Math Jams

copeland 2016-02-26 19:03:53
Welcome to the 2016 AMC 10B/12B Math Jam!

copeland 2016-02-26 19:03:54
I'm Jeremy Copeland, and I'll be leading our discussion tonight.

copeland 2016-02-26 19:04:02
I'm the school director here at AoPS. I like math.

copeland 2016-02-26 19:04:06
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.

copeland 2016-02-26 19:04:08
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.

copeland 2016-02-26 19:04:25
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.

copeland 2016-02-26 19:04:31
There are bunches 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. 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!!

copeland 2016-02-26 19:04:39
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the prerequisite 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!

copeland 2016-02-26 19:04:50
We have 4 (yep, 4!) assistants tonight:

copeland 2016-02-26 19:05:01
Kaitlin Maile (kmaile)
Andrew Jiang (andrewjjiang97)
Ben Stone (hailstone)
Shyan Akmal (Naysh).

copeland 2016-02-26 19:05:11
Kaitlin has been working as a grader for AOPS since November 2014, when she was a senior in high school. While in high school, Kaitlin participated in math team, marching band, and various sports. Now, she is a college student at the University of Minnesota, majoring in Biomedical Engineering. When not working for AOPS or studying for classes, she is usually listening to music of any genre or enjoying the outdoors.

copeland 2016-02-26 19:05:17
Shyan is currently a student at Harvey Mudd college, where he plans on majoring in math and computer science. In high school he was an avid participant in science and math competitions, qualifying for the USAMO and USAJMO, being an ARML high scorer two years in a row, and attending the national competitions for Science Bowl and Science Olympiad. His experiences have motivated him to become (more) suspicious of all sciences not based off pure mathematics. Beyond teaching and math, he enjoys reading novels (especially those by Agatha Christie and Terry Pratchett), playing guitar, and ultimate frisbee.

copeland 2016-02-26 19:05:18
Ben has a B.S. in Math from Carnegie Mellon University, with minors in Music, Philosophy, and Film Studies, and has an M.S. in Math from the University of Washington. He has spent almost as many hours working on his juggling and unicycling.

copeland 2016-02-26 19:05:24
And Andrew:
I am currently a freshman at Harvard College, considering concentrating in economics or applied mathematics. In high school, I participated in a lot of math competitions: AMC, AIME, ARML, and the USAMO in 2015. In my free time, I like playing chess, writing and reading poetry, and running. I've also taken some courses on AoPS before including WOOT! A fun fact about me: I type with on a Dvorak-formatted keyboard, instead of the standard QWERTY style.

hailstone 2016-02-26 19:05:39
Hello!

kmaile32 2016-02-26 19:05:44
Hi!

andrewjjiang97 2016-02-26 19:05:47
Hi everyone!

Naysh 2016-02-26 19:05:57
Hello.

dvdmkr 2016-02-26 19:06:00
hello

Python54 2016-02-26 19:06:00
Hi!

suika1234 2016-02-26 19:06:00
hi

kingofmathcounts 2016-02-26 19:06:00
hi

Brainiac2 2016-02-26 19:06:00
hi!

giraffe123 2016-02-26 19:06:00
Hi!

GeneralCobra19 2016-02-26 19:06:00
Hiya!

computernerd 2016-02-26 19:06:00
hi!

copeland 2016-02-26 19:06:09
We can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the incredibly large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.

copeland 2016-02-26 19:06:14
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems, and not to merely present the answers. "Working through the solutions" 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, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be posted.

copeland 2016-02-26 19:06:18
We will work the last 5 problems from the AMC 10B, then the last 5 problems from the AMC 12B.

copeland 2016-02-26 19:06:26
Everyone set? Should we get started?

elitechicken 2016-02-26 19:06:42
yo

Lovemath2015 2016-02-26 19:06:42
hi yah!

nc_sam 2016-02-26 19:06:42
hi

mj7777 2016-02-26 19:06:42
hello

AOPS12142015 2016-02-26 19:06:42
Yes

avn 2016-02-26 19:06:42
yes

temp8909 2016-02-26 19:06:42
yes!

Python54 2016-02-26 19:06:42
Yes!

Jyzhang12 2016-02-26 19:06:42
YESSS

Rotack00 2016-02-26 19:06:42
yes

Brainiac2 2016-02-26 19:06:42
yes!

abccsparta 2016-02-26 19:06:42
yes

GeneralCobra19 2016-02-26 19:06:42
Yep

IQMathlete 2016-02-26 19:06:42
yes

math101010 2016-02-26 19:06:42
yeah

copeland 2016-02-26 19:06:44
OK, let's rock.

temp8909 2016-02-26 19:06:47
will there be any overlap?

copeland 2016-02-26 19:06:57
Strangely, there isn't any overlap in these 10 problems.

copeland 2016-02-26 19:07:05
Usually there are 1-3 repeats.

copeland 2016-02-26 19:07:27
21. What is the area of the region enclosed by the graph of the equation $x^2 + y^2 = |x| + |y|$?
$\phantom{10B:21}$
(A) $\pi + \sqrt2$ (B) $\pi + 2$ (C) $\pi + 2\sqrt2$ (D) $2\pi + \sqrt2$ (E) $2\pi + 2\sqrt2$

copeland 2016-02-26 19:07:38
(This was also #18 on the 12B.)

copeland 2016-02-26 19:07:41
How can we work with those absolute value signs?

copeland 2016-02-26 19:08:03
There we go.

copeland 2016-02-26 19:08:07
Now, about those absolute values. . .

MSTang 2016-02-26 19:08:29
Casework

IQMathlete 2016-02-26 19:08:29
Casework

physangel 2016-02-26 19:08:29
casework

math101010 2016-02-26 19:08:29
casework

ninjataco 2016-02-26 19:08:29
casework based on signs of x and y

temp8909 2016-02-26 19:08:29
divide into cases

SK200 2016-02-26 19:08:29
cases

amyhu910 2016-02-26 19:08:29
casework

copeland 2016-02-26 19:08:33
Casework is a good idea.

copeland 2016-02-26 19:08:37
What can you say about the cases, though?

joey8189681 2016-02-26 19:09:27
Confine it to one quadrant with symmetry

Verjok 2016-02-26 19:09:27
Assume that x and y are positive and multiply final answer by 4

SK200 2016-02-26 19:09:27
symmetric

bguo 2016-02-26 19:09:27
x is positive results in the same thing as when it is negative

ninjataco 2016-02-26 19:09:27
symmetry

AkashD 2016-02-26 19:09:27
They are symmetric, so we can focus on one quadrant of the graph

DeathLlama9 2016-02-26 19:09:27
They're all the same! (Symmetry)

temp8909 2016-02-26 19:09:27
symmetric!!!

Benq 2016-02-26 19:09:27
All essentially the same.

copeland 2016-02-26 19:09:29
The cases are all symmetric! Replacing $x$ with $-x$ doesn't change anything in the equation. Nor does replacing $y$ with $-y$.

copeland 2016-02-26 19:09:35
The region is the same in each of the four quadrants of the coordinate plane.

copeland 2016-02-26 19:09:37
So we can just compute the region in the first quadrant, where $x \ge 0$ and $y \ge 0$, and then multiply by 4.

copeland 2016-02-26 19:09:38
In the first quadrant, what do we have?

Verjok 2016-02-26 19:10:13
x^2+y^2=x+y

DeathLlama9 2016-02-26 19:10:13
$x^2 + y^2 = x + y$

Maths4J 2016-02-26 19:10:13
x^2+y^2=x+y

Jyzhang12 2016-02-26 19:10:13
$x^2+y^2=x+y$

physangel 2016-02-26 19:10:13
x^2 + y^2=x+y

speck 2016-02-26 19:10:13
$x^2 + y^2 = x + y$

copeland 2016-02-26 19:10:16
We just have $x^2 + y^2 = x + y$.

copeland 2016-02-26 19:10:16
How can we work with this?

GeneralCobra19 2016-02-26 19:10:44
I dunno(completing the square) but it looks like a circle

bguo 2016-02-26 19:10:44
complete the square

MSTang 2016-02-26 19:10:44
Complete the square

ninjataco 2016-02-26 19:10:44
complete the square

DeathLlama9 2016-02-26 19:10:44
Subtract $x + y$ from both sides and complete the square

suika1234 2016-02-26 19:10:44
complete the square

temp8909 2016-02-26 19:10:44
complete the square

baseballcat 2016-02-26 19:10:44
complet the square to get a circle

matthias12345 2016-02-26 19:10:44
make a circle equation

GeneralCobra19 2016-02-26 19:10:44
Subtract x and y, then complete the square to get the formula of a circle!

ghghghghghghghgh 2016-02-26 19:10:44
complete the square

copeland 2016-02-26 19:10:49
We can bring all terms to one side, and complete the square.

copeland 2016-02-26 19:10:51
That is, we have $x^2 - x + y^2 - y = 0$. How do we complete the square from here?

swagger21 2016-02-26 19:11:28
i'm having technical problems; the messages aren't sending for me, i have to refresh

baseballcat 2016-02-26 19:11:35
add 1/2 to both sides

bguo 2016-02-26 19:11:35
add 1/2 to both sides

DeathLlama9 2016-02-26 19:11:35
$(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}$

idkanymath 2016-02-26 19:11:35
(x-1/2)^2 + (y-1/2)^2=1/2

GeneralCobra19 2016-02-26 19:11:35
(s-1/2)^2+(y-1/2)^2=1/2

speck 2016-02-26 19:11:35
$\left ( x - \dfrac{1}{2} \right)^2 + \left( x - \dfrac{1}{2} \right)^2 = \dfrac{1}{2}$

ghghghghghghghgh 2016-02-26 19:11:35
(x-1/2)^2+(y-1/2)^2=1/2

physangel 2016-02-26 19:11:35
add 1/2 to both sides

copeland 2016-02-26 19:11:39
We must add $\frac14$ twice to the left side, and thus we add $\frac12$ to the right side:
\[
x^2 - x + \frac14 + y^2 - y + \frac14 = \frac12.
\]

copeland 2016-02-26 19:11:40
This is just \[ \left(x-\frac12\right)^2 + \left(y-\frac12\right)^2 = \frac12 = \left(\frac{1}{\sqrt2}\right)^2.\]

copeland 2016-02-26 19:11:41
What does this curve look like?

Verjok 2016-02-26 19:12:03
A circle

fclvbfm934 2016-02-26 19:12:03
circle

emhyr 2016-02-26 19:12:03
A circle!

copeland 2016-02-26 19:12:06
What circle?

math101010 2016-02-26 19:12:28
circle with radius sqrt(2)/2 and center (1/2,1/2)

ptes77 2016-02-26 19:12:28
Circle centered at (1/2, 1/2) with a radius of sqrt2/2

Caprylic1207 2016-02-26 19:12:28
a circle centered at (1/2,1/2) with radius \sqrt{2}/2

DeathLlama9 2016-02-26 19:12:28
It's a circle with radius $\frac{1}{\sqrt{2}}$ centered at $(\frac{1}{2}, \frac{1}{2})$.

baseballcat 2016-02-26 19:12:28
a circle with center at 1/2,1/2 and radius 1/sqrt(2)

copeland 2016-02-26 19:12:30
It's a circle centered at $\left(\frac12,\frac12\right)$ and with radius $\frac{1}{\sqrt2}$.

copeland 2016-02-26 19:12:41

copeland 2016-02-26 19:12:44
Except. . .

bguo 2016-02-26 19:13:17
isocels right triangle+semicircle

DeathLlama9 2016-02-26 19:13:17
Part of it is cut off though

speck 2016-02-26 19:13:17
First quadrant only

math101010 2016-02-26 19:13:17
first quadrant

idkanymath 2016-02-26 19:13:17
Only in 1st quadrant

GeneralCobra19 2016-02-26 19:13:17
The extras are not on the quadrant

DeathLlama9 2016-02-26 19:13:17
Part of it isn't in the first quadrant

temp8909 2016-02-26 19:13:17
only the part in the first quadrant

baseballcat 2016-02-26 19:13:17
we only want it when x and y are positive

IQMathlete 2016-02-26 19:13:17
everything below or to the left of x and y axis

copeland 2016-02-26 19:13:23

copeland 2016-02-26 19:13:33
We only want the part in the firs quadrant:

copeland 2016-02-26 19:13:34

copeland 2016-02-26 19:13:37
What's the area of that region?

emhyr 2016-02-26 19:13:59
Typo @copeland

copeland 2016-02-26 19:14:08
Sorry, firs is Sweedish for first.

phi_ftw1618 2016-02-26 19:14:45
1/2 plus 1/4 pi

joey8189681 2016-02-26 19:14:45
pi/4 + 1/2

amyhu910 2016-02-26 19:14:45
1/2+pi/4

temp8909 2016-02-26 19:14:45
$\frac{1}{2}+\frac{\pi}{4}$

Verjok 2016-02-26 19:14:45
Semicircle + triangle = $\frac{\pi}{4} + \frac{1}{2}$

amyhu910 2016-02-26 19:14:45
(1/2)+(pi/4)

IQMathlete 2016-02-26 19:14:45
1/4 pi + 1/2

emhyr 2016-02-26 19:14:45
$\frac{1}{2}+\frac{\pi}{4}$

Pisensei 2016-02-26 19:14:45
1/4pi + 1/2

HVishy 2016-02-26 19:14:45
$1/2+1/4pi$

copeland 2016-02-26 19:14:47
It's a right triangle with legs of length 1, plus half of a circle with radius $\dfrac{1}{\sqrt2}$.

copeland 2016-02-26 19:14:49
Thus, the area is $\dfrac12 + \dfrac14\pi$.

copeland 2016-02-26 19:14:49
So what's our final answer?

