2017 AIME I Discussion
Go back to the Math Jam ArchiveAoPS Instructors discuss all 15 problems on the 2017 AIME I.
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Facilitator: Jeremy Copeland
copeland
2017-03-09 19:02:47
Welcome to the 2017 AIME I Math Jam!
Welcome to the 2017 AIME I Math Jam!
copeland
2017-03-09 19:02:48
I'm Jeremy Copeland, and I'll be leading our discussion tonight.
I'm Jeremy Copeland, and I'll be leading our discussion tonight.
copeland
2017-03-09 19:02:56
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.
copeland
2017-03-09 19:02:58
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.
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
2017-03-09 19:03:04
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.
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
2017-03-09 19:03:17
Who doesn't want to be kept organized, amirite?
Who doesn't want to be kept organized, amirite?
copeland
2017-03-09 19:03:24
Notice that this is a lot like one of our classes except there are a lot more of you and the same number of me. I won't be able to post all of your comments all of the time. It's not personal! Here's a secret though: I prefer to pass clear, well-written comments to the room.
Notice that this is a lot like one of our classes except there are a lot more of you and the same number of me. I won't be able to post all of your comments all of the time. It's not personal! Here's a secret though: I prefer to pass clear, well-written comments to the room.
copeland
2017-03-09 19:03:34
And of course, a lot of mathematics appears on the AIME. We don't have the chance to teach all the techniques, but you should take notes of the things you think are cool and want to research more and look into those things later. There are some great conversations going on about this year's AIME over in our community:
https://artofproblemsolving.com/community/c5t183f5_aime
If you want to know more about any of the problems, try there.
And of course, a lot of mathematics appears on the AIME. We don't have the chance to teach all the techniques, but you should take notes of the things you think are cool and want to research more and look into those things later. There are some great conversations going on about this year's AIME over in our community:
https://artofproblemsolving.com/community/c5t183f5_aime
If you want to know more about any of the problems, try there.
copeland
2017-03-09 19:03:43
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the necessary 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 always to try do so in our regular online 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!
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the necessary 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 always to try do so in our regular online 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
2017-03-09 19:03:54
We do have two teaching assistants with us tonight to help answer your questions:
We do have two teaching assistants with us tonight to help answer your questions:
copeland
2017-03-09 19:03:55
Henrik Boecken (henrikjb): Henrik is an undergraduate at the Massachusetts Institute of Technology studying economics and, of course, math. After college, he plans on teaching at his old high school for a few years. Math contests were the backbone of his high school career, and now he hopes to give back. In his free time, Henrik enjoys Ultimate Frisbee, card games, and reading the Japanese manga One Piece.
Henrik Boecken (henrikjb): Henrik is an undergraduate at the Massachusetts Institute of Technology studying economics and, of course, math. After college, he plans on teaching at his old high school for a few years. Math contests were the backbone of his high school career, and now he hopes to give back. In his free time, Henrik enjoys Ultimate Frisbee, card games, and reading the Japanese manga One Piece.
copeland
2017-03-09 19:03:56
William Wang (willwang123): William is a freshman at the University of Pennsylvania. He is a 4-time USA(J)MO qualifier and a USA Physics Olympiad qualifier. In his spare time, he enjoys playing tennis, clarinet, and StarCraft.
William Wang (willwang123): William is a freshman at the University of Pennsylvania. He is a 4-time USA(J)MO qualifier and a USA Physics Olympiad qualifier. In his spare time, he enjoys playing tennis, clarinet, and StarCraft.
henrikjb
2017-03-09 19:04:03
Hi guys!
Hi guys!
willwang123
2017-03-09 19:04:04
Hi everyone!
Hi everyone!
zac15SCASD
2017-03-09 19:04:21
PA ftw!!!
PA ftw!!!
quanhui868
2017-03-09 19:04:21
Hi
Hi
Radio2
2017-03-09 19:04:21
Hi!
Hi!
tdeng
2017-03-09 19:04:21
Hey!
Hey!
swagger
2017-03-09 19:04:21
hi
hi
samuel
2017-03-09 19:04:21
Hi!
Hi!
GeneralCobra19
2017-03-09 19:04:21
Hi. Glad to be here.
Hi. Glad to be here.
Liopleurodon
2017-03-09 19:04:21
Hello!
Hello!
MrMXS
2017-03-09 19:04:21
hello henrikjb and willwang123
hello henrikjb and willwang123
copeland
2017-03-09 19:04:24
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the 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.
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the 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
2017-03-09 19:04:30
Please also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest. 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 acknowledged.
Please also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest. 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 acknowledged.
copeland
2017-03-09 19:04:42
Before we get started, here's a question for those of you who took the test: What was your favorite question on the test?
Before we get started, here's a question for those of you who took the test: What was your favorite question on the test?
MSTang
2017-03-09 19:05:21
13
13
swagger
2017-03-09 19:05:21
#2
#2
EasyAs_Pi
2017-03-09 19:05:21
question number 16!
question number 16!
pieater314159
2017-03-09 19:05:21
Problem 14. I like logarithms.
Problem 14. I like logarithms.
samuel
2017-03-09 19:05:21
#15
#15
linqaszayi
2017-03-09 19:05:21
#14 maybe
#14 maybe
zihang
2017-03-09 19:05:21
#11
#11
mossie
2017-03-09 19:05:21
The 13th!
The 13th!
GeronimoStilton
2017-03-09 19:05:21
Problem 13, although I didn't get to it in time.
Problem 13, although I didn't get to it in time.
amzhao
2017-03-09 19:05:21
2, it was pretty simple
2, it was pretty simple
ilovemath04
2017-03-09 19:05:21
#1
#1
NewbieGamer
2017-03-09 19:05:21
#13
#13
hjia17
2017-03-09 19:05:21
1
1
reelmathematician
2017-03-09 19:05:21
Problem 15 because 15 is prime. jk
Problem 15 because 15 is prime. jk
ethanliu247
2017-03-09 19:05:21
number 8,even if i got it wrong
number 8,even if i got it wrong
Jyzhang12
2017-03-09 19:05:21
1
1
copeland
2017-03-09 19:05:27
Two of you said 14, huh?
Two of you said 14, huh?
copeland
2017-03-09 19:05:29
OK, then.
OK, then.
copeland
2017-03-09 19:05:32
Let's get started! We're going to work through all 15 problems from the 2017 AIME I, in order.
Let's get started! We're going to work through all 15 problems from the 2017 AIME I, in order.
copeland
2017-03-09 19:05:37
1. Fifteen distinct points are designated on $\triangle ABC{:}$ the 3 vertices $A,$ $B,$ and $C;$ 3 other point on side $\overline{AB};$ 4 other points on side $\overline{BC};$ and 5 other points on side $\overline{CA}$. Find the number of triangles with positive area whose vertices are among these 15 points.
1. Fifteen distinct points are designated on $\triangle ABC{:}$ the 3 vertices $A,$ $B,$ and $C;$ 3 other point on side $\overline{AB};$ 4 other points on side $\overline{BC};$ and 5 other points on side $\overline{CA}$. Find the number of triangles with positive area whose vertices are among these 15 points.
copeland
2017-03-09 19:05:43
I lorve semicolons.
I lorve semicolons.
tdeng
2017-03-09 19:06:06
Lorve?
Lorve?
Smileyklaws
2017-03-09 19:06:06
lorve
lorve
Liopleurodon
2017-03-09 19:06:06
lorve
lorve
copeland
2017-03-09 19:06:07
It's French.
It's French.
copeland
2017-03-09 19:06:17
(A romance language.)
(A romance language.)
copeland
2017-03-09 19:06:26
What's the trick here?
What's the trick here?
celestialphoenix3768
2017-03-09 19:06:51
complementary counting
complementary counting
GeneralCobra19
2017-03-09 19:06:51
Complementary Counting!
Complementary Counting!
SomethingNeutral
2017-03-09 19:06:51
complementary counting
complementary counting
strategos21
2017-03-09 19:06:51
Complimentary counting
Complimentary counting
ilikepie2003
2017-03-09 19:06:51
use complementary counting!
use complementary counting!
espeon12
2017-03-09 19:06:51
complementary counting
complementary counting
sxu
2017-03-09 19:06:51
complementary counting: subtract degenerate triangles from total
complementary counting: subtract degenerate triangles from total
copeland
2017-03-09 19:06:54
If we want to see how many triangles there are with positive area we can find the number of triangles that have area zero and subtract.
If we want to see how many triangles there are with positive area we can find the number of triangles that have area zero and subtract.
richuw
2017-03-09 19:07:09
Draw a diagram!
Draw a diagram!
oiler8
2017-03-09 19:07:09
A diagram
A diagram
copeland
2017-03-09 19:07:11
copeland
2017-03-09 19:07:13
How many total triples are there?
How many total triples are there?
MaxTplusAMSP
2017-03-09 19:07:37
15 choose 3
15 choose 3
espeon12
2017-03-09 19:07:37
15C3
15C3
EulerMacaroni
2017-03-09 19:07:37
$\binom{15}{3}$
$\binom{15}{3}$
DemonPlat4
2017-03-09 19:07:37
15 choose 3
15 choose 3
islander7
2017-03-09 19:07:37
15C3
15C3
ninjataco
2017-03-09 19:07:37
15C3
15C3
Deathranger999
2017-03-09 19:07:37
15 choose 3
15 choose 3
W.Sun
2017-03-09 19:07:37
15 choose 3
15 choose 3
letsgomath
2017-03-09 19:07:37
15 choose 3 = 455
15 choose 3 = 455
fdas
2017-03-09 19:07:37
15 choose 3
15 choose 3
gabrielsui
2017-03-09 19:07:37
15 nCr 3
15 nCr 3
copeland
2017-03-09 19:07:40
There are $\dbinom{15}3$ total triples of points.
There are $\dbinom{15}3$ total triples of points.
copeland
2017-03-09 19:07:41
How many of those points give us "triangles" of area zero?
How many of those points give us "triangles" of area zero?
mossie
2017-03-09 19:08:19
$\binom{15}{3}-\binom{7}{3}-\binom{6}{3}-\binom{5}{3}=390$
$\binom{15}{3}-\binom{7}{3}-\binom{6}{3}-\binom{5}{3}=390$
zihang
2017-03-09 19:08:19
7C3+6C3+5C3
7C3+6C3+5C3
sxu
2017-03-09 19:08:19
5C3+6C3+7C3
5C3+6C3+7C3
abvenkgoo
2017-03-09 19:08:19
7C3+6C3+5C3
7C3+6C3+5C3
EulerMacaroni
2017-03-09 19:08:19
$\binom{5}{3}+\binom{6}{3}+\binom{7}{3}$
$\binom{5}{3}+\binom{6}{3}+\binom{7}{3}$
jonzli123
2017-03-09 19:08:19
5C3 + 6C3 + 7C3
5C3 + 6C3 + 7C3
vvluo
2017-03-09 19:08:19
7 choose 3 plus 5 choose 3 plus 6 choose 3
7 choose 3 plus 5 choose 3 plus 6 choose 3
tdeng
2017-03-09 19:08:19
$\binom{5}{3}+\binom{6}{3}+\binom{7}{3}$
$\binom{5}{3}+\binom{6}{3}+\binom{7}{3}$
cwechsz
2017-03-09 19:08:19
6C3 + 5C3 + 7C3
6C3 + 5C3 + 7C3
copeland
2017-03-09 19:08:22
The points must be collinear to have area zero. There are $\dbinom53$ triples along $\overline{AB}$, $\dbinom63$ triples along $\overline{BC},$ and $\dbinom73$ triples along $\overline{CA}$.
The points must be collinear to have area zero. There are $\dbinom53$ triples along $\overline{AB}$, $\dbinom63$ triples along $\overline{BC},$ and $\dbinom73$ triples along $\overline{CA}$.
copeland
2017-03-09 19:08:22
And the final answer?
And the final answer?
math101010
2017-03-09 19:08:40
390
390
Pandasareamazing.
2017-03-09 19:08:40
390
390
nukelauncher
2017-03-09 19:08:40
390!
390!
fireflame241
2017-03-09 19:08:40
390
390
DarthRen
2017-03-09 19:08:40
390
390
dhruv
2017-03-09 19:08:40
390
390
Mrkiller
2017-03-09 19:08:40
390
390
jonzli123
2017-03-09 19:08:40
390!!!!!!
390!!!!!!
ethanliu247
2017-03-09 19:08:40
390
390
copeland
2017-03-09 19:08:42
There are
\begin{align*}
\binom{15}3-\binom53-\binom63-\binom73
&=5\cdot7\cdot13-5\cdot2-5\cdot4-7\cdot5\\
&=5(91-2-4-7)\\
&=5\cdot78=\boxed{390}
\end{align*} total triangles of positive area.
There are
\begin{align*}
\binom{15}3-\binom53-\binom63-\binom73
&=5\cdot7\cdot13-5\cdot2-5\cdot4-7\cdot5\\
&=5(91-2-4-7)\\
&=5\cdot78=\boxed{390}
\end{align*} total triangles of positive area.
copeland
2017-03-09 19:08:44
Great.
Great.
copeland
2017-03-09 19:08:47
Everyone warmed up?
Everyone warmed up?
First
2017-03-09 19:09:01
Yes!
Yes!
BLCRAFT
2017-03-09 19:09:01
ya...
ya...
SomethingNeutral
2017-03-09 19:09:01
yes
yes
Dude03
2017-03-09 19:09:01
yes
yes
Jyzhang12
2017-03-09 19:09:01
yea
yea
MountainHeight
2017-03-09 19:09:01
yes
yes
cooleybz2013
2017-03-09 19:09:01
yep
yep
lsh0589
2017-03-09 19:09:01
yeah
yeah
EasyAs_Pi
2017-03-09 19:09:01
Yes!
Yes!
tdeng
2017-03-09 19:09:01
Yeah!
Yeah!
copeland
2017-03-09 19:09:02
Cool.
Cool.
copeland
2017-03-09 19:09:14
Incidentally, we used the same trick on the AMC10A this year:
Incidentally, we used the same trick on the AMC10A this year:
copeland
2017-03-09 19:09:19
23. How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?
23. How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?
copeland
2017-03-09 19:09:29
(We're not solving that problem.)
(We're not solving that problem.)
copeland
2017-03-09 19:09:35
We are going to solve. . .
We are going to solve. . .
ethanliu247
2017-03-09 19:09:42
THisone
THisone
SomethingNeutral
2017-03-09 19:09:42
I remember that one...
I remember that one...
lego101
2017-03-09 19:09:42
oh i remember that question -_-
oh i remember that question -_-
gabrielsui
2017-03-09 19:09:42
i was there!
i was there!
Funnybunny5246
2017-03-09 19:09:48
On to #2
On to #2
Liopleurodon
2017-03-09 19:09:48
NUMBER 2
NUMBER 2
W.Sun
2017-03-09 19:09:48
Number 2!
Number 2!
copeland
2017-03-09 19:09:56
2. When each of $702,$ $787,$ and $855$ is divided by the positive integer $m,$ the remainder is always the positive integer $r.$ When each of $412,$ $722,$ and $815$ is divided by the positive integer $n,$ the remainder is always the positive integer $s\neq r.$ Find $m+n+r+s.$
2. When each of $702,$ $787,$ and $855$ is divided by the positive integer $m,$ the remainder is always the positive integer $r.$ When each of $412,$ $722,$ and $815$ is divided by the positive integer $n,$ the remainder is always the positive integer $s\neq r.$ Find $m+n+r+s.$
copeland
2017-03-09 19:10:00
Where should we start?
Where should we start?
noodlemaster
2017-03-09 19:10:37
analyze differences
analyze differences
quanhui868
2017-03-09 19:10:37
find the difference between the numbers
find the difference between the numbers
jonzli123
2017-03-09 19:10:37
find the difference between the numbers
find the difference between the numbers
DemonPlat4
2017-03-09 19:10:37
differences of the numbers
differences of the numbers
smartpgp
2017-03-09 19:10:37
find the gcf of the differences
find the gcf of the differences
Liopleurodon
2017-03-09 19:10:37
subtract each number from each other, and find their GCD
subtract each number from each other, and find their GCD
rapturt9
2017-03-09 19:10:37
subtract each pair of the 3 numbers
subtract each pair of the 3 numbers
vvluo
2017-03-09 19:10:37
differences
differences
copeland
2017-03-09 19:10:40
If two numbers have the same remainder when divided by $m$ then their difference is a multiple of $m$. Here we have
If two numbers have the same remainder when divided by $m$ then their difference is a multiple of $m$. Here we have
copeland
2017-03-09 19:10:41
\begin{align*}
702&=am+r\\
787&=bm+r\\
855&=cm+r
\end{align*}
\begin{align*}
702&=am+r\\
787&=bm+r\\
855&=cm+r
\end{align*}
copeland
2017-03-09 19:10:42
When we subtract them pairwise we get
\begin{align*}
m&\mid 787-702=85\\
m&\mid 855-787=68\\
\end{align*}
When we subtract them pairwise we get
\begin{align*}
m&\mid 787-702=85\\
m&\mid 855-787=68\\
\end{align*}
copeland
2017-03-09 19:10:47
[Remember that we say $m\mid n$ if $n$ is a multiple of $m$, so $3\mid 6$ and $71\mid71$ and $431\mid0.$]
[Remember that we say $m\mid n$ if $n$ is a multiple of $m$, so $3\mid 6$ and $71\mid71$ and $431\mid0.$]
copeland
2017-03-09 19:10:53
So what do we know about $m?$
So what do we know about $m?$
tdeng
2017-03-09 19:11:39
Only common factor is 17
Only common factor is 17
MSTang
2017-03-09 19:11:39
m = 17
m = 17
legolego
2017-03-09 19:11:39
factor of 17
factor of 17
espeon12
2017-03-09 19:11:39
it's 17
it's 17
swagger
2017-03-09 19:11:39
17
17
nukelauncher
2017-03-09 19:11:39
m ≠ 1, so m = 17
m ≠ 1, so m = 17
DarthRen
2017-03-09 19:11:39
m|17
m|17
Liopleurodon
2017-03-09 19:11:39
m is larger than 1
m is larger than 1
MrMXS
2017-03-09 19:11:39
$m=17$
$m=17$
Pandasareamazing.
2017-03-09 19:11:39
so it is 17
so it is 17
amzhao
2017-03-09 19:11:39
m is 17
m is 17
MathTechFire
2017-03-09 19:11:39
it is 17
it is 17
copeland
2017-03-09 19:11:42
Since $m$ divides both of these numbers, it divides the GCD, which is 17. (Nicely, the first step in the Euclidean Algorithm here is $85-68=17$.)
Since $m$ divides both of these numbers, it divides the GCD, which is 17. (Nicely, the first step in the Euclidean Algorithm here is $85-68=17$.)
copeland
2017-03-09 19:11:44
In order for $r$ to be positive we need $m>1,$ so $m=17$.
In order for $r$ to be positive we need $m>1,$ so $m=17$.
copeland
2017-03-09 19:11:45
So what is $r?$
So what is $r?$
tdeng
2017-03-09 19:12:06
5
5
swagger
2017-03-09 19:12:06
5
5
letsgomath
2017-03-09 19:12:06
5
5
legolego
2017-03-09 19:12:06
5
5
ilovemath04
2017-03-09 19:12:06
r=5
r=5
Funnybunny5246
2017-03-09 19:12:06
It is 5
It is 5
copeland
2017-03-09 19:12:08
I like looking at 855. Since
\[855=850+5=17\cdot50+5,\] we see that $r=5$.
I like looking at 855. Since
\[855=850+5=17\cdot50+5,\] we see that $r=5$.
copeland
2017-03-09 19:12:14
OK, now for the other three numbers. Does anyone see what $n$ is?
OK, now for the other three numbers. Does anyone see what $n$ is?
Radio2
2017-03-09 19:12:37
31
31
GeronimoStilton
2017-03-09 19:12:37
$31$
$31$
mshanmugam
2017-03-09 19:12:37
31
31
Radio2
2017-03-09 19:12:37
n=31
n=31
SomethingNeutral
2017-03-09 19:12:37
n is 31
n is 31
strategos21
2017-03-09 19:12:37
31
31
skiboy32
2017-03-09 19:12:37
31
31
copeland
2017-03-09 19:12:39
Subtracting pairwise again we get
\begin{align*}
n\mid722-412&=310\\
n\mid 815-722&=93.
\end{align*}
Subtracting pairwise again we get
\begin{align*}
n\mid722-412&=310\\
n\mid 815-722&=93.
\end{align*}
copeland
2017-03-09 19:12:47
Both of those numbers are obviously divisible by 31: we have $310=31\cdot10$ and $93=31\cdot3$. Since the quotients are coprime, the GCD is $n=31$.
Both of those numbers are obviously divisible by 31: we have $310=31\cdot10$ and $93=31\cdot3$. Since the quotients are coprime, the GCD is $n=31$.
copeland
2017-03-09 19:12:50
And what is $s?$
And what is $s?$
quanhui868
2017-03-09 19:13:26
9
9
stan23456
2017-03-09 19:13:26
9
9
rapturt9
2017-03-09 19:13:26
9
9
MrMXS
2017-03-09 19:13:26
$9$
$9$
DemonPlat4
2017-03-09 19:13:26
9
9
cooleybz2013
2017-03-09 19:13:26
9
9
SomethingNeutral
2017-03-09 19:13:26
actually 9
actually 9
copeland
2017-03-09 19:13:29
Let's start with 722. We know 620 is a multiple of 31, so if we start by subtracting 620 we get
\[722\equiv 102\equiv71\equiv40\equiv9\pmod{31}.\]
Let's start with 722. We know 620 is a multiple of 31, so if we start by subtracting 620 we get
\[722\equiv 102\equiv71\equiv40\equiv9\pmod{31}.\]
copeland
2017-03-09 19:13:30
And the answer?
And the answer?
nukelauncher
2017-03-09 19:13:54
062
062
math_noob
2017-03-09 19:13:54
62
62
samuel
2017-03-09 19:13:54
062
062
ijava
2017-03-09 19:13:54
62
62
ilikepie2003
2017-03-09 19:13:54
62
62
Pandasareamazing.
2017-03-09 19:13:54
62=31+9+17+5
62=31+9+17+5
ilikepie2003
2017-03-09 19:13:54
062
062
samuel
2017-03-09 19:13:54
$\boxed{062}$
$\boxed{062}$
DemonPlat4
2017-03-09 19:13:54
31+9+17+5 = 062
31+9+17+5 = 062
zihang
2017-03-09 19:13:54
62
62
thinkinavi
2017-03-09 19:13:54
062
062
copeland
2017-03-09 19:13:56
\[m+r+n+s=17+5+31+9=\boxed{062}.\]
\[m+r+n+s=17+5+31+9=\boxed{062}.\]
copeland
2017-03-09 19:14:01
I haven't figured out what $s\neq r$ does in this problem. My guess is that in the original version of the problem one of those GCDs was not prime and there was an extra step. If so, then they lightened the problem up by making them both prime and the $s\neq r$ constraint is vestigial.
I haven't figured out what $s\neq r$ does in this problem. My guess is that in the original version of the problem one of those GCDs was not prime and there was an extra step. If so, then they lightened the problem up by making them both prime and the $s\neq r$ constraint is vestigial.
copeland
2017-03-09 19:14:08
I once knew this kid, Willie, who had 11 toes. Who's ready for the next problem?
I once knew this kid, Willie, who had 11 toes. Who's ready for the next problem?
thomas9549
2017-03-09 19:14:26
i am
i am
jam10307
2017-03-09 19:14:26
Me1!!!
Me1!!!
swagger
2017-03-09 19:14:26
me!
me!
Pandasareamazing.
2017-03-09 19:14:26
me!!
me!!
W.Sun
2017-03-09 19:14:26
ME
ME
lego101
2017-03-09 19:14:26
okay! me!
okay! me!
