Math Jams

Who Wants to Be a Mathematician, Qualifying Round

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AoPS instructor Dave Patrick will discuss the problems on Round 1 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 2015-10-07 19:30:24
Welcome to the 2015-16 Who Wants to Be a Mathematician Round 1 Math Jam!
DPatrick 2015-10-07 19:30:37
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 back before I started working at AoPS, when Regis was still the host. (No, I didn't win the million bucks.)
TemporalFuzz 2015-10-07 19:30:54
Awww
gamjawon 2015-10-07 19:30:54
Aww
Bonami2014 2015-10-07 19:31:06
teampeeta12301 2015-10-07 19:31:11
How far did you get?
DPatrick 2015-10-07 19:31:26
Yeah, I was a little sad too. But I did win $64000, which (after taxes) paid for a new car.
DPatrick 2015-10-07 19:31:38
Joining us tonight are the co-creators of WWTBAM, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
DPatrick 2015-10-07 19:31:48
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.
DPatrick 2015-10-07 19:32:19
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 2015-10-07 19:33:13
Here's how things will work tonight. As you've probably noticed, this 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 2015-10-07 19:33:33
This helps keep the session organized and on track.
DPatrick 2015-10-07 19:34:00
There are a lot of students here! So only a fraction of you're collective 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 2015-10-07 19:34:21
Also, we won't be going through all 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 2015-10-07 19:34:34
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 2015-10-07 19:34:50
Round 1 of the national contest was held last month, and 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 2015-10-07 19:35:13
We'll cover all 10 problems, but 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 2015-10-07 19:35:33
And several of the questions have interesting sidetracks, so we'll also stop and view some of the scenery along the way.
DPatrick 2015-10-07 19:35:57
If you have questions about the how the contest works, please save them for the end, where I, Mike, and/or Bill will try to answer them.
DPatrick 2015-10-07 19:36:09
For next, let's start with the problems!
DPatrick 2015-10-07 19:36:21
Like a lot of math contests, WWTBAM starts off with a problem that most people would probably describe as "routine" -- it's just like an exercise that you would do in an algebra class.
DPatrick 2015-10-07 19:36:24
1. Solve for $x$: $2x^2-x=15.$
DPatrick 2015-10-07 19:36:39
(You'll notice that I always put the current problem under discussion at the top of the window.)
stan23456 2015-10-07 19:36:59
Subtract 15 on both sides and factor
checkmatetang 2015-10-07 19:36:59
Move all terms to one side then solve the quadratic.
Rubaiya 2015-10-07 19:36:59
bring it all to the rhs then factor
szhang7853 2015-10-07 19:36:59
2x^2-x-15=0 then solve using quadratic formula or by factoring
DPatrick 2015-10-07 19:37:07
Right, we can write this more traditionally as a quadratic equation as $2x^2-x-15 = 0.$
trumpeter 2015-10-07 19:37:25
quadratic formula
mathwhiz16 2015-10-07 19:37:25
You can use the quadratic formula or try to factor it.
mjlove 2015-10-07 19:37:25
factor
Einsteinhead 2015-10-07 19:37:32
Then plug into the quadratic formula!
Bonami2014 2015-10-07 19:37:32
now use quadratic formula
DPatrick 2015-10-07 19:37:39
At this point, there's always the quadratic formula: $$x = \frac{1 \pm \sqrt{1 - 4(2)(-15)}}{4} = \frac{1 \pm \sqrt{121}}{4}.$$
trish 2015-10-07 19:38:04
So the answers are 3 and -5/2
Natsu_Dragneel 2015-10-07 19:38:04
3 or -5/2
phi_ftw1618 2015-10-07 19:38:04
That simplifies to 12/4 and -10/4
DPatrick 2015-10-07 19:38:10
At this point we notice that $\sqrt{121} = 11$, so our solutions are $x = \dfrac{1 \pm 11}{4}.$
skiboy32 2015-10-07 19:38:18
3, -5/2
adas1 2015-10-07 19:38:18
3 or -5/2
teampeeta12301 2015-10-07 19:38:18
3 and -5/2
DPatrick 2015-10-07 19:38:20
That gives us $\boxed{x=3}$ and $\boxed{x=-\frac52}$ as our solutions.
Einsteinhead 2015-10-07 19:38:39
Factoring is much nicer
stan23456 2015-10-07 19:38:39
(2x+5)(x-3)=0, so x=3,-5/2
Liopleurodon 2015-10-07 19:38:39
we could factor it as well
mihirb 2015-10-07 19:38:39
or $(2x+5)(x-3) = 0$
goodbear 2015-10-07 19:38:39
$(x-3)(2x+5)=0$
winnertakeover 2015-10-07 19:38:39
Factor (2x+5)(x-3)
teampeeta12301 2015-10-07 19:38:39
We could have just factored
DPatrick 2015-10-07 19:38:50
Certainly, we could have solved this quadratic by factoring: $$(2x^2-x-15) = (2x+5)(x-3).$$ So $2x+5 = 0$, giving $x=-\frac52$, or $x-3=0$, giving $x=3$.
DPatrick 2015-10-07 19:39:08
Either way, as a check of our solutions, we can use Vieta's Formulas for a quadratic. What do these formulas tell us?
phi_ftw1618 2015-10-07 19:39:32
Sum and product of solutions?
winnertakeover 2015-10-07 19:39:32
sum and product of roots
Natsu_Dragneel 2015-10-07 19:39:32
-b/a is the sum of the roots, and c/a is the product of them
mjlove 2015-10-07 19:39:32
sum and product of roots
goodbear 2015-10-07 19:39:32
the product and sum of the numbers
Bonami2014 2015-10-07 19:39:32
the sum and product of the roots
DPatrick 2015-10-07 19:39:49
Exactly. In general, Vieta's Formulas tell us that in the quadratic $ax^2 + bx + c = 0$, the two roots must sum to $-\frac{b}{a}$ and have a product of $\frac{c}{a}$.
DPatrick 2015-10-07 19:40:04
So for our quadratic, the sum of the roots should be $\frac12$, and the product of the roots should be $-\frac{15}{2}$.
DPatrick 2015-10-07 19:40:18
Sure enough, this holds for the two roots that we found, so we can be sure that they're correct.
DPatrick 2015-10-07 19:40:34
Onwards to #2!
DPatrick 2015-10-07 19:40:42
2. What is the ones digit of $2017^{2015}?$
DPatrick 2015-10-07 19:40:58
What the first simplification that we want to make?
trumpeter 2015-10-07 19:41:15
get rid of the 201
cyborg108 2015-10-07 19:41:15
7^2015
Rubaiya 2015-10-07 19:41:15
same as $7^{2015}$
gamjawon 2015-10-07 19:41:15
$7^{2015}$
mihirb 2015-10-07 19:41:15
$7^{2015}$
Natsu_Dragneel 2015-10-07 19:41:15
You just do 7^2015
DarkPikachu 2015-10-07 19:41:15
7^2015 is the same
letsgomath 2015-10-07 19:41:15
7^2015
jxiao 2015-10-07 19:41:15
It is the same as 7^2015
DivideBy0 2015-10-07 19:41:15
same as 7^2015
DPatrick 2015-10-07 19:41:20
We know that the units digit of a product (or a power, which is a special type of product) depends only on the units digits of what we're starting with.
DPatrick 2015-10-07 19:41:29
So $2017^{2015}$ has the same units digit as $7^{2015}$.