DeathLlama9 2016-02-26 19:15:33
$\pi + 2$

temp8909 2016-02-26 19:15:33
$2+\pi$

Caprylic1207 2016-02-26 19:15:33
2+pi

Jyzhang12 2016-02-26 19:15:33
$\pi+2$

Rotack00 2016-02-26 19:15:33
$2+\pi$

mathelp 2016-02-26 19:15:33
pi + 2

AOPS12142015 2016-02-26 19:15:33
2 + pi

speck 2016-02-26 19:15:33
B

emhyr 2016-02-26 19:15:33
(B)

GeneralCobra19 2016-02-26 19:15:33
B

Maths4J 2016-02-26 19:15:33
multiply by 4 to get $2+\pi$

HVishy 2016-02-26 19:15:33
$\pi+2$

SK200 2016-02-26 19:15:33
$\pi + 2$

ninjataco 2016-02-26 19:15:33
pi + 2 (B)

mathelp 2016-02-26 19:15:33
(B) !!!

AkashD 2016-02-26 19:15:33
multiply by 4 and get B

SineWarrior 2016-02-26 19:15:33
2+pi

copeland 2016-02-26 19:15:36
We have four copies of this region (one in each quadrant), so our final answer is 4 times this, or $\boxed{2 + \pi}$. Answer (B).

copeland 2016-02-26 19:15:36

copeland 2016-02-26 19:15:41
Cool.

copeland 2016-02-26 19:15:43
Everyone warmed up?

temp8909 2016-02-26 19:16:01
yes!

Firstdream 2016-02-26 19:16:01
Yes

Rotack00 2016-02-26 19:16:01
yup

DeathLlama9 2016-02-26 19:16:01
Yes!

phi_ftw1618 2016-02-26 19:16:01
yeah

GeneralCobra19 2016-02-26 19:16:01
Of course

mathelp 2016-02-26 19:16:01
yap

yingyangbu 2016-02-26 19:16:01
ya

Lovemath2015 2016-02-26 19:16:01
yes

Python54 2016-02-26 19:16:01
yes!

mewtwomew 2016-02-26 19:16:01
yeah!

ninjataco 2016-02-26 19:16:01
yesh

sam15 2016-02-26 19:16:01
yep. Interesting problem

AOPS12142015 2016-02-26 19:16:01
Yeah!

avn 2016-02-26 19:16:01
yes

physangel 2016-02-26 19:16:01
YES!!!

copeland 2016-02-26 19:16:15
Isn't that the Girl Scouts logo?

Lovemath2015 2016-02-26 19:16:34
yeah, the four leaf clover

GeneralCobra19 2016-02-26 19:16:34
is it? not another girl scouts math problem!!!

copeland 2016-02-26 19:16:35
I think they want us to eat more cookies.

copeland 2016-02-26 19:16:40
We're on to you, AMC.

MSTang 2016-02-26 19:16:43
//cdn.artofproblemsolving.com/images/5/b/a/5bae59890ef7f7118d413fdf31c85fc95a92f4ae.png

copeland 2016-02-26 19:16:47
OK, so maybe not.

copeland 2016-02-26 19:16:51
22. A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won 10 games and lost 10 games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A$?
$\phantom{10A:22}$
(A) 385 (B) 665 (C) 945 (D) 1140 (E) 1330

copeland 2016-02-26 19:17:13
What are you thinking about this one?

DeathLlama9 2016-02-26 19:17:39
Complementary counting!

baseballcat 2016-02-26 19:17:39
complementary counting

fclvbfm934 2016-02-26 19:17:39
complementary counting

sas4 2016-02-26 19:17:39
complementary counting

suika1234 2016-02-26 19:17:39
complimentary counting

Superwiz 2016-02-26 19:17:39
Complimentary counting?

physangel 2016-02-26 19:17:39
complementary counting

copeland 2016-02-26 19:17:51
Interesting. That's not the game I like, but let's try it.

copeland 2016-02-26 19:18:01
Complementary counting means we count all sets of three teams, and then subtract those that don't satisfy the condition.

copeland 2016-02-26 19:18:03
How many sets of three teams are there in total?

fclvbfm934 2016-02-26 19:18:35
$\binom{21}{3}$

temp8909 2016-02-26 19:18:35
21 C 3

physangel 2016-02-26 19:18:35
1330

DeathLlama9 2016-02-26 19:18:35
$\binom{21}{3} = 1330$

idkanymath 2016-02-26 19:18:35
1330

ninjataco 2016-02-26 19:18:35
21C3 = 1330

ghghghghghghghgh 2016-02-26 19:18:35
1330

mathelp 2016-02-26 19:18:35
1330

physangel 2016-02-26 19:18:35
21 choose 3=1330

AOPS12142015 2016-02-26 19:18:35
21 choose 3

copeland 2016-02-26 19:18:38
Each team played 20 games and played each other team once. Therefore, there are 21 terms, and hence $\dbinom{21}{3} = 1330$ triples of teams.

copeland 2016-02-26 19:18:42
How many of these do not satisfy the condition in the problem?

copeland 2016-02-26 19:18:47
If $\{A,B,C\}$ don't work, what do we know?

DeathLlama9 2016-02-26 19:19:08
One of the teams beat both of the other teams

phi_ftw1618 2016-02-26 19:19:08
one team beat the other two

fclvbfm934 2016-02-26 19:19:08
One of the players beat the other two

MSTang 2016-02-26 19:19:08
somebody beat the other two teams

baseballcat 2016-02-26 19:19:08
one beat the other 2

ghghghghghghghgh 2016-02-26 19:19:08
one of the teams beat both of the others

aq1048576 2016-02-26 19:19:08
one team beat both others in the set

copeland 2016-02-26 19:19:12
We know that one of the teams must have beaten both of the other two.

copeland 2016-02-26 19:19:13
How can we count how often this happens?

copeland 2016-02-26 19:19:41
If $A$ is one of the teams, how many $\{B,C\}$ are there such that $A$ beats both $B$ and $C$?

IQMathlete 2016-02-26 19:20:15
for each team, 10c2

sicilianfan 2016-02-26 19:20:15
10C2

temp8909 2016-02-26 19:20:15
$\binom{10}{2}$

Maths4J 2016-02-26 19:20:15
10C2=45

yingyangbu 2016-02-26 19:20:15
{10 2}

84nea9tresse2 2016-02-26 19:20:15
10c2

DeathLlama9 2016-02-26 19:20:15
$\binom{10}{2}$

ghghghghghghghgh 2016-02-26 19:20:15
45

Peggy 2016-02-26 19:20:15
C (10, 2)

copeland 2016-02-26 19:20:17
$A$ beats 10 teams, so $A$ beats $\dbinom{10}{2} = 45$ pairs of teams.

copeland 2016-02-26 19:20:19
Thus, for any given $A$, there are 45 sets $\{A,B,C\}$ in which $A$ beats both $B$ and $C$.

copeland 2016-02-26 19:20:20
So how many such sets in total?

temp8909 2016-02-26 19:20:56
$21\times 45=945$

ghghghghghghghgh 2016-02-26 19:20:56
21*45=945

sicilianfan 2016-02-26 19:20:56
21*45

84nea9tresse2 2016-02-26 19:20:56
45*21

sam15 2016-02-26 19:20:56
945

IQMathlete 2016-02-26 19:20:56
945

DeathLlama9 2016-02-26 19:20:56
$21 \times 45 = 945$

copeland 2016-02-26 19:20:59
There are 21 choices for $A$. So there are $21 \cdot 45 = 945$ sets of three teams that don't work.

copeland 2016-02-26 19:21:03
Thus our final answer, the number of triples that do work, is $1330 - 945 = \boxed{385}$. Answer (A).

copeland 2016-02-26 19:21:18
Now, did anyone else solve this problem differently?

mathelp 2016-02-26 19:21:59
mee

AkashD 2016-02-26 19:21:59
graph theory?

sicilianfan 2016-02-26 19:21:59
graph theory

swagger21 2016-02-26 19:22:03
count the number of cases that do work?

copeland 2016-02-26 19:22:09
Hm, I LOVE pictures.

copeland 2016-02-26 19:22:21
What should we draw?

Pisensei 2016-02-26 19:22:59
the girl's scout logo

GeneralCobra19 2016-02-26 19:22:59
Another girl scouts logo? Jk

copeland 2016-02-26 19:23:06
You guys are killing me with these cookies.

copeland 2016-02-26 19:23:15
Let's get to the end and eat some. I assume you have some ready?

ThorJames 2016-02-26 19:23:21
A triangle with team a and teams that lost and teams that won

copeland 2016-02-26 19:23:40
Many people are assuming that we want to draw 21 vertices, but that is a lot.

copeland 2016-02-26 19:23:43
Suppose $A$ is any team. Let $X$ be the teams that $A$ beat, and $Y$ be the teams that beat $A$. There are 10 teams in each set.

copeland 2016-02-26 19:23:51
The teams in $X$ and those in $Y$ are similar in some way.

copeland 2016-02-26 19:23:52
We can draw a little diagram to keep track of this, where an arrow points from a winner to a loser.

copeland 2016-02-26 19:23:53

copeland 2016-02-26 19:23:54
When do we get a triple $\{A,B,C\}$?

phi_ftw1618 2016-02-26 19:24:35
1 in x beats one in y

GeneralCobra19 2016-02-26 19:24:35
When X beats Y

SK200 2016-02-26 19:24:35
X beats Y

ghghghghghghghgh 2016-02-26 19:24:35
when a team in X beats a team in Y

sillyd 2016-02-26 19:24:35
when a team in X beats a team in Y

amyhu910 2016-02-26 19:24:35
when x beats y?

copeland 2016-02-26 19:24:36
We need a team in $X$ to beat a team in $Y$.

copeland 2016-02-26 19:24:45
Let's see how many times that happens.

copeland 2016-02-26 19:24:47
Each team in $X$ has 10 wins and 10 losses, so they have a total of 100 wins and 100 losses.

copeland 2016-02-26 19:25:21
How many of those edges are inside $X$?

copeland 2016-02-26 19:26:30
How many wins does X have against X?

sicilianfan 2016-02-26 19:26:52
45

MSTang 2016-02-26 19:26:52
45

Peggy 2016-02-26 19:26:52
45

copeland 2016-02-26 19:26:53
We know that they all play each other. That results in $\binom{10}{2} = 45$ wins and 45 losses.

copeland 2016-02-26 19:26:57

copeland 2016-02-26 19:27:23
So how many times does a team in X beat a team in Y?

avn 2016-02-26 19:27:47
55

amyhu910 2016-02-26 19:27:47
55

vatatmaja 2016-02-26 19:27:47
55

AkashD 2016-02-26 19:27:47
55

JayJuly 2016-02-26 19:27:47
55

SineWarrior 2016-02-26 19:27:47
55

vatatmaja 2016-02-26 19:27:47
100-45

copeland 2016-02-26 19:27:57
That leaves 55 wins and 45 losses, which must be against the teams in $Y$.

copeland 2016-02-26 19:27:58

copeland 2016-02-26 19:28:03
Again, as a check: there are 100 arrows in total leaving from items in $X$: 45 within $X$, and 55 from $X$ to $Y$. And, there are 100 arrows pointing towards items in $X$: 45 within $X$, 45 from $Y$, and 10 from $A$. (We could do the same check for $Y$ too.)

copeland 2016-02-26 19:28:13
In particular, there are 55 pairs of teams $\{B,C\}$ with $B$ in $X$ and $C$ in $Y$, where $B$ beats $C$. This gives us our set $\{A,B,C\}$.

copeland 2016-02-26 19:28:16
So since there are 21 choices for $A$, that makes $21 \cdot 55 = 1155$ triples, right?

MSTang 2016-02-26 19:28:49
divide by 3!

phi_ftw1618 2016-02-26 19:28:49
overcounting

W.Sun 2016-02-26 19:28:49
You have to divide by 3

temp8909 2016-02-26 19:28:49
divide by 3

sicilianfan 2016-02-26 19:28:49
triple count

SK200 2016-02-26 19:28:49
divide by 3

fclvbfm934 2016-02-26 19:28:49
divide by 3

mathelp 2016-02-26 19:28:49
No!! divide by 3

gradysocool 2016-02-26 19:28:49
Over counting by a factor of 3

ninjataco 2016-02-26 19:28:49
no, divide by 3 because overcounting

copeland 2016-02-26 19:28:51
No! (That's not an answer choice, for one thing.) Each triple is counted three times, once for each team.

copeland 2016-02-26 19:28:51
So our final answer is $\dfrac{1155}{3} = \boxed{385}$. Answer (A).

copeland 2016-02-26 19:28:59
Great.

copeland 2016-02-26 19:29:00
Next problem?

avn 2016-02-26 19:29:24
yup

amyhu910 2016-02-26 19:29:24
bring on the heat!!!!

GeneralCobra19 2016-02-26 19:29:24
Sure enough(if it isn't hard)

mathelp 2016-02-26 19:29:24
si

SK200 2016-02-26 19:29:24
yeah

W.Sun 2016-02-26 19:29:24
Yes!

W.Sun 2016-02-26 19:29:24
Onward!

copeland 2016-02-26 19:29:33
What I liked about that one is that it is independent of the graph.

copeland 2016-02-26 19:29:41
Not all graphs of this form are isomorphic.

copeland 2016-02-26 19:29:46
Anyway. . .

copeland 2016-02-26 19:29:47
23. In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$, respectively, so that lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?
$\phantom{10A:23}$
(A) $\dfrac13$ (B) $\dfrac{10}{27}$ (C) $\dfrac{11}{27}$ (D) $\dfrac49$ (E) $\dfrac{13}{27}$.

copeland 2016-02-26 19:29:51
(You might note that the answer choices are consecutive multiples of $\dfrac{1}{27}$. That might help.)

copeland 2016-02-26 19:29:52
I have a feeling that if we just draw an accurate picture, the rest will be easy.

copeland 2016-02-26 19:29:55
What should we do?

temp8909 2016-02-26 19:30:10
DIAGRAM!!!!