W.Sun
2017-03-09 19:14:26
Moiii
Moiii
GeronimoStilton
2017-03-09 19:14:26
Me!
Me!
amackenzie1
2017-03-09 19:14:26
I am!
I am!
copeland
2017-03-09 19:14:28
3. For a positive integer $n,$ let $d_n$ be the units digit of \[1+2+3+\cdots+n.\]Find the remainder when \[\sum_{n=1}^{2017}d_n\] is divided by 1000.
3. For a positive integer $n,$ let $d_n$ be the units digit of \[1+2+3+\cdots+n.\]Find the remainder when \[\sum_{n=1}^{2017}d_n\] is divided by 1000.
copeland
2017-03-09 19:14:37
Raise your hand if your first answer was way too big on this problem.
Raise your hand if your first answer was way too big on this problem.
ilovetynker
2017-03-09 19:15:18
me!
me!
First
2017-03-09 19:15:18
Me!
Me!
BLCRAFT
2017-03-09 19:15:18
*raise*
*raise*
lego101
2017-03-09 19:15:18
*raises hand emphatically*
*raises hand emphatically*
jam10307
2017-03-09 19:15:18
RAISE
RAISE
W.Sun
2017-03-09 19:15:18
Me lol
Me lol
quanhui868
2017-03-09 19:15:18
*raises hand*
*raises hand*
sxu
2017-03-09 19:15:18
*raises hand
*raises hand
jonzli123
2017-03-09 19:15:18
*raises hand*
*raises hand*
thomas9549
2017-03-09 19:15:18
me!
me!
ilovetynker
2017-03-09 19:15:18
me
me
W.Sun
2017-03-09 19:15:18
*raises hand*
*raises hand*
cooleybz2013
2017-03-09 19:15:18
me
me
jack74
2017-03-09 19:15:18
hand raise
hand raise
samuel
2017-03-09 19:15:18
me me me
me me me
strategos21
2017-03-09 19:15:20
my hand is raised
my hand is raised
copeland
2017-03-09 19:15:21
Yeah, me too. I had that, "Jeremy, what are you doing?" moment later than I should admit.
Yeah, me too. I had that, "Jeremy, what are you doing?" moment later than I should admit.
copeland
2017-03-09 19:15:25
Every $d_n$ is a digit. That's important. Half of the mistakes on this problem are from forgetting that. We're adding 2017 digits. If you ever add a number that's more than 9, you're doing it wrong.
Every $d_n$ is a digit. That's important. Half of the mistakes on this problem are from forgetting that. We're adding 2017 digits. If you ever add a number that's more than 9, you're doing it wrong.
copeland
2017-03-09 19:15:35
So we want the units digits of the triangle numbers. I don't think I've ever seen that. But what do you expect?
So we want the units digits of the triangle numbers. I don't think I've ever seen that. But what do you expect?
nukelauncher
2017-03-09 19:16:05
theres a repeating pattern
theres a repeating pattern
ilovemath04
2017-03-09 19:16:05
a pattern
a pattern
DemonPlat4
2017-03-09 19:16:05
find the pattern
find the pattern
GeronimoStilton
2017-03-09 19:16:05
A pattern!
A pattern!
espeon12
2017-03-09 19:16:05
i expected it to be a cycle
i expected it to be a cycle
Deathranger999
2017-03-09 19:16:05
Pattern
Pattern
Liopleurodon
2017-03-09 19:16:05
A pattern
A pattern
tdeng
2017-03-09 19:16:05
It will repeat
It will repeat
gabrielsui
2017-03-09 19:16:05
pattern
pattern
Mrkiller
2017-03-09 19:16:05
It to repeat
It to repeat
dhruv
2017-03-09 19:16:05
pattern?
pattern?
Shri333
2017-03-09 19:16:05
a pattern
a pattern
abvenkgoo
2017-03-09 19:16:05
a pattern in the units digits
a pattern in the units digits
copeland
2017-03-09 19:16:08
I expect the sequence $d_n$ to be periodic. This problem is going to be really hard if it's not.
I expect the sequence $d_n$ to be periodic. This problem is going to be really hard if it's not.
copeland
2017-03-09 19:16:09
Do you see the period off-hand?
Do you see the period off-hand?
Mrkiller
2017-03-09 19:16:37
20
20
tdeng
2017-03-09 19:16:37
It has a period of 20
It has a period of 20
legolego
2017-03-09 19:16:37
period of 20
period of 20
gabrielsui
2017-03-09 19:16:37
no
no
shenmaster88
2017-03-09 19:16:37
evey 20
evey 20
atmchallenge
2017-03-09 19:16:37
no, so let's write out some terms
no, so let's write out some terms
Pandasareamazing.
2017-03-09 19:16:37
not really
not really
EulerMacaroni
2017-03-09 19:16:37
20
20
GeronimoStilton
2017-03-09 19:16:37
No. Let's write it out to find the pattern!
No. Let's write it out to find the pattern!
ilikepie2003
2017-03-09 19:16:37
nope
nope
copeland
2017-03-09 19:16:40
You might see that it's 20. Since the units digits of $n$ repeat every 10, the period has to be a multiple of 10. Since $d_0=0$ and $d_{10}=5$ is the units digit of 55, the period isn't 10. But $d_{20}=0$ is the units digit of $210$. The period is 20.
You might see that it's 20. Since the units digits of $n$ repeat every 10, the period has to be a multiple of 10. Since $d_0=0$ and $d_{10}=5$ is the units digit of 55, the period isn't 10. But $d_{20}=0$ is the units digit of $210$. The period is 20.
copeland
2017-03-09 19:16:44
You might not see that. It's not even a big deal. Why?
You might not see that. It's not even a big deal. Why?
LittleChimp
2017-03-09 19:17:35
We have to write out terms anyways
We have to write out terms anyways
jonzli123
2017-03-09 19:17:35
you write out the terms
you write out the terms
espeon12
2017-03-09 19:17:35
we can just write out some terms
we can just write out some terms
LearningMath
2017-03-09 19:17:35
Because we want the last 3 digits
Because we want the last 3 digits
Pudentane
2017-03-09 19:17:35
cause we want the sum of the units digits
cause we want the sum of the units digits
copeland
2017-03-09 19:17:37
Well, we actually need to sum these things so we need to figure out what they are. Once we list them we'll know exactly what the period is.
Well, we actually need to sum these things so we need to figure out what they are. Once we list them we'll know exactly what the period is.
copeland
2017-03-09 19:17:43
OK, so let's list them. What are the first 10? (You can even just list 10 digits in a row!)
OK, so let's list them. What are the first 10? (You can even just list 10 digits in a row!)
Liopleurodon
2017-03-09 19:18:24
$1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0$
$1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0$
SomethingNeutral
2017-03-09 19:18:24
1360518655681500
1360518655681500
EasyAs_Pi
2017-03-09 19:18:24
yes, $1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\cdots$
yes, $1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\cdots$
vishwathganesan
2017-03-09 19:18:24
1360518655
1360518655
SomethingNeutral
2017-03-09 19:18:24
1360518655
1360518655
pican
2017-03-09 19:18:24
1, 3, 6, 0, 5, 1, 8, 6, 5, 5
1, 3, 6, 0, 5, 1, 8, 6, 5, 5
GeneralCobra19
2017-03-09 19:18:24
1 3 6 0 5 1 8 6 5 5
1 3 6 0 5 1 8 6 5 5
Funnybunny5246
2017-03-09 19:18:24
1360518655
1360518655
KYang
2017-03-09 19:18:24
1 3 6 0 5 1 8 6 5 5
1 3 6 0 5 1 8 6 5 5
copeland
2017-03-09 19:18:28
\[\begin{array}{c|ccccc|ccccc}
n &1&2&3&4&5&6&7&8&9&10\\
\hline
d_n&1&3&6&0&5&1&8&6&5&5\\
\end{array}\]
\[\begin{array}{c|ccccc|ccccc}
n &1&2&3&4&5&6&7&8&9&10\\
\hline
d_n&1&3&6&0&5&1&8&6&5&5\\
\end{array}\]
copeland
2017-03-09 19:18:30
Here are the next 10:
Here are the next 10:
copeland
2017-03-09 19:18:32
\[\begin{array}{c|ccccc|ccccc|ccccc|ccccc}
n &1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20\\
\hline
d_n&1&3&6&0&5&1&8&6&5&5&6&8&1&5&0&6&3&1&0&0\\
\end{array}\]
\[\begin{array}{c|ccccc|ccccc|ccccc|ccccc}
n &1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20\\
\hline
d_n&1&3&6&0&5&1&8&6&5&5&6&8&1&5&0&6&3&1&0&0\\
\end{array}\]
copeland
2017-03-09 19:18:33
Hm, it's symmetric. We don't care (except that it helps us double-check our arithmetic).
Hm, it's symmetric. We don't care (except that it helps us double-check our arithmetic).
copeland
2017-03-09 19:18:36
What's the sum?
What's the sum?
Liopleurodon
2017-03-09 19:19:16
$70$
$70$
nosaj
2017-03-09 19:19:16
70
70
Funnybunny5246
2017-03-09 19:19:16
70
70
LittleChimp
2017-03-09 19:19:16
70
70
rapturt9
2017-03-09 19:19:16
70
70
math_noob
2017-03-09 19:19:16
70
70
SomethingNeutral
2017-03-09 19:19:16
70
70
Radio2
2017-03-09 19:19:16
70
70
legolego
2017-03-09 19:19:16
70
70
letsgomath
2017-03-09 19:19:16
70
70
copeland
2017-03-09 19:19:19
The sum is $70$. That means the average value over any block of 20 consecutive numbers is $\dfrac72$.
The sum is $70$. That means the average value over any block of 20 consecutive numbers is $\dfrac72$.
copeland
2017-03-09 19:19:25
So how do we compute the sum of the first 2017 of these $d_n?$
So how do we compute the sum of the first 2017 of these $d_n?$
LittleChimp
2017-03-09 19:20:31
the sum of the first 2000 is just 7000
the sum of the first 2000 is just 7000
rapturt9
2017-03-09 19:20:31
70*50 for the first 2000 and then sum the first 17
70*50 for the first 2000 and then sum the first 17
EasyAs_Pi
2017-03-09 19:20:31
compute the sum of the first 2000 then add the last 17...
compute the sum of the first 2000 then add the last 17...
vishwathganesan
2017-03-09 19:20:31
find the sum of the first 101 cycles and subtract 1+0+0
find the sum of the first 101 cycles and subtract 1+0+0
legolego
2017-03-09 19:20:31
first 2020 : 7070 - 0 - 0 - 1 = 7069 -> 69
first 2020 : 7070 - 0 - 0 - 1 = 7069 -> 69
ilovemath04
2017-03-09 19:20:31
the first 2000 is 70*100
the first 2000 is 70*100
spicyray
2017-03-09 19:20:31
20*101-1
20*101-1
amzhao
2017-03-09 19:20:31
2000*(7/2) then add first 17
2000*(7/2) then add first 17
owm
2017-03-09 19:20:31
100*7/2+69
100*7/2+69
MathTechFire
2017-03-09 19:20:31
7/2 x 100 + next 17
7/2 x 100 + next 17
Funnybunny5246
2017-03-09 19:20:31
1000(70) + first 17 digits
1000(70) + first 17 digits
copeland
2017-03-09 19:20:33
We can start with the first 2020 of them.
We can start with the first 2020 of them.
copeland
2017-03-09 19:20:36
Since the average value is $\dfrac72$, the sum of the first 2020 is 7070.
Since the average value is $\dfrac72$, the sum of the first 2020 is 7070.
copeland
2017-03-09 19:20:42
Then we subtract $d_{20}+d_{19}+d_{18}.$ What's the final answer?
Then we subtract $d_{20}+d_{19}+d_{18}.$ What's the final answer?
Radio2
2017-03-09 19:21:29
069
069
legolego
2017-03-09 19:21:29
69
69
sxu
2017-03-09 19:21:29
069
069
swagger
2017-03-09 19:21:29
069
069
MrMXS
2017-03-09 19:21:29
$069$
$069$
StellarG
2017-03-09 19:21:29
7069
7069
MountainHeight
2017-03-09 19:21:29
069
069
amackenzie1
2017-03-09 19:21:29
$069$
$069$
GeneralCobra19
2017-03-09 19:21:29
69
69
copeland
2017-03-09 19:21:34
We want $7070-0-0-1=7069.$ The answer is $\boxed{069}$.
We want $7070-0-0-1=7069.$ The answer is $\boxed{069}$.
copeland
2017-03-09 19:21:37
Notice that there were two separate tactics suggested above. Tackling the first 2000 or the first 2020. Both work just fine.
Notice that there were two separate tactics suggested above. Tackling the first 2000 or the first 2020. Both work just fine.
copeland
2017-03-09 19:22:08
What next?
What next?
copeland
2017-03-09 19:22:12
Should we keep going?
Should we keep going?
First
2017-03-09 19:22:22
#4
#4
GeneralCobra19
2017-03-09 19:22:22
#4
#4
Shri333
2017-03-09 19:22:22
QUESTION 4!!!
QUESTION 4!!!
nukelauncher
2017-03-09 19:22:22
YEs
YEs
stronto
2017-03-09 19:22:22
YEAH
YEAH
quanhui868
2017-03-09 19:22:22
Yeah
Yeah
owm
2017-03-09 19:22:22
Yep!
Yep!
Shri333
2017-03-09 19:22:22
YES
YES
GeorgeAvocados
2017-03-09 19:22:22
sure
sure
Liopleurodon
2017-03-09 19:22:23
Number 4
Number 4
copeland
2017-03-09 19:22:26
4. A pyramid has a triangular base with side lengths 20, 20, and 24. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length 25. The volume of the pyramid is $m\sqrt n,$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$
4. A pyramid has a triangular base with side lengths 20, 20, and 24. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length 25. The volume of the pyramid is $m\sqrt n,$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$
copeland
2017-03-09 19:22:29
Hm, OK. Do we like 20-20-24 triangles?
Hm, OK. Do we like 20-20-24 triangles?
winnertakeover
2017-03-09 19:22:45
yes\
yes\
Deathranger999
2017-03-09 19:22:45
Yes.
Yes.
cwechsz
2017-03-09 19:22:45
yes
yes
NewbieGamer
2017-03-09 19:22:45
yes
yes
copeland
2017-03-09 19:22:48
Why?
Why?
Deathranger999
2017-03-09 19:23:14
It's two 3-4-5 in disguise.
It's two 3-4-5 in disguise.
nosaj
2017-03-09 19:23:14
they're two 12-16-20's taped together
they're two 12-16-20's taped together
abvenkgoo
2017-03-09 19:23:14
it's 2 12-16-20 triangles together
it's 2 12-16-20 triangles together
Funnybunny5246
2017-03-09 19:23:14
turns into 12-16-20
turns into 12-16-20
ethanliu247
2017-03-09 19:23:14
two 20-16-12
two 20-16-12
jonzli123
2017-03-09 19:23:14
forms two 20-16-12 triangles
forms two 20-16-12 triangles
math101010
2017-03-09 19:23:14
12-16-20
12-16-20
Picroft
2017-03-09 19:23:14
3-4-5 triangles when you draw altitude
3-4-5 triangles when you draw altitude
math.fever
2017-03-09 19:23:14
3-4-5 triangles!
3-4-5 triangles!
Wuna
2017-03-09 19:23:14
3-4-5 based!
3-4-5 based!
copeland
2017-03-09 19:23:19
It's isosceles anyway. We can even compute the altitude to the base:
It's isosceles anyway. We can even compute the altitude to the base:
copeland
2017-03-09 19:23:21
copeland
2017-03-09 19:23:21
Oh, right. That's 3-4-5:
Oh, right. That's 3-4-5:
copeland
2017-03-09 19:23:22
copeland
2017-03-09 19:23:24
How do we compute the volume?
How do we compute the volume?
vishwathganesan
2017-03-09 19:23:53
1/3Bh
1/3Bh
GeronimoStilton
2017-03-09 19:23:53
We need the height.
We need the height.
EasyAs_Pi
2017-03-09 19:23:53
$\frac{bh}{3}$
$\frac{bh}{3}$
tdeng
2017-03-09 19:23:53
Now we just need the height.
Now we just need the height.
Pudentane
2017-03-09 19:23:53
1/3*B*h
1/3*B*h
gabrielsui
2017-03-09 19:23:53
1/3 * B * h
1/3 * B * h
stronto
2017-03-09 19:23:53
1/3 [ABC] h
1/3 [ABC] h
curry3030
2017-03-09 19:23:53
1/3 * b * h
1/3 * b * h
young_3141
2017-03-09 19:23:53
1/3 * b * h
1/3 * b * h
islander7
2017-03-09 19:23:53
find height to this face
find height to this face
espeon12
2017-03-09 19:23:53
bh/3
bh/3
LearningMath
2017-03-09 19:23:53
The base's area times height divided by 3
The base's area times height divided by 3
copeland
2017-03-09 19:23:56
To compute the volume, we need the area of the base and the height. Toss in a $\dfrac13$ and stir. Let's save the variable $h$ for the height of the pyramid.
To compute the volume, we need the area of the base and the height. Toss in a $\dfrac13$ and stir. Let's save the variable $h$ for the height of the pyramid.
copeland
2017-03-09 19:23:57
What's the area of the base?
What's the area of the base?
SDMKM
2017-03-09 19:24:15
192
192
ninjataco
2017-03-09 19:24:15
192
192
Pandasareamazing.
2017-03-09 19:24:15
192
192
FlamingDragon_9000
2017-03-09 19:24:15
192
192
ilovemath04
2017-03-09 19:24:15
192
192
vvluo
2017-03-09 19:24:15
192
192
SomethingNeutral
2017-03-09 19:24:15
16*12 = 192
16*12 = 192
dhruv
2017-03-09 19:24:15
192
192
copeland
2017-03-09 19:24:16
The area of the base is $12\cdot16.$
The area of the base is $12\cdot16.$
copeland
2017-03-09 19:24:24
Looks like you guys like multiplying more than I do.
Looks like you guys like multiplying more than I do.
copeland
2017-03-09 19:24:27
That's cool.
That's cool.
copeland
2017-03-09 19:24:31
And, qualitatively, what do we know about the altitude to this base?
And, qualitatively, what do we know about the altitude to this base?
a1b2
2017-03-09 19:25:15
It contains the circumcenter
It contains the circumcenter
richuw
2017-03-09 19:25:15
Touches the Circumcenter
Touches the Circumcenter
sxu
2017-03-09 19:25:15
it's above the circumcenter
it's above the circumcenter
copeland
2017-03-09 19:25:28
The point we care about is the circumcenter of the base.
The point we care about is the circumcenter of the base.
copeland
2017-03-09 19:25:35
That makes sense geometrically: every tetrahedron has a circumsphere. Since the points on the base are equidistant from the top vertex, the plane of their circumcircle is perpendicular to the diameter through the upper vertex, and, um, nevermind. . .
That makes sense geometrically: every tetrahedron has a circumsphere. Since the points on the base are equidistant from the top vertex, the plane of their circumcircle is perpendicular to the diameter through the upper vertex, and, um, nevermind. . .
copeland
2017-03-09 19:25:41
Maybe you were thinking something else.
Maybe you were thinking something else.
copeland
2017-03-09 19:25:48
Why is it the circumcenter again?
Why is it the circumcenter again?
richuw
2017-03-09 19:26:09
Equi-distant from each vertex
Equi-distant from each vertex
jonzli123
2017-03-09 19:26:09
same distance from each vertex
same distance from each vertex
stronto
2017-03-09 19:26:09
The top is equidistant from all 3 vertices
The top is equidistant from all 3 vertices
linqaszayi
2017-03-09 19:26:09
equidistance
equidistance
First
2017-03-09 19:26:09
Center of equidistance?
Center of equidistance?
copeland
2017-03-09 19:26:11
So?
So?
sxu
2017-03-09 19:27:12
well since altitude is perp to base, it's perp to every line thru foot, so congruent triangles (that's how I convinced myself)
well since altitude is perp to base, it's perp to every line thru foot, so congruent triangles (that's how I convinced myself)
alifenix-
2017-03-09 19:27:12
Because all the lines upwards have the same length. The height is the same, thus the distance must be the same and so it is the definition of the circumcenter.
Because all the lines upwards have the same length. The height is the same, thus the distance must be the same and so it is the definition of the circumcenter.
richuw
2017-03-09 19:27:12
well R^2 + Altitude^2 = 625
well R^2 + Altitude^2 = 625
copeland
2017-03-09 19:27:17
Yeah! There are right triangles in here. We're just using HL similarity, or the Pythagorean Theorem or something.
Yeah! There are right triangles in here. We're just using HL similarity, or the Pythagorean Theorem or something.
copeland
2017-03-09 19:27:44
Where's the circumcenter of this triangle?
Where's the circumcenter of this triangle?
copeland
2017-03-09 19:27:48
Deathranger999
2017-03-09 19:28:12
On the altitude.
On the altitude.
vvluo
2017-03-09 19:28:12
on altitude
on altitude
MathTechFire
2017-03-09 19:28:12
on the 16 line
on the 16 line
MrMXS
2017-03-09 19:28:12
on the altitute
on the altitute
Jyzhang12
2017-03-09 19:28:12
on the height 16
on the height 16
Radio2
2017-03-09 19:28:12
on the median with length 116
on the median with length 116
Radio2
2017-03-09 19:28:12
on the median with length 16
on the median with length 16
copeland
2017-03-09 19:28:23
We know that the distances from the vertices of this triangle to the fourth vertex are each 25.
We know that the distances from the vertices of this triangle to the fourth vertex are each 25.
copeland
2017-03-09 19:28:23
We also know by symmetry that the vertex lies over the triangle altitude that we drew in our picture. Let's plop the base of our altitude onto the picture.
We also know by symmetry that the vertex lies over the triangle altitude that we drew in our picture. Let's plop the base of our altitude onto the picture.
copeland
2017-03-09 19:28:28
copeland
2017-03-09 19:28:37
If $a^2+h^2=25^2$, then $a$ is the distance from our point to any of the vertices.
If $a^2+h^2=25^2$, then $a$ is the distance from our point to any of the vertices.
copeland
2017-03-09 19:28:39
Let's plop $a$ on there, shall we?
Let's plop $a$ on there, shall we?
copeland
2017-03-09 19:28:42
copeland
2017-03-09 19:28:49
How can we get $a$ from here?
How can we get $a$ from here?
tdeng
2017-03-09 19:29:40
(16-a)^2+12^2=a^2 by the Pythagorean Theorem
(16-a)^2+12^2=a^2 by the Pythagorean Theorem
Jyzhang12
2017-03-09 19:29:40
pythagorean theorem
pythagorean theorem
raxu
2017-03-09 19:29:40
$(16-a)^2+12^2=a^2$ using Pythagorean Theorem
$(16-a)^2+12^2=a^2$ using Pythagorean Theorem
legolego
2017-03-09 19:29:40
12^2 + (16-a)^2 = a^2
12^2 + (16-a)^2 = a^2
mathguy623
2017-03-09 19:29:40
$12^2+(16-a)^2=a^2$
$12^2+(16-a)^2=a^2$
Deathranger999
2017-03-09 19:29:40
(16 - a)^2 + 12^2 = a^2
(16 - a)^2 + 12^2 = a^2
Radio2
2017-03-09 19:29:40
Draw a perpendicular to a side with length 20 and use similarity
Draw a perpendicular to a side with length 20 and use similarity
curry3030
2017-03-09 19:29:40
drop perp from altitude to side with length 20
drop perp from altitude to side with length 20
copeland
2017-03-09 19:29:42
Oh, here are two solutions. We can either go with Pythagoras or similarity.
Oh, here are two solutions. We can either go with Pythagoras or similarity.
copeland
2017-03-09 19:29:44
Which do you want?
Which do you want?
