 2 is not prime.
DPatrick 2017-09-27 19:54:44
So we circle 2, and we cross out all the numbers that are multiples of 2.
DPatrick 2017-09-27 19:54:47
Generic_Username 2017-09-27 19:55:00
go to the first unmarked number and call it prime, mark all multiples, repeat
bubblestick3 2017-09-27 19:55:00
first remove the multiples of 2, then 3, then 5, then keep doing it with other primes
awesomemaths 2017-09-27 19:55:02
then 3
DPatrick 2017-09-27 19:55:16
Right. Now the smallest unmarked number is 3. It's prime.
DPatrick 2017-09-27 19:55:20
So we circle it and cross out multiples of 3:
DPatrick 2017-09-27 19:55:25
prajna1225 2017-09-27 19:55:34
then 5
Unlimited 2017-09-27 19:55:34
Yup and we keep doing that
DPatrick 2017-09-27 19:55:39
We can continue in this way: we circle the smallest remaining unmarked number (which must be prime, as it's not a multiple of a smaller number), and cross out its multiples.
DPatrick 2017-09-27 19:55:48
So we next do this for 5:
DPatrick 2017-09-27 19:55:54
DPatrick 2017-09-27 19:56:00
And then we do it for 7:
DPatrick 2017-09-27 19:56:04
pokemonduel 2017-09-27 19:56:11
just keep crossing out multiples of primes till you reach 100 and you dont have to go more than 10
Unlimited 2017-09-27 19:56:11
Until 10
DPatrick 2017-09-27 19:56:26
Nothing more need be done. Any number less than 100 that's composite must be a multiple of 2, 3, 5, or 7, because those are all the primes less than $\sqrt{100} = 10$. So all the numbers that remain uncrossed are prime. (I won't bother circling them.)
DPatrick 2017-09-27 19:56:41
So we see that the primes less than 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
DPatrick 2017-09-27 19:57:10
Eratosthenes is also particularly known for a couple of other insights. First, he was one of the first people known to attempt to accurately measure the circumference of the Earth. He did this by comparing the noon shadow at two different locations and using the difference in their angles. There is some scholarly dispute over the units that he used, but his calculation is believed to be within 10-15% of the actual circumference of the Earth.
DPatrick 2017-09-27 19:57:30
That's pretty impressive for over 2000 years ago.
DPatrick 2017-09-27 19:57:39
Second, Eratosthenes was one of the first people to recognize the need for leap years. He was able to determine that the length of a year was approximately 365.25 days, so that every 4th year we would need a leap day in the calendar.
DPatrick 2017-09-27 19:57:59
Next:
DPatrick 2017-09-27 19:58:00
5. Write $\dfrac{\log_5 9}{\log_{25} 3}$ as a rational number.
Pi_3.14_Squared 2017-09-27 19:58:22
Use change of base.
awang11 2017-09-27 19:58:22
Use log properties
rick101 2017-09-27 19:58:22
Use change of base formula
DPatrick 2017-09-27 19:58:36
You might know some log identities that can help you out, but they're easy to mess up if you're not careful.
DPatrick 2017-09-27 19:58:49
I prefer to introduce variables because it's way easier (for me) to keep track.
reddragon644 2017-09-27 19:58:57
5 is the base and 9 is the answer to the top so do $5^x=9$
DPatrick 2017-09-27 19:59:07
Exactly. Let's set $x = \log_5 9$ and $y = \log_{25} 3$, so that the quantity we want to find is $\dfrac{x}{y}$.
DPatrick 2017-09-27 19:59:27
So then we have $5^x = 9$ and $25^y = 3$ by the definition of log.
DarkRunner 2017-09-27 20:00:07
Oh, and rewrite 25 as 5^
DarkRunner 2017-09-27 20:00:07
5^2*
DPatrick 2017-09-27 20:00:19
Yes, good idea: they're easier to compare if they have the same base.
DPatrick 2017-09-27 20:00:31
So let's write the second one as $(5^2)^y = 3$.