Liopleurodon 2015-10-07 19:41:49
Find a pattern!
teampeeta12301 2015-10-07 19:41:49
Look for a pattern
eveningstarandlion 2015-10-07 19:41:49
Find a pattern in the unit digit with smaller powers
rt03 2015-10-07 19:41:49
Make a pattern of the units digit
atmchallenge 2015-10-07 19:41:49
Now let's find a pattern!
Rubaiya 2015-10-07 19:41:49
now look for a patttern
phi_ftw1618 2015-10-07 19:41:49
Then we need to find the cycle of 7^n
DPatrick 2015-10-07 19:41:55
Good idea. We can look at small powers of 7, and see if we can figure out the pattern.
gamjawon 2015-10-07 19:42:19
7, 9, 3, 1 is the pattern of the increasing powers
cyborg108 2015-10-07 19:42:19
so find the pattern (7,9,3,1)
szhang7853 2015-10-07 19:42:19
The units digits of the powers of 7 cycle every 4, so it's 7 9 3 1 7 9 3 1 and on and on
TheLuckyAngelo 2015-10-07 19:42:19
the cycle is 1,7,9,3,1,...
vsny23 2015-10-07 19:42:19
3 b/c 7,9,3,1...
jxiao 2015-10-07 19:42:19
The pattern is 7, 9, 3, 1, repeat.
Einsteinhead 2015-10-07 19:42:19
It repeats: 7, 9, 3, 1, 7...
gamjawon 2015-10-07 19:42:19
7, 9, 3, 1 is the pattern
mihirb 2015-10-07 19:42:19
$7,9,3,1,7,9,3,1$
Flash12 2015-10-07 19:42:19
7, 9, 3, 1, 7, 9, 3, 1.... etc
TheLuckyAngelo 2015-10-07 19:42:19
The cycle repeats by four!
DPatrick 2015-10-07 19:42:24
\begin{align*}
\text{units digit of } 7^1 &= 7, \\
\text{units digit of } 7^2 &= 9, \\
\text{units digit of } 7^3 &= 3, \\
\text{units digit of } 7^4 &= 1, \\
\text{units digit of } 7^5 &= 7, \\
&\vdots
\end{align*}
DPatrick 2015-10-07 19:42:31
We can see that the pattern repeats for every power of 4.
DPatrick 2015-10-07 19:42:37
So we just need to figure out where $7^{2015}$ fits in the pattern.
mjlove 2015-10-07 19:42:59
divide 2015 by 4
MathStudent2002 2015-10-07 19:42:59
2015 divided by 4 leaves a remainder of 3
Natsu_Dragneel 2015-10-07 19:42:59
So you divide 2015 by 4, giving a remainder of 3.
Destructio 2015-10-07 19:42:59
2015mod4
WL0410 2015-10-07 19:42:59
2015 is 3 mod 4
Ramanan369 2015-10-07 19:42:59
2015 mod 4
mathwhiz16 2015-10-07 19:42:59
It is 3 mod 4
Flash12 2015-10-07 19:42:59
2015 mod 4 = 3, so the ones digit is 3
DPatrick 2015-10-07 19:43:20
Right. To be a little more formal, we can write $2015 = 4 \cdot 503 + 3$.
DPatrick 2015-10-07 19:43:28
So $7^{2015} = 7^{4 \cdot 503 + 3} = (7^4)^{503} \cdot 7^3$ has units digit $1 \cdot 3 = \boxed{3}$.
mihirb 2015-10-07 19:43:45
or you could use euler's totient function but this way is sometimes faster
DPatrick 2015-10-07 19:43:56
Indeed: a more general way to approach problems of this type is to use Euler's Theorem, which states that if $a$ and $n$ are relatively prime positive integers, then $$a^{\phi(n)} \equiv 1 \pmod{n},$$
where $\phi(n)$ is the Euler totient function: the number of positive integers less than $n$ that are relatively prime to $n$.
DPatrick 2015-10-07 19:44:12
(Remember that name Euler. We'll see it again a little later tonight!)
DPatrick 2015-10-07 19:44:29
In our problem, $a=7$, $n=10$, and $\phi(n) = 4$ (since 1, 3, 7, and 9 are the numbers less than 10 that are relatively prime to 10). So we have $7^4 \equiv 1 \pmod{10}$, which tells us exactly that the units digit of $7^4$ is 1.
DPatrick 2015-10-07 19:44:53
From there, we can proceed as we did above: $7^{2015} = 7^{4 \cdot 503 + 3} = (7^4)^{503} \cdot 7^3$ has units digit $1 \cdot 3 = \boxed{3}$.
DPatrick 2015-10-07 19:45:25
A special case of Euler's Theorem is Fermat's Little Theorem, which is the case where $n$ is prime (and we relabel it $p$).
DPatrick 2015-10-07 19:45:29
What's $\phi(p)$ if $p$ is prime?
High 2015-10-07 19:45:50
p-1
phi_ftw1618 2015-10-07 19:45:50
p-1?
mjlove 2015-10-07 19:45:50
p-1
nosaj 2015-10-07 19:45:50
$p-1$
ImpossibleCube 2015-10-07 19:45:50
p-1
goodbear 2015-10-07 19:45:50
p-1
DPatrick 2015-10-07 19:45:55
Right. Every positive integer less than $p$ is relatively prime to $p$ if $p$ is prime. So $\phi(p) = p-1$.
DPatrick 2015-10-07 19:46:07
And then Fermat's Little Theorem says that, provided $a$ is not a multiple of $p$, \[a^{p-1} \equiv 1 \pmod{p}.\]
DPatrick 2015-10-07 19:46:34
This, of course, is more advanced math than you'd need to solve this problem -- but I said we'd be stopping and looking at some scenery along the way.
DPatrick 2015-10-07 19:46:50
On to #3!
mikebreen 2015-10-07 19:46:52
Love the scenic route!
DPatrick 2015-10-07 19:46:57
3. How many solutions are there in $[0,2\pi]$ to the equation $\sin x = \cos x?$
DPatrick 2015-10-07 19:47:09
How would/did you solve this problem?
rt03 2015-10-07 19:47:42
Maybe make a graph?
mikhailgromov 2015-10-07 19:47:42
Draw a graph
nosaj 2015-10-07 19:47:42
I solved this problem by graphing and finding intersections.
DarkPikachu 2015-10-07 19:47:42
graph it
goldentail141 2015-10-07 19:47:42
Graph sketching
High 2015-10-07 19:47:42
graph the lines
DPatrick 2015-10-07 19:47:49
Sure, one way is to sketch the graphs $y=\sin x$ and $y=\cos x$:
DPatrick 2015-10-07 19:47:54
DPatrick 2015-10-07 19:48:02
Quick quiz: which is which?
szhang7853 2015-10-07 19:48:24
red is cosx
atmchallenge 2015-10-07 19:48:24
red=$\cos x$
jxiao 2015-10-07 19:48:24
The blue is sin
atmchallenge 2015-10-07 19:48:24
blue= $\sin x$
DPatrick 2015-10-07 19:48:29
That's right: cosine is in red and sine is in blue.