EasyAs_Pi
2017-03-09 19:30:42
similairiry
similairiry
BuddyS
2017-03-09 19:30:42
similarity
similarity
leonlzg
2017-03-09 19:30:42
similarity
similarity
gabrielsui
2017-03-09 19:30:42
Similarity!!!
Similarity!!!
letsgomath
2017-03-09 19:30:42
Pythagoras
Pythagoras
J1618
2017-03-09 19:30:42
I want to see how similarity can work here
I want to see how similarity can work here
QuestForKnowledge
2017-03-09 19:30:42
Pyhatgoras
Pyhatgoras
Mathaddict11
2017-03-09 19:30:42
Pythagoras
Pythagoras
nukelauncher
2017-03-09 19:30:42
pythagoras ftw
pythagoras ftw
LearningMath
2017-03-09 19:30:42
Pythagoras
Pythagoras
copeland
2017-03-09 19:30:47
Wow, OK.
Wow, OK.
copeland
2017-03-09 19:30:56
Similarity is a little faster. Let's do that.
Similarity is a little faster. Let's do that.
copeland
2017-03-09 19:31:02
Tell me about this other altitude:
Tell me about this other altitude:
copeland
2017-03-09 19:31:04
GeronimoStilton
2017-03-09 19:31:59
It bisects because $a = a$.
It bisects because $a = a$.
GeneralCobra19
2017-03-09 19:31:59
The side it bisects is cut into two sides of 10
The side it bisects is cut into two sides of 10
amackenzie1
2017-03-09 19:31:59
triangle formed by it is similar to half of the big triangle
triangle formed by it is similar to half of the big triangle
EasyAs_Pi
2017-03-09 19:31:59
it creates similar triangles
it creates similar triangles
amzhao
2017-03-09 19:31:59
similar trianlg
similar trianlg
copeland
2017-03-09 19:32:01
It bisects that other side, so those edges have length 10:
It bisects that other side, so those edges have length 10:
copeland
2017-03-09 19:32:02
copeland
2017-03-09 19:32:02
So what is $a?$
So what is $a?$
AlisonH
2017-03-09 19:32:34
a/10=20/16
a/10=20/16
MrMXS
2017-03-09 19:32:34
using the new right triangle gives $\cfrac{a}{20}=\cfrac{10}{12}=\cfrac{5}{6}$
using the new right triangle gives $\cfrac{a}{20}=\cfrac{10}{12}=\cfrac{5}{6}$
jonzli123
2017-03-09 19:32:34
25/2
25/2
math101010
2017-03-09 19:32:34
25/2
25/2
KYang
2017-03-09 19:32:34
25/2
25/2
MountainHeight
2017-03-09 19:32:34
25/2
25/2
Funnybunny5246
2017-03-09 19:32:34
25/2
25/2
copeland
2017-03-09 19:32:36
Now $a$ is the hypotenuse of a $3-4-5$ triangle. The long leg has length 10, so the hypotenuse solves $\dfrac a5=\dfrac{10}4.$ Therefore $a=\dfrac{25}2.$
Now $a$ is the hypotenuse of a $3-4-5$ triangle. The long leg has length 10, so the hypotenuse solves $\dfrac a5=\dfrac{10}4.$ Therefore $a=\dfrac{25}2.$
copeland
2017-03-09 19:32:56
Incidentally, the Pythagoran approach, gone-plaid, looks like this:
Incidentally, the Pythagoran approach, gone-plaid, looks like this:
copeland
2017-03-09 19:32:57
copeland
2017-03-09 19:33:01
$a+b=16.$
$a+b=16.$
copeland
2017-03-09 19:33:07
And $12^2+b^2=a^2$.
And $12^2+b^2=a^2$.
copeland
2017-03-09 19:33:13
We have
\begin{align*}
12^2&=a^2-b^2\\
16&=a+b
\end{align*}
Dividing these gives \[a-b=\frac{12^2}{4^2}=9.\] Adding that to the second equation gives\[2a=16+9=25.\]
We have
\begin{align*}
12^2&=a^2-b^2\\
16&=a+b
\end{align*}
Dividing these gives \[a-b=\frac{12^2}{4^2}=9.\] Adding that to the second equation gives\[2a=16+9=25.\]
copeland
2017-03-09 19:33:26
Either way,
Either way,
copeland
2017-03-09 19:33:27
\[a=\frac{25}2.\]
\[a=\frac{25}2.\]
copeland
2017-03-09 19:33:31
What is the height?
What is the height?
nosaj
2017-03-09 19:34:03
so the height is 25*sqrt(3)/2
so the height is 25*sqrt(3)/2
islander7
2017-03-09 19:34:03
25rt3/2
25rt3/2
vishwathganesan
2017-03-09 19:34:03
25sqrt3/2
25sqrt3/2
vvluo
2017-03-09 19:34:03
25 sqrt(3)/2
25 sqrt(3)/2
legolego
2017-03-09 19:34:03
25sqrt3/2
25sqrt3/2
KYang
2017-03-09 19:34:03
{25/2}*sqrt{3}
{25/2}*sqrt{3}
NeeNeeMath
2017-03-09 19:34:03
25/2sqrt(3)
25/2sqrt(3)
algebra_star1234
2017-03-09 19:34:03
$25\sqrt{3}/2$
$25\sqrt{3}/2$
rapturt9
2017-03-09 19:34:03
25sqrt(3)/2
25sqrt(3)/2
copeland
2017-03-09 19:34:05
The height solves\[h^2=25^2-\left(\frac{25}2\right)^2=25^2\cdot\frac34.\] So $h=25\sqrt{\dfrac34}=\dfrac{25\sqrt3}2$.
The height solves\[h^2=25^2-\left(\frac{25}2\right)^2=25^2\cdot\frac34.\] So $h=25\sqrt{\dfrac34}=\dfrac{25\sqrt3}2$.
copeland
2017-03-09 19:34:07
And the volume?
And the volume?
KYang
2017-03-09 19:34:37
800 sqrt{3}
800 sqrt{3}
algebra_star1234
2017-03-09 19:34:37
$800\sqrt{3}$
$800\sqrt{3}$
ilovemath04
2017-03-09 19:34:37
800\sqrt3
800\sqrt3
a1b2
2017-03-09 19:34:37
$800\sqrt3$
$800\sqrt3$
sxu
2017-03-09 19:34:37
$800\sqrt{3}$
$800\sqrt{3}$
stronto
2017-03-09 19:34:37
800\sqrt{3}
800\sqrt{3}
vishwathganesan
2017-03-09 19:34:37
800sqrt3
800sqrt3
GeronimoStilton
2017-03-09 19:34:37
$800\sqrt{3}$
$800\sqrt{3}$
EasyAs_Pi
2017-03-09 19:34:37
$800\sqrt3$
$800\sqrt3$
rapturt9
2017-03-09 19:34:37
800sqrt3
800sqrt3
copeland
2017-03-09 19:34:39
The volume is
\begin{align*}
V&=\frac13(12\cdot16)\cdot\dfrac{25\sqrt3}2\\
&=4\cdot8\cdot25\sqrt3=800\sqrt3.
\end{align*}
The volume is
\begin{align*}
V&=\frac13(12\cdot16)\cdot\dfrac{25\sqrt3}2\\
&=4\cdot8\cdot25\sqrt3=800\sqrt3.
\end{align*}
copeland
2017-03-09 19:34:41
And the final answer?
And the final answer?
GeronimoStilton
2017-03-09 19:34:57
$803$
$803$
Picroft
2017-03-09 19:34:57
803
803
dhruv
2017-03-09 19:34:57
803
803
DylanX22
2017-03-09 19:34:57
803
803
swagger
2017-03-09 19:34:57
803
803
owm
2017-03-09 19:34:57
803
803
BooBooTM
2017-03-09 19:34:57
803
803
First
2017-03-09 19:34:57
803
803
winnertakeover
2017-03-09 19:34:57
803
803
alifenix-
2017-03-09 19:34:57
803
803
copeland
2017-03-09 19:34:59
The final answer is $800+3=\boxed{803}.$
The final answer is $800+3=\boxed{803}.$
copeland
2017-03-09 19:35:29
Incidentally, several of you suggested an even faster way to the circumradius:
Incidentally, several of you suggested an even faster way to the circumradius:
pican
2017-03-09 19:35:32
We could use the fact that the circumradius is equal to $\dfrac{abc}{4A}$, where $a$, $b$, $c$ are the sides and $A$ is the area
We could use the fact that the circumradius is equal to $\dfrac{abc}{4A}$, where $a$, $b$, $c$ are the sides and $A$ is the area
DemonPlat4
2017-03-09 19:35:32
A = abc/4R
A = abc/4R
atmchallenge
2017-03-09 19:35:32
the circumradius, so $abc/4A$.
the circumradius, so $abc/4A$.
First
2017-03-09 19:35:32
We could have just used the formula $[ABC]=\frac{abc}{4R}$
We could have just used the formula $[ABC]=\frac{abc}{4R}$
copeland
2017-03-09 19:36:06
Often it's nice to work from first principles, but there just are a few formulas that you should know.
Often it's nice to work from first principles, but there just are a few formulas that you should know.
copeland
2017-03-09 19:36:12
Time for. . .
Time for. . .
awesomemaths
2017-03-09 19:37:01
problem 5
problem 5
lego101
2017-03-09 19:37:01
#5!!!!!
#5!!!!!
cooleybz2013
2017-03-09 19:37:01
5th problem!!!!
5th problem!!!!
curry3030
2017-03-09 19:37:01
question 5
question 5
lego101
2017-03-09 19:37:01
Number five, it's alive!
Number five, it's alive!
Reef334
2017-03-09 19:37:01
#5
#5
copeland
2017-03-09 19:37:04
5. A rational number written in base eight is $\underline a \underline b.\underline c\underline d,$ where all digits are nonzero. The same number in base twelve is $\underline b \underline b.\underline b\underline a.$ Find the base-ten number $\underline a \underline b\underline c.$
5. A rational number written in base eight is $\underline a \underline b.\underline c\underline d,$ where all digits are nonzero. The same number in base twelve is $\underline b \underline b.\underline b\underline a.$ Find the base-ten number $\underline a \underline b\underline c.$
copeland
2017-03-09 19:37:27
OK, great. We have a rational number that's not an integer.
OK, great. We have a rational number that's not an integer.
copeland
2017-03-09 19:37:28
Which of the two expressions is going to be more fun to look at?
Which of the two expressions is going to be more fun to look at?
richuw
2017-03-09 19:38:12
bb.ba
bb.ba
espeon12
2017-03-09 19:38:12
the second one?
the second one?
a1b2
2017-03-09 19:38:12
$bb.ba_{12}$
$bb.ba_{12}$
ilovemath04
2017-03-09 19:38:12
bb.ba
bb.ba
MrMXS
2017-03-09 19:38:12
the second?
the second?
ethanliu247
2017-03-09 19:38:12
second one
second one
legolego
2017-03-09 19:38:12
second one with only two distinct digits
second one with only two distinct digits
rapturt9
2017-03-09 19:38:12
the second only 2 variables
the second only 2 variables
BuddyS
2017-03-09 19:38:12
base 12
base 12
vvluo
2017-03-09 19:38:12
bb.ba becase lez pozibilideez
bb.ba becase lez pozibilideez
Pandasareamazing.
2017-03-09 19:38:12
bb.ba
bb.ba
copeland
2017-03-09 19:38:15
The base-12 number has a lot of repeated digits. Let's think about it first. Can you write $\underline b \underline b.\underline b\underline a$ in a more meaningful way?
The base-12 number has a lot of repeated digits. Let's think about it first. Can you write $\underline b \underline b.\underline b\underline a$ in a more meaningful way?
vishwathganesan
2017-03-09 19:39:08
12b+b+b/12+a/144
12b+b+b/12+a/144
pi_Plus_45x23
2017-03-09 19:39:08
$13b+\frac{b}{12}+\frac{a}{144}$
$13b+\frac{b}{12}+\frac{a}{144}$
lego101
2017-03-09 19:39:08
12b + b + b/12 + a/144
12b + b + b/12 + a/144
DylanX22
2017-03-09 19:39:08
12b + b + b/12 + a/144
12b + b + b/12 + a/144
math101010
2017-03-09 19:39:08
12b+b+b/12+a/144
12b+b+b/12+a/144
Deathranger999
2017-03-09 19:39:08
12b + b + b/12 + a/144
12b + b + b/12 + a/144
nukelauncher
2017-03-09 19:39:08
13b + b/12 + a/144
13b + b/12 + a/144
GeneralCobra19
2017-03-09 19:39:08
12b+b+b/12+a/144
12b+b+b/12+a/144
WhaleVomit
2017-03-09 19:39:08
12b+b+b/12+a/144
12b+b+b/12+a/144
J1618
2017-03-09 19:39:08
12b +b +b/12 +a/144
12b +b +b/12 +a/144
copeland
2017-03-09 19:39:11
We can kind-of turn it into a "mixed number":\[12b+b+\frac{12b+a}{12^2}.\]
We can kind-of turn it into a "mixed number":\[12b+b+\frac{12b+a}{12^2}.\]
copeland
2017-03-09 19:39:20
As a "mixed number" the base-8 dude becomes \[8a+b+\frac{8c+d}{8^2}.\]
As a "mixed number" the base-8 dude becomes \[8a+b+\frac{8c+d}{8^2}.\]
copeland
2017-03-09 19:39:22
\[12b+b+\frac{12b+a}{12^2}=8a+b+\frac{8c+d}{8^2}\]
\[12b+b+\frac{12b+a}{12^2}=8a+b+\frac{8c+d}{8^2}\]
copeland
2017-03-09 19:39:23
What else can we say about this equation?
What else can we say about this equation?
NeeNeeMath
2017-03-09 19:40:08
loook at the integere parts first
loook at the integere parts first
espeon12
2017-03-09 19:40:08
the fractions are equivalent
the fractions are equivalent
EulerMacaroni
2017-03-09 19:40:08
integer and fractional parts are respectively equal
integer and fractional parts are respectively equal
hodori01
2017-03-09 19:40:08
use integer and fractional parts
use integer and fractional parts
vishwathganesan
2017-03-09 19:40:08
equate the integral parts and the fractional parts
equate the integral parts and the fractional parts
mathguy623
2017-03-09 19:40:08
fractional parts and integer parts are equal
fractional parts and integer parts are equal
DemonPlat4
2017-03-09 19:40:08
12b = 8a
12b = 8a
lego101
2017-03-09 19:40:08
we know the integer parts are equal and the fractional parts are equal too since a, b, c, d are all integers
we know the integer parts are equal and the fractional parts are equal too since a, b, c, d are all integers
rapturt9
2017-03-09 19:40:08
the fracs equal each other
the fracs equal each other
copeland
2017-03-09 19:40:10
Since these are base-12 and base-8 numbers, we know that the fractions are less than 1. So the integer parts and fractional parts are equal.
Since these are base-12 and base-8 numbers, we know that the fractions are less than 1. So the integer parts and fractional parts are equal.
copeland
2017-03-09 19:40:12
The fractional part is nasty. The integer part is cute:\[12b+b=8a+b.\]
The fractional part is nasty. The integer part is cute:\[12b+b=8a+b.\]
copeland
2017-03-09 19:40:13
What's that say?
What's that say?
pican
2017-03-09 19:40:45
$3b=2a$
$3b=2a$
tdeng
2017-03-09 19:40:45
3b=2a
3b=2a
letsgomath
2017-03-09 19:40:45
3b=2a
3b=2a
treemath
2017-03-09 19:40:45
3b=2a
3b=2a
LearningMath
2017-03-09 19:40:45
3b = 2a
3b = 2a
rapturt9
2017-03-09 19:40:45
a=2b/3
a=2b/3
ninjataco
2017-03-09 19:40:45
3b = 2a
3b = 2a
QuestForKnowledge
2017-03-09 19:40:45
3b=2a
3b=2a
islander7
2017-03-09 19:40:45
3b=2a
3b=2a
stronto
2017-03-09 19:40:45
a = 3/2 b
a = 3/2 b
bomb427006
2017-03-09 19:40:45
$2a=3b$
$2a=3b$
KYang
2017-03-09 19:40:45
3b=2a
3b=2a
copeland
2017-03-09 19:40:47
This says that $a=\dfrac32b$.
This says that $a=\dfrac32b$.
copeland
2017-03-09 19:40:48
What are all the possible values for $\underline a\underline b?$
What are all the possible values for $\underline a\underline b?$
legolego
2017-03-09 19:41:34
32, 64
32, 64
quanhui868
2017-03-09 19:41:34
32, 64, 96
32, 64, 96
J1618
2017-03-09 19:41:34
32 and 64
32 and 64
jonzli123
2017-03-09 19:41:34
32 & 64
32 & 64
cooljoseph
2017-03-09 19:41:34
64, 32
64, 32
LearningMath
2017-03-09 19:41:34
64, 32
64, 32
ezhao02
2017-03-09 19:41:34
32, 64
32, 64
amackenzie1
2017-03-09 19:41:34
32, 64
32, 64
copeland
2017-03-09 19:41:37
Since these are both base-8 digits, they're less than 8. Therefore $b$ is either 2 or 4 and the corresponding $a$ is 3 or 6.
Since these are both base-8 digits, they're less than 8. Therefore $b$ is either 2 or 4 and the corresponding $a$ is 3 or 6.
copeland
2017-03-09 19:41:39
Therefore the base-12 number is either $\underline 2 \underline 2.\underline 2\underline 3$ or $\underline 4 \underline 4.\underline 4\underline 6.$
Therefore the base-12 number is either $\underline 2 \underline 2.\underline 2\underline 3$ or $\underline 4 \underline 4.\underline 4\underline 6.$
copeland
2017-03-09 19:41:43
Let's convert $\underline 2 \underline 2.\underline 2\underline 3_{12}$ to a base-8 number. We know the integer part is $\underline 3\underline 2_{8}$. What is $.\underline 2\underline3_{12}$ when we convert it to base 8?
Let's convert $\underline 2 \underline 2.\underline 2\underline 3_{12}$ to a base-8 number. We know the integer part is $\underline 3\underline 2_{8}$. What is $.\underline 2\underline3_{12}$ when we convert it to base 8?
Funnybunny5246
2017-03-09 19:43:00
14
14
Pandasareamazing.
2017-03-09 19:43:00
.14
.14
letsgomath
2017-03-09 19:43:00
0.14
0.14
vvluo
2017-03-09 19:43:00
.14
.14
mathdragon2000
2017-03-09 19:43:00
.14
.14
jonzli123
2017-03-09 19:43:00
.14
.14
ethanliu247
2017-03-09 19:43:00
.14
.14
islander7
2017-03-09 19:43:00
.14
.14
copeland
2017-03-09 19:43:03
\[.\underline 2\underline3_{12}=\frac{27}{144}=\frac3{16}=\frac{12}{64}=.\underline1\underline4_{8}.\]
\[.\underline 2\underline3_{12}=\frac{27}{144}=\frac3{16}=\frac{12}{64}=.\underline1\underline4_{8}.\]
copeland
2017-03-09 19:43:12
Therefore $\underline 2 \underline 2.\underline 2\underline 3_{12} = \underline 3 \underline 2.\underline 1\underline 4_{8}$.
Therefore $\underline 2 \underline 2.\underline 2\underline 3_{12} = \underline 3 \underline 2.\underline 1\underline 4_{8}$.
copeland
2017-03-09 19:43:14
And what is $\underline 4 \underline 4.\underline 4\underline 6_{12}$ when converted to base 8?
And what is $\underline 4 \underline 4.\underline 4\underline 6_{12}$ when converted to base 8?
Radio2
2017-03-09 19:44:11
64.28
64.28
gabrielsui
2017-03-09 19:44:11
64.28
64.28
ezhao02
2017-03-09 19:44:11
64.30
64.30
Funnybunny5246
2017-03-09 19:44:11
64.30
64.30
Mrkiller
2017-03-09 19:44:11
64.30
64.30
IsaacZ123
2017-03-09 19:44:11
64.3
64.3
amzhao
2017-03-09 19:44:11
64.28
64.28
MountainHeight
2017-03-09 19:44:11
64.3
64.3
DemonPlat4
2017-03-09 19:44:11
64.30
64.30
mathguy623
2017-03-09 19:44:11
64.30
64.30
First
2017-03-09 19:44:11
64.3
64.3
copeland
2017-03-09 19:44:19
This is twice the previous number!\[\underline 4 \underline 4.\underline 4\underline 6_{12}=2(\underline 2 \underline 2.\underline 2\underline 3_{12}) = 2(\underline 3 \underline 2.\underline 1\underline 4_{8})=\underline 6 \underline 4.\underline 3\underline 0_{8}.\]
This is twice the previous number!\[\underline 4 \underline 4.\underline 4\underline 6_{12}=2(\underline 2 \underline 2.\underline 2\underline 3_{12}) = 2(\underline 3 \underline 2.\underline 1\underline 4_{8})=\underline 6 \underline 4.\underline 3\underline 0_{8}.\]
copeland
2017-03-09 19:44:26
Notice that base 8 you need to carry the 8.
Notice that base 8 you need to carry the 8.
copeland
2017-03-09 19:44:30
So are there two answers?
So are there two answers?
sxu
2017-03-09 19:44:55
no since nonzero
no since nonzero
BooBooTM
2017-03-09 19:44:55
No!!!!
No!!!!
legolego
2017-03-09 19:44:55
you can't have 0
you can't have 0
DemonPlat4
2017-03-09 19:44:55
NO - nonzero digits
NO - nonzero digits
vvluo
2017-03-09 19:44:55
but there is a 0........!!!!!
but there is a 0........!!!!!
Pandasareamazing.
2017-03-09 19:44:55
non zero digits
non zero digits
jonzli123
2017-03-09 19:44:55
NO! All the digits have to be nonzero
NO! All the digits have to be nonzero
Funnybunny5246
2017-03-09 19:44:55
One has a 0
One has a 0
mingxu
2017-03-09 19:44:55
nonzero!
nonzero!
owm
2017-03-09 19:44:55
No, the digits are nonzero
No, the digits are nonzero
AlisonH
2017-03-09 19:44:55
all digits nonzero
all digits nonzero
GeronimoStilton
2017-03-09 19:44:55
All the digits are nonzero.
All the digits are nonzero.
copeland
2017-03-09 19:44:57
No! The digits have to be positive. So what's the answer?
No! The digits have to be positive. So what's the answer?
legolego
2017-03-09 19:45:28
321
321
AlisonH
2017-03-09 19:45:28
321
321
Picroft
2017-03-09 19:45:28
321
321
SomethingNeutral
2017-03-09 19:45:28
321
321
jonzli123
2017-03-09 19:45:28
321
321
vvluo
2017-03-09 19:45:28
321
321
mingxu
2017-03-09 19:45:28
321
321
smartpgp
2017-03-09 19:45:28
521
521
MrMXS
2017-03-09 19:45:28
$321$
$321$
ilovemath04
2017-03-09 19:45:28
321
321
Deathranger999
2017-03-09 19:45:28
321
321
math101010
2017-03-09 19:45:28
321
321
smartpgp
2017-03-09 19:45:28
531
531
Gamabyte
2017-03-09 19:45:28
321
321
winnertakeover
2017-03-09 19:45:28
321
321
copeland
2017-03-09 19:45:30
The only viable pair is $\underline 2 \underline 2.\underline 2\underline 3_{12} = \underline 3 \underline 2.\underline 1\underline 4_{8}$. Therefore $\underline a\underline b\underline c=\boxed{321}.$
The only viable pair is $\underline 2 \underline 2.\underline 2\underline 3_{12} = \underline 3 \underline 2.\underline 1\underline 4_{8}$. Therefore $\underline a\underline b\underline c=\boxed{321}.$
copeland
2017-03-09 19:45:35
Also, the second digit after the decimal point in a base-12 number, $\underline 0.\underline0\underline1_{12},$ is the "grossths" place.