DPatrick 2017-09-27 20:00:35
This simplifies to $5^{2y} = 3$.
DPatrick 2017-09-27 20:00:50
So now we have $5^x = 9$ and $5^{2y} = 3$.
neelbhalla 2017-09-27 20:01:06
square 5^2y so they are both 9
DPatrick 2017-09-27 20:01:13
Good idea too.
DPatrick 2017-09-27 20:01:19
If we square it, we get $(5^{2y})^2 = 9$. This simplifies to $5^{4y} = 9$.
rick101 2017-09-27 20:01:32
x=4y
sgadekar 2017-09-27 20:01:32
then x=4y
DPatrick 2017-09-27 20:01:50
Right. We have $5^x = 9$ and $5^{4y} = 9$. So $x = 4y$.
ThorJames 2017-09-27 20:02:00
So the answer is $4$
kz1356 2017-09-27 20:02:00
x/y=4
prajna1225 2017-09-27 20:02:00
x=4y, ans is 4y/y=4
DPatrick 2017-09-27 20:02:07
And to finish, $\dfrac{x}{y} = \boxed{4}$ is our answer.
mikebreen 2017-09-27 20:02:28
Nice solution.
DPatrick 2017-09-27 20:02:46
Like I said, you can certainly do this more slickly with log-identities. But this way I'm pretty sure I didn't mess anything up!
DPatrick 2017-09-27 20:03:02
6. For which one of the following choices for $m$ are the base $m$ numbers $25_m$ and $27_m$ both prime?
a. 8 b. 9 c. 11 d. 12 e. 13
DPatrick 2017-09-27 20:03:36
We could brute-force check them all, but what can we notice that may make things easier?
ThorJames 2017-09-27 20:03:50
write into 2m+5 and 2m+7
reddragon644 2017-09-27 20:03:50
you put the number in terms of m
pianoman24 2017-09-27 20:03:50
Rewrite them as 2m+5 and 2m+7 and test
Generic_Username 2017-09-27 20:03:50
they are twin primes
shootingstar8 2017-09-27 20:03:50
they are twin primes
DPatrick 2017-09-27 20:03:58
Right: we need to find which $m$ makes $2m+5$ and $2m+7$ both prime.
DPatrick 2017-09-27 20:04:11
And it's clear that these are consecutive odd numbers.
DPatrick 2017-09-27 20:04:26
If they're both prime, what does that say about $2m + 3$?
pianoman24 2017-09-27 20:04:43
It isn't prime
reddragon644 2017-09-27 20:04:43
it is not prime
DPatrick 2017-09-27 20:04:52
Indeed, but more specifically...?
awesomemaths 2017-09-27 20:04:58
it has to be divisble by 3
DPatrick 2017-09-27 20:05:19
Right. If we've got three consecutive prime numbers, and two of them are prime, then the other one has to be a multiple of 3.
DPatrick 2017-09-27 20:05:30
So $2m + 3$ must be a multiple of 3. What does that say about $m$?
GeronimoStilton 2017-09-27 20:05:53
It's a multiple of $3$!
Mudkipswims42 2017-09-27 20:05:53
multiple of 3
math-family 2017-09-27 20:05:53
It has to be a multiple of 3
ShreyJ 2017-09-27 20:05:53
it is divisible by 3
yali306 2017-09-27 20:05:53
It's a multiple of 3
DPatrick 2017-09-27 20:06:09
Right. For $2m+3$ to be a multiple of 3, $m$ must be a multiple of 3 too.
pianoman24 2017-09-27 20:06:15
We only have to check 9 or 12
rick101 2017-09-27 20:06:15
We've eliminated 3 choices.
DPatrick 2017-09-27 20:06:24
Right: the answer must be (b) 9 or (d) 12.
DPatrick 2017-09-27 20:06:37
And at this point we can just check the two possibilities.
pianoman24 2017-09-27 20:06:58
9 gives 23 and 25, and 25 isn't prime, so it has to be 12 (we only even have to check 1 by process of elimination!)