AOPS12142015 2016-02-26 19:30:10
Draw a diagram

AlcumusGuy 2016-02-26 19:30:10
draw a picture!

soccerchess 2016-02-26 19:30:10
draw

vatatmaja 2016-02-26 19:30:10
Draw a Diagram!

copeland 2016-02-26 19:30:12

copeland 2016-02-26 19:30:15
OK, here's a start.

copeland 2016-02-26 19:30:25
What next? Should we draw more stuff?

phi_ftw1618 2016-02-26 19:30:56
yeah, divide the hexagon into 6 equilateral triangles

u2d2lrlrba_Ajia 2016-02-26 19:30:56
Would drawing diagonals help?

math101010 2016-02-26 19:30:56
we should draw equilateral triangles

copeland 2016-02-26 19:31:01
We can draw this:

copeland 2016-02-26 19:31:04

copeland 2016-02-26 19:31:17
In fact, here I see lots of little equilateral triangles.

copeland 2016-02-26 19:31:37
One thing we could do is color and then throw math at it:

copeland 2016-02-26 19:31:39

copeland 2016-02-26 19:31:47
Is there something simpler we could do?

yrnsmurf 2016-02-26 19:32:20
trisect all edges besides AB and DE

AOPS12142015 2016-02-26 19:32:20
Yes, we should draw three lines that are heights since they are equally spaced

phi_ftw1618 2016-02-26 19:32:20
and then divide up those equilateral triangles

avn 2016-02-26 19:32:20
split them up more

vatatmaja 2016-02-26 19:32:20
Draw a bunch of smaller sized ones!.

copeland 2016-02-26 19:32:32
We can break this up even more, right?

copeland 2016-02-26 19:32:46
Let's cut the top half into triangles all of the same size:

copeland 2016-02-26 19:32:47

copeland 2016-02-26 19:32:58
What's the answer to the problem?

phi_ftw1618 2016-02-26 19:33:47
then we can just count them

math101010 2016-02-26 19:33:47
C

sam15 2016-02-26 19:33:47
11/27

yrnsmurf 2016-02-26 19:33:47
11/27

kkkittykat 2016-02-26 19:33:47
11/27

temp8909 2016-02-26 19:33:47
$\frac{11}{27}$

AOPS12142015 2016-02-26 19:33:47
11/27

ninjataco 2016-02-26 19:33:47
11/27

Rotack00 2016-02-26 19:33:47
11/27, C

phi_ftw1618 2016-02-26 19:33:47
numerator is clearly 11 so $\frac{11]{27}$

sicilianfan 2016-02-26 19:33:47
C

yrnsmurf 2016-02-26 19:33:47
$\dfrac{11}{27}$

vatatmaja 2016-02-26 19:33:47
C!

vatatmaja 2016-02-26 19:33:47
11/27

Brainiac2 2016-02-26 19:33:47
c

Maths4J 2016-02-26 19:33:47
(B) $\frac{11}{27}$

AC2014 2016-02-26 19:33:47
11/27

Jayjayliu 2016-02-26 19:33:47
11/27

jk23541 2016-02-26 19:33:47
C

JayJuly 2016-02-26 19:33:47
C

W.Sun 2016-02-26 19:33:47
11/27

W.Sun 2016-02-26 19:33:47
11/27 Is the answer!

Math1331Math 2016-02-26 19:33:47
11/27

IQMathlete 2016-02-26 19:33:47
C

mathelp 2016-02-26 19:33:47
11/27!!

sam15 2016-02-26 19:33:47
11/27

8Invalid8 2016-02-26 19:33:47
11/27!!!!

copeland 2016-02-26 19:33:49
Aha! Now we can just count them. There are 11 green triangles and there are 27 total triangles. The ratio is 11/27.

vatatmaja 2016-02-26 19:33:56
This was a pretty easy 23!

copeland 2016-02-26 19:34:09
One could argue that the problems were not exactly in order on this test.

phi_ftw1618 2016-02-26 19:34:32
Would a construction have helped with a more accurate diagram?

copeland 2016-02-26 19:34:40
So usually that's true, just as a statement.

copeland 2016-02-26 19:34:44
I don't think it helps much here.

mssmath 2016-02-26 19:35:11
An AoPS person admits the MAA made a mistake? What!!

copeland 2016-02-26 19:35:12
No. . . I don't think they promised the problems would be in any order.

copeland 2016-02-26 19:35:13
What's next?

jk23541 2016-02-26 19:35:21
any tips for being able to see this sort of clever splitting up? Like is there a hint in the problem that gives us this idea?

copeland 2016-02-26 19:35:24
The 27.

copeland 2016-02-26 19:35:30
That's all I got.

copeland 2016-02-26 19:35:38
Plus, it just looks like a graph paper problem.

ThorJames 2016-02-26 19:35:42
24!

copeland 2016-02-26 19:35:45
24. How many four-digit positive integers $abcd$, with $a \not= 0$, have the property that the three two-digit integers $ab < bc < cd$ form an increasing arithmetic sequence? One such number is 4692, where $a=4$, $b=6$, $c=9$, and $d=2$.
$\phantom{10A:24}$
(A) 9 (B) 15 (C) 16 (D) 17 (E) 20

copeland 2016-02-26 19:35:49
How can we write this property in a form that we can work with?

MSTang 2016-02-26 19:36:41
(10a+b) + (10c+d) = 2(10b+c)

ninjataco 2016-02-26 19:36:41
10a + b + 10c + d = 20b + 2c

vatatmaja 2016-02-26 19:36:41
10x+y=xy

temp8909 2016-02-26 19:36:41
$10a+b,10b+c,10c+d$

AlcumusGuy 2016-02-26 19:36:41
10b + c - (10a + b) = 10c + d - (10b + c)

IQMathlete 2016-02-26 19:36:41
bc is average of ab and cd

Math1331Math 2016-02-26 19:36:41
10a+10c+b+d=20b+2c

yrnsmurf 2016-02-26 19:36:41
ab+cd=2bc

copeland 2016-02-26 19:36:43
Note that what they're calling "ab" is really the quantity $10a+b$, etc.

copeland 2016-02-26 19:36:45
So the numbers $10a + b$, $10b + c$ and $10c + d$ must be an increasing arithmetic sequence.

copeland 2016-02-26 19:36:48
Three numbers form an arithmetic sequence if and only if the middle number is the average of the other two.

copeland 2016-02-26 19:36:51
In our case, that means that
\[ (10a + b) + (10c + d) = 2(10b + c). \]

copeland 2016-02-26 19:36:55
This simplifies to $10a - 19b + 8c + d = 0$.

copeland 2016-02-26 19:37:00
How can we work with this to count the solutions?

copeland 2016-02-26 19:37:28
What do you notice in that big old expression?

copeland 2016-02-26 19:38:12
Bee bee bumble bee, I see, um, something that looks like a 2-digit number again.

vatatmaja 2016-02-26 19:38:40
10a+d=ad!!!

ein 2016-02-26 19:38:40
10a+d

ThorJames 2016-02-26 19:38:40
10a + d

SherlockHolmes7 2016-02-26 19:38:40
10a+d

GeneralCobra19 2016-02-26 19:38:40
10a+d!

Rotack00 2016-02-26 19:38:40
10a+d?

copeland 2016-02-26 19:38:50
OK, maybe the 10a+d in there is useful.

copeland 2016-02-26 19:38:55
We might notice is that $10a + d$ is the two-digit number $\overline{ad}$.

copeland 2016-02-26 19:38:57
So we might think to isolate it: \[ 10a + d = 19b - 8c. \]

copeland 2016-02-26 19:39:02
At this point, we need to count the number of ways to choose $(b,c)$ to ensure that the right side is a two-digit number that's at least 10 (since $a \not= 0$), and that $ab < bc < cd$.

copeland 2016-02-26 19:39:09
Also, the right side only depends on $b$ and $c$. Those are our middle two digits. Is there a more elegant way to express $19b-8c$ that looks like the problem at hand?

copeland 2016-02-26 19:39:59
In particular, can we use the same trick again somehow?

MSTang 2016-02-26 19:40:17
(10b+c) + 9(b-c)

AlcumusGuy 2016-02-26 19:40:17
10b + c + 9(b-c)

copeland 2016-02-26 19:40:23
Ooh, that's interesting.

copeland 2016-02-26 19:40:25
The middle number is $\overline{bc}=10b+c$. We can rewrite our equation as \[10a+d=(10b+c)+9b-9c\]or

copeland 2016-02-26 19:40:26
\[\overline{ad}=\overline{bc}-9(c-b).\]

copeland 2016-02-26 19:40:31
As a check, the example they gave us in the problem is $\overline{abcd}=\overline{4692}.$ What does our equation say in this case?

AlcumusGuy 2016-02-26 19:41:19
42 = 69 - 9(9-6)

ninjataco 2016-02-26 19:41:19
42 = 69 - 9(9-6)

swagger21 2016-02-26 19:41:19
42 = 69 -9(9-6), which is true

Math1331Math 2016-02-26 19:41:19
It works

phi_ftw1618 2016-02-26 19:41:19
42=69-27

Dominater76 2016-02-26 19:41:19
42=69-27

Rotack00 2016-02-26 19:41:19
42=69-27

mathmass 2016-02-26 19:41:19
42 = 69 - 9(7)

yrnsmurf 2016-02-26 19:41:19
42=69-9(9-6)

copeland 2016-02-26 19:41:22
The equation says $42=69-9(9-6)$. That's true.

copeland 2016-02-26 19:41:25
Now we're reduced to finding all the solutions to this equation that
$\bullet$ create four-digit numbers
$\bullet$ create increasing arithmetic sequences.

copeland 2016-02-26 19:41:27
What do we need to force $\overline{abcd}$ to be a 4-digit number?

DavidUsa 2016-02-26 19:41:36
What does the bar over ad and bc mean?

copeland 2016-02-26 19:41:46
I'm writing that to mean we put the digits together to make a number.

swagger21 2016-02-26 19:42:01
as opposed to multiplying a*b*c*d

copeland 2016-02-26 19:42:02
Exactly.

physangel 2016-02-26 19:42:23
a is not equal to 0

ThorJames 2016-02-26 19:42:23
a not equal 0

vatatmaja 2016-02-26 19:42:23
a=1,2,3,4...9

Peggy 2016-02-26 19:42:23
a can't be 0

thebobdbuilder 2016-02-26 19:42:23
0<a

Rotack00 2016-02-26 19:42:23
a=/=0

sicilianfan 2016-02-26 19:42:23
a=/=0

copeland 2016-02-26 19:42:25
We need $a\geq1$. However, let's stick with thinking about these as two-digit numbers. Then we need $\overline{ad}\geq10$.

copeland 2016-02-26 19:42:28
\[10\leq\overline{ad}=\overline{bc}-9(c-b).\]

copeland 2016-02-26 19:42:37
Now the sequence is guaranteed to be arithmetic because that's where we got the equation. Let's worry about "increasing" along the way.

copeland 2016-02-26 19:42:44
How should we go about finding all the solutions?

avn 2016-02-26 19:43:07
casework

Tapeman 2016-02-26 19:43:07
casework

copeland 2016-02-26 19:43:10
Casework on what?

copeland 2016-02-26 19:44:07
A lot of people are suggesting casework on $a$, but that's a little unsettling. Notice that since $\overline{ad}$ is just any old number, that's EASY. We should focus on the hard constraint.

Darn 2016-02-26 19:44:24
$c-b$

swagger21 2016-02-26 19:44:24
c-b?

copeland 2016-02-26 19:44:31
Let's proceed via casework on $c-b$. First of all, can $c-b$ be negative?

Rotack00 2016-02-26 19:45:00
no

mathelp 2016-02-26 19:45:00
no

Pisensei 2016-02-26 19:45:00
no

temp8909 2016-02-26 19:45:00
no

swagger21 2016-02-26 19:45:00
no

AOPS12142015 2016-02-26 19:45:00
No

Maths4J 2016-02-26 19:45:00
no, then cd<bc

ghghghghghghghgh 2016-02-26 19:45:00
no

amyhu910 2016-02-26 19:45:00
no?

physangel 2016-02-26 19:45:00
no

copeland 2016-02-26 19:45:02
No! If $c<b$ then $\overline{cd}<\overline{bc}$ which contradicts our increasing assumption.

copeland 2016-02-26 19:45:04
Case: $c-b=0$.

copeland 2016-02-26 19:45:04
Then what does our number look like?

Maths4J 2016-02-26 19:45:30
b=c

GeneralCobra19 2016-02-26 19:45:30
c=b

suika1234 2016-02-26 19:45:30
c is equal to b

copeland 2016-02-26 19:45:31
and. . .

temp8909 2016-02-26 19:45:49
bbbb

MSTang 2016-02-26 19:45:49
ad = bc, so we would have a number like bbbb

SK200 2016-02-26 19:45:49
same digiy

PwnageBeans 2016-02-26 19:45:49
a=b=c=d?

Tapeman 2016-02-26 19:45:49
and a=b=c=d

copeland 2016-02-26 19:45:53
Our equation reads $\overline{ad}=\overline{bb}$, so $a=d$ as well and the number is $aaaa$. This gives an arithmetic sequence but it is not increasing.

copeland 2016-02-26 19:45:58
There are no solutions with the two middle number equal.

copeland 2016-02-26 19:45:59
Case: $c-b=1$

copeland 2016-02-26 19:46:00
The equation reads $\overline{ad}=\overline{bc}-9$ with $c=b+1$. So $\overline{bc}$ is one of 12, 23, 34, 45, 56, 67, 78, or 89.

copeland 2016-02-26 19:46:06
Which don't work?

vatatmaja 2016-02-26 19:46:50
12

baseballcat 2016-02-26 19:46:50
12

gradysocool 2016-02-26 19:46:50
12

ghghghghghghghgh 2016-02-26 19:46:50
12

physangel 2016-02-26 19:46:50
12

soccerchess 2016-02-26 19:46:50
12

Peggy 2016-02-26 19:46:50
12

copeland 2016-02-26 19:46:53
If $\overline{bc}=12$ then $\overline{ad}=12-9<10$ so that fails. Otherwise we get solutions:\[
1234\quad
2345\quad
3456\quad
4567\quad
5678\quad
6789\quad
8890.\]

copeland 2016-02-26 19:46:54
(Note that 8890 is one that a lot of people miss in most approaches to this problem!)

copeland 2016-02-26 19:47:01
We had to eliminate 0123 from the list because we failed the inequality $10\leq\overline{bc}-9(c-b)$.

copeland 2016-02-26 19:47:03
There are 7 values in this case.

copeland 2016-02-26 19:47:04
Case: $c-b=2$

copeland 2016-02-26 19:47:06
What is the smallest $\overline{bc}$ that solves $10\leq \overline{bc}-9(c-b)$ now?

ghghghghghghghgh 2016-02-26 19:47:51
35

MSTang 2016-02-26 19:47:51
35

yrnsmurf 2016-02-26 19:47:51
35

Pisensei 2016-02-26 19:47:51
35

Darn 2016-02-26 19:47:51
$35$

Jayjayliu 2016-02-26 19:47:51
35

Rotack00 2016-02-26 19:47:51
35

copeland 2016-02-26 19:47:54
We want $10\leq\overline{bc}-18,$ so $28\leq\overline{bc}$. The first number that solves this is 35. How many valid 4-digit numbers are in this case?

copeland 2016-02-26 19:48:04
The possible values for $\overline{bc}$ are 35, 46, 57, 68, 79. There are 5 solutions\[
1357\quad
2468\quad
3579\quad
5680\quad
6791.\]

copeland 2016-02-26 19:48:10
Case: $c-b=3$

copeland 2016-02-26 19:48:10
What are all possible values of $\overline{bc}$ in this case?

gradysocool 2016-02-26 19:48:23
bash out the other cases now?

copeland 2016-02-26 19:48:26
Yes. . .