Also, the second digit after the decimal point in a base-12 number, $\underline 0.\underline0\underline1_{12},$ is the "grossths" place.
copeland
2017-03-09 19:45:39
I just made that up.
I just made that up.
copeland
2017-03-09 19:45:40
Next?
Next?
GeronimoStilton
2017-03-09 19:45:59
Problem 6!
Problem 6!
espeon12
2017-03-09 19:45:59
number 6
number 6
IsaacZ123
2017-03-09 19:45:59
number 6
number 6
vishwathganesan
2017-03-09 19:45:59
number 6 yay!!
number 6 yay!!
lego101
2017-03-09 19:45:59
Number six, pick up sticks, and the pace too.
Number six, pick up sticks, and the pace too.
amzhao
2017-03-09 19:45:59
numero 6
numero 6
MathTechFire
2017-03-09 19:45:59
6!!!!!!!!
6!!!!!!!!
gabrielsui
2017-03-09 19:45:59
6!
6!
copeland
2017-03-09 19:46:03
6. A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x.$ Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\dfrac{14}{25}.$ Find the difference between the largest and smallest possible values of $x.$
6. A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x.$ Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\dfrac{14}{25}.$ Find the difference between the largest and smallest possible values of $x.$
copeland
2017-03-09 19:46:06
OK, here's a picture:
OK, here's a picture:
copeland
2017-03-09 19:46:07
copeland
2017-03-09 19:46:08
When does the chord intersect the triangle?
When does the chord intersect the triangle?
IsaacZ123
2017-03-09 19:46:53
when the two end points are from two different arc segments
when the two end points are from two different arc segments
lego101
2017-03-09 19:46:53
When the two points are on different sides of the triangle
When the two points are on different sides of the triangle
a1b2
2017-03-09 19:46:53
If the two chosen points are on different arcs
If the two chosen points are on different arcs
winnertakeover
2017-03-09 19:46:53
when 2 separte arcs are chosen
when 2 separte arcs are chosen
vishwathganesan
2017-03-09 19:46:53
when both points are in different arcs of the circle
when both points are in different arcs of the circle
J1618
2017-03-09 19:46:53
When the two endpoints are on two different arcs
When the two endpoints are on two different arcs
DemonPlat4
2017-03-09 19:46:53
if the points are in 2 different of the 3 "regions" on the circle
if the points are in 2 different of the 3 "regions" on the circle
QuestForKnowledge
2017-03-09 19:46:53
when not on same section
when not on same section
GeneralCobra19
2017-03-09 19:46:53
When the points are across different "parts" of the circle divided by the triangle.
When the points are across different "parts" of the circle divided by the triangle.
alifenix-
2017-03-09 19:46:53
when the points are on two different arcs out of the three made by the points on the triangle
when the points are on two different arcs out of the three made by the points on the triangle
copeland
2017-03-09 19:46:56
The triangle cuts off three segments of the circle. A chord intersects the triangle when its vertices lie on the arcs of different segments.
The triangle cuts off three segments of the circle. A chord intersects the triangle when its vertices lie on the arcs of different segments.
copeland
2017-03-09 19:47:21
I think it's likely that we're going to want to use complementary probability on this problem, since the chords that do not intersect the triangle are easier to think about.
I think it's likely that we're going to want to use complementary probability on this problem, since the chords that do not intersect the triangle are easier to think about.
copeland
2017-03-09 19:47:23
Let's pick two points at random on the circle. What's the probability that the two points are in the top right (red) region below?
Let's pick two points at random on the circle. What's the probability that the two points are in the top right (red) region below?
copeland
2017-03-09 19:47:24
ninjataco
2017-03-09 19:48:05
(x/180)^2
(x/180)^2
tdeng
2017-03-09 19:48:05
(2x/360)^2
(2x/360)^2
rapturt9
2017-03-09 19:48:05
x^2/(180^2)
x^2/(180^2)
Reef334
2017-03-09 19:48:05
x^2/180^2
x^2/180^2
Funnybunny5246
2017-03-09 19:48:05
$(x/180)^2$
$(x/180)^2$
rapturt9
2017-03-09 19:48:05
(x/180)^2
(x/180)^2
copeland
2017-03-09 19:48:09
That region is $\dfrac{2x}{360}=\dfrac{x}{180}$ of the circle. The probability that both points are in that region is $\dfrac{x^2}{180^2}$. Let's call that $p^2$.
That region is $\dfrac{2x}{360}=\dfrac{x}{180}$ of the circle. The probability that both points are in that region is $\dfrac{x^2}{180^2}$. Let's call that $p^2$.
copeland
2017-03-09 19:48:10
What about the blue region?
What about the blue region?
IsaacZ123
2017-03-09 19:48:40
same as p^2
same as p^2
NeeNeeMath
2017-03-09 19:48:40
same
same
rapturt9
2017-03-09 19:48:40
(x/180)^2, the same thing
(x/180)^2, the same thing
GeronimoStilton
2017-03-09 19:48:40
$\frac{x^2}{180^2}$
$\frac{x^2}{180^2}$
IsaacZ123
2017-03-09 19:48:40
same as the red region
same as the red region
Peggy
2017-03-09 19:48:40
same
same
smartpgp
2017-03-09 19:48:40
p^2
p^2
copeland
2017-03-09 19:48:42
Since the triangle is isosceles, the blue and red regions are congruent, so the probability that both points lie in the blue region is also $p^2$.
Since the triangle is isosceles, the blue and red regions are congruent, so the probability that both points lie in the blue region is also $p^2$.
copeland
2017-03-09 19:48:43
What's the probability that both points are in the green region?
What's the probability that both points are in the green region?
Picroft
2017-03-09 19:50:33
(1-2p)^2
(1-2p)^2
legolego
2017-03-09 19:50:33
(1-2p)^2
(1-2p)^2
ezhao02
2017-03-09 19:50:33
$(1-p)^2$
$(1-p)^2$
GeronimoStilton
2017-03-09 19:50:33
$\frac{(180-2x)^2}{180^2}$
$\frac{(180-2x)^2}{180^2}$
ezhao02
2017-03-09 19:50:33
$(1-2p)^2$
$(1-2p)^2$
amzhao
2017-03-09 19:50:33
((180-2x)/180)^2
((180-2x)/180)^2
rapturt9
2017-03-09 19:50:33
(180-2x)^2/(180)^2
(180-2x)^2/(180)^2
MathTechFire
2017-03-09 19:50:33
(180-2x/180)^2
(180-2x/180)^2
copeland
2017-03-09 19:50:45
The probability that a point is in the green region is $1-2p$, so the probability that both points are in the green region is $(1-2p)^2=1-4p+4p^2$.
The probability that a point is in the green region is $1-2p$, so the probability that both points are in the green region is $(1-2p)^2=1-4p+4p^2$.
copeland
2017-03-09 19:50:48
And in terms of $p$, what is the probability that the chord intersects the triangle?
And in terms of $p$, what is the probability that the chord intersects the triangle?
cooljoseph
2017-03-09 19:52:08
$1-(6p^2-4p+1)$
$1-(6p^2-4p+1)$
dandyq
2017-03-09 19:52:08
1 - (1 - 4p + 4p^2 + 2p^2)
1 - (1 - 4p + 4p^2 + 2p^2)
GeronimoStilton
2017-03-09 19:52:08
$4p - 6p^2$
$4p - 6p^2$
alifenix-
2017-03-09 19:52:08
$1 - 2p^2 - 1 + 4p - 4p^2 = 4p - 6p^2$
$1 - 2p^2 - 1 + 4p - 4p^2 = 4p - 6p^2$
QuestForKnowledge
2017-03-09 19:52:08
1-(6p^2-4p+1)
1-(6p^2-4p+1)
MathTechFire
2017-03-09 19:52:08
1-4p+6p^2
1-4p+6p^2
ezhao02
2017-03-09 19:52:08
$1-(1-2p)^2-p^2-p^2$
$1-(1-2p)^2-p^2-p^2$
copeland
2017-03-09 19:52:11
The probability that the chord intersects the probability is complementary to the probability that both points are in one of the three regions above, so it is \[1-p^2-p^2-(1-4p+4p^2)=4p-6p^2.\]
The probability that the chord intersects the probability is complementary to the probability that both points are in one of the three regions above, so it is \[1-p^2-p^2-(1-4p+4p^2)=4p-6p^2.\]
copeland
2017-03-09 19:52:13
We are trying to solve\[4p-6p^2=\dfrac{14}{25}.\]
We are trying to solve\[4p-6p^2=\dfrac{14}{25}.\]
copeland
2017-03-09 19:52:15
What are the solutions to this equation?
What are the solutions to this equation?
GeneralCobra19
2017-03-09 19:52:40
Is there a nicer way without quadratic formula?
Is there a nicer way without quadratic formula?
copeland
2017-03-09 19:52:47
For sure. Shall I?
For sure. Shall I?
MrMXS
2017-03-09 19:53:00
go ahead
go ahead
MegaProblemSolver
2017-03-09 19:53:00
yes
yes
MathTechFire
2017-03-09 19:53:00
definitely
definitely
awesomemaths
2017-03-09 19:53:00
yep
yep
Pandasareamazing.
2017-03-09 19:53:00
yess!
yess!
Ani10
2017-03-09 19:53:00
yas
yas
jfmath04
2017-03-09 19:53:00
yup
yup
amackenzie1
2017-03-09 19:53:02
yes please!
yes please!
winnertakeover
2017-03-09 19:53:02
yes
yes
copeland
2017-03-09 19:53:04
When we multiply by 25 we get \[150p^2-100p+14=0.\]Dividing by 2 gives\[75p^2-50p+7=0.\]
When we multiply by 25 we get \[150p^2-100p+14=0.\]Dividing by 2 gives\[75p^2-50p+7=0.\]
copeland
2017-03-09 19:53:13
I see $5p$ in there.
I see $5p$ in there.
copeland
2017-03-09 19:53:15
In terms of $5p$ we have \[3(5p)^2-10(5p)+7=0.\] Since $3+7=10$, we see $5p=1$ as an immediate solution, so we're set up to factor. This quadratic factors as
\[\left((3(5p)-7\right)\left((5p)-1\right)=0.\]
In terms of $5p$ we have \[3(5p)^2-10(5p)+7=0.\] Since $3+7=10$, we see $5p=1$ as an immediate solution, so we're set up to factor. This quadratic factors as
\[\left((3(5p)-7\right)\left((5p)-1\right)=0.\]
copeland
2017-03-09 19:53:24
The solutions are $p=\dfrac{7}{15}$ and $p=\dfrac15$.
The solutions are $p=\dfrac{7}{15}$ and $p=\dfrac15$.
copeland
2017-03-09 19:53:26
But what do we need now?
But what do we need now?
MrMXS
2017-03-09 19:53:52
$x$
$x$
jonzli123
2017-03-09 19:53:52
x
x
islander7
2017-03-09 19:53:52
value of x
value of x
Deathranger999
2017-03-09 19:53:52
The value of x
The value of x
Atg
2017-03-09 19:53:52
Multiply p by 180
Multiply p by 180
MathTechFire
2017-03-09 19:53:52
x
x
Funnybunny5246
2017-03-09 19:53:52
x
x
vishwathganesan
2017-03-09 19:53:52
we want x in degrees, not p's
we want x in degrees, not p's
fdas
2017-03-09 19:53:54
the angles
the angles
guoziyanglovemath
2017-03-09 19:53:54
work out x
work out x
copeland
2017-03-09 19:53:56
We want things in terms of $x$. What are the two possible values of $x?$
We want things in terms of $x$. What are the two possible values of $x?$
algebra_star1234
2017-03-09 19:54:20
84 and 36
84 and 36
vvluo
2017-03-09 19:54:20
36,84
36,84
Deathranger999
2017-03-09 19:54:20
x = 84 and x = 36.
x = 84 and x = 36.
cooljoseph
2017-03-09 19:54:20
36 and 84
36 and 84
Radio2
2017-03-09 19:54:20
36 and 84
36 and 84
Reef334
2017-03-09 19:54:20
84 and 36
84 and 36
GeronimoStilton
2017-03-09 19:54:20
$36$ and $84$
$36$ and $84$
copeland
2017-03-09 19:54:22
And the answer?
And the answer?
mathman3880
2017-03-09 19:54:39
180(7/15 - 1/5) = 48
180(7/15 - 1/5) = 48
First
2017-03-09 19:54:39
48 is the answer
48 is the answer
stronto
2017-03-09 19:54:39
48
48
ninjataco
2017-03-09 19:54:39
048
048
pythonsquared
2017-03-09 19:54:39
48
48
math101010
2017-03-09 19:54:39
048
048
owm
2017-03-09 19:54:39
48
48
a1b2
2017-03-09 19:54:39
$\boxed{048}$
$\boxed{048}$
DemonPlat4
2017-03-09 19:54:39
84 - 36 = 048 :<
84 - 36 = 048 :<
cooleybz2013
2017-03-09 19:54:40
48
48
copeland
2017-03-09 19:54:43
Since $p=\dfrac{x}{180}$, these two probabilities correspond to $x=84$ and $x=36$.
Since $p=\dfrac{x}{180}$, these two probabilities correspond to $x=84$ and $x=36$.
copeland
2017-03-09 19:54:47
The difference between these two values of $x$ is $84-36=\boxed{048}.$
The difference between these two values of $x$ is $84-36=\boxed{048}.$
copeland
2017-03-09 19:54:55
Alright, not halfway there but almost. What's next?
Alright, not halfway there but almost. What's next?
letsgomath
2017-03-09 19:55:15
#7
#7
alphamom
2017-03-09 19:55:15
7!
7!
IsaacZ123
2017-03-09 19:55:15
number 7
number 7
GeorgeAvocados
2017-03-09 19:55:15
number 7 oh thank heaven
number 7 oh thank heaven
Jyzhang12
2017-03-09 19:55:15
numero 7
numero 7
copeland
2017-03-09 19:55:18
7. For nonnegative integers $a$ and $b$ with $a+b\leq6,$ let $T(a,b)=\dbinom6a\dbinom6b\dbinom6{a+b}.$ Let $S$ denote the sum of all $T(a,b),$ where $a$ and $b$ are nonnegative integers with $a+b\leq6.$ Find the remainder when $S$ is divided by 1000.
7. For nonnegative integers $a$ and $b$ with $a+b\leq6,$ let $T(a,b)=\dbinom6a\dbinom6b\dbinom6{a+b}.$ Let $S$ denote the sum of all $T(a,b),$ where $a$ and $b$ are nonnegative integers with $a+b\leq6.$ Find the remainder when $S$ is divided by 1000.
copeland
2017-03-09 19:55:19
I like the smell of this problem. There just has to be something really elegant going on here.
I like the smell of this problem. There just has to be something really elegant going on here.
copeland
2017-03-09 19:55:20
Let's think about how we construct a combinatorial argument. First, what is $\dbinom6a$?
Let's think about how we construct a combinatorial argument. First, what is $\dbinom6a$?
legolego
2017-03-09 19:55:45
choose a objects out of 6 objects
choose a objects out of 6 objects
IsaacZ123
2017-03-09 19:55:45
number of ways to choose a people from 6
number of ways to choose a people from 6
First
2017-03-09 19:55:45
I have 6 people and I need to pick a committee of a people
I have 6 people and I need to pick a committee of a people
Mathaddict11
2017-03-09 19:55:45
choosing a people from a group of 6 people
choosing a people from a group of 6 people
espeon12
2017-03-09 19:55:45
choosing a things from 6 things
choosing a things from 6 things
MrMXS
2017-03-09 19:55:45
the number of ways to choose $a$ objects from $6$ objects
the number of ways to choose $a$ objects from $6$ objects
letsgomath
2017-03-09 19:55:45
number of ways to choose a people out of 6
number of ways to choose a people out of 6
WhaleVomit
2017-03-09 19:55:45
number of ways to choose a people to murder out of 6 people
number of ways to choose a people to murder out of 6 people
amyhu910
2017-03-09 19:55:45
the number of ways to choose a items out of 6
the number of ways to choose a items out of 6
copeland
2017-03-09 19:55:48
This is the number of ways to choose $a$ elements from a set of size 6. For example, we could be picking $a$ rabbits to put hats on.
This is the number of ways to choose $a$ elements from a set of size 6. For example, we could be picking $a$ rabbits to put hats on.
copeland
2017-03-09 19:55:54
Of course there are a lot of equally nice stories that we can write, from block-walking to organizing strings of letters, etc., but they all boil down to the rabbit thing.
Of course there are a lot of equally nice stories that we can write, from block-walking to organizing strings of letters, etc., but they all boil down to the rabbit thing.
copeland
2017-03-09 19:55:56
So here we have a set of 6 things and we're picking $a$ of them.
So here we have a set of 6 things and we're picking $a$ of them.
copeland
2017-03-09 19:56:00
We also have another set of 6 things and we're picking $b$ of them. (Maybe it's the same set. Keep that option in mind.)
We also have another set of 6 things and we're picking $b$ of them. (Maybe it's the same set. Keep that option in mind.)
copeland
2017-03-09 19:56:04
Finally we have a third set and we're picking $a+b$ of them.
Finally we have a third set and we're picking $a+b$ of them.
copeland
2017-03-09 19:56:07
So we have 6 rabbits and we're putting $a$ hats on them. We have 6 coyotes and we're putting $b$ hats on them. We have 6 snowmen and we're putting $a+b$ hats on them.
So we have 6 rabbits and we're putting $a$ hats on them. We have 6 coyotes and we're putting $b$ hats on them. We have 6 snowmen and we're putting $a+b$ hats on them.
copeland
2017-03-09 19:56:15
Snowmen love hats.
Snowmen love hats.
copeland
2017-03-09 19:56:20
I don't see it yet. Is there some nice trick we can play with these sets?
I don't see it yet. Is there some nice trick we can play with these sets?
lego101
2017-03-09 19:56:35
wouldnt the coyotes eat the rabbits
wouldnt the coyotes eat the rabbits
copeland
2017-03-09 19:56:36
It depends on $b-a$.
It depends on $b-a$.
ninjataco
2017-03-09 19:57:18
6C(a+b) = 6C(6-a-b)
6C(a+b) = 6C(6-a-b)
amackenzie1
2017-03-09 19:57:18
6 choose a + b = 6 choose 6 -(a + b)
6 choose a + b = 6 choose 6 -(a + b)
GeronimoStilton
2017-03-09 19:57:18
We're choosing $6-a-b$ snowmen not to put hats on.
We're choosing $6-a-b$ snowmen not to put hats on.
legolego
2017-03-09 19:57:18
6 choose (a+b) = 6 choose (6 - a - b)
6 choose (a+b) = 6 choose (6 - a - b)
vishwathganesan
2017-03-09 19:57:18
wait, we can also choose 6-a-b snowmen not to put hats on
wait, we can also choose 6-a-b snowmen not to put hats on
copeland
2017-03-09 19:57:23
The ways to pick $a+b$ snowmen from a set of 6 are the same as the ways to pick $6-a-b$ snowmen from a set of 6.
The ways to pick $a+b$ snowmen from a set of 6 are the same as the ways to pick $6-a-b$ snowmen from a set of 6.
copeland
2017-03-09 19:57:26
So this counts the ways to pick $a$ rabbits and then pick $b$ coyotes and then pick $6-a-b$ snowmen. How is that better?
So this counts the ways to pick $a$ rabbits and then pick $b$ coyotes and then pick $6-a-b$ snowmen. How is that better?
IsaacZ123
2017-03-09 19:58:21
we are giving a total of 6 hats
we are giving a total of 6 hats
vishwathganesan
2017-03-09 19:58:21
pick 6 out of 18 total
pick 6 out of 18 total
legolego
2017-03-09 19:58:21
our sum equals 18 choose 6
our sum equals 18 choose 6
GeronimoStilton
2017-03-09 19:58:21
We're picking $6$ members out of a set with $18$ elements!
We're picking $6$ members out of a set with $18$ elements!
amackenzie1
2017-03-09 19:58:21
basically you're putting 6 hats on to 18 things
basically you're putting 6 hats on to 18 things
cooljoseph
2017-03-09 19:58:21
Well, this is the same as choosing 6 things out of 18 to put hats on.
Well, this is the same as choosing 6 things out of 18 to put hats on.
IsaacZ123
2017-03-09 19:58:21
we are choosing a total of 6 out of 18
we are choosing a total of 6 out of 18
espeon12
2017-03-09 19:58:21
if you combine all the snowmen, coyotes, and rabbits, you're choosing 6 to put hats on
if you combine all the snowmen, coyotes, and rabbits, you're choosing 6 to put hats on
islander7
2017-03-09 19:58:21
18 objects choose 6 of them
18 objects choose 6 of them
sxu
2017-03-09 19:58:21
6 hats, give a to rabbits, b to coyotes, rest to snowmen
6 hats, give a to rabbits, b to coyotes, rest to snowmen
copeland
2017-03-09 19:58:34
Now we're counting ways to pick 6 critters out of a full set of 18 critters. Say that more precisely for me.
Now we're counting ways to pick 6 critters out of a full set of 18 critters. Say that more precisely for me.
awesomemaths
2017-03-09 19:58:59
18 choose 6
18 choose 6
MrMXS
2017-03-09 19:58:59
$\binom{18}{6}$
$\binom{18}{6}$
GeronimoStilton
2017-03-09 19:58:59
We want $\dbinom{18}{6}$
We want $\dbinom{18}{6}$
DemonPlat4
2017-03-09 19:58:59
18 choose 6
18 choose 6
alifenix-
2017-03-09 19:58:59
thats $\binom{18}{6}$, wish I realized this lol
thats $\binom{18}{6}$, wish I realized this lol
nukelauncher
2017-03-09 19:58:59
18 choose 6
18 choose 6
IsaacZ123
2017-03-09 19:58:59
18C6
18C6
QuestForKnowledge
2017-03-09 19:58:59
18 chose 6
18 chose 6
copeland
2017-03-09 19:59:01
Since we are summing over all possible $a$ and $b$, we are looking at all the ways to pick some number of bunnies, $a$, out of the first set, some other number of coyotes $b$ out of the second set, and then pulling whatever else is needed, $6-a-b$ snowmen, from the third set to create a set of 6 critters.
Since we are summing over all possible $a$ and $b$, we are looking at all the ways to pick some number of bunnies, $a$, out of the first set, some other number of coyotes $b$ out of the second set, and then pulling whatever else is needed, $6-a-b$ snowmen, from the third set to create a set of 6 critters.
copeland
2017-03-09 19:59:07
That's just $\dbinom{18}6$.
That's just $\dbinom{18}6$.
copeland
2017-03-09 19:59:08
\[\binom{18}6=\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13}{6\cdot5\cdot4\cdot3\cdot2\cdot1}.\]
\[\binom{18}6=\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13}{6\cdot5\cdot4\cdot3\cdot2\cdot1}.\]
copeland
2017-03-09 19:59:09
Anybody know what that is modulo 1000?
Anybody know what that is modulo 1000?
legolego
2017-03-09 19:59:46
564
564
thinkinavi
2017-03-09 19:59:46
564
564
IsaacZ123
2017-03-09 19:59:46
564
564
samuel
2017-03-09 19:59:46
564?
564?
math101010
2017-03-09 19:59:46
564
564
awesomemaths
2017-03-09 19:59:46
too lazy to compute
too lazy to compute
algebra_star1234
2017-03-09 19:59:46
564!!!!
564!!!!
letsgomath
2017-03-09 19:59:46
564
564
GeronimoStilton
2017-03-09 19:59:46
It's congruent to $564$ modulo 1000.
It's congruent to $564$ modulo 1000.
GeneralCobra19
2017-03-09 19:59:46
My calculator says 564
My calculator says 564
ilikepie2003
2017-03-09 19:59:46
no.
no.
