Math Jams

Who Wants to Be a Mathematician, Round 1

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AoPS instructor David Patrick will discuss the problems on Round 1 of qualifying for the 2019-20 Who Wants to Be a Mathematician Championship. We will be joined by Mike Breen and Bill Butterworth, the creators of the game. Mike is also the host of the Championship finals, to be held in Denver in January 2020.

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Facilitator: Dave Patrick

DPatrick 2019-09-30 19:30:04
Welcome to the 2019-20 Who Wants to Be a Mathematician Championship Round 1 Math Jam!
DPatrick 2019-09-30 19:30:15
I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens (probably hundreds!) of AoPS classes over the past 15 years, and I've written or co-written a few of our textbooks.
DPatrick 2019-09-30 19:30:31
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 2019-09-30 19:30:35
https://www.washington.edu/alumni/columns/march00/images/patrick.jpg
DPatrick 2019-09-30 19:30:39
Photo Credit: Maria Melin, copyright 1999 ABC Television.
Epic_Dabber 2019-09-30 19:30:51
how much did you iwn
DPatrick 2019-09-30 19:30:59
I didn't win the million bucks, but I did win enough to buy a new car!
kvedula2004 2019-09-30 19:31:02
how old were u when u competed?
DPatrick 2019-09-30 19:31:09
About 20 years younger than I am now.
DPatrick 2019-09-30 19:31:16
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 2019-09-30 19:31:33
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
DPatrick 2019-09-30 19:31:47
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 2019-09-30 19:32:00
There are a lot of students here! As I said, 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.
DPatrick 2019-09-30 19:32:15
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!
DPatrick 2019-09-30 19:32:18
So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!
DPatrick 2019-09-30 19:32:36
Who Wants to Be a Mathematician (or WWTBAM for short) is conducted by the American Mathematical Society (AMS).
DPatrick 2019-09-30 19:32:39
https://www.ams.org/images/content/logo.png
DPatrick 2019-09-30 19:32:49
The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
DPatrick 2019-09-30 19:33:00
The AMS is one of the two big professional mathematics organizations in the U.S. (The other is the Mathematical Association of America, or MAA, which as you may know runs the AMC/AIME/USA(J)MO contests.)
DPatrick 2019-09-30 19:33:11
Joining us tonight are the co-creators of WWTBAM, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
DPatrick 2019-09-30 19:33:19
Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill began Who Wants to Be a Mathematician for the American Mathematical Society in 2001. The first national game was in 2010. Mike has been on Jeopardy! and Wheel of Fortune (if you want to know if he won lots of money on either show, note that he is still working for a living) and may be the only person ever to cut his hand on the wheel. Who Wants to Be a Mathematician has so far been much safer.
mikebreen 2019-09-30 19:33:27
Hi, everyone. Thanks for being here.
DPatrick 2019-09-30 19:33:41
Bill earned an undergraduate degree in mathematics from Santa Clara University and a Ph.D. from Northwestern University, and is currently an associate professor and associate chair of mathematics at DePaul University. He shares a life-long interest in game shows with colleague Mike Breen, with whom he works as the not-so-lovely assistant on the mathematics game show Who Wants to Be a Mathematician. In addition to authoring articles and presenting talks related to game-show mathematics, Bill served as mathematics consultant to the CBS television show The Price is Right from 1997 to 2009. (Hence, his username.)
TPiR 2019-09-30 19:34:04
Welcome everyone. Thanks for attending tonight
DPatrick 2019-09-30 19:34:17
As you can see, we have a lot of game show background here tonight!
DPatrick 2019-09-30 19:34:26
We also have an assistant here to help out tonight: Jeffery Yu, or AlcumusGuy. Jeffery is an undergraduate student at MIT studying mathematics. His mathematical interests span the entire range from pure, abstract math, to its applications in physics and computer science. He was an avid participant of math competitions in high school, and still enjoys keeping up with new contest problems today. Besides math, he also loves music, having played piano and violin since elementary school. In his free time, he enjoys playing Euro-style board games and card games.
DPatrick 2019-09-30 19:34:46
Jeffery 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 2019-09-30 19:35:01
Tonight we'll be talking about Round 1 of the Championship contest, which concluded last week.
DPatrick 2019-09-30 19:35:18
Round 1 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 access was permitted during the contest.
DPatrick 2019-09-30 19:35:35
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 2019-09-30 19:35:51
Let's get started!
DPatrick 2019-09-30 19:35:57
1. What is the perimeter (in inches) of a square that has area 9 square inches?
$\phantom{hi!}$
(a) 8 (b) 10 (c) 12 (d) 16
DPatrick 2019-09-30 19:36:17
The current question will always be posted to the top of the window. You can resize that top area as needed.
Allen31415 2019-09-30 19:36:32
Side length is 3, P=12
amcha9 2019-09-30 19:36:32
sqrt(9) * 4 = 12
Dragonite42 2019-09-30 19:36:32
c. sqrt(9)=3 so the side length is 3 so the perimeter is 12
PandaForLife 2019-09-30 19:36:40
sware root 9 = 3 and then x4 = 12
priyankaadhikari 2019-09-30 19:36:40
side is each 3, and because its a square 3*4 = c) 12 inches
smartninja2000 2019-09-30 19:36:40
Letting the side length of the square being n, we find that n^2=9. This means that n=3 since we want the positive solution. The perimeter is 4n=3*4=12
Toinfinity 2019-09-30 19:36:40
$s=\sqrt{9}=3\Rightarrow 4s=\boxed{12}$
DPatrick 2019-09-30 19:36:52
Right -- this question was pretty straightforward.
DPatrick 2019-09-30 19:36:58
Its area is $9$, so it has side length $\sqrt9 = 3$.
DPatrick 2019-09-30 19:37:06
And then four sides of length $3$ gives a perimeter of $4 \cdot 3 = 12$. Answer $\boxed{\text{(c)}}$.
Allen31415 2019-09-30 19:37:15
Are these problems very trivial like that one?
DPatrick 2019-09-30 19:37:23
No. That was the warmup problem.
DPatrick 2019-09-30 19:37:33
If you didn't get this correct, a likely reason is that you didn't read the problem carefully!
DPatrick 2019-09-30 19:37:50
Indeed, 95.3% of students taking the contest electronically got this correct. (I only have the stats for students who took the test electronically, which was about three-fourths of all students in Round 1.)
DPatrick 2019-09-30 19:38:08
2. The graph of which of the following is perpendicular to the line $y + 3x =7$?
$\phantom{hi!}$
(a) $y - 3x = 7$ (b) $x - 3y = 7$
(c) $3x - y = 7$ (d) $x + 3y = 7$
DPatrick 2019-09-30 19:38:21
How do we tell when two lines on the coordinate plane are perpendicular?
NASA1234 2019-09-30 19:38:47
slopes multiply to -1
kvedula2004 2019-09-30 19:38:47
slopes multiply to -1
newton2023 2019-09-30 19:38:47
perpendicular lines have negative reciprocal slopes
sbans 2019-09-30 19:38:47
negative recripocal slopes
aryabhata000 2019-09-30 19:38:47
If the product of their slopes is -1
colin_aops 2019-09-30 19:38:47
negative reciprocal slope
coolmath2017 2019-09-30 19:38:47
the slope is the negative reciprocal
kc5170 2019-09-30 19:38:47
the slopes are negative reciprocals of each other
luckysw 2019-09-30 19:38:47
when the slop multiplied is -1
DPatrick 2019-09-30 19:38:53