DPatrick 2015-10-07 19:48:39
If we're reasonably confident that these graphs are sufficiently accurate, then there are clearly 2 intersection points, which means that there are $\boxed{2}$ values of $x$ at which $\sin x = \cos x$.
trumpeter 2015-10-07 19:49:01
$\tan x=1$
jxiao 2015-10-07 19:49:01
Look when tan x = 1
mjlove 2015-10-07 19:49:01
divide cosx, find tanx=1
EulerMacaroni 2015-10-07 19:49:01
Think $\tan{x}=1$
chezbgone 2015-10-07 19:49:01
Divide by $\cos x$ to get $\tan x$
EulerMacaroni 2015-10-07 19:49:01
$\tan{x}=1$, then you look at $x=\frac{\pi}{4}$ and $x=\frac{5\pi}{4}$
DPatrick 2015-10-07 19:49:07
Right, another option is to divide both sides of the original equation by $\cos x$, giving $\tan x = 1$.
DPatrick 2015-10-07 19:49:18
Now we use our knowledge of the tangent function to note that $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ are the solutions for $0 \le x \le 2\pi$, so again we have $\boxed{2}$ solutions (and in fact we know what those solutions are!).
DPatrick 2015-10-07 19:49:32
However: what do we have to be careful about whenever we divide an equation like this?
phi_ftw1618 2015-10-07 19:49:51
Division by 0?
EulerMacaroni 2015-10-07 19:49:51
dividing by zero
mathwhiz16 2015-10-07 19:49:51
cos(x)=0
Knin2820 2015-10-07 19:49:51
Division by 0
szhang7853 2015-10-07 19:49:51
when cosx=0
WL0410 2015-10-07 19:49:51
Dividing by 0
stronto 2015-10-07 19:49:51
it doesnt equal 0
guluguluga 2015-10-07 19:49:51
can't divide by 0
DPatrick 2015-10-07 19:49:56
We have to be careful not to divide by 0. Otherwise we might lose a solution.
DPatrick 2015-10-07 19:50:02
For example, if there were a value of $x$ for which $\cos x = 0$ that solved the equation, then we'd lose that solution when we divide by $\cos x$.
Liopleurodon 2015-10-07 19:50:14
but it's impossible!
EulerMacaroni 2015-10-07 19:50:17
doesn't matter since $\sin$ and $\cos$ are never zero at the same time
DPatrick 2015-10-07 19:50:28
Right you all. Because $\sin^2x + \cos^2x = 1$ always, we can't have them both be 0 at the same time.
DPatrick 2015-10-07 19:50:53
On to #4!
DPatrick 2015-10-07 19:50:57
One of the unique and (in my opinion) fun aspects of WWTBAM is that there is always a question about the history of mathematics:
DPatrick 2015-10-07 19:51:00
4. George Boole, who developed the logic upon which computers operate, was born in which of the following countries?
(a) Austria (b) England (c) France (d) Germany
DPatrick 2015-10-07 19:51:12
If you think it'll help, here's a picture of Mr. Boole:
DPatrick 2015-10-07 19:51:17

//cdn.artofproblemsolving.com/images/5/9/6/596453d2acf31a15fca0e9ad7f9d246f2b4c575b.jpg
Liopleurodon 2015-10-07 19:51:40
England!
szhang7853 2015-10-07 19:51:40
his name sounds englishh so b
EulerMacaroni 2015-10-07 19:51:40
His name is George so it's clearly England
Liopleurodon 2015-10-07 19:51:40
His name sounds english lol
szhang7853 2015-10-07 19:51:40
george is an english name
eveningstarandlion 2015-10-07 19:51:45
England actually
joeyusa2013 2015-10-07 19:51:45
b. England
DPatrick 2015-10-07 19:51:51
George Boole was born on November 2, 1815, in Lincoln, Lincolnshire, England. So the answer is $\boxed{(\text{b})}$. (However, Boole spent much of his adult life in Ireland, where he was a mathematics professor.)
mikebreen 2015-10-07 19:52:01
Boole making all this possible.
DPatrick 2015-10-07 19:52:16
Indeed! Although Boole did research in a variety of mathematical topics, he is best known for developing the foundations of what we today call Boolean algebra, which is a mathematical description of logical operations, and the foundation of much computer programming.
DPatrick 2015-10-07 19:52:26
In fact, many computer programming languages using the term boolean to refer to a variable that can take the values "true" or "false". These variables can then be manipulated using Boolean algebra operations, as in this snippet from the Python programming language:
DPatrick 2015-10-07 19:52:30
DPatrick 2015-10-07 19:52:40
Go ahead and click "Run"!
DPatrick 2015-10-07 19:53:13
By the way, my source for the photo and the biographical information is the page http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Boole.html, which is part of the great MacTutor History of Mathematics archive at the University of St. Andrews in Scotland. The main page is http://www-groups.dcs.st-and.ac.uk/~history. They have biographies of hundreds of mathematicians, from Asger Aaboe to Eustachy Zylinski. (No, I didn't make those names up.)
DPatrick 2015-10-07 19:53:47
And a shameless plug: if you'd even like to learn the Python programming language, a snippet of which is above, AoPS has Python classes!
DPatrick 2015-10-07 19:54:03
Anyway, on to problem #5:
DPatrick 2015-10-07 19:54:06
5. How many subsets (including the set itself) of a four-element set have at least two elements?
nosaj 2015-10-07 19:54:36
We can do casework on the number of elements in the subset!
mihirb 2015-10-07 19:54:36
Do case work on the number of elements
rt03 2015-10-07 19:54:36
Divide into cases based on the number of elements in the subset
DPatrick 2015-10-07 19:54:46
One way to approach this problem is to use casework. Specifically, if a subset of a 4-element set has at least two elements, then it has either 2, 3, or 4 elements.
phi_ftw1618 2015-10-07 19:55:07
Isn't it just $4C2+4C3+4C4$ = $6+4+1$=11?
gamjawon 2015-10-07 19:55:07
4 choose 2 + 4 choose 3 + 4 choose 4
mihirb 2015-10-07 19:55:13
there are ${4 \choose 2}$ 2 element ones ${4 \choose 3}$ 3 element ones and ${4 \choose 4}$ 4 element ones
DPatrick 2015-10-07 19:55:17
Exactly.
vsny23 2015-10-07 19:55:21
11 b/c 4C2 plus 4C3 plus 4C4
adas1 2015-10-07 19:55:21
4C2 + 4C3 + 4C4 = 6+4+1=11
DPatrick 2015-10-07 19:55:28
To count 2-element subsets, we have to choose two of the four elements to form the subset. One way to express this is as the combination $\binom42$, which is 6.
DPatrick 2015-10-07 19:55:38
Another way to think of it is that there are 4 choices for the first element of the subset, and then 3 choices for the second element of the subset. So that's $4 \cdot 3 = 12$ ways to choose two elements.
DPatrick 2015-10-07 19:55:44
But subsets don't care about order, so this method of choosing counts each subset twice: once for either order of choosing the two elements. Thus, when counting 12, we've double-counted, and the actual number of subsets is $12/2 = 6$.
DPatrick 2015-10-07 19:56:04
Similarly, we can count $\binom43 = 4$ ways to choose 3 elements out of 4. (But it's easier to just think of which element we decide to leave out: we have 4 choices for the missing element.)
DPatrick 2015-10-07 19:56:18
And just 1 four-element subset: the set itself.
DPatrick 2015-10-07 19:56:27
So our final answer is that there are $6 + 4 + 1 = \boxed{11}$ subsets with 2 or more elements.
Einsteinhead 2015-10-07 19:56:41
Use complementary counting!
atmchallenge 2015-10-07 19:56:41
Complementary counting!
mikhailgromov 2015-10-07 19:56:41
Complementary counting
KevTu 2015-10-07 19:56:41
Complementary counting
WL0410 2015-10-07 19:56:41
Complementary counting
axue 2015-10-07 19:56:41
or complementary counting if casework is too complicated
Rubaiya 2015-10-07 19:56:41
complementary
Knin2820 2015-10-07 19:56:41
Or complementary counting, no?