DemonPlat4
2017-03-09 19:59:46
*cricket-chirp*
*cricket-chirp*
brainiac1
2017-03-09 19:59:46
564, clearly
564, clearly
Lance57
2017-03-09 19:59:46
564
564
copeland
2017-03-09 19:59:49
\begin{align*}
\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13}{6\cdot5\cdot4\cdot3\cdot2\cdot1}
&=17\cdot2\cdot3\cdot14\cdot13\\
&=102\cdot182\\
&=18{,}200+364\\
&=18{,}564.
\end{align*}
\begin{align*}
\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13}{6\cdot5\cdot4\cdot3\cdot2\cdot1}
&=17\cdot2\cdot3\cdot14\cdot13\\
&=102\cdot182\\
&=18{,}200+364\\
&=18{,}564.
\end{align*}
copeland
2017-03-09 19:59:56
The answer is $\boxed{564}$.
The answer is $\boxed{564}$.
copeland
2017-03-09 20:00:01
This is a case of the Vandermonde's Identity, which is the general name for all of the identities that you get by writing $\dbinom nm$ as a sum by breaking $n$ into subsets like this.
This is a case of the Vandermonde's Identity, which is the general name for all of the identities that you get by writing $\dbinom nm$ as a sum by breaking $n$ into subsets like this.
copeland
2017-03-09 20:00:13
Now we're definitely almost halfway there maybe.
Now we're definitely almost halfway there maybe.
First
2017-03-09 20:00:36
#8!
#8!
GeronimoStilton
2017-03-09 20:00:36
Problem 8!
Problem 8!
MathTechFire
2017-03-09 20:00:36
POUND 8
POUND 8
ethanliu247
2017-03-09 20:00:36
Great! number 8 next!
Great! number 8 next!
IsaacZ123
2017-03-09 20:00:36
number 8 is not half of 15
number 8 is not half of 15
jkittykitkat
2017-03-09 20:00:36
number 8
number 8
Pandasareamazing.
2017-03-09 20:00:36
On to number 8 we go!
On to number 8 we go!
copeland
2017-03-09 20:00:46
8. Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0,75).$ Let $O$ and $P$ be two points in the plane with $OP=200.$ Let $Q$ and $R$ be points on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b,$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR\leq 100$ is equal to $\dfrac mn,$ where $m$ and $N$ are relatively prime positive integers. Find $m+n$.
8. Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0,75).$ Let $O$ and $P$ be two points in the plane with $OP=200.$ Let $Q$ and $R$ be points on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b,$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR\leq 100$ is equal to $\dfrac mn,$ where $m$ and $N$ are relatively prime positive integers. Find $m+n$.
copeland
2017-03-09 20:00:49
Where should we start?
Where should we start?
IsaacZ123
2017-03-09 20:01:10
drawing a picture
drawing a picture
NewbieGamer
2017-03-09 20:01:10
Draw
Draw
mathman3880
2017-03-09 20:01:10
diagram
diagram
jonzli123
2017-03-09 20:01:10
draw a diagram!!!
draw a diagram!!!
awesomemaths
2017-03-09 20:01:10
diagram
diagram
Pandasareamazing.
2017-03-09 20:01:10
draw a diagram
draw a diagram
ninjataco
2017-03-09 20:01:10
diagram
diagram
stronto
2017-03-09 20:01:10
DIAGRAM
DIAGRAM
copeland
2017-03-09 20:01:16
Let's draw a picture! What does the right angle constraint on $\angle OQP$ and $\angle ORP$ tell us?
Let's draw a picture! What does the right angle constraint on $\angle OQP$ and $\angle ORP$ tell us?
fdas
2017-03-09 20:02:11
Draw a circle with OP as a diameter
Draw a circle with OP as a diameter
jkittykitkat
2017-03-09 20:02:11
It makes a circle
It makes a circle
DemonPlat4
2017-03-09 20:02:11
on a semicircle
on a semicircle
ethanliu247
2017-03-09 20:02:11
in a circle
in a circle
tdeng
2017-03-09 20:02:11
They are on a semicircle
They are on a semicircle
WhaleVomit
2017-03-09 20:02:11
they are on circle with diameter OP
they are on circle with diameter OP
islander7
2017-03-09 20:02:11
on semicircle with diameter pq
on semicircle with diameter pq
fdas
2017-03-09 20:02:11
they are in a circle
they are in a circle
GeronimoStilton
2017-03-09 20:02:11
$Q$ and $R$ are on a semicircle.
$Q$ and $R$ are on a semicircle.
copeland
2017-03-09 20:02:13
If $\angle OQP=90^\circ$ then $Q$ lies on the circle with diameter $OP.$ So we're picking two points on this arc at random (for some notion of "random").
If $\angle OQP=90^\circ$ then $Q$ lies on the circle with diameter $OP.$ So we're picking two points on this arc at random (for some notion of "random").
copeland
2017-03-09 20:02:15
copeland
2017-03-09 20:02:16
So, the center is $\heartsuit$. It's not my fault; the problem-writer already took $O.$
So, the center is $\heartsuit$. It's not my fault; the problem-writer already took $O.$
copeland
2017-03-09 20:02:21
OK, so when is $QR<100?$
OK, so when is $QR<100?$
dt800298
2017-03-09 20:03:31
when the minor arc QR is less than 60 degrees
when the minor arc QR is less than 60 degrees
AlisonH
2017-03-09 20:03:31
angle Q(heart)R is less than 60
angle Q(heart)R is less than 60
Root01
2017-03-09 20:03:31
When arc formed by QR is less than 60 degrees
When arc formed by QR is less than 60 degrees
tdeng
2017-03-09 20:03:31
When $\angle Q\heart R \leq 60$
When $\angle Q\heart R \leq 60$
Celebrated
2017-03-09 20:03:31
arc QR is less than or equal to 60
arc QR is less than or equal to 60
WhaleVomit
2017-03-09 20:03:31
when <Q<3R < 60 degrees
when <Q<3R < 60 degrees
rapturt9
2017-03-09 20:03:31
When angle QheartR is <60 degrees
When angle QheartR is <60 degrees
Radio2
2017-03-09 20:03:31
When $\angle \heartsuit > 60^\circ$
When $\angle \heartsuit > 60^\circ$
copeland
2017-03-09 20:03:33
(Props for <Q<3R<60.)
(Props for <Q<3R<60.)
copeland
2017-03-09 20:03:36
The radius of the circle is $100$, so if $QR=100$ then $\triangle\heartsuit QR$ is equilateral.
The radius of the circle is $100$, so if $QR=100$ then $\triangle\heartsuit QR$ is equilateral.
copeland
2017-03-09 20:03:38
What is $\angle Q\heartsuit R$ in terms of $a$ and $b?$
What is $\angle Q\heartsuit R$ in terms of $a$ and $b?$
vishwathganesan
2017-03-09 20:04:52
when |a-b| < 30
when |a-b| < 30
stronto
2017-03-09 20:04:52
2a-2b
2a-2b
WhaleVomit
2017-03-09 20:04:52
|2a-2b|
|2a-2b|
ninjataco
2017-03-09 20:04:52
|2a-2b|
|2a-2b|
guoziyanglovemath
2017-03-09 20:04:52
2|a-b|
2|a-b|
vvluo
2017-03-09 20:04:52
a-b<=30
a-b<=30
jonzli123
2017-03-09 20:04:52
|2b-2a|
|2b-2a|
rapturt9
2017-03-09 20:04:52
The absolute value of 2a-2b
The absolute value of 2a-2b
copeland
2017-03-09 20:04:54
Since $\angle POQ=a$ the central angle is $\angle P\heartsuit Q=2a$. Likewise, $\angle P\heartsuit R=2b.$ Therefore $\angle Q\heartsuit R=|2a-2b|.$
Since $\angle POQ=a$ the central angle is $\angle P\heartsuit Q=2a$. Likewise, $\angle P\heartsuit R=2b.$ Therefore $\angle Q\heartsuit R=|2a-2b|.$
copeland
2017-03-09 20:04:57
We want to find the probability that two angles $a$ and $b$ chosen uniformly from $[0,75]$ satisfy $|2a-2b|<60.$
We want to find the probability that two angles $a$ and $b$ chosen uniformly from $[0,75]$ satisfy $|2a-2b|<60.$
copeland
2017-03-09 20:05:00
This is a tool for. . .
This is a tool for. . .
IsaacZ123
2017-03-09 20:05:43
geometric probability
geometric probability
ninjataco
2017-03-09 20:05:43
geometric probability
geometric probability
First
2017-03-09 20:05:43
geometric probability
geometric probability
stronto
2017-03-09 20:05:43
geometric probability
geometric probability
a1b2
2017-03-09 20:05:43
Geometric probability
Geometric probability
pythonsquared
2017-03-09 20:05:43
geometric probability
geometric probability
tdeng
2017-03-09 20:05:43
Geometric probability
Geometric probability
WhaleVomit
2017-03-09 20:05:43
geometric probability
geometric probability
fdas
2017-03-09 20:05:43
geometric probability
geometric probability
Radio2
2017-03-09 20:05:43
geometric probability
geometric probability
copeland
2017-03-09 20:05:45
Geometric probability! First we draw a rectangle representing the possible values for $a$ and $b$:
Geometric probability! First we draw a rectangle representing the possible values for $a$ and $b$:
copeland
2017-03-09 20:05:46
copeland
2017-03-09 20:05:47
And what does the success region look like?
And what does the success region look like?
IsaacZ123
2017-03-09 20:06:38
two triangles in the corners
two triangles in the corners
tdeng
2017-03-09 20:06:38
A hexagon
A hexagon
vishwathganesan
2017-03-09 20:06:38
it is a hexagon
it is a hexagon
andsun19
2017-03-09 20:06:38
diagonal
diagonal
brainiac1
2017-03-09 20:06:38
a strip along the long diagonal
a strip along the long diagonal
NewbieGamer
2017-03-09 20:06:38
Square minus two triangles
Square minus two triangles
rapturt9
2017-03-09 20:06:38
a wide diagonal line
a wide diagonal line
Blue_Whale
2017-03-09 20:06:38
diagonal stripe from bottomleft to topright?
diagonal stripe from bottomleft to topright?
J1618
2017-03-09 20:06:38
A diagonal strip with horizontal width 30
A diagonal strip with horizontal width 30
ninjataco
2017-03-09 20:06:38
chop off two triangles in the upper left and lower right corners
chop off two triangles in the upper left and lower right corners
legolego
2017-03-09 20:06:38
a stripe
a stripe
vvluo
2017-03-09 20:06:38
stripe with 2 isosceles right triangles being unsuccessful
stripe with 2 isosceles right triangles being unsuccessful
J1618
2017-03-09 20:06:38
A diagonal strip from b=30 and a=30
A diagonal strip from b=30 and a=30
copeland
2017-03-09 20:06:41
The success region is where $a$ and $b$ are within 30 of one another. We want to draw the lines $b=a+30$ and $b=a-30,$ since those are the boundaries of the successful region.
The success region is where $a$ and $b$ are within 30 of one another. We want to draw the lines $b=a+30$ and $b=a-30,$ since those are the boundaries of the successful region.
copeland
2017-03-09 20:06:43
The whole diagonal is in the success region, so we want to color between the boundary lines.
The whole diagonal is in the success region, so we want to color between the boundary lines.
copeland
2017-03-09 20:06:44
copeland
2017-03-09 20:06:49
What do we want to compute?
What do we want to compute?
IsaacZ123
2017-03-09 20:07:23
the area of the shaded region/total area
the area of the shaded region/total area
cooleybz2013
2017-03-09 20:07:23
the area ratio
the area ratio
Celebrated
2017-03-09 20:07:23
shaded region in terms of whole thig
shaded region in terms of whole thig
math129
2017-03-09 20:07:23
area of stripe/area of square
area of stripe/area of square
Radio2
2017-03-09 20:07:23
the shaded divided by the total
the shaded divided by the total
mathboy4
2017-03-09 20:07:23
the shaded place
the shaded place
stronto
2017-03-09 20:07:23
shaded region/total area
shaded region/total area
copeland
2017-03-09 20:07:30
We want to compute the proportion of the square that is gray.
We want to compute the proportion of the square that is gray.
copeland
2017-03-09 20:07:32
What should we do first?
What should we do first?
copeland
2017-03-09 20:07:40
Eh. That's a bad question.
Eh. That's a bad question.
vishwathganesan
2017-03-09 20:07:50
scale it!!!!!!!1
scale it!!!!!!!1
copeland
2017-03-09 20:07:56
But that's a good answer:
But that's a good answer:
copeland
2017-03-09 20:07:57
It's fine to rescale everything in sight by 15, since that won't change the proportions. Here's a better picture:
It's fine to rescale everything in sight by 15, since that won't change the proportions. Here's a better picture:
copeland
2017-03-09 20:07:58
copeland
2017-03-09 20:08:22
What's the probability that $QR$ is less than 100?
What's the probability that $QR$ is less than 100?
letsgomath
2017-03-09 20:08:54
16/25
16/25
sxu
2017-03-09 20:08:54
16/25
16/25
jkittykitkat
2017-03-09 20:08:54
25-9=16. 16/25
25-9=16. 16/25
bogstop320
2017-03-09 20:08:54
16/25
16/25
mathman3880
2017-03-09 20:08:54
16/25
16/25
Reef334
2017-03-09 20:08:54
16/25
16/25
islander7
2017-03-09 20:08:54
16/25
16/25
Ani10
2017-03-09 20:08:54
16/25
16/25
MountainHeight
2017-03-09 20:08:54
16/25
16/25
owm
2017-03-09 20:08:54
16/25
16/25
copeland
2017-03-09 20:08:58
The white area can be shoved together to get a $3\times3$ square, so its area is 9.
The white area can be shoved together to get a $3\times3$ square, so its area is 9.
copeland
2017-03-09 20:08:59
The full square has area 25 and the grey region has area $25-9=16$. Therefore the probability of success is $\dfrac{16}{25}.$ The answer is $16+25=\boxed{041}.$
The full square has area 25 and the grey region has area $25-9=16$. Therefore the probability of success is $\dfrac{16}{25}.$ The answer is $16+25=\boxed{041}.$
copeland
2017-03-09 20:09:02
Theorem: the probability that a geometric probability problem uses some version of this diagram is also $\dfrac{16}{25}.$
Theorem: the probability that a geometric probability problem uses some version of this diagram is also $\dfrac{16}{25}.$
copeland
2017-03-09 20:09:04
copeland
2017-03-09 20:09:14
Ready for the next one?
Ready for the next one?
Celebrated
2017-03-09 20:09:34
prob 9
prob 9
Celebrated
2017-03-09 20:09:34
yea
yea
IsaacZ123
2017-03-09 20:09:34
wait what
wait what
MegaProblemSolver
2017-03-09 20:09:34
yeah!
yeah!
jonzli123
2017-03-09 20:09:34
yeah!
yeah!
jfmath04
2017-03-09 20:09:34
yas
yas
yrnsmurf
2017-03-09 20:09:34
yes #9
yes #9
IsaacZ123
2017-03-09 20:09:34
what was the theorem u just stated
what was the theorem u just stated
lego101
2017-03-09 20:09:34
Number nine, lookin fine.
Number nine, lookin fine.
jkittykitkat
2017-03-09 20:09:34
Time to get another one wrong
Time to get another one wrong
copeland
2017-03-09 20:09:52
9. Let $a_{10}=10,$ and for each integer $n>10$ let $a_n=100a_{n-1}+n.$ Find the least $n>10$ such that $a_n$ is a multiple of 99.
9. Let $a_{10}=10,$ and for each integer $n>10$ let $a_n=100a_{n-1}+n.$ Find the least $n>10$ such that $a_n$ is a multiple of 99.
copeland
2017-03-09 20:09:58
So we have "is a multiple of 99", which means. . .
So we have "is a multiple of 99", which means. . .
dhruv
2017-03-09 20:10:41
mods
mods
sxu
2017-03-09 20:10:41
mod 99
mod 99
First
2017-03-09 20:10:41
$\pmod{99}$
$\pmod{99}$
GeronimoStilton
2017-03-09 20:10:41
Express it modulo $99$!
Express it modulo $99$!
ninjataco
2017-03-09 20:10:41
a_n == 0 (mod 99)
a_n == 0 (mod 99)
Mrkiller
2017-03-09 20:10:41
congruent to 0 mod 99
congruent to 0 mod 99
yojan_sushi
2017-03-09 20:10:41
0 mod 99
0 mod 99
jkittykitkat
2017-03-09 20:10:41
0 mod 99
0 mod 99
pythonsquared
2017-03-09 20:10:41
modular arithmetic
modular arithmetic
jonzli123
2017-03-09 20:10:41
mod 99
mod 99
copeland
2017-03-09 20:10:49
Consider everything modulo 99. What's the first simplification of the problem?
Consider everything modulo 99. What's the first simplification of the problem?
copeland
2017-03-09 20:11:00
(Also, nobody loves the CRT more than I do, but let's hold off until we need it.)
(Also, nobody loves the CRT more than I do, but let's hold off until we need it.)
Mrkiller
2017-03-09 20:11:41
so 100 becomes 1
so 100 becomes 1
QuestForKnowledge
2017-03-09 20:11:41
an=a(n-1)+n MUCH nicer
an=a(n-1)+n MUCH nicer
Reef334
2017-03-09 20:11:41
a_n = a_{n-1)+n
a_n = a_{n-1)+n
LittleChimp
2017-03-09 20:11:41
change the 100 to a 1
change the 100 to a 1
sxu
2017-03-09 20:11:41
$a_n=a_{n-1}+n$
$a_n=a_{n-1}+n$
amackenzie1
2017-03-09 20:11:41
$100{a_{n-1}} = {a_{n-1}}$.
$100{a_{n-1}} = {a_{n-1}}$.
jkittykitkat
2017-03-09 20:11:41
an=a(n-1)+n
an=a(n-1)+n
MountainHeight
2017-03-09 20:11:41
a_n = a_(n-1) + n
a_n = a_(n-1) + n
copeland
2017-03-09 20:11:43
Well, $100\equiv1\pmod{99},$ so we should think of the recursion $a_n=a_{n-1}+n$.
Well, $100\equiv1\pmod{99},$ so we should think of the recursion $a_n=a_{n-1}+n$.
copeland
2017-03-09 20:11:45
What's that the recursion for?
What's that the recursion for?
yrnsmurf
2017-03-09 20:12:17
triangular numbers
triangular numbers
letsgomath
2017-03-09 20:12:17
triangular numbers
triangular numbers
First
2017-03-09 20:12:17
Triangular numbers
Triangular numbers
DemonPlat4
2017-03-09 20:12:17
triangular numbers
triangular numbers
amackenzie1
2017-03-09 20:12:17
triangle numbers
triangle numbers
tdeng
2017-03-09 20:12:17
Triangular numbers
Triangular numbers
ethanliu247
2017-03-09 20:12:17
triangular numbers
triangular numbers
copeland
2017-03-09 20:12:19
That's the recursion for the triangle numbers. What's the relationship between $a_n$ and $\dfrac{n(n+1)}2?$
That's the recursion for the triangle numbers. What's the relationship between $a_n$ and $\dfrac{n(n+1)}2?$
yrnsmurf
2017-03-09 20:13:02
an=n(n+1)/2-45
an=n(n+1)/2-45
letsgomath
2017-03-09 20:13:02
missing 1+2+3+4+5+6+7+8+9
missing 1+2+3+4+5+6+7+8+9
tdeng
2017-03-09 20:13:02
a_n=n(n+1)/2-45
a_n=n(n+1)/2-45
IsaacZ123
2017-03-09 20:13:02
we have to take out 1+2+3+4+5+6+7+8+9
we have to take out 1+2+3+4+5+6+7+8+9
vishwathganesan
2017-03-09 20:13:02
the first is 45 more than the seconf
the first is 45 more than the seconf
islander7
2017-03-09 20:13:02
a_n=n(n+1)/2-45
a_n=n(n+1)/2-45
vishwathganesan
2017-03-09 20:13:05
i mean 45 less
i mean 45 less
copeland
2017-03-09 20:13:08
The difference is that we forgot to add in the first $1+2+\cdots+9=\dfrac{9(10)}2$.
The difference is that we forgot to add in the first $1+2+\cdots+9=\dfrac{9(10)}2$.
copeland
2017-03-09 20:13:12
\[a_n=\dfrac{n(n+1)}2-\dfrac{9\cdot10}2.\]
\[a_n=\dfrac{n(n+1)}2-\dfrac{9\cdot10}2.\]
copeland
2017-03-09 20:13:14
This is a polynomial in $n$. Check this trick out: Do you see any roots of this polynomial?
This is a polynomial in $n$. Check this trick out: Do you see any roots of this polynomial?
ninjataco
2017-03-09 20:13:39
9
9
math129
2017-03-09 20:13:39
n=9
n=9
amackenzie1
2017-03-09 20:13:39
9 is a root
9 is a root
IsaacZ123
2017-03-09 20:13:39
n=9
n=9
DemonPlat4
2017-03-09 20:13:39
n=9
n=9
brainiac1
2017-03-09 20:13:39
9
9
copeland
2017-03-09 20:13:43
If you set $n=9$ you get \[\dfrac{n(n+1)}2-\dfrac{9\cdot10}2=\dfrac{9\cdot10}2-\dfrac{9\cdot10}2=0.\] That makes sense since $a_9=0$ must be true if $a_{10}=10$.
If you set $n=9$ you get \[\dfrac{n(n+1)}2-\dfrac{9\cdot10}2=\dfrac{9\cdot10}2-\dfrac{9\cdot10}2=0.\] That makes sense since $a_9=0$ must be true if $a_{10}=10$.
copeland
2017-03-09 20:13:43
See the other root?
See the other root?
nukelauncher
2017-03-09 20:13:58
9 and -10
9 and -10
brainiac1
2017-03-09 20:13:58
9 and -10
9 and -10
brainiac1
2017-03-09 20:13:58
-10
-10
QuestForKnowledge
2017-03-09 20:13:58
-10
-10
DemonPlat4
2017-03-09 20:13:58
n=-10
n=-10
Funnybunny5246
2017-03-09 20:13:58
-10
-10
NeeNeeMath
2017-03-09 20:13:58
-10
-10
fdas
2017-03-09 20:13:58
-10
-10
copeland
2017-03-09 20:13:59
If you set $n=-10$ you get \[\dfrac{n(n+1)}2-\dfrac{9\cdot10}2=\dfrac{(-10)(-9)}2-\dfrac{9\cdot10}2=0.\]
If you set $n=-10$ you get \[\dfrac{n(n+1)}2-\dfrac{9\cdot10}2=\dfrac{(-10)(-9)}2-\dfrac{9\cdot10}2=0.\]
copeland
2017-03-09 20:14:00
Therefore $a_n=\dfrac12(n-9)(n+10).$
Therefore $a_n=\dfrac12(n-9)(n+10).$
copeland
2017-03-09 20:14:03
Incidentally, difference of rectangles is a thing. Try to find a geometric proof that $a(a+x)-b(b+x)=(a-b)(a+b+x)$. Difference of squares is $x=0$, and we just used the $x=1$ case.
Incidentally, difference of rectangles is a thing. Try to find a geometric proof that $a(a+x)-b(b+x)=(a-b)(a+b+x)$. Difference of squares is $x=0$, and we just used the $x=1$ case.
copeland
2017-03-09 20:14:11
But not now.
But not now.
copeland
2017-03-09 20:14:13
What are we looking for?
What are we looking for?
letsgomath
2017-03-09 20:15:10
0 mod 99
0 mod 99
QuestForKnowledge
2017-03-09 20:15:10
a n such that the equation is 0 mod 99
a n such that the equation is 0 mod 99
jkittykitkat
2017-03-09 20:15:10
it being a multiple of 99
it being a multiple of 99
a1b2
2017-03-09 20:15:10
$\equiv 0 \mod 99$
$\equiv 0 \mod 99$
amackenzie1
2017-03-09 20:15:10
a_n to be a multiple of $99$.
a_n to be a multiple of $99$.