Right. Their slopes must have product $-1$.
DPatrick 2019-09-30 19:39:01
(The exception is that a horizontal line is perpendicular to a vertical line, but the slope of a vertical line is undefined.)
DPatrick 2019-09-30 19:39:26
Before we go on with this problem, can anyone explain why this fact is true?
DPatrick 2019-09-30 19:40:09
It's a little but of an unfair question for me to ask, because you probably want to draw a picture to explain it!
yaj.jhajhria 2019-09-30 19:40:13
Slope is rise over run, and perpendicular lines have opposite run, so slopes are opposite.
DPatrick 2019-09-30 19:40:30
Yeah, this is pretty much it without a picture!
DPatrick 2019-09-30 19:40:44
Here's a quick picture that I drew:
DPatrick 2019-09-30 19:40:48
DPatrick 2019-09-30 19:40:59
The blue line has slope $\dfrac{b}{a}$.
DPatrick 2019-09-30 19:41:03
The red line has slope $-\dfrac{d}{c}$.
DPatrick 2019-09-30 19:41:13
But what else do we know if the lines are perpendicular?
origamimaster 2019-09-30 19:41:35
a/b=d/c
DPatrick 2019-09-30 19:41:52
The red triangle and the blue triangle are similar! (They're both right triangles with the same angles.)
DPatrick 2019-09-30 19:42:04
Which tells us that $\dfrac{a}{b} = \dfrac{d}{c}$.
DPatrick 2019-09-30 19:42:21
Therefore, the red line has slope $-\dfrac{d}{c} = -\dfrac{a}{b}$.
DPatrick 2019-09-30 19:42:32
And indeed, the product of the slopes is $\dfrac{b}{a} \cdot \left(-\dfrac{a}{b}\right) = -1$.
DPatrick 2019-09-30 19:42:45
And that's a nice simple proof of the fact we need for this problem!
DPatrick 2019-09-30 19:43:08
As we say a lot around AoPS, it's really useful to know how to prove facts that you might be tempted to memorize -- it can help with more difficult problems!
DPatrick 2019-09-30 19:43:16
So back to our problem...
DPatrick 2019-09-30 19:43:21
What is the slope of the given line?
DPatrick 2019-09-30 19:43:30
(I mean the line $y + 3x =7$ that we're starting with.)
Allen31415 2019-09-30 19:43:48
-3
Stellar967 2019-09-30 19:43:48
-3
priyankaadhikari 2019-09-30 19:43:48
-3
Alpha56 2019-09-30 19:43:48
-3
yoyopianow 2019-09-30 19:43:48
Our slope in our given equation is -3.
mdawn_65 2019-09-30 19:43:48
-3
JackMathReacher 2019-09-30 19:43:48
-3
RadiantCheddar 2019-09-30 19:43:48
-3
vincent_wang 2019-09-30 19:43:48
-3
Hyo 2019-09-30 19:43:48
-3
r0518 2019-09-30 19:43:48
-3
bluelinfish 2019-09-30 19:43:48
-3
Epic_Dabber 2019-09-30 19:43:56
-3 because you can move the 3x to the other side
DPatrick 2019-09-30 19:44:00
Right. In slope-intercept form, the equation is $y = -3x + 7$.
DPatrick 2019-09-30 19:44:04
So its slope is $-3$.
DPatrick 2019-09-30 19:44:12
So now we know what slope we're looking for in the answer...
slinky 2019-09-30 19:44:33
1/3
MathClassStudent 2019-09-30 19:44:33
1/3
hp77 2019-09-30 19:44:33
1/3
box.jellyfish712 2019-09-30 19:44:33
1/3!
DPatrick 2019-09-30 19:44:39
We're looking for the answer choice with a slope that, when multiplied by $-3$, gives $-1$. So we're looking for the line with slope $\dfrac13$.
DPatrick 2019-09-30 19:44:47
Which one is it?
NASA1234 2019-09-30 19:45:10
b
amcha9 2019-09-30 19:45:10
B
himani.musku 2019-09-30 19:45:10
b
ogbooger 2019-09-30 19:45:10
b
jcf8c1203 2019-09-30 19:45:10
B
Toinfinity 2019-09-30 19:45:10
(b) $x-3y=7$
Fjmt 2019-09-30 19:45:10
b
DPatrick 2019-09-30 19:45:15
It's $\boxed{\text{(b)}}$. If we rewrite it in slope-intercept form, we get $y = \dfrac13x - \dfrac73$.
DPatrick 2019-09-30 19:45:30
Just to double-check: the slopes of the answer choices are: (a) $3$ (b) $\dfrac13$ (c) $3$ (d) $-\dfrac13$.
DPatrick 2019-09-30 19:45:46
Note also that the lines given by (a) and (c) are parallel. (Two lines with the same slope are either parallel or the same line.) Indeed, we could multiply the equation given by (c) by $-1$ to get $y - 3x = -7$, which is more clearly parallel to (a) $y - 3x = 7$.
DPatrick 2019-09-30 19:46:05
Also note that the "7" in all these equations is irrelevant. It's only the relationship between $x$ and $y$ that determines the slope of a line!
DPatrick 2019-09-30 19:46:22
On to a trig question:
DPatrick 2019-09-30 19:46:27
3. Let $T = \sin\left(\cos\left(\dfrac{\pi}{2}\right)\right)$ (where $\dfrac\pi2$ is measured in radians). Then
$\phantom{hi!}$
(a) $T=0$ (b) $0 < T \le \dfrac12$ (c) $\dfrac12 < T < 1$ (d) $T=1$
DPatrick 2019-09-30 19:46:51
Although I really encourage you to think in radians, which is what mathematicians do, you might be more comfortable with degrees.
DPatrick 2019-09-30 19:47:00
What is $\dfrac\pi2$ radians in degrees?
logz 2019-09-30 19:47:19
90 degrees
Giakki 2019-09-30 19:47:19
90
Elppa 2019-09-30 19:47:19
90
newton2023 2019-09-30 19:47:19
90
PandaForLife 2019-09-30 19:47:19
90
sbans 2019-09-30 19:47:19
90 degrees
yaj.jhajhria 2019-09-30 19:47:19
$90$
DPatrick 2019-09-30 19:47:24
$\dfrac\pi2$ radians is the same as $90^\circ$.
DPatrick 2019-09-30 19:47:34
So how do we compute $T$?
prajna1225 2019-09-30 19:47:51
cos pi/2
kvedula2004 2019-09-30 19:47:54
start from inside out. cos(90)=0
I-_-I 2019-09-30 19:47:58
cos (90 degrees) = 0
logz 2019-09-30 19:47:59
Unit circle time
DPatrick 2019-09-30 19:48:04
We compute it from inside out.
DPatrick 2019-09-30 19:48:30
To start, what is $\cos\left(\dfrac\pi2\right)$?
bigwolfy 2019-09-30 19:48:52
cos(pi/2) = 0
Edwinyc 2019-09-30 19:48:52
cos(pi/2)=0
Mathematician1010 2019-09-30 19:48:52
cos(90 degrees) using unit circle is $0$
Allen31415 2019-09-30 19:48:52
$\cos \left( \frac{\pi}2 \right)=0$
DPatrick 2019-09-30 19:48:58
Right, it is $0$. If you don't have this fact on recall, you can see this by looking at the $x$-coordinate of the point of the unit circle corresponding to the angle $\dfrac\pi2$:
DPatrick 2019-09-30 19:49:03
DPatrick 2019-09-30 19:49:10
So now we know that $T = \sin\left(\cos\left(\dfrac{\pi}{2}\right)\right) = \sin(0)$.
DPatrick 2019-09-30 19:49:18
And what is this?
luckysw 2019-09-30 19:49:40
0
Orangestripe 2019-09-30 19:49:40
0
ProblemSolver_63 2019-09-30 19:49:40
0
dori_vinnie 2019-09-30 19:49:40
0...
pow_h_2 2019-09-30 19:49:40
0
DPatrick 2019-09-30 19:49:45
It's also $0$. It's the $y$-coordinate of the point of the unit circle corresponding to the angle $0$:
DPatrick 2019-09-30 19:49:48
DPatrick 2019-09-30 19:49:56
So $T = 0$. Answer $\boxed{\text{(a)}}$.
DPatrick 2019-09-30 19:50:22
Next, a probability problem:
DPatrick 2019-09-30 19:50:27
4. Sue rolls two fair six-sided dice (with faces numbered 1-6) and computes their sum, while Diane rolls a single fair dodecahedral die (with faces numbered 1-12). Which of the following numbers has the property that Sue and Diane are equally likely to roll that number?
$\phantom{hi!}$
(a) 8 (b) 9 (c) 10 (d) 11
Epic_Dabber 2019-09-30 19:51:07
look for something that has a 1/12th chance to be rolled with 2 dice
Allen31415 2019-09-30 19:51:07
P(n) on dodecahedron die=$\frac 1{12}$
DPatrick 2019-09-30 19:51:14
Right: that's the key observation here!
DPatrick 2019-09-30 19:51:24
Each number that Diane could roll is equally likely.
DPatrick 2019-09-30 19:51:34
So each number has the same probability, which is $\dfrac{1}{12}$.
sbans 2019-09-30 19:51:41
so we have to find a number that is 1/12 chance for Sue
DPatrick 2019-09-30 19:51:52
Exactly. We're looking for the sum that Sue rolls with probability $\dfrac{1}{12}$.
jcf8c1203 2019-09-30 19:52:00
because you need both to get to 1/12, you need the two dice to have three ways to equal x
kootrapali 2019-09-30 19:52:00
3 possible ways then
DPatrick 2019-09-30 19:52:05
Good!
DPatrick 2019-09-30 19:52:10
There are 6 (equally likely) ways for her to roll her first die.