DPatrick 2015-10-07 19:56:53
Right: another approach is that we could count all the subsets, and subtract those with fewer than 2 elements. (This is sometimes called complementary counting: we're counting what we don't want.)
Pimaster314 2015-10-07 19:57:30
2^n = number of subsets
EulerMacaroni 2015-10-07 19:57:30
or just $16-4-1=11$ by complementary counting
jxiao 2015-10-07 19:57:30
You can also have 2^4 = 16 sets in total, minus 1 empty set and 4 one-element sets to get 11.
phi_ftw1618 2015-10-07 19:57:30
So that's 16 subsets - $4C0+4C1$ = 16-5 = 11
phi_ftw1618 2015-10-07 19:57:30
Oops forgot the parenthesis
DPatrick 2015-10-07 19:57:45
Right. To start, there are $2^4 = 16$ subsets altogether. Think about how you would construct a subset: for each of the 4 elements, you have to decide whether it's in the subset or not. So you have 2 choices for each of 4 elements, and these choices are independent of each other, giving $2^4 = 16$ possibilities.
DPatrick 2015-10-07 19:58:03
And then we count how many of these 16 subsets have fewer than 2 elements.
DPatrick 2015-10-07 19:58:10
Well, there's the empty set $\emptyset$. That's 1.
DPatrick 2015-10-07 19:58:15
There's also the 4 subsets each consisting of a single element.
DPatrick 2015-10-07 19:58:20
So that's $1+4 = 5$ subsets we have to exclude, leaving us a total of $16 - 5 = \boxed{11}$ desired subsets.
DPatrick 2015-10-07 19:58:34
Counting problems can often be solved in multiple ways, and doing them in more than one way is a good check of your answer!
DPatrick 2015-10-07 19:58:52
Halfway home! On to #6:
DPatrick 2015-10-07 19:58:56
6. What is the area of the quadrilateral formed by gluing together a 3-4-5 right triangle and a 5-12-13 right triangle along their common side of length 5?
mjlove 2015-10-07 19:59:19
draw a diagram
Pimaster314 2015-10-07 19:59:19
picture
DPatrick 2015-10-07 19:59:40
This is a bit of a "trick" question in that it's much easier than it might first appear. Most of the difficulty of this problem is getting the accurate picture to be sure you're not missing anything.
DPatrick 2015-10-07 19:59:49
The side of length 5 is the hypotenuse of the smaller triangle, and the smaller leg of the larger triangle. So they glue together like this:
DPatrick 2015-10-07 19:59:53
phi_ftw1618 2015-10-07 20:00:14
Just add the triangle areas?
trumpeter 2015-10-07 20:00:14
treat the triangles as separate and add the areas
Harry0531 2015-10-07 20:00:14
The are right triangles!
mathwhiz16 2015-10-07 20:00:14
This is quite simple: just find the area of the two triangles separately!
letsgomath 2015-10-07 20:00:14
find the ares of the to triangles and just add them together
ohmcfifth 2015-10-07 20:00:14
Find the areas separately and add them together
Liopleurodon 2015-10-07 20:00:14
it's just the area of the two triangles glued together!
matl26 2015-10-07 20:00:14
Just add both areas
szhang7853 2015-10-07 20:00:14
find the areas of the two triangles and add them together
DPatrick 2015-10-07 20:00:33
Just to be careful, we note that the entire region is a quadrilateral with sides 3, 4, 12, and 13, as required in the problem statement.
DPatrick 2015-10-07 20:00:45
So its area is just the sum of the two triangles' areas.
john456852 2015-10-07 20:01:04
its 36 because (3*4/2) + (5*12/2)
phi_ftw1618 2015-10-07 20:01:04
Then we get $\frac{3(4)}{2}+\frac{5(12)}{2}$ which is equal to $6+30$= $36$
eveningstarandlion 2015-10-07 20:01:04
30+6=36
Wakashubi12 2015-10-07 20:01:04
36
int_user 2015-10-07 20:01:04
=36
mihirb 2015-10-07 20:01:04
The area of the 3-4-5 is 6 and 5-12-13 30. 30+6= 36
goodbear 2015-10-07 20:01:04
6+30=36
DPatrick 2015-10-07 20:01:13
The small triangle has area $\frac12 \cdot 3 \cdot 4 = 6$, and the large triangle has area $\frac12 \cdot 5 \cdot 12 = 30$.
DPatrick 2015-10-07 20:01:17
Thus the area of the quadrilateral is $6 + 30 = \boxed{36}$.
mikebreen 2015-10-07 20:01:33
Love watching people solve these so quickly (and so well).
DPatrick 2015-10-07 20:01:44
You might wonder if different gluings might lead to different areas. We could certainly flip the larger triangle:
DPatrick 2015-10-07 20:01:48
DPatrick 2015-10-07 20:01:58
This is a different quadrilateral, but it still has the same area, given by the sum of the two triangles' areas.
EulerMacaroni 2015-10-07 20:02:11
non-intersecting area is additive, you don't even need to draw a diagram
mathwhiz16 2015-10-07 20:02:14
Different gluings will not lead to different areas unless one triangle overlaps another.
DPatrick 2015-10-07 20:02:27
Right, we might worry a little about an overlapping gluing that still gives a quadrilateral.
DPatrick 2015-10-07 20:02:33
DPatrick 2015-10-07 20:02:43
But the resulting shape is not a quadrilateral, so this is no good.
DPatrick 2015-10-07 20:02:48
Flipping the large triangle the other way also completely covers the smaller triangle by the larger one.
DPatrick 2015-10-07 20:02:52
DPatrick 2015-10-07 20:03:18
OK, on to #7:
DPatrick 2015-10-07 20:03:25
7. How many real solutions are there to the equation $\ln(x^2+x) = \ln(x^2) + \ln(x)?$
DPatrick 2015-10-07 20:03:43
First to note some slightly more advanced math than what we've seen so far: what is $\ln$?
Rubaiya 2015-10-07 20:04:04
natural log
ninjataco 2015-10-07 20:04:04
log base e
azmath333 2015-10-07 20:04:04
natural log
stronto 2015-10-07 20:04:04
natural log or log based e
bharatputra 2015-10-07 20:04:04
log base e
winnertakeover 2015-10-07 20:04:04
Does IN mean base e?
nosaj 2015-10-07 20:04:04
It's a base $e$ logarithm.
vsny23 2015-10-07 20:04:04
log with base e
jxiao 2015-10-07 20:04:04
Log base e / natural log
Harry0531 2015-10-07 20:04:04
The inverse of $e^x$
ninjataco 2015-10-07 20:04:04
inverse of e^x
DPatrick 2015-10-07 20:04:12
$\ln$ is the natural logarithm, or the logarithm to the base $e$. ($e \approx 2.718281828\ldots$ is a very special irrational number...more about $e$ later.)
DPatrick 2015-10-07 20:04:23
So this is a problem about logarithms. What can we do?
nosaj 2015-10-07 20:04:55
Guys the domain of logs is positive numbers!!