DPatrick 2017-09-27 20:07:09
Indeed, $m=9$ doesn't work because $27_9 = 25$ isn't prime.
DPatrick 2017-09-27 20:07:25
So by process of elimination the answer must be $\boxed{\text{(d) } 12}$.
DPatrick 2017-09-27 20:07:35
And indeed, $25_{12} = 29$ and $27_{12} = 31$ are both prime.
DPatrick 2017-09-27 20:08:00
Next up:
DPatrick 2017-09-27 20:08:04
7. You and two friends decide to pay a restaurant bill by having each person flip a fair, two-sided coin. If all three coins show heads or all show tails, you will split the bill three ways. Otherwise, the person whose coin landed differently from the other two will pay the entire bill. What is the probability that you will not have to pay anything?
albertlu00 2017-09-27 20:08:29
casework
Pi_3.14_Squared 2017-09-27 20:08:29
Casework?
DPatrick 2017-09-27 20:08:59
Right. There are two cases listed in the problem: we all flip the same, or someone flips different.
DPatrick 2017-09-27 20:09:46
But if we all flip the same, I'll definitely have to pay. So we can ignore that case, because it's not part of what we want!
DPatrick 2017-09-27 20:10:06
What's the probability that we don't all flip the same?
IlinoisMathlete 2017-09-27 20:10:23
3/4
TopNotchMath 2017-09-27 20:10:23
3/4
awang11 2017-09-27 20:10:23
3/4
Plasma_Vortex 2017-09-27 20:10:23
3/4
NLMath 2017-09-27 20:10:23
3/4
DPatrick 2017-09-27 20:10:43
Right. There are $2^3 = 8$ possible outcomes for the flips, and only 2 of those (HHH and TTT) are where we all flip the same.
DPatrick 2017-09-27 20:11:01
So the other 6 outcomes, we don't all flip the same, and that occurs with probability $\dfrac68 = \dfrac34$.
DPatrick 2017-09-27 20:11:25
And if we don't flip the same, what's the probably I don't have to pay?
Hippoman 2017-09-27 20:11:49
2/3
GeronimoStilton 2017-09-27 20:11:49
$\frac{2}{3}$
awang11 2017-09-27 20:11:49
2/3
DPatrick 2017-09-27 20:12:07
Right. By symmetry, each of the 3 of us are equally likely to be the person who has to pay it all.
DPatrick 2017-09-27 20:12:24
So I have probability 1/3 that I'll have to pay it all, leaving probability $\dfrac23$ that I'll have to pay nothing.
awang11 2017-09-27 20:12:34
Then 3/4 * 2/3 = $\boxed{1/2}$
DPatrick 2017-09-27 20:13:18
Right: to summarize and finish, there's a $\dfrac34$ probability that one person will end up paying it all, and in that case there's a $\dfrac23$ probability that person won't be me.
DPatrick 2017-09-27 20:13:38
So I get a free lunch with probability $\dfrac34 \cdot \dfrac23 = \dfrac24 = \boxed{\dfrac12}$.
DPatrick 2017-09-27 20:14:15
As with most probability problems, there were certainly other ways you could have computed this.
DPatrick 2017-09-27 20:14:46
I'll leave that for you to think about on your own.
DPatrick 2017-09-27 20:14:53
Let's move on -- just 3 problems to go!
DPatrick 2017-09-27 20:14:57
8. Suppose that $m$ and $n$ are positive integers such that $5m + 3n = 41$. What is the smallest possible value for $|m^2 - n^2|$?
DaniyalQazi2 2017-09-27 20:15:22
Just bash it out. It's only positive integers.
DPatrick 2017-09-27 20:15:35
Indeed, there can't possibly be very many solutions in positive integers to this equation. Maybe we can just list them all.
DPatrick 2017-09-27 20:16:12
Is there an easy way to list all the solutions using positive integers?