Darn 2016-02-26 19:48:55
$47, 58, 69$

temp8909 2016-02-26 19:48:55
47,58,69

yrnsmurf 2016-02-26 19:48:55
47, 58, 69

ghghghghghghghgh 2016-02-26 19:48:55
47,58,69

ninjataco 2016-02-26 19:48:55
47, 58, 69

Pisensei 2016-02-26 19:48:55
47, 58, 69

Jayjayliu 2016-02-26 19:48:55
47, 58, 69

copeland 2016-02-26 19:48:57
We need $10\leq\overline{bc}-27$ or $\overline{bc}\geq37$. The possible values for $\overline{bc}$ are 47, 58, 69.

copeland 2016-02-26 19:48:59
There are 3 solutions:\[
2470\quad
3581\quad
4692.\]

Dominater76 2016-02-26 19:49:02
Well that's kinds boring /:

copeland 2016-02-26 19:49:03
Yes.

copeland 2016-02-26 19:49:07
Hopefully we're almost done.

copeland 2016-02-26 19:49:08
Case: $c-b=4$

copeland 2016-02-26 19:49:09
What are the solutions now?

yrnsmurf 2016-02-26 19:49:58
48 and 59

swagger21 2016-02-26 19:49:58
48, 59

Pisensei 2016-02-26 19:49:58
48, 59

AOPS12142015 2016-02-26 19:49:58
48,59

ghghghghghghghgh 2016-02-26 19:49:58
48,59

Darn 2016-02-26 19:49:58
$48, 59$

Brainiac2 2016-02-26 19:49:58
59,48

copeland 2016-02-26 19:50:02
We need $10\leq\overline{bc}-36$ or $\overline{bc}\geq47$. The only possible values for $\overline{bc}$ are 48 and 59.

copeland 2016-02-26 19:50:03
So there are two solutions: \[ 1482 \quad 2593.\]

copeland 2016-02-26 19:50:04
Are there any other cases?

avn 2016-02-26 19:50:28
no

agbdmrbirdyface 2016-02-26 19:50:28
no

MSTang 2016-02-26 19:50:28
no

Darn 2016-02-26 19:50:28
No, since $49$ doesn't work.

ghghghghghghghgh 2016-02-26 19:50:28
no

Rotack00 2016-02-26 19:50:28
no?

math101010 2016-02-26 19:50:28
no

yrnsmurf 2016-02-26 19:50:28
nope

PwnageBeans 2016-02-26 19:50:28
no

lkarhat 2016-02-26 19:50:28
no

AOPS12142015 2016-02-26 19:50:28
no

copeland 2016-02-26 19:50:32
There are no other cases: If $c-b\geq5$ then $\overline{bc}\geq10+45,$ but then $b\geq6$ so $c\geq11$ and that's not how digits work.

copeland 2016-02-26 19:50:38
So what's the answer?

kkkittykat 2016-02-26 19:51:04
17

avn 2016-02-26 19:51:04
17

jeremyyu 2016-02-26 19:51:04
17

Rotack00 2016-02-26 19:51:04
17

nosaj 2016-02-26 19:51:04
17!

AOPS12142015 2016-02-26 19:51:04
17 (D)

jeremyyu 2016-02-26 19:51:04
17 D

mathelp 2016-02-26 19:51:04
D (17)

avn 2016-02-26 19:51:04
d

physangel 2016-02-26 19:51:04
D,17

Darn 2016-02-26 19:51:04
$7+5+3+2=17$.

SineWarrior 2016-02-26 19:51:04
17

avn 2016-02-26 19:51:04
17

ninjataco 2016-02-26 19:51:04
17

Pisensei 2016-02-26 19:51:04
17

copeland 2016-02-26 19:51:06
There are $7+5+3+2=\boxed{17}$ solutions. The answer is D.

copeland 2016-02-26 19:51:35
Alright, time to finish off this test so we can move to the next one.

mathelp 2016-02-26 19:51:43
YAY 25 now!!

temp8909 2016-02-26 19:51:43
next one!

Darn 2016-02-26 19:51:45
That took a while.

copeland 2016-02-26 19:51:49
Yeah, 24 was pretty hard.

Maths4J 2016-02-26 19:51:55
I couldn't follow this complex casework. Isn't there a better way?

copeland 2016-02-26 19:51:56
No.

copeland 2016-02-26 19:52:01
There are a lot of different things you can do.

copeland 2016-02-26 19:52:10
In a different approach, I also got to this cool equation:

copeland 2016-02-26 19:52:22
\[10(a-2b+c)=-(b-2c+d).\]

copeland 2016-02-26 19:52:30
Now if those are digits there aren't a lot of solutions.

copeland 2016-02-26 19:52:40
Since $-18<-(b-2c+d)<18,$ we know $a-2b+c$ is -1, 0, or 1.

copeland 2016-02-26 19:53:12
For example, if those are both zero we have \[a-2b+c=0\]and\[b-2c+d=0.\]

copeland 2016-02-26 19:53:20
That means the numbers are an honest arithmetic progression.

copeland 2016-02-26 19:53:29
Unfortunately the other 2 cases are still pretty annoying.

SineWarrior 2016-02-26 19:53:39
where did the 18 come from

copeland 2016-02-26 19:53:51
20-2=18.

copeland 2016-02-26 19:53:59
Oh, no. that's not right.

copeland 2016-02-26 19:54:02
9+9=18.

copeland 2016-02-26 19:54:15
And 2*9 is also 18.

MSTang 2016-02-26 19:54:26
You could just stick with the initial equation (10a-19b+8c-d=0) and bash, right?

copeland 2016-02-26 19:54:35
Probably, but when I tried it I missed all kinds of answers.

copeland 2016-02-26 19:54:43
Problem 25:

copeland 2016-02-26 19:54:47
25. Let $f(x) = \sum_{k=2}^{10}\left(\lfloor kx \rfloor - k \lfloor x \rfloor\right)$, where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to $r$. How many distinct values does $f(x)$ assume for $x \ge 0$?
$\phantom{10A:25}$
(A) 32 (B) 36 (C) 45 (D) 46 (E) infinitely many

copeland 2016-02-26 19:54:58
How can we deal with the floor function?

math101010 2016-02-26 19:55:26
casework

legozelda 2016-02-26 19:55:26
casework?

copeland 2016-02-26 19:55:27
Who's trolling whom?

swagger21 2016-02-26 19:55:48
MAA's trolling us

GeneralCobra19 2016-02-26 19:55:48
The MAA is trolling us

gradysocool 2016-02-26 19:56:00
separate into integer and fractional parts

AlcumusGuy 2016-02-26 19:56:00
split it into the number minus its fractional part

mathguy5041 2016-02-26 19:56:00
write x as integer part + decimal part

Darn 2016-02-26 19:56:00
Let it equal $n-\{n\}$, where $\{n\}$ is the fractional part of $n$.

nosaj 2016-02-26 19:56:00
x - fractional part of x

Puzzled417 2016-02-26 19:56:00
let x = m+r where r is the remainder 0 < r < 1

copeland 2016-02-26 19:56:04
One convenient way is to split a number into its integer part and its fractional part.

copeland 2016-02-26 19:56:05
That is, for any $x$, we write $x = y + z$, where $y$ is an integer and $0 \le z < 1$. Note that $y = \lfloor x \rfloor$ and we sometimes write $z = \{ x \}$.

copeland 2016-02-26 19:56:07
Then what happens to the expression inside the sum?

copeland 2016-02-26 19:57:49
OK, so first we know $\lfloor x\rfloor =y$, so we get $\lfloor kx \rfloor - k \lfloor x \rfloor = \lfloor k(y+z) \rfloor - ky$.

copeland 2016-02-26 19:57:51
But then what?

nosaj 2016-02-26 19:59:08
it becomes $k\{x\}$

yrnsmurf 2016-02-26 19:59:08
it beomes floor(kz)

MSTang 2016-02-26 19:59:08
take out the ky from the floor

Dominater76 2016-02-26 19:59:08
ky cancels

Hyperspace01 2016-02-26 19:59:08
$ky-ky+\lfloor kz\rfloor=\lfloor kz\rfloor$

AlcumusGuy 2016-02-26 19:59:08
floor(k(y+z)) = floor(ky + kz) = ky + floor(kz)

sillyd 2016-02-26 19:59:08
ky is an integer so just take it out of the floor function

ghghghghghghghgh 2016-02-26 19:59:08
take out ky from the floor

randomdude10807 2016-02-26 19:59:08
it is floor(kz) i think

copeland 2016-02-26 19:59:16
$ky$ is an integer, so we can pull it out of the floor function: \[ \lfloor k(y+z) \rfloor = ky + \lfloor kz \rfloor.\]

copeland 2016-02-26 19:59:18
Now the $ky$ parts cancel, and we're left with $\lfloor kx \rfloor - k \lfloor x \rfloor = \lfloor kz \rfloor$.

copeland 2016-02-26 19:59:20
That means that \[f(x) = \sum_{k=2}^{10} \lfloor kz \rfloor,\] where $0 \le z < 1$ is the fractional part of $x$.

copeland 2016-02-26 19:59:23
That's great, because now we can forget about $x$ and just worry about $0 \le z < 1$. In fact, I'm going to call it $f(z)$.

copeland 2016-02-26 19:59:25
What do we know about $f(z)$?

copeland 2016-02-26 19:59:37
Let's look at each term of the sum separately. What about the $k=2$ term?

temp8909 2016-02-26 20:00:17
0 or 1

randomdude10807 2016-02-26 20:00:17
f(z) is 0 or 1

copeland 2016-02-26 20:00:18
When is it 0 and when is it 1?

agbdmrbirdyface 2016-02-26 20:00:45
it depends on the value of z: if z > 1/2 we have 1, and if z < 1/2 we have 0

epiclucario 2016-02-26 20:00:45
if z<.5, the term is 0, if z>=.5, the term is 1

mathguy5041 2016-02-26 20:00:45
x<1/2: 0

gradysocool 2016-02-26 20:00:45
0 below 1/2 1 above 1/2

WhaleVomit 2016-02-26 20:00:45
0 when z<0.5

agbdmrbirdyface 2016-02-26 20:00:45
0 when z < 1/2, 1 when z >= 1/2

copeland 2016-02-26 20:00:48
It's 0 for $0 \le z < \frac12$ and it's 1 for $\frac12 \le z < 1$.

copeland 2016-02-26 20:00:48
What about the $k=3$ term?

mathguy5041 2016-02-26 20:01:35
split into thirds

gradysocool 2016-02-26 20:01:35
0,1,2 between the thirds

soccerchess 2016-02-26 20:01:35
0,1,2

Burrito 2016-02-26 20:01:35
it is 0 or 1 or 2

ninjataco 2016-02-26 20:01:35
0 when 0<= z < 1/3, 1 when 1/3 <= z < 2/3, 2 when 2/3 <= z < 1

AlcumusGuy 2016-02-26 20:01:35
0 if $0 \le z < \frac{1}{3}$, 1 if $\frac{1}{3} \le z < \frac{2}{3}$, and 2 otherwise

agbdmrbirdyface 2016-02-26 20:01:35
0 for when z < 1/3, 1 when z is between 1/3 and 2/3, and 2 for z above 2/3

swagger21 2016-02-26 20:01:35
0 when z < 1/3, 1 when 1/3 ≤ z < 2/3, 2 when 2/3 ≤ z < 1

copeland 2016-02-26 20:01:37
It's 0 for $0 \le z < \frac13$, it's 1 for $\frac13 \le z < \frac23$, and it's 2 for $\frac23 \le z < 1$.

copeland 2016-02-26 20:01:39
See the pattern? The $k^\text{th}$ term increases by 1 whenever $z$ crosses a fraction $\frac{m}{k}$ for some $0 < m < k$.

copeland 2016-02-26 20:01:40
So the $k=2$ terms "steps up" once, the $k=3$ term "steps up" twice, etc., up to the $k=10$ term "steps up" 9 times.

copeland 2016-02-26 20:01:41
So, since we start with $f(0) = 0$ and we step up $1+2+\cdots+9 = 45$ times, is the answer 46?

Darn 2016-02-26 20:02:29
No, because you repeat some fractions.

agbdmrbirdyface 2016-02-26 20:02:29
like some cases for k = 4 and k = 2 overlap, etc.