EulerMacaroni 2015-10-07 20:04:55
$\ln(x^2)+\ln(x)=\ln(x^3)$
bguo 2015-10-07 20:04:55
log addition property
rt03 2015-10-07 20:04:55
Use logarithm addition rules
Rubaiya 2015-10-07 20:04:55
or ln x^3
ohmcfifth 2015-10-07 20:04:55
Use the sum of logs = product rule
atmchallenge 2015-10-07 20:04:55
Logarithm rules. $\ln (x^2)+\ln (x)=\ln (x^3)$
azmath333 2015-10-07 20:04:55
turn the RHS into $\ln(x^3)$ using log properties?
DPatrick 2015-10-07 20:05:14
I'd first note what nosaj noted: since logarithms are only defined for positive numbers, we must have $x > 0$. (This automatically makes $x^2 > 0$ and $x^2+x>0$.)
DPatrick 2015-10-07 20:05:34
Now we can use the identity $\ln(a) + \ln(b) = \ln(ab)$.
DPatrick 2015-10-07 20:05:44
So our equation becomes $\ln(x^2+x) = \ln(x^2 \cdot x) = \ln(x^3)$.
DPatrick 2015-10-07 20:05:51
How does this help?
zsp 2015-10-07 20:06:22
Take away the $ln$
mjlove 2015-10-07 20:06:22
raise both sides to the e power
Rubaiya 2015-10-07 20:06:22
x^2 + x = x^3
jxiao 2015-10-07 20:06:22
Now we know x^2+x=x^3.
trumpeter 2015-10-07 20:06:22
$\ln$ is injective
WL0410 2015-10-07 20:06:22
$x^3=x^2+x$
szhang7853 2015-10-07 20:06:22
if you take both sides to the power of e, we can create an equation with no logarithms
mikhailgromov 2015-10-07 20:06:22
Raise to power of e to get rid of logs
ninjataco 2015-10-07 20:06:22
x^2 + x = x^3
dtxiong 2015-10-07 20:06:22
we can set what's in the logs equal to each other
mathwhiz16 2015-10-07 20:06:22
We can now change it into the equation $x^2+x=x^3$
eveningstarandlion 2015-10-07 20:06:22
x^2+x=x^3
azmath333 2015-10-07 20:06:22
raise both sides to the $e$ power
Harry0531 2015-10-07 20:06:22
That means we can $e^x$ both sides. We get $x^3=x^2+x$
DPatrick 2015-10-07 20:06:29
Right. Logarithms are 1-to-1 functions: this means that $\ln(p) = \ln(q)$ if and only if $p = q$.
DPatrick 2015-10-07 20:06:35
So we conclude that $x^2 + x = x^3$.
DPatrick 2015-10-07 20:06:42
How do we count the solutions to this?
int_user 2015-10-07 20:07:02
now you can divide by x
mathwhiz16 2015-10-07 20:07:02
Now we can divide by x
xayy 2015-10-07 20:07:02
divide by x
szhang7853 2015-10-07 20:07:02
divide by x to create quadratic
zsp 2015-10-07 20:07:02
Divide by x
ninjataco 2015-10-07 20:07:02
divide by x since x>0 and then solve the quadratic
Rubaiya 2015-10-07 20:07:02
x(x^2 -x - 1) = 0
DPatrick 2015-10-07 20:07:15
Right. Clearly $x=0$ is a solution to our cubic equation. But we're discarding that solution, since $\ln(0)$ is undefined.
DPatrick 2015-10-07 20:07:26
So we can divide by $x$ and we have $x + 1 = x^2$. This is the quadratic equation $x^2 - x - 1 = 0$.
DPatrick 2015-10-07 20:07:38
Thus by the quadratic formula, $x = \dfrac{1 \pm \sqrt5}{2}$. So 2 solutions, right?
ninjataco 2015-10-07 20:08:03
no, x>0
xayy 2015-10-07 20:08:03
no 1-sqrt5 is negative
goodbear 2015-10-07 20:08:03
positive only
dtxiong 2015-10-07 20:08:03
no, 1-sqrt5 is negative
mjlove 2015-10-07 20:08:03
no, x cann't be negative
szhang7853 2015-10-07 20:08:03
1-sqrt5/2 is negative not in domain
mikhailgromov 2015-10-07 20:08:03
No. X>0.
john456852 2015-10-07 20:08:03
no!!
gxah 2015-10-07 20:08:03
one is negative
WL0410 2015-10-07 20:08:03
No, discard the negative one
axue 2015-10-07 20:08:03
x cannot be negative
john456852 2015-10-07 20:08:03
its one because it can only be positive!
DPatrick 2015-10-07 20:08:08
No! Remember we must have $x>0$ for the original logarithm equation to make sense!
DPatrick 2015-10-07 20:08:17
So the only solution to our original equation is $x = \dfrac{1 + \sqrt5}{2}$, and thus there is $\boxed{1}$ solution.
DPatrick 2015-10-07 20:08:39
$e$ is a really cool number: it's arguably the most important number in mathematics. (I would vote for 0 myself.)
DPatrick 2015-10-07 20:08:51
One way to define $e$ is as the infinite sum of the reciprocals of all the factorials:
\[ e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots. \]
DPatrick 2015-10-07 20:09:06
Another way to define $e$ is to do the following thought experiment:
DPatrick 2015-10-07 20:09:11
Imagine that the bank down the street has a savings account that pays $100\%$ annual interest! If I put $\$1$ in the bank at the start of the year, how much money will I have at the end of the year?
mathwhiz16 2015-10-07 20:09:27
$2.
zsp 2015-10-07 20:09:27
$2
jxiao 2015-10-07 20:09:27
$2
Destructio 2015-10-07 20:09:27
2
DPatrick 2015-10-07 20:09:34
True, if the interest is just compounded once -- that is, is just paid as a single lump sum at the end of the year -- then I'll receive $\$1$ in interest, and have a total of $\$2$ at the end of the year.
DPatrick 2015-10-07 20:09:46
But what if the interest is compounded semi-annually: that is, once after 6 months and again at the end of the year?
zsp 2015-10-07 20:10:03
$2.25
mathwhiz16 2015-10-07 20:10:03
$2.25
DPatrick 2015-10-07 20:10:09
Exactly. After 6 months, I'll get $50\%$ interest on my dollar, so I'll have $\$1.50$.
DPatrick 2015-10-07 20:10:15
Then, at the end of the year, I'll get $50\%$ interest on my $\$1.50$. So that's another $\$0.75$ of interest, and I'll have $\$2.25$ at the end of the year.
stronto 2015-10-07 20:10:32
It goes towards e ... doesn't it
quartzgirl 2015-10-07 20:10:32
this number zeroes in on e
DPatrick 2015-10-07 20:10:41
Indeed...let's see what happens if the interest is compounded monthly. Note that I get $\frac{100}{12}\%$ on my money at the end of each month. (I'm going to leave off the dollar signs, and round to 3 decimal places -- some of the totals might be off by 0.001 due to rounding.)
DPatrick 2015-10-07 20:10:46
\[\begin{array}{r|c|c}
\text{Month} & \text{Interest Earned} & \text{New Balance} \\ \hline
1 & 0.083 & 1.083 \\
2 & 0.090 & 1.174 \\
3 & 0.098 & 1.271 \\
4 & 0.106 & 1.377 \\
5 & 0.115 & 1.492 \\
6 & 0.124 & 1.616 \\
7 & 0.135 & 1.751 \\
8 & 0.146 & 1.897 \\
9 & 0.158 & 2.055 \\
10 & 0.171 & 2.226 \\
11 & 0.186 & 2.412 \\
12 & 0.201 & 2.613
\end{array}\]
DPatrick 2015-10-07 20:10:54
Now I've got a little over $\$2.61$ at the end of the year! The more often we compound, the better I make out.
jxiao 2015-10-07 20:11:16
When time goes down infinitely small, you get $e.