DaniyalQazi2 2017-09-27 20:16:34
The only possible pairs are (m,n) - (1,12), (4,7), (7,2)
pianoman24 2017-09-27 20:16:34
1 and 12, 4 and 7, 7 and 2 are our choices
Pi_3.14_Squared 2017-09-27 20:16:34
You only have to test the pairs (1,12) (4,7) (7,2)
IlinoisMathlete 2017-09-27 20:16:40
go down by multiples of 5 and see if the number is divisible by 3
randomsolver 2017-09-27 20:17:01
(m,n) = (1,12), (4,7), (7,2)
losedude 2017-09-27 20:17:01
because 5m has to end in a 5 or 0, 3m must end in 1 or 6
DPatrick 2017-09-27 20:17:21
Indeed, there are lots of ways to find the list of solutions, including just plain old brute force.
DPatrick 2017-09-27 20:17:46
What I did is essentially was losedude suggested: notice that $3n$ has to be one more than a multiple of 5.
DPatrick 2017-09-27 20:17:56
This means $n$ must be 2 more than a multiple of 5.
DPatrick 2017-09-27 20:18:13
(You can experiment and check this if you're not sure about my last assertion.)
DPatrick 2017-09-27 20:18:19
So we must have $n = 2, 7, \text{ or } 12$.
DPatrick 2017-09-27 20:18:28
And now we can just make a little chart.
DPatrick 2017-09-27 20:18:31
\[

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

m & n & m^2 - n^2 \\ \hline

7 & 2 & 45 \\

4 & 7 & -33 \\

1 & 12 & -143

\end{array}

\]
prajna1225 2017-09-27 20:18:42
we must make m close to n
bubblestick3 2017-09-27 20:18:42
but m and n have to be as close as possible, so its 7^2 - 4^2 = 33
DPatrick 2017-09-27 20:18:55
Right: we see that the smallest value of $|m^2 - n^2|$ is the middle row, $\boxed{33}$.
DPatrick 2017-09-27 20:19:20
On to geometry:
DPatrick 2017-09-27 20:19:24
9. In rectangle $ABCD$ below, $\overline{DP}$ is perpendicular to $\overline{PC}$. Find $x$, the distance from $A$ to $P$, where $P$ is between $A$ and the midpoint of segment $\overline{AB}$.
a. $3-2\sqrt2$ b. $3-\sqrt2$ c. $3-\sqrt7$ d. $3-\sqrt7/2$
DPatrick 2017-09-27 20:19:31
DPatrick 2017-09-27 20:19:36
Note that the provided picture is clearly not to scale: there's no way that $CD = 6AD$ in the drawing. So we need to be careful about "assuming" facts from the picture.
Pi_3.14_Squared 2017-09-27 20:20:09
Let AP=x, then PB=6-x
DPatrick 2017-09-27 20:20:23
Let's label $x = AP$ in the drawing (what we're trying to find), and add the other missing segments:
DPatrick 2017-09-27 20:20:28
CforSeahorse 2017-09-27 20:20:34
Similar Triangles
tworigami 2017-09-27 20:20:34
we use similar triangles
GeronimoStilton 2017-09-27 20:20:34
Let's use similar triangles!
DPatrick 2017-09-27 20:20:40
We have similar triangles!
DPatrick 2017-09-27 20:20:52
To see this, note that since $\angle DPC$ is right, the angles $\angle APD$ and $\angle BPC$ must sum to $90^\circ$.
DPatrick 2017-09-27 20:20:57
DPatrick 2017-09-27 20:21:11
But this means that $\angle ADP$ is the red angle, and $\angle BCP$ is the blue angle, since the two acute angles in a right triangle must sum to $90^\circ$.
DPatrick 2017-09-27 20:21:14
DPatrick 2017-09-27 20:21:27
In particular, $\triangle APD \sim \triangle BCP$.
ThorJames 2017-09-27 20:21:52
6-x=1/x
pianoman24 2017-09-27 20:21:52
(6-x)/1 = 1/x
IlinoisMathlete 2017-09-27 20:22:01
1/x=6-x/1?