There are 6 (equally likely) ways for her to roll her second die.



So there are $6 \cdot 6 = 36$ ways for her to roll both dice.
DPatrick 2019-09-30 19:52:35
So, if we want the probability of the sum $s$ being rolled to be $\dfrac{1}{12}$, we need it to be able to be rolled $3$ different ways, since $\dfrac{3}{36} = \dfrac{1}{12}$.
DPatrick 2019-09-30 19:52:42
Which of the answer choices works?
aryabhata000 2019-09-30 19:53:04
The answer is 10 because it can be represented in 3 ways (4+6, 5+5, 6+4)
smartninja2000 2019-09-30 19:53:04
That number is 10 because 10 can be partitioned into 5+5, 4+6, and 6+4, which means that this is 3 possible ways
Hyo 2019-09-30 19:53:04
10 has 3 ways (5 + 5, 6 + 4, 4 + 6)
MathClassStudent 2019-09-30 19:53:04
there are 3 ways out of 36 ways Sue can roll a 10 4,6 5,5 and 6,4 so answe is c
yoyopianow 2019-09-30 19:53:04
There are three ways to roll the number 10: (4, 6); (5, 5); (6, 4)
DPatrick 2019-09-30 19:53:13
Right. $10$ can be rolled exactly three different ways: 4-6, 5-5, or 6-4.
DPatrick 2019-09-30 19:53:18
So the answer is $10$. Answer $\boxed{\text{(c)}}$.
DPatrick 2019-09-30 19:53:31
More generally, here is the chart for the sum of two dice:

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

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

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

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

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

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

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

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

\end{array}$$
DPatrick 2019-09-30 19:53:46
So we can count up the number of ways to roll each sum:

$$\begin{array}{r|c}

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

2 & 1 \\

3 & 2 \\

4 & 3 \\

5 & 4 \\

6 & 5 \\

7 & 6 \\

8 & 5 \\

9 & 4 \\

10 & 3 \\

11 & 2 \\

12 & 1

\end{array}$$
DPatrick 2019-09-30 19:53:53
As we can see, $7$ is the most likely sum of two dice, and the sums get less probable the further away we get from $7$.
DPatrick 2019-09-30 19:54:14
It's helpful to know these counts: not only do they come up in contest problems a lot, but it'll help you have a better chance of winning your next game of backgammon or Monopoly! (Or any game that uses the sum of two dice.)
DPatrick 2019-09-30 19:54:33
Next:
DPatrick 2019-09-30 19:54:35
5. A positive integer is called square-free if it is not divisible by any perfect square greater than $1$. Suppose $m$ and $n$ are square-free integers greater than $1$, with $m \not= n$. Which of the following is not possible?
$\phantom{Hi!}$
(a) $mn$ is square-free and composite (b) $\frac{m}{n}$ is prime
(c) $\frac{m}{n}$ is square-free and composite (d) $\sqrt{mn}$ is rational
DPatrick 2019-09-30 19:55:04
Any suggestions for how to approach this?
smartninja2000 2019-09-30 19:55:21
We can just test out all of A,B, C, and D.
Epic_Dabber 2019-09-30 19:55:21
Just trial and error and see if any of the answer choices don't work
bigwolfy 2019-09-30 19:55:21
Go through all the options
NASA1234 2019-09-30 19:55:21
use an example
bissue 2019-09-30 19:55:28
Note that being square-free means being the product of distinct primes.
DPatrick 2019-09-30 19:55:47
I like the idea of experimentation...maybe we can find examples for the three choices that are possible.
DPatrick 2019-09-30 19:56:08
And bissue makes a nice observation: a number is square-free if and only if every prime appears at most once in its prime factorization.
DPatrick 2019-09-30 19:56:46
Because: if some prime $p$ appeared more than once, then $p^2$ would have to be a divisor of the number.
DPatrick 2019-09-30 19:57:05
So, any examples for any of the choices?
BendingLight 2019-09-30 19:57:21
if m=6 and n=3 then m/n=6/3=2, which is prime, making (b) possible
DPatrick 2019-09-30 19:57:37
Good. We can rule out (b).
ar492 2019-09-30 19:57:46
a is trivial to eliminate, because if $m=2$, $n=3$
DPatrick 2019-09-30 19:58:00
Right. $mn = 6$ in that example is square-free and composite.
DPatrick 2019-09-30 19:58:06
So (a) is possible, and we can rule it out.
kvedula2004 2019-09-30 19:58:27
(30, 3) eliminates C...
Giakki 2019-09-30 19:58:27
c, m=30, n=2
DPatrick 2019-09-30 19:58:57
Right: in both of these examples, $\dfrac{m}{n}$ is square-free and composite. So (c) is possible, and that rules out answer (c).
ajp 2019-09-30 19:59:16
which only leaves (d)
Edwinyc 2019-09-30 19:59:16
so the answer is (d)
sbans 2019-09-30 19:59:16
so D is left as our answer
Stellar967 2019-09-30 19:59:16
so only one left if d; as it makes sense by ruling out and by poe
DPatrick 2019-09-30 19:59:18
By process of elimination, (d) must be impossible.
DPatrick 2019-09-30 19:59:49
This is fine for the actual contest...but since we're exploring a little bit, can we prove that (d) is impossible?
pow_h_2 2019-09-30 20:00:04
I think it is D, there must be at least one distinct prime number in sqrt(mn), and sqrt(primenum) results in a inrational number\
Allen31415 2019-09-30 20:00:04
Call the prime factorization of $m$ $p_1p_2\cdots p_n$ and $n$ be $q_1q_2\cdots q_n.$ The only way for $\sqrt{mn}$ to be rational is for each prime $p_i$ be repeated twice. Therefore $m=n$, which contradicts our assumption.
kvedula2004 2019-09-30 20:00:04
Note that for D, the same primes must appear in the prime factorization of m and n, clearly impossible.
Hyo 2019-09-30 20:00:09
it's d because there can't be pairs of factors as m and n are different and the square root of something without even pairs are irrational
DPatrick 2019-09-30 20:00:28
Good! Let me break it down a little bit...
DPatrick 2019-09-30 20:00:46
What does it mean that $\sqrt{mn}$ is rational? It actually means that $mn$ is a perfect square (since it is an integer).
DPatrick 2019-09-30 20:00:56
And what do we know about the primes in a perfect square?
JackMathReacher 2019-09-30 20:01:14
they have even exponents
Epic_Dabber 2019-09-30 20:01:14
They have even powers
amcha9 2019-09-30 20:01:14
there are pairs of them
NASA1234 2019-09-30 20:01:14
they have even exponents
jcf8c1203 2019-09-30 20:01:28
there's two of them
priyankaadhikari 2019-09-30 20:01:28
there needs to be an even exponeent
DPatrick 2019-09-30 20:01:37
Right. Each one must appear an even number of times in the prime factorization of $mn$.
DPatrick 2019-09-30 20:01:55
But we already decided that since $m$ and $n$ are square-free, each prime can only appear in each of them once.
DPatrick 2019-09-30 20:02:20
So for $mn$ to have an even number of each prime, it has to acquire each prime once from $m$ and once from $n$.
Stellar967 2019-09-30 20:02:39
but then it would mean they are equal to each other
JackMathReacher 2019-09-30 20:02:39
so they must be equal which contradicts what was stated in the problem
DPatrick 2019-09-30 20:02:41
But that would mean that the primes of $m$ and $n$ are exactly the same!
DPatrick 2019-09-30 20:02:49
That is, $m=n$.
DPatrick 2019-09-30 20:02:53
But we were told in the problem statement that $m \not= n$.
DPatrick 2019-09-30 20:02:56
So this is impossible!
DPatrick 2019-09-30 20:03:04
Therefore, $\boxed{\text{(d)}}$ is the answer.
DPatrick 2019-09-30 20:03:21
By the way, another way to think of square-free is that $n$ being square-free means that $\sqrt{n}$ can't be simplified. (That is, there are no square factors to take outside the radical.)