Harry0531 2015-10-07 20:11:16
The limit of this is $e$
EulerMacaroni 2015-10-07 20:11:16
approaches $e$
DPatrick 2015-10-07 20:11:30
Right! Now imagine that my bank is compounding infinitely often! How much will I have at the end of the year?
DPatrick 2015-10-07 20:11:33
I'll have exactly $\$e \approx \$2.71828$ in my account if the interest is compounded infinitely often.
DPatrick 2015-10-07 20:11:49
To say this in a formula: the formula for the amount of money that I'll have at the end of the year, if it is compounded $n$ times during the year, is
\[ \text{Amount} = \left(1 + \frac{1}{n}\right)^n. \]
DPatrick 2015-10-07 20:11:59
For example, for the monthly compounding, the amount is
\[ \text{Amount} = \left(1 + \frac{1}{12}\right)^{12} = \left(\frac{13}{12}\right)^{12} \approx \frac{2.3298 \cdot 10^{13}}{8.9161 \cdot 10^{12}} \approx 2.613. \]
DPatrick 2015-10-07 20:12:15
We get $e$ by letting $n$ get arbitrarily large. The specific mathematical term for this is a limit:
\[ e = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n. \]
Harry0531 2015-10-07 20:12:43
Actually, we define $e$ to be the number where $e^x$'s derivitive is itself
DPatrick 2015-10-07 20:12:48
Right. Two other amazing facts about $e$ before we go on.
DPatrick 2015-10-07 20:12:51
First, if you know a little calculus, you'll undoubtedly know the significance of $e$: the functions $f(x) = ce^x$ (for some constant $c$) are the only real-valued functions that equal their own derivative. That is, the exponential function with base $e$, up to a constant multiple, is the only solution to the differential equation $y' = y$.
DPatrick 2015-10-07 20:13:01
(If you don't know what this means, don't worry -- you'll learn it in calculus class!)
atmchallenge 2015-10-07 20:13:09
$e^{\pi i}+1=0$
DPatrick 2015-10-07 20:13:20
Second, Euler (there he is again!) discovered the fundamental relationship between $e$, complex number, and trigonometric functions, in what is today called Euler's Formula:
\[ e^{i\theta} = \cos \theta + i \sin \theta.\]
DPatrick 2015-10-07 20:13:34
If we set $\theta$ equal to $\pi$, we get what was voted the most beautiful formula in mathematics:
\[ e^{i \pi} + 1 = 0. \]
This formula combines the five most important numbers in mathematics: 0, 1, $i$, $\pi$, and $e$.
DPatrick 2015-10-07 20:14:01
There are a gazillion more facts about $e$ I could talk about, but let's move onwards!
DPatrick 2015-10-07 20:14:10
8. A prism has an equilateral triangle as its base and top and three rectangles for its lateral sides. A sphere of radius 1 fits inside the prism, touching all five of its faces. What is the volume of the prism?
Wave-Particle 2015-10-07 20:14:32
diagram
Pimaster314 2015-10-07 20:14:32
picture
DPatrick 2015-10-07 20:14:33
Here's a crude picture of our prism (without the sphere drawn):
DPatrick 2015-10-07 20:14:36
DPatrick 2015-10-07 20:14:44
What's the formula for the volume of a prism?
Liopleurodon 2015-10-07 20:15:07
BH
trumpeter 2015-10-07 20:15:07
$Bh$
matl26 2015-10-07 20:15:07
bh
atmchallenge 2015-10-07 20:15:07
$V=bh$
runikmehrotra 2015-10-07 20:15:07
Base time height
mihirb 2015-10-07 20:15:07
Area of the base times the height
bguo 2015-10-07 20:15:07
base area*height
szhang7853 2015-10-07 20:15:07
base times height
aYummyNoodle 2015-10-07 20:15:07
base times height
nosaj 2015-10-07 20:15:07
$Bh$
eveningstarandlion 2015-10-07 20:15:07
triangle area * height
DPatrick 2015-10-07 20:15:13
Yep, it's just the area of the base times the height.
DPatrick 2015-10-07 20:15:19
Do we know either of these right away?
nosaj 2015-10-07 20:15:41
Note that the height of the prism is twice the radius of the sphere.
ninjataco 2015-10-07 20:15:41
height = 2
goodbear 2015-10-07 20:15:41
h=2
mathman2048 2015-10-07 20:15:41
the height is 2
azmath333 2015-10-07 20:15:41
$h=2$
int_user 2015-10-07 20:15:41
height is 2
rt03 2015-10-07 20:15:41
h=2r=2 because the sphere fits inside the prism
ohmcfifth 2015-10-07 20:15:41
Height = 2
DPatrick 2015-10-07 20:15:46
Right. The sphere fits snugly inside the prism and touches all 5 faces.
DPatrick 2015-10-07 20:15:50
So in particular, the diameter of the sphere is the height of the prism.
DPatrick 2015-10-07 20:16:01
But we're given that the radius is 1, so the diameter is 2, and thus the height is 2.
DPatrick 2015-10-07 20:16:07
So if we can find the area of the equilateral triangle base, then we'll be done.
trumpeter 2015-10-07 20:16:27
take a cross section
mihirb 2015-10-07 20:16:27
We should make a cross section
mihirb 2015-10-07 20:16:42
the radius of the sphere is the inradius of the triangle
nosaj 2015-10-07 20:16:42
Take a cross section parallel to the base and passing through the center of the sphere.
atmchallenge 2015-10-07 20:16:42
The great circle of the sphere is the equilateral triangle's incircle.
DPatrick 2015-10-07 20:16:51
Right. The sphere is also tangent to the three sides. So if we cut a cross-section through the middle of the prism, we'll get a circle of radius 1 that's tangent to all three sides of an equilateral triangle:
DPatrick 2015-10-07 20:16:54
DPatrick 2015-10-07 20:17:13
How can we compute the area of this triangle, given that we know the radius of the circle is 1?
Pimaster314 2015-10-07 20:17:37
inradius
letsgomath 2015-10-07 20:17:37
inradius formula
lkarhat 2015-10-07 20:17:37
inradius formula
bguo 2015-10-07 20:17:37
incircle formula
DPatrick 2015-10-07 20:17:47
There's a formula for this, but I like to keep it simple.
mathman2048 2015-10-07 20:17:50
make a 30 60 90 with the radius and the left side
DPatrick 2015-10-07 20:18:02
Yes, I'd just draw in a 30-60-90 triangle:
DPatrick 2015-10-07 20:18:05
DPatrick 2015-10-07 20:18:14
What's the area of the blue triangle?
jxiao 2015-10-07 20:18:37
sqrt(3)/2
eveningstarandlion 2015-10-07 20:18:37
sqrt3/2
stronto 2015-10-07 20:18:37
sqrt3 / 2
ImpossibleCube 2015-10-07 20:18:37
$\sqrt{3}/2$
zsp 2015-10-07 20:18:37
$\frac{\sqrt{3}}{2}$
Alpha-aops 2015-10-07 20:18:37
sqrt3/2
mathwhiz16 2015-10-07 20:18:37
$\sqrt{3}/2$
DPatrick 2015-10-07 20:18:45
It's a right triangle with legs of length $1$ and $\sqrt3$, so its area is $\frac{\sqrt3}{2}$.
szhang7853 2015-10-07 20:18:52
1/6 of the area of the triangle
sturdyoak2012 2015-10-07 20:18:55
multiply by 6
DPatrick 2015-10-07 20:19:03
And yes, the entire triangle is just six of these smaller blue triangles:
DPatrick 2015-10-07 20:19:07
Alpha-aops 2015-10-07 20:19:20
3sqrt3
int_user 2015-10-07 20:19:20
so 3sqrt3
mathwhiz16 2015-10-07 20:19:20
so $3sqrt{3}$
eveningstarandlion 2015-10-07 20:19:28
the triangle is $3\sqrt{3}$
rt03 2015-10-07 20:19:28
Area of base is $(6)(\frac{\sqrt{3}}{2}=3\sqrt{3}$.