DPatrick 2017-09-27 20:22:05
So we have the proportion $\dfrac{x}{1} = \dfrac{1}{6-x}$.
willmathxu 2017-09-27 20:22:38
6x-x^2=1
pianoman24 2017-09-27 20:22:38
6x-x^2=1 (cross multiply)
yali306 2017-09-27 20:22:38
6x - x^2 = 1
reddragon644 2017-09-27 20:22:38
quadratics!!!
DPatrick 2017-09-27 20:22:48
Right, this simplifies to $x(6-x) = 1$, or $x^2 - 6x + 1 = 0$.
DPatrick 2017-09-27 20:23:00
And the solutions to this are $x = \dfrac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2\sqrt2$.
brainiac1 2017-09-27 20:23:32
Obviously the solution less than 3 must be the correct one
CforSeahorse 2017-09-27 20:23:32
3 + 2sqrt2 is too big
DPatrick 2017-09-27 20:23:49
Clearly only solution $\boxed{\text{a. } 3-2\sqrt2}$ works. $3 + 2\sqrt2$ isn't even a choice.
DPatrick 2017-09-27 20:23:58
And we're told $P$ is closer to $A$ than to $B$, so $x < 3$.
mikebreen 2017-09-27 20:23:58
People did much better on this problem than I would have in high school. Good for them/you!
DPatrick 2017-09-27 20:24:36
OK, as I said, numbers 2-9 were roughly equal in difficulty. Each was solved by somewhere between 24% and 42% of participants.
DPatrick 2017-09-27 20:24:49
Not so #10. Only about 7% got this one. Here we go...
DPatrick 2017-09-27 20:24:56
10. What is the largest prime factor of $2^{14} + 1$.
DPatrick 2017-09-27 20:25:40
Some of you who are very experienced in algebra may have recognized what to do right away. But let's see how we can stumble our way there.
DPatrick 2017-09-27 20:26:00
Maybe a lot of you tried brute force first -- I did.
DPatrick 2017-09-27 20:26:10
You may know that $2^{10} = 1024$. (This is a really useful fact to memorize.)
DPatrick 2017-09-27 20:26:20
So $2^{14} = 2^{10} \cdot 2^4 = 1024 \cdot 16 = 16384$.
Pi_3.14_Squared 2017-09-27 20:26:34
now add one
DPatrick 2017-09-27 20:26:41
Right. So the number we are given is $16385$.
DPatrick 2017-09-27 20:26:49
Clearly 5 is a prime factor, and we can divide to get $16385 = 5 \cdot 3277$.
DPatrick 2017-09-27 20:27:12
But now we're in a bit of a pickle. Is 3277 prime? If not, how does it factor?
DPatrick 2017-09-27 20:27:21
It's really hard to tell, and there's no clever way to check. This looks like a lot of brute force.
DPatrick 2017-09-27 20:27:35
It doable, but maybe we should back up a step -- it feels like we lost some information when we replaced $2^{14} + 1$ with $16385$.
DPatrick 2017-09-27 20:27:49
What could we do algebraically with $2^{14}+1$? It is a sum of two perfect squares -- does that help?
reddragon644 2017-09-27 20:28:00
$(2^7+1)^2-2^8$
DPatrick 2017-09-27 20:28:14
Indeed, there's always the identity $x^2 + y^2 = (x+y)^2 - 2xy$.
DPatrick 2017-09-27 20:28:21
In this case, that gives us $2^{14} + 1 = (2^7+1)^2 - 2 \cdot 2^7$.
DPatrick 2017-09-27 20:28:51
Or in other words, it's $(2^7+1)^2 - 2^8$.
reddragon644 2017-09-27 20:28:58
use differences of squares
Pi_3.14_Squared 2017-09-27 20:28:58
And this equals (2^7+1+2^4)(2^7+1-2^4)
GeronimoStilton 2017-09-27 20:28:58
Now we can factor!