AlcumusGuy 2016-02-26 20:02:29
no, because some fractions are overcounted; for example $\frac{3}{6} = \frac{1}{2}$

copeland 2016-02-26 20:02:31
No! Some of these "steps" overlap.

copeland 2016-02-26 20:02:32
For example, at $z=\frac14$, we step up the $k=4$ term, but we also step up the $k=8$ term (since $\frac14 = \frac28$). That's a total "step up" of 2, but we only want to count it once because it only produces one new value.

copeland 2016-02-26 20:02:34
So how to we count the "step ups" so that we don't overcount the overlaps?

swagger21 2016-02-26 20:03:41
count how many distinct reduced proper fractions with denominator ≤ 10 there are

copeland 2016-02-26 20:03:46
Yeah, OK.

copeland 2016-02-26 20:03:50
One way is to note that each new $k$ only produces a new "step up" at $\dfrac{m}{k}$ if $m$ is relatively prime to $m$; that is, if $\dfrac{m}{k}$ is in lowest terms. Because, if $\dfrac{m}{k}$ can be reduced to a smaller denominator, then we've already counted that "step up" when we counted the smaller denominator.

idomath12345 2016-02-26 20:03:54
Bash the whole set out.

copeland 2016-02-26 20:04:04
Ayup. I might just start writing them down now.

copeland 2016-02-26 20:04:12
In fact. . .

copeland 2016-02-26 20:04:14
\[\begin{array}{r|l|l}
k & \text{fractions} & \text{number} \\ \hline
2 & \frac12 & 1 \\
3 & \frac13,\frac23 & 2 \\
4 & \frac14,\frac34 & 2 \\
5 & \frac15,\frac25,\frac35,\frac45 & 4 \\
6 & \frac16,\frac56 & 2 \\
7 & \frac17,\frac27,\frac37,\frac47,\frac57,\frac67 & 6 \\
8 & \frac18,\frac38,\frac58,\frac78 & 4 \\
9 & \frac19,\frac29,\frac49,\frac59,\frac79,\frac89 & 6 \\
10 & \frac{1}{10},\frac{3}{10},\frac{7}{10},\frac{9}{10} & 4
\end{array}\]

copeland 2016-02-26 20:04:17
Ta-da!

Darn 2016-02-26 20:04:23
The number of common fractions with denominator $n$ is equal to $\phi(n)$, so we can sum $\phi(1)+\phi(2)+\ldots+\phi(10)$.

copeland 2016-02-26 20:04:30
Fancy.

copeland 2016-02-26 20:04:34
(If you know the Euler $\phi$-function, then you might be able to do this more quickly.)

copeland 2016-02-26 20:04:43
That's a total of 1+2+2+4+2+6+4+6+4 = 31 "step ups" on the way from $z=0$ to $z=1$. Each "step up" increases the value of $f(z)$.

greenguy03 2016-02-26 20:04:51
1+2+2+4+2+6+4+6+4=31

agbdmrbirdyface 2016-02-26 20:04:51
which yields 31

copeland 2016-02-26 20:04:55
Since we start with $f(z) = 0$, and step it up 31 times on the way to $z=1$, we get a total of $\boxed{32}$ different values. Answer (A).

copeland 2016-02-26 20:05:20
Cool.

copeland 2016-02-26 20:05:25
The 10 is down.

copeland 2016-02-26 20:05:33
30 points us. Go us!

copeland 2016-02-26 20:05:42
Should we go get 30 more points?

temp8909 2016-02-26 20:06:04
on to 12!

avn 2016-02-26 20:06:04
now the 12

Aathreyakadambi 2016-02-26 20:06:04

temp8909 2016-02-26 20:06:04
from the 12!

Tapeman 2016-02-26 20:06:04
yes

darthvader1521 2016-02-26 20:06:04
yes

ThorJames 2016-02-26 20:06:04
yes!

weeshree 2016-02-26 20:06:04
sure!

yrnsmurf 2016-02-26 20:06:04
definetly

Magikarp1 2016-02-26 20:06:04
yes

mj7777 2016-02-26 20:06:04
yes

baseballcat 2016-02-26 20:06:04
yea

copeland 2016-02-26 20:06:09
21. Let $ABCD$ be a unit square. Let $Q_1$ be the midpoint of $\overline{CD}$. For $i=1,2,\ldots$, let $P_i$ be the intersection of $\overline{AQ_i}$ and $\overline{BD}$, and let $Q_{i+1}$ be the foot of the perpendicular from $P_i$ to $\overline{CD}$. What is \[ \sum_{i=1}^\infty \text{Area of } \triangle DQ_iP_i\;?\]
(A) $\dfrac16$ (B) $\dfrac14$ (C) $\dfrac13$ (D) $\dfrac12$ (E) $1$

copeland 2016-02-26 20:06:34
Maybe this is a geometry problem. There are lots of letters. . .

phi_ftw1618 2016-02-26 20:06:46
diagram?

temp8909 2016-02-26 20:06:46
DIAGRAM!!!!

Puzzled417 2016-02-26 20:06:46
draw a picture

greenguy03 2016-02-26 20:06:46
Diagram!

TheRealDeal 2016-02-26 20:06:46
diagram

yrnsmurf 2016-02-26 20:06:46
diagram!

_--__--_ 2016-02-26 20:06:46
Diagram

copeland 2016-02-26 20:06:49
Oh, for sure.

copeland 2016-02-26 20:06:53
Now it's definitely geometry.

copeland 2016-02-26 20:06:56
Let's draw a picture on the coordinate plane. We'll start with the square. I put $A$ at the origin, mainly because I thought having one endpoint of $\overline{AQ_i}$ always be the origin would be useful, but it probably doesn't matter too much.

copeland 2016-02-26 20:06:57

copeland 2016-02-26 20:07:17
Should I add some Ps and Qs?

copeland 2016-02-26 20:07:19
Do you mind?

Puzzled417 2016-02-26 20:07:23
us coordinate geometry

copeland 2016-02-26 20:07:26
Ooh. coordinates.

Aathreyakadambi 2016-02-26 20:07:29
SURE!

TheRealDeal 2016-02-26 20:07:29
add them

copeland 2016-02-26 20:07:32

copeland 2016-02-26 20:07:37
What do you notice?

copeland 2016-02-26 20:08:15
First off, a lot of people are saying "similar triangles."

copeland 2016-02-26 20:08:18
Are those similar?

Burrito 2016-02-26 20:08:54
nope

temp8909 2016-02-26 20:08:54
no

SineWarrior 2016-02-26 20:08:54
no

Tapeman 2016-02-26 20:08:54
no

jkoj25 2016-02-26 20:08:54
no

Jyzhang12 2016-02-26 20:08:54
no

agbdmrbirdyface 2016-02-26 20:08:54
wait no

jkoj25 2016-02-26 20:08:54
the angles are different

SK200 2016-02-26 20:08:54
no

Pisensei 2016-02-26 20:08:54
nope

physangel 2016-02-26 20:08:54
sorry, no

Jayjayliu 2016-02-26 20:08:54
no

Rotack00 2016-02-26 20:08:54
no, angles are different?

copeland 2016-02-26 20:08:55
No. The angle at Q gets closer and closer to 90 degrees.

copeland 2016-02-26 20:09:03
If we let $q_i$ be the $y$-coordinate of $Q_i$, then what's the area of $\triangle DQ_iP_i$?

copeland 2016-02-26 20:09:16
It looks like we want base multiplied by height.

copeland 2016-02-26 20:09:22
What base should we use?

temp8909 2016-02-26 20:09:47
$QD$

yrnsmurf 2016-02-26 20:09:47
DQi

greenguy03 2016-02-26 20:09:47
DQi

agbdmrbirdyface 2016-02-26 20:09:47
we should use DQi

copeland 2016-02-26 20:09:54
What's that length?

copeland 2016-02-26 20:10:35
(as a function of i)

copeland 2016-02-26 20:10:59
Sorry, that question was harder than I meant it to be.

copeland 2016-02-26 20:11:14
The distance from the point (1,0) to the point $(1,q_i)$ is. .

swagger21 2016-02-26 20:11:39
q_i

temp8909 2016-02-26 20:11:39
$q_i$

agbdmrbirdyface 2016-02-26 20:11:39
qi

Dominater76 2016-02-26 20:11:39
$q_i$

ninjataco 2016-02-26 20:11:39
q_i

Jayjayliu 2016-02-26 20:11:39
q_i

greenguy03 2016-02-26 20:11:39
$qi$

copeland 2016-02-26 20:11:42
The base of $\triangle DQ_iP_i$ is $DQ_i$, which has length $q_i$.

copeland 2016-02-26 20:11:49
The height of $\triangle DQ_iP_i$ is $P_iQ_{i+1}$.

copeland 2016-02-26 20:11:54
That one is harder.

copeland 2016-02-26 20:12:02
How else can we express the length $P_iQ_{i+1}$?

copeland 2016-02-26 20:12:05
What do we know about triangle $DP_iQ_{i+1}$?

greenguy03 2016-02-26 20:12:28
It's right

yrnsmurf 2016-02-26 20:12:28
It's a right triangle

swagger21 2016-02-26 20:12:28
it's right

Rotack00 2016-02-26 20:12:28
its a right triangle?

Dominater76 2016-02-26 20:12:28
Right triangle?

_--__--_ 2016-02-26 20:12:28
right triangle

copeland 2016-02-26 20:12:30
It's more than just right. . .

copeland 2016-02-26 20:12:48
It's, like, annoying about how right it is. Like it keeps telling you how wrong you are. . .

TheRealDeal 2016-02-26 20:12:58
right isoscelese triangle

goldentail141 2016-02-26 20:12:58
45-45-90 right triangle

gradysocool 2016-02-26 20:12:58
isoceles right

swagger21 2016-02-26 20:12:58
isoceles right triangle

_--__--_ 2016-02-26 20:12:58
isosceles right?

math101010 2016-02-26 20:12:58
isosceles

ghghghghghghghgh 2016-02-26 20:12:58
isosceles right

temp8909 2016-02-26 20:12:58
isosceles right!

avn 2016-02-26 20:12:58
isoceles

copeland 2016-02-26 20:13:08
Uh, I mean, it's an isosceles right triangle! (Since $\angle BDC = 45^\circ$.)

copeland 2016-02-26 20:13:14
So $P_iQ_{i+1}$ has the same length as $DQ_{i+1}$. Thus, the height of $\triangle DQ_iP_i$ is $P_iQ_{i+1} = DQ_{i+1} = q_{i+1}$.

copeland 2016-02-26 20:13:16
So, the area of $\triangle DQ_iP_i$ is $\frac12 q_iq_{i+1}$, and the sum we want is \[ \frac12 \sum_{i=1}^\infty q_iq_{i+1}.\]

copeland 2016-02-26 20:13:25
We know $q_1 = \frac12$. How can we determine the other $q$'s?

copeland 2016-02-26 20:14:09
At some point we need to do some arithmetic on this test. . .

chenmeister22 2016-02-26 20:14:30
Coordinate Geometry

Puzzled417 2016-02-26 20:14:30
use coordinate geometry?

thepiercingarrow 2016-02-26 20:14:33
calculator!

copeland 2016-02-26 20:14:35
Weak.

copeland 2016-02-26 20:14:49
I don't mean "You are weak"

copeland 2016-02-26 20:14:53
I mean, um, something else.

DashK 2016-02-26 20:15:00
graphic calculator!

copeland 2016-02-26 20:15:02
. . .

copeland 2016-02-26 20:15:07
Who's trolling whom again?

copeland 2016-02-26 20:15:19
What line is $AQ_i$?

techguy2 2016-02-26 20:15:31
um

copeland 2016-02-26 20:15:57
What is its equation.

copeland 2016-02-26 20:16:08
I'll give you one point: $A=(0,0)$.

chenmeister22 2016-02-26 20:16:40
$y=\frac{x}{2}$

yrnsmurf 2016-02-26 20:16:40
y=x/2

mathguy5041 2016-02-26 20:16:40
y=1/2x

copeland 2016-02-26 20:16:46
OK, but we need to do more than just the first one.

copeland 2016-02-26 20:16:50
What about for general $i$?

SK200 2016-02-26 20:17:02
y=(q_i)(x)

agbdmrbirdyface 2016-02-26 20:17:04
$ y = q_i * x

copeland 2016-02-26 20:17:10
Great.

copeland 2016-02-26 20:17:13
We know that line $AQ_i$ is just the line $y = q_ix$.

copeland 2016-02-26 20:17:21
And what is line $BD?$

agbdmrbirdyface 2016-02-26 20:17:59
y = -x + 1

yrnsmurf 2016-02-26 20:17:59
x+y=1

goldentail141 2016-02-26 20:17:59
$y = 1 - x$

Jayjayliu 2016-02-26 20:17:59
y=1-x

ghghghghghghghgh 2016-02-26 20:17:59
y=-x+1

swagger21 2016-02-26 20:17:59
y = -x + 1

temp8909 2016-02-26 20:17:59
$y=1-x$

ein 2016-02-26 20:17:59
-x+1=y

mathguy5041 2016-02-26 20:17:59
$x+y=1$

greenguy03 2016-02-26 20:17:59
With the equation y = -x+1

_--__--_ 2016-02-26 20:17:59
$y = -x + 1$

weeshree 2016-02-26 20:17:59
y=-x+1

sillyd 2016-02-26 20:18:02
y=-x+1

copeland 2016-02-26 20:18:04
$\overline{BD}$ is part of the line $x+y = 1$.

copeland 2016-02-26 20:18:06
So $q_{i+1}$ is $y$-coordinate of the intersection of the lines
\begin{align*}
x+y&=1\\
y&=q_ix.
\end{align*}

copeland 2016-02-26 20:18:10
We want to eliminate $x$ so we multiply the first by $q_i$ and substitute.

copeland 2016-02-26 20:18:17
So $y=q_{i+1}$ solves \[q_ix+q_iy=q_i.\] Substituting $y$ for $q_ix$ and setting $y=q_{i+1}$ gives us the recurrence\[q_{i+1}+q_iq_{i+1}=q_{i}.\]

copeland 2016-02-26 20:18:42
Now we could solve this for $q_{i+1}$, but what's a more clever move?

copeland 2016-02-26 20:19:08
What are we trying to find?

mathwizard888 2016-02-26 20:19:57
$q_iq_{i+1}=q_i-q_{i+1}$ and we can telescope

goldentail141 2016-02-26 20:19:57
Write the equation as $q_{i}q_{i + 1} = q_{1} - q_{i + 1}.$

mathguy5041 2016-02-26 20:19:57
subtract q_i+1 from both sides

temp8909 2016-02-26 20:19:57
$q_iq_{i+1}$

legozelda 2016-02-26 20:19:57
find q(i)*q(i+1)

agbdmrbirdyface 2016-02-26 20:19:57
oh wait qi * qi+1

yrnsmurf 2016-02-26 20:19:57
subtract $(q_i+1)$ from each side

ein 2016-02-26 20:19:57
Substitute q_iq_(i+1) into the sum

mathguy5041 2016-02-26 20:20:06
telescoping sum

agbdmrbirdyface 2016-02-26 20:20:06
ohhhh we subtract q(i+1) from both sides

idomath12345 2016-02-26 20:20:06
Then cancel like mad.

copeland 2016-02-26 20:20:08
Our recurrence simplifies to $q_iq_{i+1} = q_i - q_{i+1}$.

copeland 2016-02-26 20:20:10
So our sum is a telescoping series! \[ \frac12 \sum_{i=1}^\infty (q_i - q_{i+1}).\]

copeland 2016-02-26 20:20:11
We're home free now. What does it sum to?

techguy2 2016-02-26 20:20:33
Can you like, sticky the question? that would help.

techguy2 2016-02-26 20:20:45
so we don't have to scroll.