DPatrick 2015-10-07 20:19:29
So the triangle has area $6 \cdot \frac{\sqrt3}{2} = 3\sqrt3$.
runikmehrotra 2015-10-07 20:19:48
so the volume is 3sqrt(3) *2
ninjataco 2015-10-07 20:19:48
so the volume is $6\sqrt{3}$
runikmehrotra 2015-10-07 20:19:48
or 6sqrt(3)
DPatrick 2015-10-07 20:19:54
And to finish up, the base area is $3\sqrt3$ and the height is 2, so the volume is their product, $\boxed{6\sqrt3}$.
DPatrick 2015-10-07 20:20:14
Nearing the end of the contest -- on to #9:
DPatrick 2015-10-07 20:20:18
9. Bob and Jane have three children. Given that one child is their daughter Mary, what is the probability that Bob and Jane have at least two daughters?
trumpeter 2015-10-07 20:20:36
were there two possible answers to this, depending on interpretation?
DPatrick 2015-10-07 20:20:43
Indeed...
DPatrick 2015-10-07 20:21:08
The Who Wants to Be a Mathematician staff have decided that, due to the ambiguity of the problem, they are accepting two different answers.
DPatrick 2015-10-07 20:21:33
One interpretation of this problem is as an example of conditional probability.
DPatrick 2015-10-07 20:21:51
That is, we're asked to find the probability that they have at least 2 daughters, given that they have at least 1 daughter.
mathwhiz16 2015-10-07 20:22:11
Count the number of possible cases and the number of cases with two daughters.
DPatrick 2015-10-07 20:22:31
Exactly. We can list all the possibilities for 3 children in a chart:
DPatrick 2015-10-07 20:22:34
\[\begin{array}{c|c|c}
\text{Oldest} & \text{Middle} & \text{Youngest} \\ \hline
B&B&B \\
B&B&G \\
B&G&B \\
G&B&B \\
B&G&G \\
G&B&G \\
G&G&B \\
G&G&G
\end{array}\]
DPatrick 2015-10-07 20:22:58
But because we are told that the couple has a daughter, we can eliminate the top row, leaving 7 equally-likely possibilities:
DPatrick 2015-10-07 20:23:06
\[\begin{array}{c|c|c}
\text{Oldest} & \text{Middle} & \text{Youngest} \\ \hline
B&B&G \\
B&G&B \\
G&B&B \\
B&G&G \\
G&B&G \\
G&G&B \\
G&G&G
\end{array}\]
mathwhiz16 2015-10-07 20:23:21
7 cases, 4 with two daughters.
DPatrick 2015-10-07 20:23:31
Right. The bottom four rows are the possibilities with at least two girls, so the probability of that occurring is $\boxed{\frac47}$.
DPatrick 2015-10-07 20:23:43
However...
DPatrick 2015-10-07 20:23:48
The fact of the matter is, English is a lot less precise than math. And as such, problems like this are open to different interpretations depending on how you interpret the English.
DPatrick 2015-10-07 20:23:56
A reasonable alternative interpretation of the problem would be:
DPatrick 2015-10-07 20:24:01
9. Bob and Jane have three children. Given that one child is their daughter Mary, what is the probability that at least one of their other children is a girl?
DPatrick 2015-10-07 20:24:25
Which basically reads the same way in English, but now it's not a conditional probability problem at all: it's just a "usual" probability problem with no condition.
goodbear 2015-10-07 20:24:36
3/4
Liopleurodon 2015-10-07 20:24:36
Ohhhhhh! so it would be 3/4
Knin2820 2015-10-07 20:24:36
There's a $\frac{1}{2}$ chance of having a boy. So there's a $\frac{1}{4}$ chance of having 2 boys, and no girls. Therefore, there is a $1-\frac{1}{4}=\frac{3}{4}$ chance of having at least one more girl.
DPatrick 2015-10-07 20:24:56
Right, that's one solution: they can only not have a girl (among the two other children) if the two other kids are both boys. This occurs with probability $\left(\frac12\right)^2 = \frac14$. Therefore, the probability that, among two children, at least one is a girl, is $1 - \frac14 = \boxed{\frac34}$.
DPatrick 2015-10-07 20:25:11
Or, we could make a chart for the two other children:
\[\begin{array}{c|c}
\text{Older} & \text{Younger} \\ \hline
B & B \\
B & G \\
G & B \\
G & G
\end{array}\]
DPatrick 2015-10-07 20:25:16
There are 4 equally-likely rows, and 3 of these rows have at least one girl. Therefore, the probability of at least one girl is $\boxed{\frac34}$.
EulerMacaroni 2015-10-07 20:25:26
this problem is a more general version of a well-known paradox
DPatrick 2015-10-07 20:25:36
Right you are. There's a lengthy discussion of this sort of problem on Wikipedia at
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
DPatrick 2015-10-07 20:25:46
It's an unfortunate fact of life, and of the imprecision of the English language, that problems like this are usually ambiguous and open to multiple interpretations.
DPatrick 2015-10-07 20:25:53
Though I wouldn't call it a "paradox" -- rather, it's a consequence of the imprecision of the English language.
TheLuckyAngelo 2015-10-07 20:26:00
The Monty Hall Problem!!!!
DPatrick 2015-10-07 20:26:04
This is also somewhat related to the famous "Monty Hall" problem. If you haven't heard of that before, I highly recommend looking it up.
DPatrick 2015-10-07 20:26:14
And as I said at the beginning of this problem, the Who Wants to Be a Mathematician staff have decided that, due to the ambiguity of the problem, they are accepting both $\frac47$ and $\frac34$ as correct answers.
DPatrick 2015-10-07 20:26:54
And on to the finish at #10, which was actually a bit easier (in my opinion) than some of the earlier problem:
DPatrick 2015-10-07 20:26:58
10. What integer is closest to the square root of the product of the largest two-digit prime number and the smallest three-digit prime number?
szhang7853 2015-10-07 20:27:46
the largest 2 digit prime is 97 and the smallest 3 digit prime is 101
mathman2048 2015-10-07 20:27:46
97 x 101 square rooted
runikmehrotra 2015-10-07 20:27:46
isnt it just 97 times 101
DeathLlama9 2015-10-07 20:27:46
$97 \times 101$
phi_ftw1618 2015-10-07 20:27:46
So closest to $\sqrt{97\cdot101}$?
bearytasty 2015-10-07 20:27:46
sqrt(97*101)
Liopleurodon 2015-10-07 20:27:46
97*101
Harry0531 2015-10-07 20:27:46
$97\times 101$
bearytasty 2015-10-07 20:27:46
97 and 101 are the two numbers
ninjataco 2015-10-07 20:27:46
97 and 101
DPatrick 2015-10-07 20:28:04
Right. The largest two-digit prime is 97, and the smallest three-digit prime is 101.