SirCalcsALot 2017-09-27 20:28:58
You can factor this with difference of squares!
CforSeahorse 2017-09-27 20:28:58
2^8 is a perfect square
DPatrick 2017-09-27 20:29:05
Aha! It's a difference of two perfect squares!
DPatrick 2017-09-27 20:29:13
So our number factors as $(2^7+1+2^4)(2^7+1-2^4)$.
DPatrick 2017-09-27 20:29:24
This is $145 \cdot 113$.
DPatrick 2017-09-27 20:29:34
So after factoring out the 5 as before, we get $5 \cdot 29 \cdot 113$.
pianoman24 2017-09-27 20:29:41
113 is prime
reddragon644 2017-09-27 20:29:41
so 113
mikebreen 2017-09-27 20:29:45
This is how we did it.
Pi_3.14_Squared 2017-09-27 20:29:47
113 is prime.
andrusha2 2017-09-27 20:29:47
$113$ is our answer!
yali306 2017-09-27 20:29:47
Is 113 prime...
DPatrick 2017-09-27 20:29:54
Yes, 113 is prime! (This is easy to check: since $11^2 = 121 > 113$, we only need to check divisibility by 2, 3, 5, and 7, and none of them work.)
DPatrick 2017-09-27 20:30:01
So we have the prime factorization $2^{14} + 1 = 5 \cdot 29 \cdot 113$, and thus $\boxed{113}$ is the largest prime factor.
Pi_3.14_Squared 2017-09-27 20:30:21
Sophie Germain?
DaniyalQazi2 2017-09-27 20:30:21
Sophie Germaine Identity
ShreyJ 2017-09-27 20:30:21
We could use Sophie-Germain
CforSeahorse 2017-09-27 20:30:21
Could you not also use Sophie Germain?
DPatrick 2017-09-27 20:30:32
Indeed, what we did was just a special case of the Sophie Germain Identity, which is
\[ 4x^4 + y^4 = (2x^2+y^2)^2 - 4x^2y^2 = (2x^2+y^2+2xy)(2x^2+y^2-2xy).\]
DPatrick 2017-09-27 20:30:46
In our problem, $x = 2^3$ and $y = 1$ gives the above factorization.
DPatrick 2017-09-27 20:31:22
So that's it! On the actual contest: if you got 8 or more correct, congratulations! -- you're moving on to Round 2. Your teacher should have your invitation (or should get it soon). Round 2 will be held October 9-25. And come back here for our Round 2 Math Jam on Thursday, October 26 at 7:30 pm ET / 4:30 pm PT.
DPatrick 2017-09-27 20:31:37
After Round 2 is complete, 12 students will be invited to compete in the National Finals, live in San Diego at the 2018 Joint Mathematics Meetings on January 13. Travel costs to and from San Diego will be covered by the AMS. This is especially exciting for us this year, as San Diego is also the hometown of AoPS, so we expect that many AoPS staff will be in attendance at the Finals!
DPatrick 2017-09-27 20:31:53
Here's how the 12 finalists will be determined: 10 of the 12 will be the top scorer from Round 2 in each of the following regions:
DPatrick 2017-09-27 20:31:57
http://www.ams.org/images/wwtbam-map-us-canada.jpg
DPatrick 2017-09-27 20:32:07
The other two contestants will be the top scorer in the United Kingdom and the top scorer in the San Diego metro area (or, I suppose, the second-highest scorer, if the highest scorer in Region 9 happens to be from San Diego).
DPatrick 2017-09-27 20:32:17
The National Finals are held live in front of an audience at the Joint Mathematics Meetings, and are also live streamed on the web. (You can watch the archives of past years' finals on the WWTBAM website.) Contestants will compete directly against each other in semi-final rounds, with the semi-final winners advancing to a Jeopardy!-style buzz-in final round to determine a champion.
DPatrick 2017-09-27 20:32:40
The WWTBAM web site is ams.org/wwtbam and has lots more information and past years' contests.
mikebreen 2017-09-27 20:33:13
Everyone wins something, so we hope it's not too stressful.