IsaacZ123
2017-03-09 20:15:10
looking for the next time it is 0 mod 99
looking for the next time it is 0 mod 99
guoziyanglovemath
2017-03-09 20:15:10
multiple of 99
multiple of 99
DjokerNole
2017-03-09 20:15:10
find the least a_n is a multiple of 99
find the least a_n is a multiple of 99
First
2017-03-09 20:15:10
0 \mod 99
0 \mod 99
vvluo
2017-03-09 20:15:10
multiple of 99
multiple of 99
copeland
2017-03-09 20:15:14
Since 99 is odd, we're looking for the first $n>10$ such that $99\mid (n-9)(n+10)$.
Since 99 is odd, we're looking for the first $n>10$ such that $99\mid (n-9)(n+10)$.
copeland
2017-03-09 20:15:17
Can we simplify this a little?
Can we simplify this a little?
copeland
2017-03-09 20:16:29
We could take a swing at CRT. Do we need to?
We could take a swing at CRT. Do we need to?
letsgomath
2017-03-09 20:16:46
9 and 11
9 and 11
NeeNeeMath
2017-03-09 20:16:46
yeah crt
yeah crt
IsaacZ123
2017-03-09 20:16:46
99=9*11
99=9*11
NeeNeeMath
2017-03-09 20:16:46
divisible b 9 and 11
divisible b 9 and 11
jkittykitkat
2017-03-09 20:16:46
do the case for 11 and 9 seperately
do the case for 11 and 9 seperately
Radio2
2017-03-09 20:16:46
no
no
IsaacZ123
2017-03-09 20:16:46
no
no
NeeNeeMath
2017-03-09 20:16:46
noooo
noooo
DjokerNole
2017-03-09 20:16:46
no
no
copeland
2017-03-09 20:17:10
Let's just start searching for the answer.
Let's just start searching for the answer.
copeland
2017-03-09 20:17:22
I want to make life easier by letting $m=n-9$ then we want the first $m>1$ such that $99\mid m(m+19)$.
I want to make life easier by letting $m=n-9$ then we want the first $m>1$ such that $99\mid m(m+19)$.
copeland
2017-03-09 20:17:30
Now what?
Now what?
Dr4gon39
2017-03-09 20:18:00
BASH
BASH
copeland
2017-03-09 20:18:02
Yeah, there had to be a point where we bash. This is the AIME after all.
Yeah, there had to be a point where we bash. This is the AIME after all.
fdas
2017-03-09 20:18:08
Start listing ms that create a multiple of 11
Start listing ms that create a multiple of 11
legolego
2017-03-09 20:18:12
either factor must have a factor of 11
either factor must have a factor of 11
copeland
2017-03-09 20:18:18
One of $m$ and $m+19$ is a multiple of 11. Let's start listing multiples of 11:
One of $m$ and $m+19$ is a multiple of 11. Let's start listing multiples of 11:
copeland
2017-03-09 20:18:20
\[\begin{array}{c|c|c}
x&x-19&x+19\\
\hline
11&-8&30\\
22&3&41\\
33&14&52\\
44&25&63\\
55&36&74\\
66&47&85\\
77&58&96\\
\end{array}\]
\[\begin{array}{c|c|c}
x&x-19&x+19\\
\hline
11&-8&30\\
22&3&41\\
33&14&52\\
44&25&63\\
55&36&74\\
66&47&85\\
77&58&96\\
\end{array}\]
copeland
2017-03-09 20:18:23
See any winners on that list?
See any winners on that list?
jkittykitkat
2017-03-09 20:19:10
55
55
GeronimoStilton
2017-03-09 20:19:10
$x = 44$
$x = 44$
islander7
2017-03-09 20:19:10
55
55
legolego
2017-03-09 20:19:10
55!
55!
GeronimoStilton
2017-03-09 20:19:10
$x = 55$
$x = 55$
amackenzie1
2017-03-09 20:19:10
55 - 36
55 - 36
yrnsmurf
2017-03-09 20:19:10
63 and 36 for 44 and 55
63 and 36 for 44 and 55
GeronimoStilton
2017-03-09 20:19:10
$x = 44,55$
$x = 44,55$
QuestForKnowledge
2017-03-09 20:19:10
x=44
x=44
Funnybunny5246
2017-03-09 20:19:10
36
36
vishwathganesan
2017-03-09 20:19:10
63?
63?
amyhu910
2017-03-09 20:19:10
x=44
x=44
IsaacZ123
2017-03-09 20:19:10
55
55
copeland
2017-03-09 20:19:12
The pairs $44\cdot63$ and $36\cdot55$ both work.
The pairs $44\cdot63$ and $36\cdot55$ both work.
copeland
2017-03-09 20:19:13
So what's $m?$
So what's $m?$
copeland
2017-03-09 20:20:16
Remember that $m$ is the smallest number such that $m(m+19)$ is a multiple of 99.
Remember that $m$ is the smallest number such that $m(m+19)$ is a multiple of 99.
legolego
2017-03-09 20:20:28
36
36
GeronimoStilton
2017-03-09 20:20:28
$36$
$36$
fdas
2017-03-09 20:20:28
36 because it is lower
36 because it is lower
Pandasareamazing.
2017-03-09 20:20:28
36
36
stronto
2017-03-09 20:20:28
36
36
guoziyanglovemath
2017-03-09 20:20:28
36
36
yrnsmurf
2017-03-09 20:20:28
36
36
Reef334
2017-03-09 20:20:28
36
36
islander7
2017-03-09 20:20:28
36
36
Jyzhang12
2017-03-09 20:20:28
36
36
copeland
2017-03-09 20:20:32
Since there aren't going to be any numbers less than 36 "lower" on the table, the first multiple of 99 comes with $m=36.$
Since there aren't going to be any numbers less than 36 "lower" on the table, the first multiple of 99 comes with $m=36.$
copeland
2017-03-09 20:20:32
And what's the final answer?
And what's the final answer?
nukelauncher
2017-03-09 20:21:01
045
045
brainiac1
2017-03-09 20:21:01
45
45
brainiac1
2017-03-09 20:21:01
45
45
strategos21
2017-03-09 20:21:01
045
045
rapturt9
2017-03-09 20:21:01
$45$
$45$
fdas
2017-03-09 20:21:01
45
45
mshanmugam
2017-03-09 20:21:01
45
45
stronto
2017-03-09 20:21:01
45
45
LittleChimp
2017-03-09 20:21:01
045
045
Pandasareamazing.
2017-03-09 20:21:01
36+9=45
36+9=45
thinkinavi
2017-03-09 20:21:01
045
045
copeland
2017-03-09 20:21:04
The first multiple of 99 comes when $n=9+36=\boxed{045}$, when $a_{45}=\dfrac{36\cdot55}2.$
The first multiple of 99 comes when $n=9+36=\boxed{045}$, when $a_{45}=\dfrac{36\cdot55}2.$
copeland
2017-03-09 20:21:10
Great.
Great.
copeland
2017-03-09 20:21:37
I'm sad we didn't get to CRT, but maybe we'll get to later.
I'm sad we didn't get to CRT, but maybe we'll get to later.
Lance57
2017-03-09 20:22:13
Question 10!
Question 10!
Pandasareamazing.
2017-03-09 20:22:13
now we go to #10
now we go to #10
lego101
2017-03-09 20:22:13
next problem?
next problem?
MathTechFire
2017-03-09 20:22:13
zen 10
zen 10
copeland
2017-03-09 20:22:22
CRT is the Chinese Remainder Theorem.
CRT is the Chinese Remainder Theorem.
copeland
2017-03-09 20:22:27
10. Let $z_1=18+83i,$ $z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that \[\frac{z_3-z_1}{z_2-z_1}\cdot\frac{z-z_2}{z-z_3}\] is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$.
10. Let $z_1=18+83i,$ $z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that \[\frac{z_3-z_1}{z_2-z_1}\cdot\frac{z-z_2}{z-z_3}\] is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$.
copeland
2017-03-09 20:22:30
OK, so that's a lot of fancy complex numbers stuff. I think this is probably a geometry problem. Think for a second where those three points are in the complex plane and I'll draw them for you.
OK, so that's a lot of fancy complex numbers stuff. I think this is probably a geometry problem. Think for a second where those three points are in the complex plane and I'll draw them for you.
copeland
2017-03-09 20:22:45
copeland
2017-03-09 20:22:49
Let's focus on the hard part.
Let's focus on the hard part.
copeland
2017-03-09 20:22:51
We have this crazy expression, \[\frac{z_3-z_1}{z_2-z_1}\cdot\frac{z-z_2}{z-z_3},\]and we want to know when it's real. What does being real tell us about the geometry here?
We have this crazy expression, \[\frac{z_3-z_1}{z_2-z_1}\cdot\frac{z-z_2}{z-z_3},\]and we want to know when it's real. What does being real tell us about the geometry here?
ninjataco
2017-03-09 20:23:28
on the x-axis
on the x-axis
a1b2
2017-03-09 20:23:28
It is on the real axis
It is on the real axis
espeon12
2017-03-09 20:23:28
it's on teh x axis?
it's on teh x axis?
Ani10
2017-03-09 20:23:28
lies on the x axis
lies on the x axis
tdeng
2017-03-09 20:23:28
It's on the real axis
It's on the real axis
JJShan26
2017-03-09 20:23:28
it's on the x axis
it's on the x axis
NeeNeeMath
2017-03-09 20:23:28
x axis
x axis
copeland
2017-03-09 20:23:29
So a number is real when it's on the "$x$-axis".
So a number is real when it's on the "$x$-axis".
copeland
2017-03-09 20:23:37
When have we seen a product be real?
When have we seen a product be real?
IsaacZ123
2017-03-09 20:24:14
two conjugates
two conjugates
First
2017-03-09 20:24:14
cojugates
cojugates
vishwathganesan
2017-03-09 20:24:14
conjugates
conjugates
ethanliu247
2017-03-09 20:24:14
when the complex number are conjugates
when the complex number are conjugates
sxu
2017-03-09 20:24:14
complex conjugates?
complex conjugates?
GeneralCobra19
2017-03-09 20:24:14
conjugates
conjugates
copeland
2017-03-09 20:24:15
Fabulous! Multiplying conjugates gives us a real number.
Fabulous! Multiplying conjugates gives us a real number.
copeland
2017-03-09 20:24:43
Complex conjugates have an angle relationship and a length relationship. Which one does reality care about?
Complex conjugates have an angle relationship and a length relationship. Which one does reality care about?
richuw
2017-03-09 20:25:34
Degree measures add up to be 180n
Degree measures add up to be 180n
islander7
2017-03-09 20:25:34
angle
angle
tree3
2017-03-09 20:25:34
angle
angle
a1b2
2017-03-09 20:25:34
Angle
Angle
IsaacZ123
2017-03-09 20:25:34
angle relationshiop
angle relationshiop
fishy15
2017-03-09 20:25:34
angle
angle
samuel
2017-03-09 20:25:34
angle!
angle!
dr3463
2017-03-09 20:25:34
angle
angle
samuel
2017-03-09 20:25:36
length doesn't matter
length doesn't matter
copeland
2017-03-09 20:25:37
If we change the lengths, it doesn't change reality. This is a statement about angles.
If we change the lengths, it doesn't change reality. This is a statement about angles.
copeland
2017-03-09 20:26:11
We've been talking about angles that add to zero. How can we convert that to a statement like, "angles are equal?"
We've been talking about angles that add to zero. How can we convert that to a statement like, "angles are equal?"
dr3463
2017-03-09 20:26:50
reflection
reflection
copeland
2017-03-09 20:26:58
Cool, we could reflect one of the numbers.
Cool, we could reflect one of the numbers.
copeland
2017-03-09 20:27:10
How do you reflect a number across the real axis?
How do you reflect a number across the real axis?
a1b2
2017-03-09 20:27:50
Complex Conjugate
Complex Conjugate
MountainHeight
2017-03-09 20:27:50
take its conjugate
take its conjugate
ninjataco
2017-03-09 20:27:50
take its conjugate
take its conjugate
espeon12
2017-03-09 20:27:50
graph its conjugate
graph its conjugate
copeland
2017-03-09 20:27:56
That's good. Anything else?
That's good. Anything else?
yrnsmurf
2017-03-09 20:28:09
take its reciprocal
take its reciprocal
MSTang
2017-03-09 20:28:09
Reciprocate
Reciprocate
copeland
2017-03-09 20:28:24
Great. The argument of $\dfrac1z$ is the negative of the argument of $z.$
Great. The argument of $\dfrac1z$ is the negative of the argument of $z.$
copeland
2017-03-09 20:28:26
If we have a pair of complex numbers $u$ and $v$ then $\dfrac u v$ is real when $u$ and $v$ can be written with the same argument. That is, if $u=re^{i\theta}$ and $v=se^{i\phi}$ then in order for $\dfrac uv$ to be real, we have to have $\theta=\phi$ (or possibly $\theta=\phi+\pi$).
If we have a pair of complex numbers $u$ and $v$ then $\dfrac u v$ is real when $u$ and $v$ can be written with the same argument. That is, if $u=re^{i\theta}$ and $v=se^{i\phi}$ then in order for $\dfrac uv$ to be real, we have to have $\theta=\phi$ (or possibly $\theta=\phi+\pi$).
copeland
2017-03-09 20:28:34
So in order for $\dfrac{z_3-z_1}{z_2-z_1}\cdot\dfrac{z-z_2}{z-z_3}$ to be real, we would need $\dfrac{z_3-z_1}{z_2-z_1}$ and $\dfrac{z-z_3}{z-z_2}$ to have the same argument.
So in order for $\dfrac{z_3-z_1}{z_2-z_1}\cdot\dfrac{z-z_2}{z-z_3}$ to be real, we would need $\dfrac{z_3-z_1}{z_2-z_1}$ and $\dfrac{z-z_3}{z-z_2}$ to have the same argument.
copeland
2017-03-09 20:28:44
Where can I find the argument of $\dfrac{z_3-z_1}{z_2-z_1}$ on this diagram?
Where can I find the argument of $\dfrac{z_3-z_1}{z_2-z_1}$ on this diagram?
copeland
2017-03-09 20:28:53
copeland
2017-03-09 20:30:03
Where can I find $z_3-z_1?$
Where can I find $z_3-z_1?$
IsaacZ123
2017-03-09 20:30:28
the line connecting z_3 and z_1
the line connecting z_3 and z_1
GeronimoStilton
2017-03-09 20:30:28
The line from $z_3$ to $z_1$.
The line from $z_3$ to $z_1$.
copeland
2017-03-09 20:30:30
That's the "vector" that points from $z_1$ to $z_3.$
That's the "vector" that points from $z_1$ to $z_3.$
copeland
2017-03-09 20:31:19
And $z_2-z_1$ is the "vector" pointing from $z_1$ to $z_2$.
And $z_2-z_1$ is the "vector" pointing from $z_1$ to $z_2$.
copeland
2017-03-09 20:31:20
What is the angle of their quotient?
What is the angle of their quotient?
IsaacZ123
2017-03-09 20:31:35
draw $z_3z_1z_2$
draw $z_3z_1z_2$
KYang
2017-03-09 20:31:35
the angle z2 z1 z3 ??
the angle z2 z1 z3 ??
IsaacZ123
2017-03-09 20:31:35
the angle of $z_2z_1z_3$
the angle of $z_2z_1z_3$
yrnsmurf
2017-03-09 20:31:35
the angle of <z3z1z2
the angle of <z3z1z2
GeronimoStilton
2017-03-09 20:31:35
The angle $Z_2Z_1Z_3$?
The angle $Z_2Z_1Z_3$?
brainiac1
2017-03-09 20:31:35
the angle from z2 to z3 through z1
the angle from z2 to z3 through z1
copeland
2017-03-09 20:31:37
$z_3-z_1$ is the "vector" that points from $z_1$ to $z_3$ and $z_2-z_1$ is the "vector" that points from $z_1$ to $z_2$. The argument of their quotient is the angle at $z_1$ from $z_3$ to $z_2$.
$z_3-z_1$ is the "vector" that points from $z_1$ to $z_3$ and $z_2-z_1$ is the "vector" that points from $z_1$ to $z_2$. The argument of their quotient is the angle at $z_1$ from $z_3$ to $z_2$.
copeland
2017-03-09 20:31:39
copeland
2017-03-09 20:31:51
And if we have a $z$ such that $\dfrac{z-z_3}{z-z_2}$ has the same argument, what does that say?
And if we have a $z$ such that $\dfrac{z-z_3}{z-z_2}$ has the same argument, what does that say?
islander7
2017-03-09 20:32:12
same angle
same angle
vishwathganesan
2017-03-09 20:32:12
similar triangles?
similar triangles?
copeland
2017-03-09 20:32:13
So?
So?
cjquines0
2017-03-09 20:32:44
it's concyclic with $z1, z2, z3$!
it's concyclic with $z1, z2, z3$!
yrnsmurf
2017-03-09 20:32:44
it lies on the circle with z1z2z3
it lies on the circle with z1z2z3
MSTang
2017-03-09 20:32:44
cyclic quad!!
cyclic quad!!
sxu
2017-03-09 20:32:44
z lies on the same arc?
z lies on the same arc?
mathman3880
2017-03-09 20:32:44
the quadrilateral is cyclic
the quadrilateral is cyclic
Root01
2017-03-09 20:32:44
On the circumcircle of z1,z2,z3
On the circumcircle of z1,z2,z3
Mrkiller
2017-03-09 20:32:44
a semicircle?
a semicircle?
copeland
2017-03-09 20:32:51
We can rewrite $\dfrac{z_3-z}{z_2-z}=\dfrac{z-z_3}{z-z_2}$. This is the same expression that we had for $z_1$ above. That means that the angle at $z$ from $z_3$ to $z_2$ is the same as the corresponding angle at $z_1$, so $\angle z_3z_1z_2=\angle z_3zz_2$.
We can rewrite $\dfrac{z_3-z}{z_2-z}=\dfrac{z-z_3}{z-z_2}$. This is the same expression that we had for $z_1$ above. That means that the angle at $z$ from $z_3$ to $z_2$ is the same as the corresponding angle at $z_1$, so $\angle z_3z_1z_2=\angle z_3zz_2$.
copeland
2017-03-09 20:32:57
The set of $z$ that we care about are the points on the circumcircle of $\triangle z_3z_1z_2$.
The set of $z$ that we care about are the points on the circumcircle of $\triangle z_3z_1z_2$.
copeland
2017-03-09 20:32:58
copeland
2017-03-09 20:33:02
[If you want to space out at this picture more later, think about this: The function $f(z)=\dfrac{z_3-z_1}{z_2-z_1}\cdot\dfrac{z-z_2}{z-z_3}$ takes points on this circle and spits out real numbers. That means the circle is identified with the real line. Where is 0? Where is $\infty?$ What is the difference between the positive and negatives? Where are the integers?]
[If you want to space out at this picture more later, think about this: The function $f(z)=\dfrac{z_3-z_1}{z_2-z_1}\cdot\dfrac{z-z_2}{z-z_3}$ takes points on this circle and spits out real numbers. That means the circle is identified with the real line. Where is 0? Where is $\infty?$ What is the difference between the positive and negatives? Where are the integers?]
copeland
2017-03-09 20:33:13
Back to the problem. What are we trying to find?
Back to the problem. What are we trying to find?
richuw
2017-03-09 20:34:01
when the imaginary part is as great as possible (The peak of the circle)
when the imaginary part is as great as possible (The peak of the circle)
First
2017-03-09 20:34:01
the highest from the real axis
the highest from the real axis
Dr4gon39
2017-03-09 20:34:01
mazimize the imaginary part and then give the real part of the complex number with the optimized imaginary part
mazimize the imaginary part and then give the real part of the complex number with the optimized imaginary part
Mrkiller
2017-03-09 20:34:01
the z with the biggest imaginary part
the z with the biggest imaginary part
yrnsmurf
2017-03-09 20:34:01
the point with the greatest y
the point with the greatest y
brainiac1
2017-03-09 20:34:01
the maximum possible imaginary part for z (basically the top of the circle)
the maximum possible imaginary part for z (basically the top of the circle)
gabrielsui
2017-03-09 20:34:01
greatest possible imaginary part
greatest possible imaginary part
GeronimoStilton
2017-03-09 20:34:01
The $z$ such that $\operatorname{Im} z$ is the greatest possible.
The $z$ such that $\operatorname{Im} z$ is the greatest possible.
yrnsmurf
2017-03-09 20:34:01
the x coordinate of the point with greatest y
the x coordinate of the point with greatest y
IsaacZ123
2017-03-09 20:34:01
the greatest imaginary part of z
the greatest imaginary part of z
copeland
2017-03-09 20:34:03
We want the point with largest imaginary part and we want to figure out what its real part is. That point is definitely way up at the top:
We want the point with largest imaginary part and we want to figure out what its real part is. That point is definitely way up at the top:
copeland
2017-03-09 20:34:05
copeland
2017-03-09 20:34:09
Now I think it's time for some numbers. What should our goal be now?
Now I think it's time for some numbers. What should our goal be now?
yrnsmurf
2017-03-09 20:34:59
the circumcenter of z1z2z3
the circumcenter of z1z2z3
ninjataco
2017-03-09 20:34:59
find circumradius and center of the circle
find circumradius and center of the circle
gabrielsui
2017-03-09 20:34:59
find the center of the circle
find the center of the circle
brainiac1
2017-03-09 20:34:59
find the center of this circle, since it will have the same x-value as our desired point
find the center of this circle, since it will have the same x-value as our desired point
vishwathganesan
2017-03-09 20:34:59
find center and radius
find center and radius
fdas
2017-03-09 20:34:59
Find the center of the circle
Find the center of the circle
copeland
2017-03-09 20:35:01
The top of the circle has the same real part as the center of the circle. So if we can find the center of this circle then we're done.
The top of the circle has the same real part as the center of the circle. So if we can find the center of this circle then we're done.
copeland
2017-03-09 20:35:06
Let's just turn these complex coordinates into vectors. (That's the first and last time in my life I'll ever say that.)
Let's just turn these complex coordinates into vectors. (That's the first and last time in my life I'll ever say that.)
copeland
2017-03-09 20:35:08
We want to find the $x$-coordinate of the center of the circle. What's its $y$-coordinate?
We want to find the $x$-coordinate of the center of the circle. What's its $y$-coordinate?
copeland
2017-03-09 20:35:57
vishwathganesan
2017-03-09 20:36:38
61
61
brainiac1
2017-03-09 20:36:38
61
61
cjquines0
2017-03-09 20:36:38
midway between $39$ and $83$, so $61$
midway between $39$ and $83$, so $61$
sxu
2017-03-09 20:36:38
61
61
strategos21
2017-03-09 20:36:38
61
61
copeland
2017-03-09 20:36:41
Since there is a vertical chord between $z_1=(18,83)$ and $z_2=(18,39)$, we know that the horizontal diameter bisects this chord. So the horizontal diameter of the circle is $y=\dfrac{83+39}2=61$.
Since there is a vertical chord between $z_1=(18,83)$ and $z_2=(18,39)$, we know that the horizontal diameter bisects this chord. So the horizontal diameter of the circle is $y=\dfrac{83+39}2=61$.
copeland
2017-03-09 20:36:42
This is our circle:
This is our circle:
copeland
2017-03-09 20:36:43
copeland
2017-03-09 20:36:52
Now what?
Now what?
liuh008
2017-03-09 20:37:34
Distance between 2 points?
Distance between 2 points?
IsaacZ123
2017-03-09 20:37:34
we can create a set of equations using the radius of the circle
we can create a set of equations using the radius of the circle
fdas
2017-03-09 20:37:34
Draw right triangles
Draw right triangles
J1618
2017-03-09 20:37:34
Find x using a right triangle
Find x using a right triangle
copeland
2017-03-09 20:37:37
We could use the distance formula. That would be nice. Stare at the coordinates of those three points, though. Do you see anything nice?