DPatrick 2015-10-07 20:28:08
So we're looking for the integer that's closest to $\sqrt{97 \cdot 101}$.
Alpha-aops 2015-10-07 20:28:42
99
djduan 2015-10-07 20:28:42
99
EulerMacaroni 2015-10-07 20:28:42
so like 99
SilverPersian1 2015-10-07 20:28:42
99
rt03 2015-10-07 20:28:42
Start around 99 and estimate numbers that are $\sqrt{97 \times 101}$.
WL0410 2015-10-07 20:28:49
$\sqrt{(99+2)(99-2)}=\sqrt{99^2-4}$
trumpeter 2015-10-07 20:28:49
$97\cdot101=99^2-4$
DPatrick 2015-10-07 20:28:54
It's very tempting to say 99.
DPatrick 2015-10-07 20:28:59
Note that if $x = 99$, then our quantity is $\sqrt{(x-2)(x+2)}$, which simplifies to $\sqrt{x^2-4}$.
DPatrick 2015-10-07 20:29:09
It seems very likely that if $x$ is a large enough integer, then $\sqrt{x^2 - 4}$ is closest to the integer $\sqrt{x^2} = x$. Is 99 large enough for this to happen?
zsp 2015-10-07 20:29:29
Yes
letsgomath 2015-10-07 20:29:29
yes
dtxiong 2015-10-07 20:29:29
yes
eveningstarandlion 2015-10-07 20:29:29
Yes
Harry0531 2015-10-07 20:29:29
Itwould seem so...
DPatrick 2015-10-07 20:29:33
Absolutely!
DPatrick 2015-10-07 20:29:38
We just need $x^2-4$ to be between $\left(x-\frac12\right)^2$ and $x^2$.
DPatrick 2015-10-07 20:29:46
But $\left(x-\frac12\right)^2 = x^2 - x + \frac14$.
DPatrick 2015-10-07 20:29:56
So as long as $x^2 - x + \frac14 < x^2 - 4$, then $\sqrt{x^2-4}$ will round to $x$.
DPatrick 2015-10-07 20:30:16
Thus all we need is $-x < -\frac{17}{4}$, or $x > \frac{17}{4}$. That's clearly the case if $x=99$, so we're done.
DeathLlama9 2015-10-07 20:30:22
So $x > \frac{15}{4}$, and since $99$ is a lot bigger than $3.75$ it works
DPatrick 2015-10-07 20:30:28
Right. The answer is $\boxed{99}$.
DPatrick 2015-10-07 20:30:36
That's it for Round 1!
DPatrick 2015-10-07 20:30:45
If you officially participated in the contest, and you got at least 8 of these 10 problems correct, then you'll move on to Round 2 later this month. Official notification of this should go out to your teacher later this week.
DPatrick 2015-10-07 20:30:55
We will have a Round 2 Math Jam on Tuesday, November 3 at 7:30 PM Eastern / 4:30 PM Pacific to discuss the Round 2 problems and solutions.
DPatrick 2015-10-07 20:31:09
They'll be, on average, a bit harder.
DPatrick 2015-10-07 20:31:20
Then, the top scorer on Round 2 in each of 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 in January. Perhaps I'll see you there!
szhang7853 2015-10-07 20:31:38
what are the regions?
Liopleurodon 2015-10-07 20:31:38
why is the Seattle region special?
DPatrick 2015-10-07 20:31:49
There's a map of the regions on the website (which is coming in a moment).
DPatrick 2015-10-07 20:31:56
Seattle is special because that's where the live finals will be.
DPatrick 2015-10-07 20:32:01
Visit http://www.ams.org/programs/students/wwtbam/wwtbam to learn more about the game and to visit the archive of past years' contests.
DPatrick 2015-10-07 20:32:19
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. Check out the above website for more info.
mikebreen 2015-10-07 20:32:23
If you didn't participate in this year's test, we invite all of your teachers to email us at paoffice@ams.org and we'll put you on next year's mailing list.
Snowie 2015-10-07 20:32:36
Will there be a transcript of this math jam?
DPatrick 2015-10-07 20:32:45
Yes, the transcript will be available later this evening on the AoPS website.
DPatrick 2015-10-07 20:33:04
That's it for tonight's Math Jam. Thanks for coming!
jxiao 2015-10-07 20:35:17
If the top scorer ties, what is the tiebreaker?
DPatrick 2015-10-07 20:35:41
Give me just a second -- this info is on the AMS website and I'll cut-and-paste it over here.
DPatrick 2015-10-07 20:36:02
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
DPatrick 2015-10-07 20:36:33
And here's the regional map:
DPatrick 2015-10-07 20:36:36

//cdn.artofproblemsolving.com/images/1/a/9/1a96f9d75b5b00544cc7f488315169b2b7a177ff.jpg
Rubaiya 2015-10-07 20:36:48
How do you register? Is it done by teams or individuals?
DPatrick 2015-10-07 20:37:06
It's through your school. Your teacher should email paoffice@ams.org for information.
1-1 is 3 2015-10-07 20:37:48
Where is a good place to learn all the history stuff for the contest?
DPatrick 2015-10-07 20:38:01
There are a lot of websites devoted to mathematical history.
DPatrick 2015-10-07 20:38:28
My favorite is the one I mentioned earlier: the MacTutor History of Mathematics archive at the University of St. Andrews in Scotland. The main page is http://www-groups.dcs.st-and.ac.uk/~history.
guluguluga 2015-10-07 20:38:33
How many history questions are there, usually?
DPatrick 2015-10-07 20:38:38
Just 1 out of the 10.
nosaj 2015-10-07 20:38:45
One student from each region gets to go to semifinals in seattle?
DPatrick 2015-10-07 20:38:52
That's right. Plus one student from the Seattle area.
guluguluga 2015-10-07 20:39:40
Does our teacher have to administer the test to us in a specific date?
DPatrick 2015-10-07 20:40:06
There is a 2-week (I think) window, which for Round 1 was the last two weeks.
DPatrick 2015-10-07 20:40:16
Round 2 is 10/17-11/2.
mikebreen 2015-10-07 20:40:27
Yes, any time in the two-week window.
Rubaiya 2015-10-07 20:40:48
Are there options for homeschooled students?
DPatrick 2015-10-07 20:40:57
I'll let Mike answer this because I'm not sure.
mikebreen 2015-10-07 20:41:26
Homeschooled students are eligible. We only require that the test is administered by whomever does the teaching at home.
kungfugirl 2015-10-07 20:41:34
Does the admission process require anything fancy?
DPatrick 2015-10-07 20:41:52
Nope -- just write to the email address that we provided a few minutes ago.
mikebreen 2015-10-07 20:42:14
No registration fee either.
Ciscool 2015-10-07 20:42:52
How do you check score?
DPatrick 2015-10-07 20:43:03
I'll defer to Mike on the scoring -- he's the one in charge of it.
mikebreen 2015-10-07 20:43:34
Teachers can see their students' scores online, via Maple TA.
DPatrick 2015-10-07 20:44:34
OK, I think we're going to close down for the evening, which means we'll be closing the classroom very shortly. Thanks again for coming -- I hope you enjoyed it!
mikebreen 2015-10-07 20:45:23
Loved it! Thanks.
TPiR 2015-10-07 20:45:53
Terrific. Thanks.

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