GeronimoStilton 2017-09-27 20:33:56
Even the people who get knocked out in round 1?
DPatrick 2017-09-27 20:34:04
No, I think he was talking about the 12 finalists.
DPatrick 2017-09-27 20:34:21
It can be stressful to be in front of a live audience and being live-streamed on the web at the same time.
mikebreen 2017-09-27 20:34:23
Yes.
DPatrick 2017-09-27 20:34:31
But everyone among the final 12 will win prizes just for making it to the National Finals.
TheBoredKid 2017-09-27 20:35:45
What if there are ties for the top in each region?
DPatrick 2017-09-27 20:35:50
There are tiebreaking procedures.
mikebreen 2017-09-27 20:35:57
We have a tie-breaking question in Round 2.
DPatrick 2017-09-27 20:36:36
For example, last year's tiebreaker question was:
DPatrick 2017-09-27 20:36:41
10. What is the smallest (minimum) value of $n$ such that $(n+1)!$ has four more digits than $n!$?
Note: this is the tie-breaking question. In the event of a tie for high scores in a region, the winner is the person with the closest to the correct answer without going over.
DPatrick 2017-09-27 20:37:05
This is almost impossible to get exactly without a computer. But with a little ingenuity you can make a pretty good estimate.
DaniyalQazi2 2017-09-27 20:37:30
what are the prizes for making the finals?
DPatrick 2017-09-27 20:37:41
It's $10000 to the winner.
DPatrick 2017-09-27 20:37:57
(I think that's split halfway between the student and the school.)
DPatrick 2017-09-27 20:38:12
Plus other cash prizes and prizes from some of the other sponsors to the other students.
mikebreen 2017-09-27 20:38:19
Yes. Everyone gets at least $500 plus matching amount for math dept. of school.
Fluxed 2017-09-27 20:39:14
Why is the website so hard to navigate? Do you have web developers?
GeronimoStilton 2017-09-27 20:39:34
Where are the past WWTBAM problems?
DPatrick 2017-09-27 20:40:13
Those are the summaries of past years.
mikebreen 2017-09-27 20:40:19
Videos linked from http://www.ams.org/wwtbam. Past tests linked to from there, too.
DPatrick 2017-09-27 20:40:24
http://ams.org/programs/students/wwtbam/tests
DPatrick 2017-09-27 20:40:39
...has all the past tests. I see 2018 Round One is already posted there.
DPatrick 2017-09-27 20:41:03
(Which is the contest we just discussed. The finals are in January 2018.)
Fluxed 2017-09-27 20:41:09
Who won last year?
mikebreen 2017-09-27 20:41:26
Graham O'Donnell from FL.
DPatrick 2017-09-27 20:42:01
//cdn.artofproblemsolving.com/images/6/8/f/68f17ee474e439bb57b9dda966daa39ee47700a5.jpg
DPatrick 2017-09-27 20:42:52
And here's Graham with a giant check at his school:
DPatrick 2017-09-27 20:43:00
//cdn.artofproblemsolving.com/images/d/e/3/de314b19dc09d107f0f7c2bdccd0fcbe44da3f21.jpg
mikebreen 2017-09-27 20:43:42
He and his family drove there under threat of ice storm. One contestant drove from Chicago (because of ice storm)!
DPatrick 2017-09-27 20:44:15
That's right, I remember: last year in Atlanta at the finals there was a significant ice storm the night before. It was hard to get in and out of town, but everyone made it!
DPatrick 2017-09-27 20:44:26
I can guarantee that this will not be a problem this year here in San Diego.
mikebreen 2017-09-27 20:44:58
We're holding you to that.
DPatrick 2017-09-27 20:45:08
I think with that, we'll call it a night. Thanks for coming, and hopefully we'll see you on October 26 for our Round 2 Math Jam!

Copyright © 2020 AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this file with others.

Invalid username
Sign In