We could use the distance formula. That would be nice. Stare at the coordinates of those three points, though. Do you see anything nice?
copeland
2017-03-09 20:38:24
We have $z_2=(18,39)$ and $z_3=(78,99)$. Those have the same ones digits. . .
We have $z_2=(18,39)$ and $z_3=(78,99)$. Those have the same ones digits. . .
brainiac1
2017-03-09 20:38:37
find the perpendicular bisector of z2z3, which happens to have slope -1
find the perpendicular bisector of z2z3, which happens to have slope -1
yrnsmurf
2017-03-09 20:38:37
line through z2 and z3 has slope 1
line through z2 and z3 has slope 1
sxu
2017-03-09 20:38:37
oh the x and y coords both differ by 60
oh the x and y coords both differ by 60
duck_master
2017-03-09 20:38:37
z1 and z3 are 45 degrees to the x-axis!
z1 and z3 are 45 degrees to the x-axis!
vishwathganesan
2017-03-09 20:38:37
oops z_2z_3 slope is 1
oops z_2z_3 slope is 1
copeland
2017-03-09 20:38:39
The difference between $z_3$ and $z_2$ is $z_3-z_2=(78,99)-(18,39)=(60,60).$
The difference between $z_3$ and $z_2$ is $z_3-z_2=(78,99)-(18,39)=(60,60).$
copeland
2017-03-09 20:38:42
That means the line through $z_2$ and $z_3$ has slope $+1$.
That means the line through $z_2$ and $z_3$ has slope $+1$.
copeland
2017-03-09 20:38:45
So its perpendicular bisector has slope $-1!$ It's $x+y=k$. What is $k?$
So its perpendicular bisector has slope $-1!$ It's $x+y=k$. What is $k?$
brainiac1
2017-03-09 20:39:53
117
117
samuel
2017-03-09 20:39:53
$k=117$
$k=117$
vishwathganesan
2017-03-09 20:39:53
117
117
sxu
2017-03-09 20:39:53
117
117
duck_master
2017-03-09 20:39:53
117?
117?
cjquines0
2017-03-09 20:39:53
it passes through the midpoint $(48, 69)$, so it's $48+69=117$
it passes through the midpoint $(48, 69)$, so it's $48+69=117$
islander7
2017-03-09 20:39:53
117
117
copeland
2017-03-09 20:39:55
The perpendicular bisector contains the midpoint $(48,69)$, so its equation is $x+y=117.$
The perpendicular bisector contains the midpoint $(48,69)$, so its equation is $x+y=117.$
copeland
2017-03-09 20:39:56
What is the $x$-coordinate of our center?
What is the $x$-coordinate of our center?
Buddy03
2017-03-09 20:40:37
56
56
IsaacZ123
2017-03-09 20:40:37
56
56
Reef334
2017-03-09 20:40:37
56
56
legolego
2017-03-09 20:40:37
56
56
strategos21
2017-03-09 20:40:37
56
56
pythonsquared
2017-03-09 20:40:37
56
56
DemonPlat4
2017-03-09 20:40:37
56
56
vishwathganesan
2017-03-09 20:40:37
56
56
brainiac1
2017-03-09 20:40:37
56
56
copeland
2017-03-09 20:40:38
The center lies on the lines
\begin{align*}
x+y&=117\\
y&=61.
\end{align*}
The center lies on the lines
\begin{align*}
x+y&=117\\
y&=61.
\end{align*}
copeland
2017-03-09 20:40:41
That makes $x=117-61=\boxed{056}.$
That makes $x=117-61=\boxed{056}.$
copeland
2017-03-09 20:40:48
Cool.
Cool.
copeland
2017-03-09 20:40:58
When I was a kid, people used to say "Cool beans."
When I was a kid, people used to say "Cool beans."
copeland
2017-03-09 20:41:03
I wonder why that went out of style.
I wonder why that went out of style.
copeland
2017-03-09 20:41:04
What's next?
What's next?
QuestForKnowledge
2017-03-09 20:41:22
palindrome .... 11
palindrome .... 11
letsgomath
2017-03-09 20:41:22
#11!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
#11!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
lego101
2017-03-09 20:41:22
Number eleven, we're in heaven bc we're doing math
Number eleven, we're in heaven bc we're doing math
NewbieGamer
2017-03-09 20:41:22
q 11
q 11
sxu
2017-03-09 20:41:22
11
11
MathTechFire
2017-03-09 20:41:22
#11
#11
copeland
2017-03-09 20:41:25
Palindrome, huh?
Palindrome, huh?
copeland
2017-03-09 20:41:50
(You know most of them have been palindromes already.)
(You know most of them have been palindromes already.)
brainiac1
2017-03-09 20:42:05
the only palindromic prime with an even number of digits
the only palindromic prime with an even number of digits
copeland
2017-03-09 20:42:10
OK, I did not know that.
OK, I did not know that.
copeland
2017-03-09 20:42:11
11. Consider arrangements of the 9 numbers $1,2,3,\ldots,9$ in a $3\times3$ array. For each such arrangement, let $a_1,$ $a_2,$ and $a_3$ be the medians of the numbers in rows 1, 2, and 3, respectively, and let $m$ be the median of $\{a_1,a_2,a_3\}$. Let $Q$ be the number of arrangements for which $m=5.$ Find the remainder when $Q$ is divided by 1000.
11. Consider arrangements of the 9 numbers $1,2,3,\ldots,9$ in a $3\times3$ array. For each such arrangement, let $a_1,$ $a_2,$ and $a_3$ be the medians of the numbers in rows 1, 2, and 3, respectively, and let $m$ be the median of $\{a_1,a_2,a_3\}$. Let $Q$ be the number of arrangements for which $m=5.$ Find the remainder when $Q$ is divided by 1000.
copeland
2017-03-09 20:42:20
What does it mean for 5 to be the median of the medians?
What does it mean for 5 to be the median of the medians?
checkmatetang
2017-03-09 20:43:08
m is median of row
m is median of row
checkmatetang
2017-03-09 20:43:08
5 is median of a row
5 is median of a row
IsaacZ123
2017-03-09 20:43:08
one median is larger and the other is smaller
one median is larger and the other is smaller
vishwathganesan
2017-03-09 20:43:08
it is the median of its row
it is the median of its row
GeronimoStilton
2017-03-09 20:43:08
It means that one of the medians is greater than $5$ and one is less.
It means that one of the medians is greater than $5$ and one is less.
mshanmugam
2017-03-09 20:43:08
one of the medians is 5
one of the medians is 5
copeland
2017-03-09 20:43:12
5 must be one of the medians. One of the medians is smaller than 5 and the other median is larger than 5.
5 must be one of the medians. One of the medians is smaller than 5 and the other median is larger than 5.
copeland
2017-03-09 20:43:14
Well, 5 is definitely in some row. What's the probability that it's the median of its own row?
Well, 5 is definitely in some row. What's the probability that it's the median of its own row?
brainiac1
2017-03-09 20:44:59
4/7
4/7
JJShan26
2017-03-09 20:44:59
4/7
4/7
DemonPlat4
2017-03-09 20:44:59
$\frac{4}{7}$?
$\frac{4}{7}$?
strategos21
2017-03-09 20:44:59
4/7
4/7
IsaacZ123
2017-03-09 20:44:59
4/7
4/7
copeland
2017-03-09 20:45:03
5 is the median of its row when one of its neighbors is less than 5 and the other neighbor is more than 5. We have 4 little numbers and 4 big numbers in the array.
5 is the median of its row when one of its neighbors is less than 5 and the other neighbor is more than 5. We have 4 little numbers and 4 big numbers in the array.
copeland
2017-03-09 20:45:06
We need the probability that 5 is in a row with one little number and one big number. Without loss of generality, assume 5 is in the middle. If the left number is little, then $\dfrac47$ of the time the right number will be big. If the left number is big then $\dfrac47$ of the time the right number will be little.
We need the probability that 5 is in a row with one little number and one big number. Without loss of generality, assume 5 is in the middle. If the left number is little, then $\dfrac47$ of the time the right number will be big. If the left number is big then $\dfrac47$ of the time the right number will be little.
copeland
2017-03-09 20:45:10
Either way, there is a $\dfrac47$ chance that 5 is the median of one of the rows.
Either way, there is a $\dfrac47$ chance that 5 is the median of one of the rows.
copeland
2017-03-09 20:45:18
Or this:
Or this:
owm
2017-03-09 20:45:20
(8C2-4C2-4C2)/8C2?
(8C2-4C2-4C2)/8C2?
copeland
2017-03-09 20:45:26
If 5 is the median of one of the rows then we have one little and one big number in the row with 5. How can we fill the array such that 5 is the median of the medians?
If 5 is the median of one of the rows then we have one little and one big number in the row with 5. How can we fill the array such that 5 is the median of the medians?
copeland
2017-03-09 20:45:49
How can we fill in the rest of such an array, that is?
How can we fill in the rest of such an array, that is?
IsaacZ123
2017-03-09 20:47:39
wait that's what the problem is asking for basically because if it is the median of its own row it has to be the median of the set
wait that's what the problem is asking for basically because if it is the median of its own row it has to be the median of the set
a1b2
2017-03-09 20:47:39
Any arrangement
Any arrangement
yrnsmurf
2017-03-09 20:47:39
any way you want
any way you want
strategos21
2017-03-09 20:47:39
The rest of the array doesn't matter; just calculate the number of arrangements.
The rest of the array doesn't matter; just calculate the number of arrangements.
brainiac1
2017-03-09 20:47:39
every other arrangement has to work
every other arrangement has to work
zihang
2017-03-09 20:47:39
6!=720
6!=720
a1b2
2017-03-09 20:47:39
It is impossible for both other medians to be more than 5 so any arrangement is valid.
It is impossible for both other medians to be more than 5 so any arrangement is valid.
Dr4gon39
2017-03-09 20:47:39
ANY WAY works as long as five is already the median of its own row
ANY WAY works as long as five is already the median of its own row
copeland
2017-03-09 20:47:42
We have 3 little numbers and 3 big numbers to put in the grid.
We have 3 little numbers and 3 big numbers to put in the grid.
copeland
2017-03-09 20:47:43
If all 3 little numbers go in the same row then that row's median will be smaller than 5 and the other row will have larger median and 5 is the median of the medians.
If all 3 little numbers go in the same row then that row's median will be smaller than 5 and the other row will have larger median and 5 is the median of the medians.
copeland
2017-03-09 20:47:44
If one row has 2 little numbers and 1 big then the median of that row will be little. Symmetrically, the other row will have a big median and 5 is still the median of the medians.
If one row has 2 little numbers and 1 big then the median of that row will be little. Symmetrically, the other row will have a big median and 5 is still the median of the medians.
copeland
2017-03-09 20:47:50
So if 5 is the median of its row, it's the median of the medians of all the rows.
So if 5 is the median of its row, it's the median of the medians of all the rows.
copeland
2017-03-09 20:47:53
No matter what, $\dfrac47$ of the time 5 is the median of the medians.
No matter what, $\dfrac47$ of the time 5 is the median of the medians.
copeland
2017-03-09 20:47:55
How many total ways are there to fill the array?
How many total ways are there to fill the array?
Buddy03
2017-03-09 20:48:49
9!
9!
vishwathganesan
2017-03-09 20:48:49
9!
9!
brainiac1
2017-03-09 20:48:49
9!
9!
IsaacZ123
2017-03-09 20:48:49
362880 or 9!
362880 or 9!
GeronimoStilton
2017-03-09 20:48:49
$362880 = 9!$
$362880 = 9!$
Kiola
2017-03-09 20:48:49
9!
9!
checkmatetang
2017-03-09 20:48:49
9!
9!
Einsteinhead
2017-03-09 20:48:49
9!
9!
yrnsmurf
2017-03-09 20:48:49
9!
9!
mathcrazymar
2017-03-09 20:48:49
9!
9!
copeland
2017-03-09 20:48:56
Who knows 9!? You guys are crazy.
Who knows 9!? You guys are crazy.
copeland
2017-03-09 20:48:59
There are $9!$ total ways to fill the array.
There are $9!$ total ways to fill the array.
copeland
2017-03-09 20:49:00
How many arrays have 5 as the median of the medians?
How many arrays have 5 as the median of the medians?
Buddy03
2017-03-09 20:49:41
9!*(4/7)
9!*(4/7)
a1b2
2017-03-09 20:49:41
$9!*\frac{4}{7}$
$9!*\frac{4}{7}$
IsaacZ123
2017-03-09 20:49:41
9!*4/7
9!*4/7
awesomemaths
2017-03-09 20:49:41
4/7*9!
4/7*9!
Mrkiller
2017-03-09 20:49:41
4/7 of them so
4/7 of them so
Einsteinhead
2017-03-09 20:49:41
(4/7)*9!
(4/7)*9!
checkmatetang
2017-03-09 20:49:41
207360
207360
a1b2
2017-03-09 20:49:41
$\frac{4}{7}*9!=207360$
$\frac{4}{7}*9!=207360$
letsgomath
2017-03-09 20:49:41
207360
207360
strategos21
2017-03-09 20:49:41
4/7 of them
4/7 of them
mathcrazymar
2017-03-09 20:49:41
4/7 * 9!
4/7 * 9!
Buddy03
2017-03-09 20:49:41
207360=9!*(4/7)
207360=9!*(4/7)
sxu
2017-03-09 20:49:41
9*8*6*5*4*4*3*2
9*8*6*5*4*4*3*2
Pot
2017-03-09 20:49:41
207360
207360
Mrkiller
2017-03-09 20:49:41
207360
207360
DemonPlat4
2017-03-09 20:49:41
$\frac{4}{7}\cdot362880$
$\frac{4}{7}\cdot362880$
ninjataco
2017-03-09 20:49:41
4*9!/7
4*9!/7
cjquines0
2017-03-09 20:49:41
$\frac47$ of those, so $207360$
$\frac47$ of those, so $207360$
copeland
2017-03-09 20:49:45
Since $\dfrac47$ of the arrays have this property, there are $\dfrac47\cdot9!$ total arrays.
Since $\dfrac47$ of the arrays have this property, there are $\dfrac47\cdot9!$ total arrays.
copeland
2017-03-09 20:49:47
\[\frac47\cdot9!=9\cdot8\cdot7\cdot6!=9\cdot8\cdot4\cdot720.\]
\[\frac47\cdot9!=9\cdot8\cdot7\cdot6!=9\cdot8\cdot4\cdot720.\]
copeland
2017-03-09 20:49:53
And what's the final answer?
And what's the final answer?
letsgomath
2017-03-09 20:50:12
360
360
Pot
2017-03-09 20:50:12
360
360
IsaacZ123
2017-03-09 20:50:12
360
360
stronto
2017-03-09 20:50:12
360
360
KYang
2017-03-09 20:50:12
360
360
GeronimoStilton
2017-03-09 20:50:12
$360$
$360$
QuestForKnowledge
2017-03-09 20:50:12
360
360
AlisonH
2017-03-09 20:50:12
360
360
MountainHeight
2017-03-09 20:50:12
360
360
legolego
2017-03-09 20:50:12
360
360
amgmflannigan
2017-03-09 20:50:12
360
360
math.fever
2017-03-09 20:50:12
360
360
Damalone
2017-03-09 20:50:12
360
360
copeland
2017-03-09 20:50:13
\begin{align*}
9\cdot8\cdot4\cdot720
&\equiv9\cdot8\cdot880\\
&\equiv9\cdot40\\
&\equiv\boxed{360}\\
\end{align*}
\begin{align*}
9\cdot8\cdot4\cdot720
&\equiv9\cdot8\cdot880\\
&\equiv9\cdot40\\
&\equiv\boxed{360}\\
\end{align*}
copeland
2017-03-09 20:50:18
OK, great.
OK, great.
copeland
2017-03-09 20:50:44
11 down. 4 to go. I bet the rest are easy. Should we skip them?
11 down. 4 to go. I bet the rest are easy. Should we skip them?
awesomemaths
2017-03-09 20:51:05
problem 12
problem 12
awesomemaths
2017-03-09 20:51:05
problem 12 yay!!!
problem 12 yay!!!
cjquines0
2017-03-09 20:51:05
totally
totally
legolego
2017-03-09 20:51:05
noo
noo
DemonPlat4
2017-03-09 20:51:05
no
no
First
2017-03-09 20:51:05
No!
No!
MountainHeight
2017-03-09 20:51:05
NO
NO
amackenzie1
2017-03-09 20:51:05
Haha no!
Haha no!
tdeng
2017-03-09 20:51:05
No.
No.
awesomemaths
2017-03-09 20:51:05
nope
nope
strategos21
2017-03-09 20:51:05
nah, you should just humor us
nah, you should just humor us
yrnsmurf
2017-03-09 20:51:05
nah
nah
lego101
2017-03-09 20:51:10
Number twelve, as deeper into the world of mathematica shall we delve.
Number twelve, as deeper into the world of mathematica shall we delve.
lego101
2017-03-09 20:51:10
NO POWER THROUGH
NO POWER THROUGH
copeland
2017-03-09 20:51:16
POWER THROUGH, in caps. I like it.
POWER THROUGH, in caps. I like it.
copeland
2017-03-09 20:51:18
12. Call a set $S$ product-free if there do not exist $a,b,c\in S$ (not necessarily distinct) such that $ab=c.$ For example, the empty set and the set $\{16,20\}$ are product-free, whereas the sets $\{4,16\}$ and $\{2,8,16\}$ are not product-free. Find the number of product-free subsets of the set $\{1,2,3,4,5,6,7,8,9,10\}.$
12. Call a set $S$ product-free if there do not exist $a,b,c\in S$ (not necessarily distinct) such that $ab=c.$ For example, the empty set and the set $\{16,20\}$ are product-free, whereas the sets $\{4,16\}$ and $\{2,8,16\}$ are not product-free. Find the number of product-free subsets of the set $\{1,2,3,4,5,6,7,8,9,10\}.$
copeland
2017-03-09 20:51:32
I'm thinking more people should use caps. It would make the classroom much more interesting.
I'm thinking more people should use caps. It would make the classroom much more interesting.
copeland
2017-03-09 20:51:36
Also, that's a joke.
Also, that's a joke.
copeland
2017-03-09 20:51:40
Don't do that.
Don't do that.
Pot
2017-03-09 20:51:46
OKAY
OKAY
reelmathematician
2017-03-09 20:51:46
CAPS FOR THE WIN
CAPS FOR THE WIN
IsaacZ123
2017-03-09 20:51:46
ARE YOU SURE IT WAS A JOKE
ARE YOU SURE IT WAS A JOKE
MSTang
2017-03-09 20:51:46
I WON'T!
I WON'T!
copeland
2017-03-09 20:51:48
Alright, what do you see immediately?
Alright, what do you see immediately?
vishwathganesan
2017-03-09 20:52:46
anything with 1 in it fails
anything with 1 in it fails
GeronimoStilton
2017-03-09 20:52:46
$1$ Can't be a member of any of the sets!
$1$ Can't be a member of any of the sets!
Mrkiller
2017-03-09 20:52:46
1 can be in any of them
1 can be in any of them
ninjataco
2017-03-09 20:52:46
1 can't be in the subset
1 can't be in the subset
Einsteinhead
2017-03-09 20:52:46
Set cannot contain 1
Set cannot contain 1
tdeng
2017-03-09 20:52:46
We can't have 1
We can't have 1
duck_master
2017-03-09 20:52:46
DON'T PUT 1 THERE OR U WON'T BE ABLE TO PUT ANYTHING ELSE
DON'T PUT 1 THERE OR U WON'T BE ABLE TO PUT ANYTHING ELSE
mathfever
2017-03-09 20:52:46
that they can't have 1
that they can't have 1
amackenzie1
2017-03-09 20:52:46
can;t have a 1
can;t have a 1
strategos21
2017-03-09 20:52:46
1 doesn't fit anywhere
1 doesn't fit anywhere
amyhu910
2017-03-09 20:52:46
1 cannot be in the set
1 cannot be in the set
KYang
2017-03-09 20:52:46
you can't have 1
you can't have 1
MountainHeight
2017-03-09 20:52:46
we can ignore 1
we can ignore 1
copeland
2017-03-09 20:52:49
I see that 1 can't be in the set since $1\cdot1=1$.
I see that 1 can't be in the set since $1\cdot1=1$.
copeland
2017-03-09 20:52:53
Are there a lot of products to avoid or just a few?
Are there a lot of products to avoid or just a few?
IsaacZ123
2017-03-09 20:54:11
a few
a few
EdwinNational
2017-03-09 20:54:11
Just a few
Just a few
amgmflannigan
2017-03-09 20:54:11
2 and 3
2 and 3
QuestForKnowledge
2017-03-09 20:54:11
few
few
DemonPlat4
2017-03-09 20:54:11
few-ish
few-ish
andsun19
2017-03-09 20:54:11
2*2=4, 2*3=6, 2*4=8, 2*5=10, 3*3=9
2*2=4, 2*3=6, 2*4=8, 2*5=10, 3*3=9
Mathaddict11
2017-03-09 20:54:11
really only a few and we can quickly list them out
really only a few and we can quickly list them out
rraj
2017-03-09 20:54:11
just a few
just a few
Metal_Bender19
2017-03-09 20:54:11
a few
a few
itxxc12
2017-03-09 20:54:11
just a few
just a few
ilikepie2003
2017-03-09 20:54:11
a few
a few
copeland
2017-03-09 20:54:14
There are only a few products we have to avoid. Every product we could make (omitting 1) has either a 2 or a 3 in it.
There are only a few products we have to avoid. Every product we could make (omitting 1) has either a 2 or a 3 in it.
copeland
2017-03-09 20:54:16
Actually, here are all the products just so we see them:
Actually, here are all the products just so we see them:
copeland
2017-03-09 20:54:16
\[\begin{array}{c}
2\cdot2=4&2\cdot3=6&2\cdot4=8&2\cdot5=10\\
3\cdot2=6&3\cdot3=9.\\
\end{array}\]
\[\begin{array}{c}
2\cdot2=4&2\cdot3=6&2\cdot4=8&2\cdot5=10\\
3\cdot2=6&3\cdot3=9.\\
\end{array}\]
copeland
2017-03-09 20:54:18
I put $2\cdot3=3\cdot2$ on there twice because it kinda fits twice.
I put $2\cdot3=3\cdot2$ on there twice because it kinda fits twice.
copeland
2017-03-09 20:54:22
So how should we organize our count?
So how should we organize our count?
a1b2
2017-03-09 20:55:24
By twos and threes
By twos and threes
legolego
2017-03-09 20:55:24
has a 2, has a 3, both, or neither
has a 2, has a 3, both, or neither
tdeng
2017-03-09 20:55:24
{Has 2, no 3}, {has 3, no 2}, {has 2 and 3}, {no 2, no 3}
{Has 2, no 3}, {has 3, no 2}, {has 2 and 3}, {no 2, no 3}
copeland
2017-03-09 20:55:26
Let's use casework on whether we have 2 or 3 in the set.
Let's use casework on whether we have 2 or 3 in the set.
copeland
2017-03-09 20:55:28
If we don't have 2 or 3 in the set, how many possible sets can we make?
If we don't have 2 or 3 in the set, how many possible sets can we make?
IsaacZ123
2017-03-09 20:55:56
2^7
2^7
tdeng
2017-03-09 20:55:56
2^7
2^7
legolego
2017-03-09 20:55:56
2^7
2^7
IsaacZ123
2017-03-09 20:55:56
128
128
letsgomath
2017-03-09 20:55:56
128
128
Ericaops
2017-03-09 20:55:56
128
128
DemonPlat4
2017-03-09 20:55:56
128
128
GeronimoStilton
2017-03-09 20:55:56
$2^7 = 128$
$2^7 = 128$
brainiac1
2017-03-09 20:55:56
2^7=128
2^7=128
copeland
2017-03-09 20:55:58
If we don't have 2 or 3 in the set then we want a subset of $\{4,5,6,7,8,9,10\}.$ Any numbers can be in our set, so there are $2^7$ total sets that miss 2 and 3.