W.Sun 2016-02-26 20:20:53
1/4

Jayjayliu 2016-02-26 20:20:53
1/4

mathguy5041 2016-02-26 20:20:53
1/4

Readingrocks88 2016-02-26 20:20:53
1/4

W.Sun 2016-02-26 20:20:53
Telescoping!!!! 1/4!! :spidey:

idomath12345 2016-02-26 20:20:53
JUST 1/2*q1?

goldentail141 2016-02-26 20:20:53
$\frac{1}{4}$

gradysocool 2016-02-26 20:20:53
1/2 qi=1/4

temp8909 2016-02-26 20:20:53
$\frac{1}{2}q_1=\frac{1}{4}$

yrnsmurf 2016-02-26 20:20:53
$\dfrac14$

copeland 2016-02-26 20:20:55
The answer is just $\dfrac12 q_1 = \boxed{\dfrac14}$. Answer (B).

copeland 2016-02-26 20:21:09
Alright, any questions?

mathguy5041 2016-02-26 20:21:35
On to 22!

gradysocool 2016-02-26 20:22:10
Is there a purely geometric way of doing the last one?

copeland 2016-02-26 20:22:14
That's a fantastic question.

copeland 2016-02-26 20:22:16
I'd love to see one.

copeland 2016-02-26 20:22:26
I wouldn't be too surprised, now that you mention it.

copeland 2016-02-26 20:22:52
I'd probably start by thinking about the respective areas and then try to find equal areas that fill, maybe, the big red triangle.

copeland 2016-02-26 20:23:15
Like, maybe the red triangles are the right areas and then you're done? Give it a whirl. . .

copeland 2016-02-26 20:23:25
Oh, that is true.

copeland 2016-02-26 20:23:56
$(q_i-q_{i+1})/2$ is the area of one of those triangles. Neat.

copeland 2016-02-26 20:24:00
I'd love to see the rest.

steveshaff 2016-02-26 20:24:08
There is...check out my solution on the contest forum

copeland 2016-02-26 20:24:15
Apparently it exists out there in the interwebs.

copeland 2016-02-26 20:24:21
Problem 22?

copeland 2016-02-26 20:24:23
22. For a certain positive integer $n$ less than 1000, the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$, a repeating decimal of period 6, and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$, a repeating decimal of period 4. In which interval does $n$ lie?
$\phantom{12B:22}$
(A) [1,200] (B) [201,400] (C) [401,600] (D) [601,800] (E) [801,999]

copeland 2016-02-26 20:24:24
How can we work with repeating decimals?

Dominater76 2016-02-26 20:25:59
With lots of 9s

ptes77 2016-02-26 20:25:59
Turn into fraction form

weeshree 2016-02-26 20:25:59
express as fractions

yrnsmurf 2016-02-26 20:25:59
it is equal to $\dfrac{\overline{abcdef}}{999999}$

jkoj25 2016-02-26 20:25:59
999999

copeland 2016-02-26 20:26:09
A repeating decimal can be written as a fraction with a bunch of 9's in the denominator. For example, $0.22222\ldots = \dfrac29$ and $0.137137137\ldots = \dfrac{137}{999}$.

copeland 2016-02-26 20:26:15
So we can write $\dfrac{1}{n} =\dfrac{r}{999999}$ and $\dfrac{1}{n+6} = \dfrac{s}{9999}$, where $r$ is the number $abcdef$ and $s$ is the number $wxyz$.

copeland 2016-02-26 20:26:17
How does that help?

MSTang 2016-02-26 20:27:25
$n$ divides 999999 and $n+6$ divides 9999

temp8909 2016-02-26 20:27:25
n is a factor of 999999 and n+6 is a factor of 9999

macandcheese 2016-02-26 20:27:25
n must divide 999999, n+6 must divide 9999

SK200 2016-02-26 20:27:25
$rn=999999$&$s(n+6)=9999$

summitwei 2016-02-26 20:27:25
n is a factor of 999999 and n+6 is a factor of 9999

yrnsmurf 2016-02-26 20:27:25
n is divisor of 999999 while n+6 is divisor of 9999

copeland 2016-02-26 20:27:27
These give $999999 = rn$ and $9999 = s(n+6)$. In particular, $n$ is a factor of 999999 and $n+6$ is a factor of 9999.

copeland 2016-02-26 20:27:29
Which of these is likely to be easier to work with?

mathguy5041 2016-02-26 20:28:38
9999

greenguy03 2016-02-26 20:28:38
The n+6

summitwei 2016-02-26 20:28:38
9999 because it's smaller

SherlockHolmes7 2016-02-26 20:28:38
9999

sillyd 2016-02-26 20:28:38
9999

mathguy5041 2016-02-26 20:28:38
since it has less factors, 9999

yrnsmurf 2016-02-26 20:28:38
9999 because 101 and 11

illogical_21 2016-02-26 20:28:39
9999=s(n+6)

copeland 2016-02-26 20:28:41
9999 is smaller, so it probably has fewer factors.

copeland 2016-02-26 20:28:42
Indeed, $9999 = 99 \cdot 101 = 9 \cdot 11 \cdot 101$.

copeland 2016-02-26 20:28:44
However, what else do we know about $n+6$?

copeland 2016-02-26 20:28:50
Can it be any of those factors?

thkim1011 2016-02-26 20:30:15
therefore n+6 is one of 101, 303, 909

thkim1011 2016-02-26 20:30:15
it must contain a 101 else the period is not 4

temp8909 2016-02-26 20:30:15
it can't divide 9 or 99

MSTang 2016-02-26 20:30:15
99 = 9 * 11, so n+6 can't also divide 99 (4 is the minimal period)

copeland 2016-02-26 20:30:22
It can't be a factor of $99$.

copeland 2016-02-26 20:30:23
If it were, then we could write $\dfrac{1}{n+6} = \dfrac{t}{99}$ for some $t$, and in particular the decimal $\dfrac{1}{n+6}$ would only have period 2 (or 1).

copeland 2016-02-26 20:30:25
So we must have $n+6 = 101k$ for some $k$ that divides 99.

copeland 2016-02-26 20:30:27
Since we're looking for $n < 1000$, that only leaves $n+6 \in \{101, 303, 909\}$ as possibilities.

copeland 2016-02-26 20:30:28
That is, $n \in \{95, 297, 903\}$.

copeland 2016-02-26 20:30:30
How can we tell which one works?

temp8909 2016-02-26 20:31:21
divide them into 999999

Couper 2016-02-26 20:31:21
3 possibilities so just check the cases

MSTang 2016-02-26 20:31:21
try dividing into 999999

agbdmrbirdyface 2016-02-26 20:31:21
we have to go over to the other equation?

macandcheese 2016-02-26 20:31:21
297 divides 999999

thkim1011 2016-02-26 20:31:25
297 = 27 * 11 so has period 3 * 2

copeland 2016-02-26 20:31:33
We need $n$ to be a factor of 999999. So we can check them all.

copeland 2016-02-26 20:31:34
Clearly 95 is not a factor of 999999, since 5 doesn't divide 999999.

copeland 2016-02-26 20:31:35
297 works! $297 = 3 \cdot 99$, and thus $999999 \div 297 = 10101 \div 3$ is an integer.

copeland 2016-02-26 20:31:38
Finally, $903 = 3 \cdot 301 = 3 \cdot 7 \cdot 43$ doesn't work, since 43 is not a factor of 999999.

copeland 2016-02-26 20:31:44
So $n = 297$ must be the number we're looking for. This is in the interval $\boxed{[201,400]}$. Answer (B).

mathguy5041 2016-02-26 20:32:20
As a check the abcdef is 003367

mathguy5041 2016-02-26 20:32:20
ah, 3367, the magical number

copeland 2016-02-26 20:32:31
Note that we never verified that $\dfrac{1}{297}$ has period 6 -- we only learned that its period is at most 6. If 297 were a factor of a smaller number of 9's, then the period would be smaller.

copeland 2016-02-26 20:32:33
But since we checked that the other possible values of $n$ definitely didn't work, $n=297$ must be the answer. Just ask Sherlock Holmes: "Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."

copeland 2016-02-26 20:32:39
23. What is the volume of the region in three-dimensional space defined by the inequalities $|x| + |y| + |z| \le 1$ and $|x| + |y| + |z-1| \le 1$?
$\phantom{12B:23}$
(A) $\dfrac16$ (B) $\dfrac13$ (C) $\dfrac12$ (D) $\dfrac23$ (E) $1$

copeland 2016-02-26 20:32:52
This problem is vaguely reminiscent of 12B #18 / 10B #21.

copeland 2016-02-26 20:33:50
Can we use a similar trick?

idomath12345 2016-02-26 20:34:22
Yes?

copeland 2016-02-26 20:34:25
I hope so. . .

copeland 2016-02-26 20:34:29
We had some symmetry before.

copeland 2016-02-26 20:34:33
Do we have any symmetry now?

agbdmrbirdyface 2016-02-26 20:35:26
kind of, in the first equation

agbdmrbirdyface 2016-02-26 20:35:26
but not in the second one

avn 2016-02-26 20:35:26
yes

SineWarrior 2016-02-26 20:35:26
yes and no

copeland 2016-02-26 20:35:43
Let's just look at $x$. Is there any symmetry (in BOTH equations) with respect to $x$?

SherlockHolmes7 2016-02-26 20:36:48
yes

temp8909 2016-02-26 20:36:48
yes!!!

swagger21 2016-02-26 20:36:48
yes?

copeland 2016-02-26 20:36:54
So what does that tells us?

summitwei 2016-02-26 20:37:35
We only need to work with x>0 and then multiply by 2

copeland 2016-02-26 20:37:40
This is symmetric with respect to $x\mapsto -x$, so we can just restrict to $x>0$ and multiply by 2 at the end.

copeland 2016-02-26 20:37:44
Right?

copeland 2016-02-26 20:37:47
What else?

Couper 2016-02-26 20:38:19
Symmetry with respect to y

idomath12345 2016-02-26 20:38:19
y is same.

swagger21 2016-02-26 20:38:19
it is symmetric with respect to y?

yrnsmurf 2016-02-26 20:38:19
Same thing with y

macandcheese 2016-02-26 20:38:19
same with y

idomath12345 2016-02-26 20:38:19
y has symmetry.

legozelda 2016-02-26 20:38:19
same for y>0, then multiply by 2(multiply by 4 total)

weeshree 2016-02-26 20:38:19
restrict y similarly

gradysocool 2016-02-26 20:38:19
The y's as well

copeland 2016-02-26 20:38:21
This is symmetric with respect to $y\mapsto -y$, so we can just restrict to $y>0$ and multiply by 2 at the end also!

copeland 2016-02-26 20:38:27
Great. And also $z?$

legozelda 2016-02-26 20:38:54
no

temp8909 2016-02-26 20:38:54
not so much!

legozelda 2016-02-26 20:38:54
kind of?

agbdmrbirdyface 2016-02-26 20:38:54
no

W.Sun 2016-02-26 20:38:54
Not with Z

chenmeister22 2016-02-26 20:38:54
Nope

copeland 2016-02-26 20:39:01
We can just compute the region where $x$ and $y$ are positive, and multiply by 4.

copeland 2016-02-26 20:39:01
But we don't have that flexibility with $z$.

copeland 2016-02-26 20:39:03
Or do we? Is there a symmetric with $z$ that we can exploit?

copeland 2016-02-26 20:39:06
What can we replace $z$ with and not change the inequalities?

agbdmrbirdyface 2016-02-26 20:39:53
(1-z)?

temp8909 2016-02-26 20:39:53
reflection via $z=\frac{1}{2}$

mssmath 2016-02-26 20:39:53
1-z

yrnsmurf 2016-02-26 20:39:53
1-z

copeland 2016-02-26 20:39:59
We can replace $z$ with $1-z$, and vice versa. Note that $|1-z| = |z-1|$ and $|(1-z)-1| = |-z| = |z|$, so replacing $z$ with $1-z$ just interchanges the two inequalities.

copeland 2016-02-26 20:39:59
Thus there's a third symmetry.

weeshree 2016-02-26 20:40:12
z>1/2

illogical_21 2016-02-26 20:40:12
z-1/2

Couper 2016-02-26 20:40:12
z -> z-1/2

weeshree 2016-02-26 20:40:12
q = z-1/2

copeland 2016-02-26 20:40:16
Note that $z = 1-z$ is true along $z= \dfrac12$, so replacing $z$ with $1-z$ is the same as reflecting across the $z = \dfrac12$ plane.

copeland 2016-02-26 20:40:24
Thus, our solid consists of 8 identical pieces, symmetric across the $x=0$, $y=0$, and $z=\dfrac12$ planes.

copeland 2016-02-26 20:40:24
We can just compute the volume of one of them, and then multiply by 8 to get our final answer.

copeland 2016-02-26 20:40:26
So let's compute the volume of the octant in which $x \ge 0$, $y \ge 0$, and $z \ge \dfrac12$.

copeland 2016-02-26 20:40:31
We want the region in which $x + y + z \le 1$ in this octant. What does this region look like?