DPatrick 2019-09-30 20:03:43
6. Let $f(x) = 5x^3(2x+3)^4$ and $g(x) = 50x^5(8x-4)^2$. Which of the following is closest to $f(10^6) \div g(10^6)$?
$\phantom{hi!}$
(a) $0.025$ (b) $0.25$ (c) $2.5$ (d) $25$
DPatrick 2019-09-30 20:04:03
I suppose that we could try to compute $f(10^6)$ and $g(10^6)$. But those are really, really big numbers.
ogbooger 2019-09-30 20:04:27
we can simplify f(x)/g(x) first
Stellar967 2019-09-30 20:04:27
simplify!
NASA1234 2019-09-30 20:04:27
simplify the polynomial a bit?
DPatrick 2019-09-30 20:04:37
Yeah, let's try to play with some algebra first.
DPatrick 2019-09-30 20:04:41
We can write $$f(x) \div g(x) = \frac{5x^3(2x+3)^4}{50x^5(8x-4)^2}.$$
DPatrick 2019-09-30 20:04:54
Aha, some of this simplifies!
NASA1234 2019-09-30 20:05:29
cancel 5x^3
MySixSenses 2019-09-30 20:05:29
take out the 5x^3
Stellar967 2019-09-30 20:05:29
10x^2
DPatrick 2019-09-30 20:05:36
Yes, the $5x^3$ in the numerator divides into $50x^5$ in the denominator.
DPatrick 2019-09-30 20:05:41
So this leaves us with $$f(x) \div g(x) = \frac{(2x+3)^4}{10x^2(8x-4)^2}.$$
DPatrick 2019-09-30 20:05:48
And now what?
pow_h_2 2019-09-30 20:06:14
expand
bluelinfish 2019-09-30 20:06:14
expand?
DPatrick 2019-09-30 20:06:27
We could expand the binomials in the numerator and denominator...but those are big numbers too.
Epic_Dabber 2019-09-30 20:06:39
You can ignore the 3 and the -4 because it won't affect it that much
tpuhan 2019-09-30 20:06:39
we can forget about the +3 and -4 bc these are such big numbers
prajna1225 2019-09-30 20:06:39
the 3 and the 4 dont matter
mathdude2 2019-09-30 20:06:39
the plus 3 and minus 4 are negligible
DPatrick 2019-09-30 20:07:00
That's a pretty good idea, I think, but suppose I'm not convinced.
DPatrick 2019-09-30 20:07:18
How can we be more convinced that the +3 and the -4 "don't matter"?
Epic_Dabber 2019-09-30 20:07:41
because 10^6 is so big
Allen31415 2019-09-30 20:07:41
They are small compared to $10^6$
Stellar967 2019-09-30 20:07:41
Well, the answers are so big, and the question says CLOSEST to....
Hyo 2019-09-30 20:07:43
because they are very small in comparison to 10^6^4
DPatrick 2019-09-30 20:07:50
That's certainly all true.
DPatrick 2019-09-30 20:08:04
Let me try to suggest something a little more precise.
DPatrick 2019-09-30 20:08:26
What do we notice about the polynomials in the numerator and denominator?
snow_monkey 2019-09-30 20:08:44
same degree
origamimaster 2019-09-30 20:08:44
the are both at ^4
DPatrick 2019-09-30 20:08:51
Both the numerator and denominator are quartic (degree 4) polynomials.
DPatrick 2019-09-30 20:09:21
So one common way to deal with these is to divide numerator and denominator by $x^4$, like so:
DPatrick 2019-09-30 20:09:24
We get $$f(x) \div g(x) = \frac{\left(2 + \dfrac{3}{x}\right)^4}{10\left(8 - \dfrac{4}{x}\right)^2}.$$
DPatrick 2019-09-30 20:10:26
Now it's way more clear (at least to me!) that when we plug in $x = 10^6$, the result of $(2.000003)^4 / (7.999996)^2$ is going to be very very close to $2^4 / 8^2$.
Hyo 2019-09-30 20:10:37
now we can get rid of the fractions since the value of x is so small
snow_monkey 2019-09-30 20:10:37
$\frac{3}{x}$ and $\frac{4}{x}$ are extremely small since $x$ is extremely big so you can ignore them
DPatrick 2019-09-30 20:10:41
Exactly.
DPatrick 2019-09-30 20:10:50
So for all practical purposes, when $x$ is really big, we have $$f(x) \div g(x) \approx \dfrac{2^4}{10(8)^2}.$$
Stellar967 2019-09-30 20:11:16
16/640
pow_h_2 2019-09-30 20:11:16
16/640= 1/40
Mathisawesome234 2019-09-30 20:11:16
1/40
DPatrick 2019-09-30 20:11:28
So now you just have to do the arithmetic correctly!
DPatrick 2019-09-30 20:11:38
It simplifies to $\dfrac{16}{640} = \dfrac{1}{40}$.
jcf8c1203 2019-09-30 20:12:00
0.025
MySixSenses 2019-09-30 20:12:00
1/40 = .025
AoPS_Brian 2019-09-30 20:12:00
1/40 is exactly 0.025
yoyopianow 2019-09-30 20:12:00
1/4 = 0.25 => 1/4*1/10= 0.25*0.1 = 0.025
DPatrick 2019-09-30 20:12:15
Yes: I like the way yoyopianow worked it out.
DPatrick 2019-09-30 20:12:21
Since $\dfrac{1}{4} = 0.25$, we move the decimal point one to the left to get $\dfrac{1}{40} = 0.025$. Answer $\boxed{\text{(a)}}$.
kvedula2004 2019-09-30 20:12:34
How close is the real number from 0.025?
DPatrick 2019-09-30 20:12:40
No idea: I didn't compute it!
DPatrick 2019-09-30 20:13:01
7. The graph of the equation $(x+2)^2 + (y-3)^2 = 12$ contains points in all quadrants except quadrant
$\phantom{hi!}$
(a) I (b) II (c) III (d) IV
DPatrick 2019-09-30 20:13:14
What does this graph look like?
ogbooger 2019-09-30 20:13:23
this is the equation of a circle
newton2023 2019-09-30 20:13:23
it's a circle
Baabi 2019-09-30 20:13:23
circle
priyankaadhikari 2019-09-30 20:13:23
its a graph of a circle
NASA1234 2019-09-30 20:13:23
circle
montana_mathlete 2019-09-30 20:13:23
A circle.
tpuhan 2019-09-30 20:13:23
it's a circle
DPatrick 2019-09-30 20:13:26
It's a circle!
DPatrick 2019-09-30 20:13:31
More specifically, what circle?
newton2023 2019-09-30 20:13:59
with center -2,3 and radius radical 12
RockmanEX3 2019-09-30 20:13:59
Circle center (-2,3) radius sqrt(12)
r0518 2019-09-30 20:13:59
circle with center -2,3 and radius sqrt(12)
Epic_Dabber 2019-09-30 20:13:59
Center at (-2,3) and radius sqrt(12)
DPatrick 2019-09-30 20:14:03
Specifically, it's the circle with center $(-2,3)$ and radius $\sqrt{12}$.
DPatrick 2019-09-30 20:14:13
$\sqrt{12}$ is about $3.5$ (as it's about halfway between $\sqrt{9} = 3$ and $\sqrt{16} = 4$), so we can make a rough sketch:
DPatrick 2019-09-30 20:14:20
DPatrick 2019-09-30 20:14:28
Since I did this with a computer, this sketch is not so "rough".
DPatrick 2019-09-30 20:14:43
But even a rough sketch should show you which quadrants it goes through.
sbans 2019-09-30 20:15:04
Quads 1-3
ajp 2019-09-30 20:15:04
I, II, and III
DPatrick 2019-09-30 20:15:29
It clearly goes though quadrants I, II, and III. (You can just go straight up, left, down, or right to get points in those quadrants.)
luckysw 2019-09-30 20:15:47
so the answer is d?
vincent_wang 2019-09-30 20:15:47
It's not in the lower right corner
prajna1225 2019-09-30 20:15:47
no IV, so the ans is d
Mathematician1010 2019-09-30 20:15:47
It does not go through IV
DPatrick 2019-09-30 20:15:56
So this circle doesn't quite reach the lower-right quadrant, which is Quadrant IV.
AoPS_Brian 2019-09-30 20:16:02
Is there an algebraic way to confirm the answer?
DPatrick 2019-09-30 20:16:07
Sure!
DPatrick 2019-09-30 20:16:14
What characterizes points in Quadrant IV?
Allen31415 2019-09-30 20:16:38
positive x, negative y
Giakki 2019-09-30 20:16:38
x>0, y<0
DukeTipster2017 2019-09-30 20:16:38
Positive x, negative y
Bananazzz 2019-09-30 20:16:38
(positive,negative)
DPatrick 2019-09-30 20:16:41
They are points $(x,y)$ with $x>0$ and $y<0$.
DPatrick 2019-09-30 20:16:48
But if $x>0$, what do we know about $(x+2)^2$?
Allen31415 2019-09-30 20:17:19
its greater than 4.
Epic_Dabber 2019-09-30 20:17:19
Greater than 4
Baabi 2019-09-30 20:17:19
its bigger than 4
mdawn_65 2019-09-30 20:17:27
bigger than 4
BendingLight 2019-09-30 20:17:27
bigger then 4
DPatrick 2019-09-30 20:17:30
We must have $(x+2)^2 > 2^2 = 4$.
DPatrick 2019-09-30 20:17:34
And if $y<0$, what do we know about $(y-3)^2$?
Stellar967 2019-09-30 20:17:59
is has to greater than 9
JackMathReacher 2019-09-30 20:17:59
greater than 9
ILvMath 2019-09-30 20:17:59
greater than 9
Mathisawesome234 2019-09-30 20:17:59
greater than 9
DPatrick 2019-09-30 20:18:03
We must have $(y-3)^2 > 3^2 = 9$. (If $y$ is negative, then the quantity $y-3$ is "more negative" than $-3$, and thus its square is bigger.)
DPatrick 2019-09-30 20:18:16
So, combining these, what do we know about the expression on the left side of our equation?
priyankaadhikari 2019-09-30 20:18:45
greater than 13
prajna1225 2019-09-30 20:18:45
it is always greater than 13
tientien1 2019-09-30 20:18:45
greater than 13
Giakki 2019-09-30 20:18:45
it is greater than 13
mdawn_65 2019-09-30 20:18:45
greater than 13
DPatrick 2019-09-30 20:18:56
It must be greater than $4 + 9 = 13$.
DPatrick 2019-09-30 20:19:07
So there's no way it could equal $12$ for any point in quadrant IV.
MySixSenses 2019-09-30 20:19:16
So no soulutions
sbans 2019-09-30 20:19:16
so it won't work
DPatrick 2019-09-30 20:19:19
Thus, there are no points in our graph that are in Quadrant IV. Answer $\boxed{\text{(d)}}$.
DPatrick 2019-09-30 20:19:34
Three to go!
DPatrick 2019-09-30 20:19:39
8. $A$,$B$,$C$,$D$,$E$,$F$,$G$, and $H$ represent eight different digits selected from $\{1,2,\ldots,9\}$. If $\dfrac{A+B}{C+D} + \dfrac{E+F}{G+H}$ is as large as possible, which digit is not used?
$\phantom{hi!}$
(a) 1 (b) 2 (c) 4 (d) 5
DPatrick 2019-09-30 20:20:04
You might have a "gut feeling" as to what the correct answer is.
origamimaster 2019-09-30 20:20:27
5?
bluelinfish 2019-09-30 20:20:27
5
yoyopianow 2019-09-30 20:20:27
Gut 5
jcf8c1203 2019-09-30 20:20:27
it is 5 because you want the largest number in the numerator and the smallest numbers in the denominator
MySixSenses 2019-09-30 20:20:30
5 because 1234 are used as denominators and 6789 are numerators
DPatrick 2019-09-30 20:20:38
Right. My "gut feeling" is that to make this expression as large as possible, we want to make the numerators as large as possible and the denominators as small as possible.
DPatrick 2019-09-30 20:20:50
Which means that I think we want $\{1,2,3,4\}$ as the digits in the denominators, and $\{6,7,8,9\}$ as the digits in the numerators.
DPatrick 2019-09-30 20:20:57
Of course, this leaves $5$ omitted, which points towards answer (d).
DPatrick 2019-09-30 20:21:11
Strategically, since we don't have to prove our answer, if you trust your gut enough, you might mark (d) and move on.
DPatrick 2019-09-30 20:21:23
However, can we rigorously prove that (d) is the correct answer?
bissue 2019-09-30 20:22:15
Replacing any digit with 5 will decrease the total amount.
r0518 2019-09-30 20:22:15
we know that the 4 biggest have to be on top and the 4 smallest have to be on the bottom
pow_h_2 2019-09-30 20:22:15
The largest 4 would go on the numerator to make largest, smallest 4 on bottom to make smallest denominator
DPatrick 2019-09-30 20:22:40
Yes -- let me try to make the case that this must be true.
DPatrick 2019-09-30 20:22:56
In any given fraction, we certainly want the numerator's numbers to be greater than the denominator's numbers.
DPatrick 2019-09-30 20:23:02
That is, in $\dfrac{A+B}{C+D}$, we certainly want $A,B$ to each be greater than $C,D$.
DPatrick 2019-09-30 20:23:31
Because if, say, $A < C$, we could swap them and make the numerator bigger and the denominator smaller, both of which increase the fraction.
DPatrick 2019-09-30 20:24:01
So, we can assume that $A>B>C>D$ and $E>F>G>H$. (It doesn't matter whether $A$ or $B$ is bigger, since addition is commutative, so we might as well assume $A>B$.)
DPatrick 2019-09-30 20:24:24
Now, what happens if we write an expression with $4$ omitted?
DPatrick 2019-09-30 20:24:42
We must then be using all of $\{5,6,7,8,9\}$.
priyankaadhikari 2019-09-30 20:24:47
you have to use 5 instead
DPatrick 2019-09-30 20:24:56
Right.
DPatrick 2019-09-30 20:25:31
But since we're using five numbers bigger than the omitted $4$, one of them must be in one of the denominators...
DPatrick 2019-09-30 20:25:49
...and if we had used the $4$ instead, we'd have made that denominator smaller, and thus the whole expression bigger.