LetThePlatPlay 2016-02-26 20:41:30
a tetrahedron

legozelda 2016-02-26 20:41:30
triangular pyramid

agbdmrbirdyface 2016-02-26 20:41:30
it has to be a pyramid?

ein 2016-02-26 20:41:30
a triangular prism

yrnsmurf 2016-02-26 20:41:30
a pyramid

adas1 2016-02-26 20:41:30
a tetrahedron?

greenguy03 2016-02-26 20:41:30
A triangular pyramid?

copeland 2016-02-26 20:41:33
In the $z = \dfrac12$ plane, it's the region $x+y \le \dfrac12$. That's a triangle in the $x,y \ge 0$ quadrant.

copeland 2016-02-26 20:41:37
And the plane intersects the $z$-axis at $(0,0,1)$.

copeland 2016-02-26 20:41:38
So we have the following picture:

copeland 2016-02-26 20:45:07

copeland 2016-02-26 20:45:40
What is the volume?

illogical_21 2016-02-26 20:46:34
1/48

mathguy5041 2016-02-26 20:46:34
1/48

yrnsmurf 2016-02-26 20:46:34
It's still $\dfrac{1}{48}$

temp8909 2016-02-26 20:46:34
$\frac{1}{48}$

SherlockHolmes7 2016-02-26 20:46:34
1/48

copeland 2016-02-26 20:46:45
Its base has area $\dfrac18$ and its height is $\dfrac12$.

copeland 2016-02-26 20:46:46
Then, its volume is $$\frac13bh = \frac13 \cdot \frac 18 \cdot \frac12 = \frac{1}{48}.$$

copeland 2016-02-26 20:46:47
And to finish?

Couper 2016-02-26 20:47:14
Multiply by 8 to get 1/6!

sillyd 2016-02-26 20:47:14
multiply by 8 to get 1/6

illogical_21 2016-02-26 20:47:14
1/48*8=1/6

agbdmrbirdyface 2016-02-26 20:47:14
multiply by 8, since 8 fold symmetry

Dominater76 2016-02-26 20:47:14
Multiply by 8

baseballcat 2016-02-26 20:47:14
multiply by 8

mathguy5041 2016-02-26 20:47:14
Times 8?

yrnsmurf 2016-02-26 20:47:14
multiply that by 8

ninjataco 2016-02-26 20:47:14
we have 8 of these

agbdmrbirdyface 2016-02-26 20:47:14
so that yields 1/6

temp8909 2016-02-26 20:47:14
$\frac{1}{6}$

copeland 2016-02-26 20:47:16
We have eight of these things, so the total volume is $8 \cdot \dfrac{1}{48} = \boxed{\dfrac16}$. Answer (A).

copeland 2016-02-26 20:47:17
You might also have noticed that these 8 octants fit together to form an octahedron, with center $(0,0,\frac12)$ and vertices $(0,0,0)$, $(0,0,1)$, and $\left(\pm\frac12,\pm\frac12,\frac12\right)$. Here's how our first octant fits into it:

copeland 2016-02-26 20:47:21

copeland 2016-02-26 20:47:34
And I bet the perspective is wrong, too.

copeland 2016-02-26 20:47:50
Oh, I also drew you a picture for that conversation on problem 21, was it?

copeland 2016-02-26 20:47:57
We conjecture that this region:

copeland 2016-02-26 20:47:58

copeland 2016-02-26 20:48:07
Has the same area as this region:

copeland 2016-02-26 20:49:47

copeland 2016-02-26 20:50:27
I think that's a really good hint to get us there.

copeland 2016-02-26 20:50:41
All we need to do now is shear the bottom half of the triangle from D over to A.

copeland 2016-02-26 20:50:46
Now let's rock number 24:

copeland 2016-02-26 20:51:03
24. There are exactly 77,000 ordered quadruples $(a,b,c,d)$ such that $\gcd(a,b,c,d) = 77$ and $\text{lcm}(a,b,c,d) = n$ .What is the smallest possible value of $n$?
$\phantom{12B:24}$
(A) 13,860 (B) 20,790 (C) 21,560 (D) 27,720 (E) 41,580

copeland 2016-02-26 20:51:17
What can we do write at the start to simplify the problem?

copeland 2016-02-26 20:51:49
wow.

copeland 2016-02-26 20:51:55
What can we do right at the start to simplify the problem?

copeland 2016-02-26 20:52:14
Would you bye. . . text-to-speech fail?

copeland 2016-02-26 20:52:40
Sorry, homophone humor. I'm dun.

legozelda 2016-02-26 20:52:49
use the fact that lcm * gcd = product: 77 * n = abcd

LetThePlatPlay 2016-02-26 20:52:49
a,b,c,d all divisible by 77

adas1 2016-02-26 20:52:49
abcd = 77n

Couper 2016-02-26 20:52:49
Copeland... get it together man

Tapeman 2016-02-26 20:52:55
77v=a 77x=b 77y=c 77z=d

yrnsmurf 2016-02-26 20:52:55
Write a,b,c,d as 77w,77x,77y,77z

copeland 2016-02-26 20:52:57
$a$, $b$, $c$, and $d$ all have a common factor of 77. So we can divide any such quadruple by 77.

copeland 2016-02-26 20:53:13
That divdes the lcm by 77 as well.

copeland 2016-02-26 20:53:14
So we can instead solve the problem:

copeland 2016-02-26 20:53:15
24X. There are exactly 77,000 ordered quadruples $(a,b,c,d)$ such that $\gcd(a,b,c,d) = 1$ and $\text{lcm}(a,b,c,d) = m$. What is the smallest possible value of $m$?
$\phantom{12B:24X}$
(A) $\dfrac{13,860}{77}$ (B) $\dfrac{20,790}{77}$ (C) $\dfrac{21,560}{77}$ (D) $\dfrac{27,720}{77}$ (E) $\dfrac{41,580}{77}$

copeland 2016-02-26 20:53:31
There's no real point in dividing those answer choices out -- once we find $m$, it's easier to compute $77m$ to get the original answer choice.

copeland 2016-02-26 20:53:41
However, a couple of people here at AoPS HQ did this to get a quick feel for the problem. The values are:

copeland 2016-02-26 20:53:43
(A) $\dfrac{13,860}{77}$ (B) $\dfrac{20,790}{77}$ (C) $\dfrac{21,560}{77}$ (D) $\dfrac{27,720}{77}$ (E) $\dfrac{41,580}{77}$

copeland 2016-02-26 20:53:44
(A) $2^2 \cdot 3^2 \cdot 5$ (B) $2\cdot 3^3 \cdot 5$ (C) $2^3 \cdot 5 \cdot 7$ (D) $2^3 \cdot 3^2 \cdot 5$ (E) $2^2 \cdot 3^3 \cdot 5$

copeland 2016-02-26 20:53:54
Do you see any choices that are just bad?

MSTang 2016-02-26 20:55:14
B, C, and E

sillyd 2016-02-26 20:55:14
C and E are bad

agbdmrbirdyface 2016-02-26 20:55:14
actually no, scratch that, B and E

summitwei 2016-02-26 20:55:14
B, 2^3*3*5 is better, C, 2^3*3*5 is better, and E, 2^3*3^2*5 is better

googol.plex 2016-02-26 20:55:14
B, C, and E

SineWarrior 2016-02-26 20:55:14
C

SineWarrior 2016-02-26 20:55:14
c

agbdmrbirdyface 2016-02-26 20:55:14
B and E are pretty big

copeland 2016-02-26 20:55:16
Choice B is bad: $2\cdot3^3\cdot5$ is bigger than $2^3\cdot3\cdot5$ and they both have the same size solution set.

copeland 2016-02-26 20:55:28
Choice E is similarly bad.

copeland 2016-02-26 20:55:37
Choice C fails because we skipped 3.

copeland 2016-02-26 20:55:41
Therefore the answer must be A or D.

copeland 2016-02-26 20:55:46
Intriguing, AMC dudes.

copeland 2016-02-26 20:55:51
Let's keep rocking.

copeland 2016-02-26 20:55:58
Given some number $m$, how do we count quadruples that have no common factors and lcm $m$?

MSTang 2016-02-26 20:56:47
Prime factorize $m$

illogical_21 2016-02-26 20:56:47
factor m ?

SimonSun 2016-02-26 20:56:50
calculate factors

copeland 2016-02-26 20:56:52
Let's look at it one prime at a time.

copeland 2016-02-26 20:56:53
Suppose that $m$ has a $p^k$ term in its prime factorization. In how many different ways can powers of $p$ appear in the prime factorizations of $a,b,c,d$?

copeland 2016-02-26 20:56:59
To put it another way, suppose:
$a$ has a $p^w$ term
$b$ has a $p^x$ term
$c$ has a $p^y$ term
$d$ has a $p^z$ term
What are the conditions on $(w,x,y,z)$ so that the gcd is 1 and that the lcm has a $p^k$ term?

pinkrock 2016-02-26 20:58:22
w, x, y, or z is 0

summitwei 2016-02-26 20:58:22
min(w,x,y,z)=0 and max(w,x,y,z)=k

yrnsmurf 2016-02-26 20:58:22
At least 1 must be $p^k$ and 1 must be $p^0$

copeland 2016-02-26 20:58:27
We need to have $0 \le w,x,y,z \le k$, with at least one of them 0 (to force gcd 1) and at least one of them $k$ (to force the lcm to have the prime factor $p^k$).

copeland 2016-02-26 20:58:39
How do we count this?

copeland 2016-02-26 20:59:40
There's a keyword "at least" here.

copeland 2016-02-26 20:59:42
What's that tell us?

agbdmrbirdyface 2016-02-26 21:00:09
complementary counting?

W.Sun 2016-02-26 21:00:09
Complimentary counting

Tapeman 2016-02-26 21:00:09
complementary

mathguy5041 2016-02-26 21:00:09
Complementary counting?

weeshree 2016-02-26 21:00:09
Complimentary

LetThePlatPlay 2016-02-26 21:00:09
complementary counting

copeland 2016-02-26 21:00:15
Often that's the cue for complementary counting.

copeland 2016-02-26 21:00:19
It seems hard here.

copeland 2016-02-26 21:00:26
Is there another technique that we can use?

MSTang 2016-02-26 21:00:40
PIE

temp8909 2016-02-26 21:00:40
PIE and complmentary counting!!!

eveningstarandlion 2016-02-26 21:00:40
PIE?

copeland 2016-02-26 21:00:42
We can use PIE! (The "at least" is often our clue to use PIE.)

copeland 2016-02-26 21:01:21
This is a little more subtle than just these names imply. We're kind of using complementary counting and then fixing stuff here, but what is a name, really?

W.Sun 2016-02-26 21:01:29
Principle of inclusion exclusion!

copeland 2016-02-26 21:01:49
PIE is the Principle of Inclusion-Exclusion. Check this out, it's cool.

copeland 2016-02-26 21:02:00
How many ways to have $(w,x,y,z)$ with $0 \le w,x,y,z \le k$, forgetting about the other conditions?

USA 2016-02-26 21:02:41
What is PIE?

swagger21 2016-02-26 21:02:42
(k+1)^4

mathguy5041 2016-02-26 21:02:42
(k+1)^4?

illogical_21 2016-02-26 21:02:42
(k+1)^4

legozelda 2016-02-26 21:02:42
(k+1)^4

yrnsmurf 2016-02-26 21:02:42
$(k+1)^4$

SimonSun 2016-02-26 21:02:42
$(k+1)^4$

randomsolver 2016-02-26 21:02:42
(k+1)4

copeland 2016-02-26 21:02:45
Each exponent has $k+1$ choices, so there are $(k+1)^4$ ways.

copeland 2016-02-26 21:02:52
Now we have to subtract those with no 0's, and subtract those with no $k$'s.

copeland 2016-02-26 21:02:53
How many with no 0's?

mathguy5041 2016-02-26 21:03:22
k^4

agbdmrbirdyface 2016-02-26 21:03:22
K^4?

temp8909 2016-02-26 21:03:22
-k^4

pinkrock 2016-02-26 21:03:22
(k)^4

summitwei 2016-02-26 21:03:22
$k^4$

illogical_21 2016-02-26 21:03:22
k^4

thequantumguy 2016-02-26 21:03:22
k^4

copeland 2016-02-26 21:03:25
If no 0's, then we have $1 \le w,x,y,z \le k$. Each has $k$ choices, so that's $k^4$ possibilities.

copeland 2016-02-26 21:03:27
If no $k$'s, then we have $0 \le w,x,y,z \le k-1$. Again each has $k$ choices, so that's $k^4$ possibilities.

copeland 2016-02-26 21:03:28
So that gives us $(k+1)^4 - 2k^4$ as our count so far. Are we done?

ein 2016-02-26 21:04:06
double counted those with none of both

randomsolver 2016-02-26 21:04:06
(k-1)^4

sillyd 2016-02-26 21:04:06
nope add the number without k's or 0's

summitwei 2016-02-26 21:04:06
No, you still need to add the ones which don't have either of them

copeland 2016-02-26 21:04:08
No -- we've doubly subtracted the $(w,x,y,z)$ with both no 0's and no $k$'s. We only want to subtract them out once, so we need to add them back in.

copeland 2016-02-26 21:04:13
These have $1 \le w,x,y,z \le k-1$, so there are $(k-1)^4$ of them.

copeland 2016-02-26 21:04:14
Thus our final count is $(k+1)^4 - 2k^4 + (k-1)^4$.

copeland 2016-02-26 21:04:15
How does this simplify?

swagger21 2016-02-26 21:05:18
12k^2 + 2

illogical_21 2016-02-26 21:05:18
12k^2+2

agbdmrbirdyface 2016-02-26 21:05:18
oh wait it just comes down to 12k^2 + 2

Jayjayliu 2016-02-26 21:05:18
12*k^2+2

thequantumguy 2016-02-26 21:05:18
12k^2 + 2

copeland 2016-02-26 21:05:23
Rather nicely! Most of the terms cancel, and we're left with just $12k^2 + 2$.

copeland 2016-02-26 21:05:24
(Incidentally, this is the "second discrete derivative of $k^4$", so we know that $4\cdot3 k^2$ is the right leading term.)

copeland 2016-02-26 21:05:26
So, to recap, if we have $(a,b,c,d)$ with gcd 1 and a factor of $p^k$ in the lcm, then there are $12k^2 + 2$ ways to allocate powers of $p$ to the four numbers.

copeland 2016-02-26 21:05:28
How do we use this to get 77,000?