DPatrick 2019-09-30 20:26:10
The same conclusion holds if our omitted number is any of $\{1,2,3,4\}$: there will have to be a number that's $5$ or greater in one of the denominators, and we can swap this with our omitted number to make the expression bigger.
DPatrick 2019-09-30 20:26:50
The same conclusion holds if we had omitted one of $\{6,7,8,9\}$ (we can make a swap in the numerator to make the expression bigger).
DPatrick 2019-09-30 20:27:09
So, to be precise about it, what's we've proved is that if we omit any digit other than $5$, then we have a way to make the expression bigger.
DPatrick 2019-09-30 20:27:13
Therefore, when the expression is as big as possible, the omitted digit must be $5$. Answer $\boxed{\text{(d)}}$.
DPatrick 2019-09-30 20:27:38
Next, logarithms!
DPatrick 2019-09-30 20:27:43
9. Suppose $x$ is an integer satisfying $\log_3(9x) + \log_9(3x) = 7$. What is the ones digit of $x$?
$\phantom{hi!}$
(a) 3 (b) 5 (c) 7 (d) 9
DPatrick 2019-09-30 20:27:57
It might be a little hard to read on the screen. The first logarithm has base $3$ and the second logarithm has base $9$.
DPatrick 2019-09-30 20:28:19
If you haven't seen this before: $\log_b(z) = n$ means that $b^n = z$.
DPatrick 2019-09-30 20:28:53
This problem, like a lot of logarithm problems on contests, is mainly a test to see how well you know the various log manipulation rules.
Epic_Dabber 2019-09-30 20:29:02
Make the bases the same
Allen31415 2019-09-30 20:29:02
rewrite the logs with common bases.
tpuhan 2019-09-30 20:29:02
turn everything to base 3
DPatrick 2019-09-30 20:29:10
Good idea! Usually logs are easier to work with if they all have the same base.
Stellar967 2019-09-30 20:29:35
You would convert log 9 to log 3
DPatrick 2019-09-30 20:30:04
OK. If we want to rewrite the $\log_9(3x)$ term using a log with base $3$, what do we get?
Allen31415 2019-09-30 20:30:42
$\frac{\log_3 (3x)}2$
jonathanfan 2019-09-30 20:30:42
1/2 * log3(3x)
sbans 2019-09-30 20:30:49
log_3 (3x)/2
hp77 2019-09-30 20:30:49
1/2*log3x base 3
DPatrick 2019-09-30 20:30:58
Good -- this is one that is really easy to mess up.
DPatrick 2019-09-30 20:31:10
Like a lot of log rules, it can be helpful to just plug in some numbers to help you remember the rules.
DPatrick 2019-09-30 20:31:26
For example, if we plug in $x=3$, we get $\log_9(3x) = \log_9(9) = 1$.
DPatrick 2019-09-30 20:31:44
So since $\log_3(9) = 2$, we need to multiply it by a factor of $\dfrac12$ to be equal.
DPatrick 2019-09-30 20:32:06
That is an example of the more general, correct $\log_9(3x) = \frac12\log_3(3x)$.
DPatrick 2019-09-30 20:32:32
So now the equation is $\log_3(9x) + \frac12\log_3(3x) = 7$.
amcha9 2019-09-30 20:32:50
1/2*log_3(3x) = log_3(3x^1/2)
prajna1225 2019-09-30 20:32:50
log(ab) = log(a) + log(b)
DPatrick 2019-09-30 20:33:05
Yes, we can do those two steps in succession!
DPatrick 2019-09-30 20:33:18
The $\frac12$ comes inside the log as an exponent.
DPatrick 2019-09-30 20:33:36
So we have $\log_3(9x) + \log_3((3x)^\frac12) = 7$.
DPatrick 2019-09-30 20:33:53
And then we can add the logs as a product.
DPatrick 2019-09-30 20:34:24
Careful...the 3 is inside the square root too!
DPatrick 2019-09-30 20:34:35
We get $\log_3(9x \cdot (3x)^\frac12) = 7$.
DPatrick 2019-09-30 20:34:46
Now what?
mdawn_65 2019-09-30 20:34:56
remove log
DPatrick 2019-09-30 20:35:03
Now we can remove the log altogether.
Creator 2019-09-30 20:35:08
we take 3^7
Epic_Dabber 2019-09-30 20:35:08
set the thing inside the log as 3^7
DPatrick 2019-09-30 20:35:19
We get $9x \cdot (3x)^\frac12 = 3^7$.
RagingWar 2019-09-30 20:35:45
square it
serichaoo 2019-09-30 20:35:45
square both sides
wertguk 2019-09-30 20:35:47
square both sides and cancel factors of 3
DPatrick 2019-09-30 20:35:55
Yeah, I don't like fractional exponents. Let's square.
DPatrick 2019-09-30 20:36:03
$81x^2 \cdot 3x = 3^{14}$.
DPatrick 2019-09-30 20:36:38
So that's $3^5 \cdot x^3 = 3^{14}$ if I did my arithmetic right!
Mathisawesome234 2019-09-30 20:36:48
x cubed equals 3 to the 9th power
amcha9 2019-09-30 20:36:55
x^3=3^9
pratyaypandey 2019-09-30 20:36:55
So it is x^3 = 3^9
DPatrick 2019-09-30 20:36:59
So $x^3 = 3^9$.
GrizzyProblemSolver79c 2019-09-30 20:37:10
x = 27
sbans 2019-09-30 20:37:10
x=27
Giakki 2019-09-30 20:37:10
x=27
DPatrick 2019-09-30 20:37:15
And thus $x = 3^3 = 27$.
DPatrick 2019-09-30 20:37:18
This has units digit $7$. Answer $\boxed{\text{(c)}}$.
DPatrick 2019-09-30 20:38:05
When I did this, I converted the logs to base $9$ at the beginning. Having done tons of log problems in my life, I knew that would avoid the fractional exponents by converting to the higher base.
DPatrick 2019-09-30 20:38:12
And another good idea is:
kvedula2004 2019-09-30 20:38:16
Let $x=3^a$ to remove logarithms.
DPatrick 2019-09-30 20:38:50
Right: we can pretty much guess that $x$ is a power of $3$, so making the substitution $x = 3^a$ at the beginning lets us get rid of the logs at the start.
DPatrick 2019-09-30 20:39:23
This #9 was the 2nd hardest problem: only about 1/3 of the participants got it.
DPatrick 2019-09-30 20:39:31
And to the last problem, which was statistically the hardest:
DPatrick 2019-09-30 20:39:36
10. How many ordered pairs of positive integers $(m,n)$ are there such that $m^2n^5 = 20^{20}$?
$\phantom{hi!}$
(a) 15 (b) 20 (c) 30 (d) 40
mathtiger6 2019-09-30 20:40:11
factor 20
wertguk 2019-09-30 20:40:11
factor 20
NASA1234 2019-09-30 20:40:19
prime factor?
DPatrick 2019-09-30 20:40:41
Good idea: when looking at a number theory problem like this one, prime factorization is probably going to be helpful!
programjames1 2019-09-30 20:40:54
Factor first:

$m^2n^5 = 2^{40}5^{20}$
yoyopianow 2019-09-30 20:40:54
2^40*5^20
DPatrick 2019-09-30 20:41:16
Right: since $20 = 2^2 \cdot 5$, we have $20^{20} = 2^{40} \cdot 5^{20}$.
DPatrick 2019-09-30 20:41:32
So the equation is $m^2n^5 = 2^{40}5^{20}$.
DPatrick 2019-09-30 20:41:34
Now what?
programjames1 2019-09-30 20:42:08
Now divide it separately by the $2$'s and the $5$'s.
Creator 2019-09-30 20:42:08
look at the prime factorization and then look for factors of 5 and 2 in each exponent
Orangestripe 2019-09-30 20:42:08
Find out how many powers of 5 and 2 go into 40 and 50?
DPatrick 2019-09-30 20:42:47
Right...we want to know how we can get the correct number of $2$'s and $5$'s in the product of $m^2n^5$.
DPatrick 2019-09-30 20:42:57
So we should probably look at the $2$'s and $5$'s in $m$ and $n$.
DPatrick 2019-09-30 20:44:33
Suppose we say $m = 2^a5^b$ and $n = 2^c5^d$.
HAUNEBU-IV 2019-09-30 20:44:59
2a+5c=40
bfan0805 2019-09-30 20:44:59
2a + 5c = 40 and 2b + 5d = 20
HAUNEBU-IV 2019-09-30 20:44:59
2b+5d=20
DPatrick 2019-09-30 20:45:03
Right!
DPatrick 2019-09-30 20:45:23
We have $(2^a5^b)^2 \cdot (2^c5^d)^5 = 2^{40}5^{20}$.
DPatrick 2019-09-30 20:45:49
So when we collect and match exponents, we get $2a + 5c = 40$ from the powers of $2$, and $2b + 5d = 20$ from the powers of $5$.
DPatrick 2019-09-30 20:46:13
So we need to count the solutions to each of these equations.
DPatrick 2019-09-30 20:46:19
How many solutions to $2a + 5c = 40$?
DPatrick 2019-09-30 20:46:27
Remember that $a$ and $c$ are nonnegative integers!
serichaoo 2019-09-30 20:46:50
5
Giakki 2019-09-30 20:46:50
5 solutions
Hyo 2019-09-30 20:46:50
5
AoPS_Brian 2019-09-30 20:46:50
c must be even for a to be a possible solution
DPatrick 2019-09-30 20:47:06
Right. Since $2a$ and $40$ are both even, $5c$ must be even too, so $c$ must be even.
DPatrick 2019-09-30 20:47:20
So $c$ must be $0,2,4,6,8$ and then we can solve for $a$.
DPatrick 2019-09-30 20:47:26
This gives 5 solutions to the first equation.
DPatrick 2019-09-30 20:47:34
How many solutions to $2b + 5d = 20$?
Epic_Dabber 2019-09-30 20:48:02
3 solutions
AoPS_Brian 2019-09-30 20:48:02
like the other problem, d must be even
mdawn_65 2019-09-30 20:48:02
3
Allen31415 2019-09-30 20:48:02
3
AoPS_Brian 2019-09-30 20:48:08
so 0, 2, 4 which gives 3 solutions
SuperAndrew6 2019-09-30 20:48:08
3 solutions
DPatrick 2019-09-30 20:48:30
Right: it's the same logic. $d$ must be even, so $d$ must be $0,2,4$.
DPatrick 2019-09-30 20:48:39
That gives 3 solutions to the second equation.
Stellar967 2019-09-30 20:48:44
15!
priyankaadhikari 2019-09-30 20:48:44
so multiply 5*3 = 15 ordered pairs
DPatrick 2019-09-30 20:49:04
Yeah: together that gives $5 \cdot 3 = 15$ solutions to the system.
DPatrick 2019-09-30 20:49:08
Hence there are $15$ ordered pairs that work. Answer $\boxed{\text{(a)}}$.
Hyo 2019-09-30 20:49:25
how many people got this problem right?
DPatrick 2019-09-30 20:49:31
About 20%.
DPatrick 2019-09-30 20:49:43
Which is less than random guessing would have expected to give!
DPatrick 2019-09-30 20:50:11
So that was Round 1 of Who Wants to Be a Mathematician for 2019-20!
DPatrick 2019-09-30 20:50:17
On the actual contest: if you got 7 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 in October. And come back here for our Round 2 Math Jam on Monday, October 14 at 7:30 pm ET / 4:30 pm PT. (That's two weeks from today, same Bat-time, same Bat-channel.)
amcha9 2019-09-30 20:50:45
What happens after round 2?
DPatrick 2019-09-30 20:50:49
I'm glad you asked!
DPatrick 2019-09-30 20:50:52
After Round 2 is complete, 12 students will be invited to compete in the Championship Finals, live in Denver at the 2020 Joint Mathematics Meetings in January. Travel costs to and from Denver will be covered by the AMS.
DPatrick 2019-09-30 20:51:07
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 2019-09-30 20:51:10
http://www.ams.org/images/wwtbam-map-us-canada.jpg
DPatrick 2019-09-30 20:51:24
(Yes, Region 10 is Canada!)
DPatrick 2019-09-30 20:51:38
The other two contestants will be the top scorer in the United Kingdom and the top scorer in the Denver metro area (or, I suppose, the second-highest scorer, if the highest scorer in Region 8 happens to be from Denver).
AoPS_Brian 2019-09-30 20:51:53
How to attend this competition? Where to sign up?
DPatrick 2019-09-30 20:52:07
Unfortunately, you've missed it for 2019-20 -- it was earlier this month.
DPatrick 2019-09-30 20:52:22
The Championship 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.
skywalker321 2019-09-30 20:52:38
what happens if you and someone else tie on round 2 for the highest score
DPatrick 2019-09-30 20:52:56
One of the questions will be used as a tie-breaker.
DPatrick 2019-09-30 20:53:09
The Round 2 instructions that you get will explain in more detail.
priyankaadhikari 2019-09-30 20:53:20
is it a lot harder than round 1
DPatrick 2019-09-30 20:53:32
The questions on Round 2 are definitely harder than Round 1.
r0518 2019-09-30 20:53:38
what percent of people got a perfect score on round 1
DPatrick 2019-09-30 20:53:50
About 200 out of 3700 got a perfect score.
DPatrick 2019-09-30 20:54:02
So a bit over 5%.
DPatrick 2019-09-30 20:54:26
If you want more information about the contest (for next year), here are a couple of sources.
DPatrick 2019-09-30 20:54:40
You can go read the transcript of the information Math Jam that we held in early September.
DPatrick 2019-09-30 20:55:01
It's on the AoPS website here: https://artofproblemsolving.com/school/mathjams-transcripts?id=498
DPatrick 2019-09-30 20:55:14
Or, you can get information directly from the AMS website.
amcha9 2019-09-30 20:55:33
What happens if you win round 3?
DPatrick 2019-09-30 20:55:41
All of the finalists will receive prizes (in addition to the free trip to Denver!), and the champion will win $10,000.
DPatrick 2019-09-30 20:55:55
And you'll get to meet me and Mike and Bill in person!
Hyo 2019-09-30 20:56:15
how to we come back to this class after 2 weeks?
DPatrick 2019-09-30 20:56:35
Just do the same as you did tonight. We'll be here at the same time in the same place. And it's free, just like tonight!
NASA1234 2019-09-30 20:56:55
how will our math teacher be notified if their students are moving on to round 2?
NASA1234 2019-09-30 20:56:55
is round 2 multiple choice
mikebreen 2019-09-30 20:57:06
Not multiple choice.
DPatrick 2019-09-30 20:57:06
I believe they will be notified by email.
DPatrick 2019-09-30 20:57:12
Round 2 is short answer.
DPatrick 2019-09-30 20:57:43
Again, if you have questions about the contest itself, please go to one of the two sources that I just posted.
Creator 2019-09-30 20:57:53
will this transcript be avaliable for us?
DPatrick 2019-09-30 20:58:00
Hopefully within the next 30 minutes.
DPatrick 2019-09-30 20:58:15
In fact, I'm going to end the Math Jam now so that I can prepare and post the transcript.
mikebreen 2019-09-30 20:58:20
Thanks, everyone. Very good job working out the problems.
DPatrick 2019-09-30 20:58:23
Thanks for participating tonight. See you in 2 weeks for Round 2!
TPiR 2019-09-30 20:58:31
So long, everyone. Great job!

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