SherlockHolmes7 2016-02-26 21:06:26
set it equal to 77000

agbdmrbirdyface 2016-02-26 21:06:26
set it equal to 77000?

copeland 2016-02-26 21:06:39
Almost, except what are we forgetting?

yrnsmurf 2016-02-26 21:07:16
$(12k^2+2)(12l^2+2)(12m^2+2)$etc.=77000

swagger21 2016-02-26 21:07:16
theres more than one prime factor most likely

copeland 2016-02-26 21:07:20
Ah.

copeland 2016-02-26 21:07:22
We do this separately for each prime power in the lcm, and then multiply them together.

copeland 2016-02-26 21:07:24
So we're looking for values of $k_1,k_2,\ldots$ (possibly repeated) that we can multiply together the $(12k_i^2 + 2)\text{'s}$ to get 77,000.

copeland 2016-02-26 21:07:26
It might help to compute some values of $12k^2 + 2$, and compare to the factors of 77,000.

copeland 2016-02-26 21:07:31
First, note that $77000 = 77 \cdot 1000 = 2^3 \cdot 5^3 \cdot 7 \cdot 11$.

copeland 2016-02-26 21:07:33
Then, we have, for $1 \le k \le 5$:
\[\begin{array}{r||c|c|c|c|c}
k & 1 & 2 & 3 & 4 & 5 & \\ \hline
12k^2 + 2 & 14 & 50 & 110 & 194 & 302
\end{array}\]

copeland 2016-02-26 21:07:36
Can we see any combination of these that multiply to give 77000?

summitwei 2016-02-26 21:08:34
You need one power of 1, one of 2, and one of 3, because $(12*1^2+2)(12*2^2+2)(12*3^2+2)=77000

mathwizard888 2016-02-26 21:08:34
14*50*110

summitwei 2016-02-26 21:08:34
14*50*110=77000

ninjataco 2016-02-26 21:08:34
14*50*110

agbdmrbirdyface 2016-02-26 21:08:34
110, 14, and 50

ein 2016-02-26 21:08:34
14, 50. 110

sillyd 2016-02-26 21:08:34
14*50*110=77,000

illogical_21 2016-02-26 21:08:34
110,14,50

LetThePlatPlay 2016-02-26 21:08:34
14*50*110?

yrnsmurf 2016-02-26 21:08:34
110 and 50 and 14

illogical_21 2016-02-26 21:08:34
110*14*50=77000

Jayjayliu 2016-02-26 21:08:34
1,2,3

SK200 2016-02-26 21:08:34
14*50*110

techguy2 2016-02-26 21:08:34
14*50*110

thequantumguy 2016-02-26 21:08:34
14,15,110

copeland 2016-02-26 21:08:36
$k=1$ is the only one so far with a factor of 7, and $k=3$ is the only one with a factor of 11, so we'll probably need those.

copeland 2016-02-26 21:08:37
And there's not enough 5's in 110, but we can't take any more 11's, so we'll need a $k=2$ term too.

copeland 2016-02-26 21:08:38
And that's enough! $14 \cdot 50 \cdot 110 = 77000$.

copeland 2016-02-26 21:08:39
So what have we concluded?

sillyd 2016-02-26 21:10:11
m=2^3 * 3^2 * 5

yrnsmurf 2016-02-26 21:10:11
$k^1l^2m^3$

thequantumguy 2016-02-26 21:10:11
theres a 3,2,1 in the exponents so u put the larger exponents on the smaller bases 2, 3, 5

copeland 2016-02-26 21:10:14
We use one prime with $k=1$, one prime with $k=2$, and one prime with $k=3$.

copeland 2016-02-26 21:10:17
That is, if $m$ is of the form $pq^2r^3$, then the number of quadruples $(a,b,c,d)$ with the desired property is exactly 77,000.

copeland 2016-02-26 21:10:18
What's the smallest this can be?

agbdmrbirdyface 2016-02-26 21:11:11
so then it has to be 2^3 3^2 * 5

Superwiz 2016-02-26 21:11:11
D

LetThePlatPlay 2016-02-26 21:11:11
D

swagger21 2016-02-26 21:11:11
360

weeshree 2016-02-26 21:11:11
360

copeland 2016-02-26 21:11:14
We want the higher powers to be as small a prime as possible.

copeland 2016-02-26 21:11:15
So we want $m = 2^3 \cdot 3^2 \cdot 5 = 8 \cdot 9 \cdot 5 = 360$.

copeland 2016-02-26 21:11:23
In the original problem, we get $n = 360 \cdot 77 = \boxed{27720}$. Answer (D).

copeland 2016-02-26 21:11:31
Alright, it looks like we have, um, 54 points.

copeland 2016-02-26 21:11:39
Anybody want to give up now?

weeshree 2016-02-26 21:11:49
60!

agbdmrbirdyface 2016-02-26 21:11:49
25 lezzgo

W.Sun 2016-02-26 21:11:49
NO

agbdmrbirdyface 2016-02-26 21:11:49
no

temp8909 2016-02-26 21:11:49
no!

eveningstarandlion 2016-02-26 21:11:49
That's not AIME qualification, so no

liuh008 2016-02-26 21:11:49
NEVER

agbdmrbirdyface 2016-02-26 21:11:49
not in the slightest

copeland 2016-02-26 21:12:01
Alright, let's rock this so-called number 25.

copeland 2016-02-26 21:12:04
25. The sequence $(a_n)$ is defined recursively by $a_0 = 1$, $a_1 = \sqrt[19]{2}$, and $a_n = a_{n-1}a_{n-2}^2$ for $n \ge 2$. What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
$\phantom{12B:25}$
(A) 17 (B) 18 (C) 19 (D) 20 (E) 21

copeland 2016-02-26 21:12:06
19th roots, ick. What makes them go away?

copeland 2016-02-26 21:13:18
Exponentiation does make them go away.

copeland 2016-02-26 21:13:19
What else?

copeland 2016-02-26 21:13:31
We can't take the 19th power since we always get integers that way.

mathwizard888 2016-02-26 21:13:40
take log2, becomes linear recurrence

LetThePlatPlay 2016-02-26 21:13:40
just express each term as 2^(1/19)^n

temp8909 2016-02-26 21:13:43
logarithms!

yrnsmurf 2016-02-26 21:13:43
Logarithms

copeland 2016-02-26 21:13:46
Let's take the logarithm base 2 of everything.

copeland 2016-02-26 21:13:47
That is, let's define $b_n = \log_2 a_n$.

copeland 2016-02-26 21:13:50
How does this rephrase the problem?

agbdmrbirdyface 2016-02-26 21:14:54
a0 = 0, a1 = 1/19

temp8909 2016-02-26 21:14:54
$b_0=1,b_1=\frac{1}{19},\text{ and }b_n=b_{n-1}+2b_{n-2}$

ninjataco 2016-02-26 21:14:54
b_1 + b_2 + ... + b_k is multiple of 19

agbdmrbirdyface 2016-02-26 21:14:54
and bn = bn-1 + 2bn-2

weeshree 2016-02-26 21:14:57
an = a_n-1 + 2a_n-2

copeland 2016-02-26 21:14:59
Now multiplication is addition!

copeland 2016-02-26 21:15:05
And our problem looks more like:

copeland 2016-02-26 21:15:06
25X. The sequence $(b_n)$ is defined recursively by $b_0 = 0$, $b_1 = \dfrac{1}{19}$, and $b_n = b_{n-1} + 2b_{n-2}$ for $n \ge 2$. What is the smallest positive integer $k$ such that the sum $b_1 + b_2 + \cdots + b_k$ is an integer?
$\phantom{12B:25X}$
(A) 17 (B) 18 (C) 19 (D) 20 (E) 21

copeland 2016-02-26 21:15:10
Note that the product of the $a$'s in the original problem is an integer if and only if the sum of the $b$'s in the new problem is an integer.

copeland 2016-02-26 21:15:12
What else would simplify things?

yrnsmurf 2016-02-26 21:15:48
Multiplying by 19

yrnsmurf 2016-02-26 21:15:48
and also mods

TheRealDeal 2016-02-26 21:15:48
multiply by 19

MSTang 2016-02-26 21:15:48
Set $c_k = 19b_k$

summitwei 2016-02-26 21:15:48
Let $b_1=x$ and look for a multiple of 19

copeland 2016-02-26 21:15:50
Everything in sight is a multiple of $\frac{1}{19}$. So we could multiply by 19, and it's necessary and sufficient for the sum to be a multiple of 19.

copeland 2016-02-26 21:15:51
That is, we can change the problem again:

copeland 2016-02-26 21:15:52
25Y. The sequence $(c_n)$ is defined recursively by $c_0 = 0$, $c_1 = 1$, and $c_n = c_{n-1} + 2c_{n-2}$ for $n \ge 2$. What is the smallest positive integer $k$ such that the sum $c_1 + c_2 + \cdots + c_k \equiv 0 \pmod{19}$?
$\phantom{12B:25Y}$
(A) 17 (B) 18 (C) 19 (D) 20 (E) 21

copeland 2016-02-26 21:15:53
Seems simple enough now.

copeland 2016-02-26 21:15:59
Now what should we do?

LetThePlatPlay 2016-02-26 21:16:43
just start plugging in

summitwei 2016-02-26 21:16:43
Start listing out some numbers

weeshree 2016-02-26 21:16:43
bash out!

TheRealDeal 2016-02-26 21:16:43
plug in numbers up to 21 XD

copeland 2016-02-26 21:16:45
Yeah, wait, those answers are really small. Let's just start computing. . .

copeland 2016-02-26 21:16:54
What do you get?

copeland 2016-02-26 21:16:56
Indeed, here's the table mod 19:
\[\begin{array}{r|c|c}
k & c_k & c_1 + \cdots + c_k \\ \hline
1 & 1 & 1 \\ \hline
2 & 1 & 2 \\ \hline
3 & 3 & 5 \\ \hline
4 & 5 & 10 \\ \hline
5 & 11 & 2 \\ \hline
6 & 2 & 4 \\ \hline
7 & 5 & 9 \\ \hline
8 & 9 & 18 \\ \hline
9 & 0 & 18 \\ \hline
10 & 18 & 17 \\ \hline
11 & 18 & 16 \\ \hline
12 & 16 & 13 \\ \hline
13 & 14 & 8 \\ \hline
14 & 8 & 16 \\ \hline
15 & 17 & 14 \\ \hline
16 & 14 & 9 \\ \hline
17 & 10 & 0
\end{array}\]

copeland 2016-02-26 21:17:09
So. . .

temp8909 2016-02-26 21:17:54
17

math101010 2016-02-26 21:17:54
17 yay

thequantumguy 2016-02-26 21:17:54
17 WORKS!!!!

SimonSun 2016-02-26 21:17:54
17 A

agbdmrbirdyface 2016-02-26 21:17:54
oh jeez copeland

lucasxia01 2016-02-26 21:17:54
17 !

SineWarrior 2016-02-26 21:17:54
so 17 is the answer?

ninjataco 2016-02-26 21:17:54
(A)

illogical_21 2016-02-26 21:17:54
17

techguy2 2016-02-26 21:17:54
17!

baseballcat 2016-02-26 21:17:54
A?

MSTang 2016-02-26 21:17:54
A, gg.

agbdmrbirdyface 2016-02-26 21:17:54
A.... but that seems kinda cheap

copeland 2016-02-26 21:17:57
We see that $k = \boxed{17}$ is the first time the right column in 0. Answer (A).

copeland 2016-02-26 21:18:03
Yeah, huh?

copeland 2016-02-26 21:18:18
We came up with a few ways to make this problem harder, but they just made it harder. . .

summitwei 2016-02-26 21:18:26
Also, $c_9=-1$ and $c_10=-2$, so you can simplify some of the calculation

copeland 2016-02-26 21:18:43
Yes, in fact, when I solved this problem I used negative numbers liberally.

techguy2 2016-02-26 21:18:47
wow, were done with.... 60 points! wow. not even honor roll.

copeland 2016-02-26 21:18:50
Killer.

copeland 2016-02-26 21:18:55
Oh well. . . maybe next year.

copeland 2016-02-26 21:19:39
So, that's it for the 2016 AMCs.

copeland 2016-02-26 21:19:49
But we'll be back!

copeland 2016-02-26 21:19:51
AoPS will hold a Math Jam after each AIME, where we'll discuss all 15 problems from the contest. The schedule is:
AIME I Math Jam: Saturday, March 5, 7 PM Eastern
AIME II Math Jam: Friday, March 18, 7 PM Eastern
We hope to see you there!

copeland 2016-02-26 21:20:43
Oh, also: AoPS is running our weekend Special AIME Problem Seminar this upcoming weekend, Feb 27 and 28, from 3:30-6:30 PM Eastern each day. More information is at http://artofproblemsolving.com/school/course/catalog/maa-aime-special.

copeland 2016-02-26 21:20:52
That is tomorrow.

SK200 2016-02-26 21:20:56
are all math jams at 7pm eastern?

copeland 2016-02-26 21:21:00
No, most are 7:30.

copeland 2016-02-26 21:21:06
The AMC and AIME ones start early.

copeland 2016-02-26 21:21:12
I'll be closing the room shortly. Thanks for coming!

thequantumguy 2016-02-26 21:21:47
goodbye

math101010 2016-02-26 21:21:47
Thank you!

SimonSun 2016-02-26 21:21:47
Thks

jenmath888 2016-02-26 21:21:47
Thanks.

ein 2016-02-26 21:21:47
Thanks!

agbdmrbirdyface 2016-02-26 21:21:47
seeya copeland

SimonSun 2016-02-26 21:21:47
K bye

DavidUsa 2016-02-26 21:21:47
THANK YOU

copeland 2016-02-26 21:21:48
Bye!

2016 AMC 10/12 B Discussion

Facilitator: Jeremy Copeland