If we don't have 2 or 3 in the set then we want a subset of $\{4,5,6,7,8,9,10\}.$ Any numbers can be in our set, so there are $2^7$ total sets that miss 2 and 3.
copeland
2017-03-09 20:56:00
What if 3 is in the set and 2 is not? (Three is in fewer products, so putting it in the set is less restrictive. That's why it's a better case to tackle next.)
What if 3 is in the set and 2 is not? (Three is in fewer products, so putting it in the set is less restrictive. That's why it's a better case to tackle next.)
legolego
2017-03-09 20:56:54
2^6
2^6
owm
2017-03-09 20:56:54
2^6
2^6
a1b2
2017-03-09 20:56:54
9 can't be in it
9 can't be in it
IsaacZ123
2017-03-09 20:56:54
64
64
mathcrazymar
2017-03-09 20:56:54
2^6
2^6
IsaacZ123
2017-03-09 20:56:54
2^6
2^6
cjquines0
2017-03-09 20:56:54
any subset of $\{4, 5, 6, 7, 8, 10\}$ so $2^6$
any subset of $\{4, 5, 6, 7, 8, 10\}$ so $2^6$
MountainHeight
2017-03-09 20:56:54
2^6
2^6
Ericaops
2017-03-09 20:56:54
cant have 9, 2^6
cant have 9, 2^6
Mathaddict11
2017-03-09 20:56:54
we can't have 3 and 9 so there are $2^6=64$ ways
we can't have 3 and 9 so there are $2^6=64$ ways
Archimedes15
2017-03-09 20:56:54
2^6 = 64
2^6 = 64
brainiac1
2017-03-09 20:56:54
2^6=64
2^6=64
copeland
2017-03-09 20:56:56
We can add anything but 9 now.
We can add anything but 9 now.
copeland
2017-03-09 20:56:57
If 3 is in the set and 2 is not then we finish the set by adding some subset of $\{4,5,6,7,8,10\}$ to the set. There are $2^6$ subsets like this.
If 3 is in the set and 2 is not then we finish the set by adding some subset of $\{4,5,6,7,8,10\}$ to the set. There are $2^6$ subsets like this.
copeland
2017-03-09 20:57:00
What if 2 is in the set and 3 is not? This is harder. What numbers can we add or not, without fear?
What if 2 is in the set and 3 is not? This is harder. What numbers can we add or not, without fear?
Inco
2017-03-09 20:57:51
6, 7, 9
6, 7, 9
owm
2017-03-09 20:57:51
we can add 6,7,and 9
we can add 6,7,and 9
copeland
2017-03-09 20:57:54
We can add any of $\{6,7,9\}$ to the set without worrying about making products with any of the other elements.
We can add any of $\{6,7,9\}$ to the set without worrying about making products with any of the other elements.
copeland
2017-03-09 20:57:55
What about 5 and 10?
What about 5 and 10?
Ericaops
2017-03-09 20:58:28
cannot have both
cannot have both
legolego
2017-03-09 20:58:28
choose one or neither
choose one or neither
yrnsmurf
2017-03-09 20:58:28
5, 10, or neither
5, 10, or neither
letsgomath
2017-03-09 20:58:28
only one of the 2
only one of the 2
QuestForKnowledge
2017-03-09 20:58:28
u cant have em both
u cant have em both
Archimedes15
2017-03-09 20:58:28
2*5 = 10, meaning we can't have 5 and 10
2*5 = 10, meaning we can't have 5 and 10
cooljoseph
2017-03-09 20:58:28
You can have one, but not the other
You can have one, but not the other
cjquines0
2017-03-09 20:58:28
we have to pick between $5$, $10$, or neither
we have to pick between $5$, $10$, or neither
Peggy
2017-03-09 20:58:28
only one of them
only one of them
mishai
2017-03-09 20:58:28
only one of them or none
only one of them or none
DemonPlat4
2017-03-09 20:58:28
one or the other, but not both
one or the other, but not both
BuddyS
2017-03-09 20:58:28
only one of 5 or 10
only one of 5 or 10
IsaacZ123
2017-03-09 20:58:28
we just can't have both
we just can't have both
MountainHeight
2017-03-09 20:58:28
they can't be in at the same time
they can't be in at the same time
copeland
2017-03-09 20:58:31
We can have either of them or none. How many ways is that to add 5 or 10?
We can have either of them or none. How many ways is that to add 5 or 10?
brainiac1
2017-03-09 20:58:48
3
3
amackenzie1
2017-03-09 20:58:48
3
3
Doink
2017-03-09 20:58:48
3
3
QuestForKnowledge
2017-03-09 20:58:48
3
3
thetank
2017-03-09 20:58:48
3
3
letsgomath
2017-03-09 20:58:48
3
3
vishwathganesan
2017-03-09 20:58:48
3
3
mishai
2017-03-09 20:58:48
3
3
copeland
2017-03-09 20:58:51
There are three choices: neither, 5, or 10.
There are three choices: neither, 5, or 10.
copeland
2017-03-09 20:58:52
What about 4 and 8?
What about 4 and 8?
IsaacZ123
2017-03-09 20:59:24
also can't we add 8 safely as well since we can't have 4 in there
also can't we add 8 safely as well since we can't have 4 in there
tdeng
2017-03-09 20:59:24
We can't have 4=2*2, so we can just add 8
We can't have 4=2*2, so we can just add 8
letsgomath
2017-03-09 20:59:24
no 4, 8 works
no 4, 8 works
DjokerNole
2017-03-09 20:59:24
only 8
only 8
Reef334
2017-03-09 20:59:24
neither or only 8
neither or only 8
IsaacZ123
2017-03-09 20:59:24
we can only have 8 because 2*2=4
we can only have 8 because 2*2=4
mishai
2017-03-09 20:59:24
no 4 but 8 can appear
no 4 but 8 can appear
cooljoseph
2017-03-09 20:59:24
You can never have 4, so you can always have 8.
You can never have 4, so you can always have 8.
owm
2017-03-09 20:59:24
there are 2 possibilities
there are 2 possibilities
a1b2
2017-03-09 20:59:24
The 8, or nothing!
The 8, or nothing!
EdwinNational
2017-03-09 20:59:24
No 4
No 4
GeronimoStilton
2017-03-09 20:59:24
We can have $8$ either way, but we can't have $4$ since $2 \cdot 2 = 4$
We can have $8$ either way, but we can't have $4$ since $2 \cdot 2 = 4$
MountainHeight
2017-03-09 20:59:24
4 can't be in but 8 can
4 can't be in but 8 can
copeland
2017-03-09 20:59:26
We can't have 4 since $2\cdot2=4$, so we can either have 8 or not.
We can't have 4 since $2\cdot2=4$, so we can either have 8 or not.
copeland
2017-03-09 20:59:29
How many sets have 2 and not 3?
How many sets have 2 and not 3?
GeronimoStilton
2017-03-09 21:00:32
$48$
$48$
mishai
2017-03-09 21:00:32
48
48
IsaacZ123
2017-03-09 21:00:32
48
48
yrnsmurf
2017-03-09 21:00:32
2*2*2*2*3=48
2*2*2*2*3=48
Tribefan
2017-03-09 21:00:32
48
48
legolego
2017-03-09 21:00:32
48
48
letsgomath
2017-03-09 21:00:32
48
48
MountainHeight
2017-03-09 21:00:32
48
48
vishwathganesan
2017-03-09 21:00:32
48
48
cjquines0
2017-03-09 21:00:32
$2^3$ ways to pick from $\{6, 7, 9\}$, $3$ ways to pick $5$ or $10$ or neither, and $2$ ways to pick either $8$ or nothing, so $48$
$2^3$ ways to pick from $\{6, 7, 9\}$, $3$ ways to pick $5$ or $10$ or neither, and $2$ ways to pick either $8$ or nothing, so $48$
islander7
2017-03-09 21:00:32
48
48
copeland
2017-03-09 21:00:33
The sets with 2 and not 3 are made by choosing from the $2^4$ subests of $\{6,7,8,9\}$ and then choosing either 5, 10, or neither (3 choices). There are $2^4\cdot3$ product-free sets that contain 2 and not 3.
The sets with 2 and not 3 are made by choosing from the $2^4$ subests of $\{6,7,8,9\}$ and then choosing either 5, 10, or neither (3 choices). There are $2^4\cdot3$ product-free sets that contain 2 and not 3.
copeland
2017-03-09 21:00:36
Now let's look at all sets with 2 and 3. What numbers can we throw in without fear?
Now let's look at all sets with 2 and 3. What numbers can we throw in without fear?
GeronimoStilton
2017-03-09 21:01:23
$7$
$7$
a1b2
2017-03-09 21:01:23
7
7
MountainHeight
2017-03-09 21:01:23
7
7
owm
2017-03-09 21:01:23
7
7
Z_Math404
2017-03-09 21:01:23
7
7
letsgomath
2017-03-09 21:01:23
7
7
copeland
2017-03-09 21:01:25
Only 7. All the other numbers appear in products with 2 or 3.
Only 7. All the other numbers appear in products with 2 or 3.
copeland
2017-03-09 21:01:34
Oh, wait, there's one more we just learned goes well with 2.
Oh, wait, there's one more we just learned goes well with 2.
IsaacZ123
2017-03-09 21:01:48
7 and 8
7 and 8
vishwathganesan
2017-03-09 21:01:48
7, 8
7, 8
GeronimoStilton
2017-03-09 21:01:48
$7$ and $8$
$7$ and $8$
Funnybunny5246
2017-03-09 21:01:48
7,8
7,8
yrnsmurf
2017-03-09 21:01:48
7 and 8
7 and 8
tdeng
2017-03-09 21:01:48
7,8
7,8
vishwathganesan
2017-03-09 21:01:48
also 8 though
also 8 though
IsaacZ123
2017-03-09 21:01:48
wait but 8 can be in there by the same reasoning as the last case
wait but 8 can be in there by the same reasoning as the last case
Tribefan
2017-03-09 21:01:48
8
8
amackenzie1
2017-03-09 21:01:48
8
8
letsgomath
2017-03-09 21:01:48
8
8
copeland
2017-03-09 21:02:00
We can add 7 or 8 without any issues.
We can add 7 or 8 without any issues.
copeland
2017-03-09 21:02:03
What other numbers can we easily make sense of?
What other numbers can we easily make sense of?
copeland
2017-03-09 21:02:11
Are there any that are forbidden?
Are there any that are forbidden?
letsgomath
2017-03-09 21:02:33
no 4 or 9
no 4 or 9
Z_Math404
2017-03-09 21:02:33
no 6
no 6
pianoman24
2017-03-09 21:02:33
6
6
islander7
2017-03-09 21:02:33
4,6,9
4,6,9
vishwathganesan
2017-03-09 21:02:33
4,6,9
4,6,9
IsaacZ123
2017-03-09 21:02:33
6 9
6 9
checkmatetang
2017-03-09 21:02:33
4,6,9
4,6,9
rockyisi
2017-03-09 21:02:33
4,6,9
4,6,9
wrc400
2017-03-09 21:02:33
4, 6, 9
4, 6, 9
treemath
2017-03-09 21:02:33
4,9
4,9
andsun19
2017-03-09 21:02:33
4,6,9
4,6,9
copeland
2017-03-09 21:02:38
We definitely can't have 6 if we already have 2 or 3. We also can't have 4 since we have 2 or 9 since we have 3.
We definitely can't have 6 if we already have 2 or 3. We also can't have 4 since we have 2 or 9 since we have 3.
copeland
2017-03-09 21:02:40
What about 5?
What about 5?
mishai
2017-03-09 21:03:24
either 5 or 10
either 5 or 10
Z_Math404
2017-03-09 21:03:24
either 5 or 10
either 5 or 10
yrnsmurf
2017-03-09 21:03:24
either 5, 10, or neither
either 5, 10, or neither
mshanmugam
2017-03-09 21:03:24
same as last time; 5 or 10
same as last time; 5 or 10
GeronimoStilton
2017-03-09 21:03:24
We can have $5$, $10$, or neither.
We can have $5$, $10$, or neither.
brainiac1
2017-03-09 21:03:24
if we have 5, we can't have 10 and vice versa
if we have 5, we can't have 10 and vice versa
MountainHeight
2017-03-09 21:03:24
we can either have 5 or 10 but not both
we can either have 5 or 10 but not both
Archimedes15
2017-03-09 21:03:24
we can only have 5 if there is no 10
we can only have 5 if there is no 10
owm
2017-03-09 21:03:24
we can have 5 or 10 but not both
we can have 5 or 10 but not both
treemath
2017-03-09 21:03:24
5 or 10, but not both
5 or 10, but not both
andsun19
2017-03-09 21:03:24
5 or 10 or neither
5 or 10 or neither
copeland
2017-03-09 21:03:26
We can have 5, 10, or neither.
We can have 5, 10, or neither.
copeland
2017-03-09 21:03:26
I think that's all. How many sets have 2 and 3?
I think that's all. How many sets have 2 and 3?
IsaacZ123
2017-03-09 21:03:57
12
12
mishai
2017-03-09 21:03:57
12
12
yrnsmurf
2017-03-09 21:03:57
12
12
vishwathganesan
2017-03-09 21:03:57
12
12
andsun19
2017-03-09 21:03:57
12
12
samuel
2017-03-09 21:03:57
12
12
copeland
2017-03-09 21:03:59
In a set with 2 and 3, we can't have any of 4, 6, or 9. We can have 7 or 8 whenever we want ($2^2$ choices). We can have either 5, 10, or neither (3 choices). There are $3\cdot2^2$ total sets with both 2 and 3.
In a set with 2 and 3, we can't have any of 4, 6, or 9. We can have 7 or 8 whenever we want ($2^2$ choices). We can have either 5, 10, or neither (3 choices). There are $3\cdot2^2$ total sets with both 2 and 3.
copeland
2017-03-09 21:04:00
How many total product-free sets are there?
How many total product-free sets are there?
GeronimoStilton
2017-03-09 21:04:24
$252$
$252$
vishwathganesan
2017-03-09 21:04:24
252
252
IsaacZ123
2017-03-09 21:04:24
252
252
brainiac1
2017-03-09 21:04:24
252
252
a1b2
2017-03-09 21:04:24
$\boxed{252}$
$\boxed{252}$
stronto
2017-03-09 21:04:24
252
252
tdeng
2017-03-09 21:04:24
252
252
copeland
2017-03-09 21:04:27
There are \[2^7+2^6+2^4\cdot3+2^2\cdot3=128+64+48+12=\boxed{252}\]total sets.
There are \[2^7+2^6+2^4\cdot3+2^2\cdot3=128+64+48+12=\boxed{252}\]total sets.
copeland
2017-03-09 21:04:39
Twelve down.
Twelve down.
lego101
2017-03-09 21:04:49
Number thirteen, by this time our scrap paper is far from clean. #POWERTHROUGH
Number thirteen, by this time our scrap paper is far from clean. #POWERTHROUGH
copeland
2017-03-09 21:04:57
Yeah. Maybe we'll get to another palindrome soon, too!
Yeah. Maybe we'll get to another palindrome soon, too!
copeland
2017-03-09 21:05:04
13. For every $m\geq2,$ let $Q(m)$ be the least positive integer with the following property: For every $n\geq Q(m),$ there is always a perfect cube $k^3$ in the range $n< k^3\leq m\cdot n.$ Find the reminader when
\[\sum_{m=2}^{2017} Q(m)\]is divided by 1000.
13. For every $m\geq2,$ let $Q(m)$ be the least positive integer with the following property: For every $n\geq Q(m),$ there is always a perfect cube $k^3$ in the range $n< k^3\leq m\cdot n.$ Find the reminader when
\[\sum_{m=2}^{2017} Q(m)\]is divided by 1000.
copeland
2017-03-09 21:05:14
This is an alphabet soup of a problem. I don't really know what's going on. What should we do to get a better grasp on the problem?
This is an alphabet soup of a problem. I don't really know what's going on. What should we do to get a better grasp on the problem?
Doink
2017-03-09 21:05:41
try small cases
try small cases
liuh008
2017-03-09 21:05:41
Small examples!
Small examples!
letsgomath
2017-03-09 21:05:41
plug in a few numbers
plug in a few numbers
sxu
2017-03-09 21:05:41
try out some stuff
try out some stuff
First
2017-03-09 21:05:41
Try an example
Try an example
amackenzie1
2017-03-09 21:05:41
try some simpler cases?
try some simpler cases?
Ericaops
2017-03-09 21:05:41
try small cases
try small cases
copeland
2017-03-09 21:05:44
Let's experiment! We probably want to start with small numbers, but I don't even know which variable to make small. There's $m$ and there's $Q$, of course, but there's also $n$ and $k$.
Let's experiment! We probably want to start with small numbers, but I don't even know which variable to make small. There's $m$ and there's $Q$, of course, but there's also $n$ and $k$.
copeland
2017-03-09 21:05:49
Let's focus on $m$ since we know for sure that $m=2$ is a small and interesting case.
Let's focus on $m$ since we know for sure that $m=2$ is a small and interesting case.
copeland
2017-03-09 21:05:53
Now we look at all these intervals
\begin{align*}
&(2,4]\\
&(3,6]\\
&(4,8]\\
&(5,10]\\
&(6,12]\\
&(7,14]\\
\end{align*}
and eventually these intervals should all have cubes in them.
Now we look at all these intervals
\begin{align*}
&(2,4]\\
&(3,6]\\
&(4,8]\\
&(5,10]\\
&(6,12]\\
&(7,14]\\
\end{align*}
and eventually these intervals should all have cubes in them.
copeland
2017-03-09 21:05:54
Which one of those first six have cubes in them?
Which one of those first six have cubes in them?
vishwathganesan
2017-03-09 21:07:10
third onwards
third onwards
IYN
2017-03-09 21:07:10
(4, 8] and the ones after
(4, 8] and the ones after
Doink
2017-03-09 21:07:10
(4,8], (5,10], (6,12], (7,14]
(4,8], (5,10], (6,12], (7,14]
ninjataco
2017-03-09 21:07:10
all except the first two
all except the first two
andsun19
2017-03-09 21:07:10
3rd, 4th, 5th, 6th
3rd, 4th, 5th, 6th
letsgomath
2017-03-09 21:07:10
(4,8],(5,10],(6,12],(7,14]
(4,8],(5,10],(6,12],(7,14]
alifenix-
2017-03-09 21:07:10
the last four
the last four
DemonPlat4
2017-03-09 21:07:10
(4,8], (5,10], (6,12], (7,14]
(4,8], (5,10], (6,12], (7,14]
a1b2
2017-03-09 21:07:23
The last 4, but that doesn't matter as $(8,16]$ doesn't have any.
The last 4, but that doesn't matter as $(8,16]$ doesn't have any.
vishwathganesan
2017-03-09 21:07:23
but then there's (8, 16] which doesn't have any
but then there's (8, 16] which doesn't have any
copeland
2017-03-09 21:07:25
These red ones have cubes in them:
\begin{align*}
&(2,4]\\
&(3,6]\\
&\color{red}{(4,8]}\\
&\color{red}{(5,10]}\\
&\color{red}{(6,12]}\\
&\color{red}{(7,14]}\\
\end{align*}
but the next interval $(8,16]$ does not contain a cube.
These red ones have cubes in them:
\begin{align*}
&(2,4]\\
&(3,6]\\
&\color{red}{(4,8]}\\
&\color{red}{(5,10]}\\
&\color{red}{(6,12]}\\
&\color{red}{(7,14]}\\
\end{align*}
but the next interval $(8,16]$ does not contain a cube.
copeland
2017-03-09 21:07:28
What's the next interval $(n,2n]$ that has a cube in it?
What's the next interval $(n,2n]$ that has a cube in it?
Picroft
2017-03-09 21:07:55
14,28
14,28
cooljoseph
2017-03-09 21:07:55
(14,28]
(14,28]
Reef334
2017-03-09 21:07:55
(14,28]
(14,28]
GeronimoStilton
2017-03-09 21:07:55
$(14,28]$
$(14,28]$
liuh008
2017-03-09 21:07:55
(14, 28]
(14, 28]
dhruv
2017-03-09 21:07:55
(14, 28]
(14, 28]
letsgomath
2017-03-09 21:07:55
(14,28]
(14,28]
QuestForKnowledge
2017-03-09 21:07:55
the 27 one, (14,28]
the 27 one, (14,28]
Inco
2017-03-09 21:07:55
(14, 28]
(14, 28]
thetank
2017-03-09 21:07:55
(14, 28]
(14, 28]
copeland
2017-03-09 21:07:57
The next available cube is 27, and the first interval that contains 27 is $(14,28]$.
The next available cube is 27, and the first interval that contains 27 is $(14,28]$.
copeland
2017-03-09 21:08:00
And what's the last interval that contains 27?
And what's the last interval that contains 27?
DemonPlat4
2017-03-09 21:08:34
(26,52]
(26,52]
stronto
2017-03-09 21:08:34
(26,52]
(26,52]
sxu
2017-03-09 21:08:34
(26, 52]
(26, 52]
yrnsmurf
2017-03-09 21:08:34
(26, 52]
(26, 52]
Doink
2017-03-09 21:08:34
(26,52]
(26,52]
mishai
2017-03-09 21:08:34
(26,52]
(26,52]
awesomemaths
2017-03-09 21:08:34
(26, 52]
(26, 52]
dantaxyz
2017-03-09 21:08:34
(26,52]
(26,52]
letsgomath
2017-03-09 21:08:34
(26,52]
(26,52]
grant.wilkins
2017-03-09 21:08:34
(26, 52]
(26, 52]
Mrkiller
2017-03-09 21:08:34
(26,52]
(26,52]
EpicCarrotMaster
2017-03-09 21:08:34
(26, 52]
(26, 52]
copeland
2017-03-09 21:08:36
The last interval that contains 27 is $(26,52]$. Does the next interval $(27,54]$ contain a cube?
The last interval that contains 27 is $(26,52]$. Does the next interval $(27,54]$ contain a cube?
curry3030
2017-03-09 21:08:55
no
no
grant.wilkins
2017-03-09 21:08:55
No
No
checkmatetang
2017-03-09 21:08:55
no
no
fdas
2017-03-09 21:08:55
No
No
cooleybz2013
2017-03-09 21:08:55
no
no
copeland
2017-03-09 21:08:57
No. What's the next interval $(n,2n]$ that does contain a cube?
No. What's the next interval $(n,2n]$ that does contain a cube?
NewbieGamer
2017-03-09 21:09:21
32,64
32,64
mishai
2017-03-09 21:09:21
(32,64]
(32,64]
cooljoseph
2017-03-09 21:09:21
32,64
32,64
Reef334
2017-03-09 21:09:21
(32,64]
(32,64]
owm
2017-03-09 21:09:21
32,64
32,64
dhruv
2017-03-09 21:09:21
(32, 64]
(32, 64]
letsgomath
2017-03-09 21:09:21
(32,64]
(32,64]
varun555
2017-03-09 21:09:21
(32, 64]
(32, 64]
liuh008
2017-03-09 21:09:21
(32, 64]
(32, 64]
EpicCarrotMaster
2017-03-09 21:09:21
(32, 64]
(32, 64]
islander7
2017-03-09 21:09:21
(32,64]
(32,64]
copeland
2017-03-09 21:09:23
The next cube is 64, so the next interval that contains a cube is $(32,64]$.
The next cube is 64, so the next interval that contains a cube is $(32,64]$.
copeland
2017-03-09 21:10:00
And the last interval that contains 64 is $(63,126]$.
And the last interval that contains 64 is $(63,126]$.
copeland
2017-03-09 21:10:01
What's the next interval $(n,2n]$ that doesn't contain a cube?
What's the next interval $(n,2n]$ that doesn't contain a cube?
Z_Math404
2017-03-09 21:10:13
none
none
vishwathganesan
2017-03-09 21:10:13
there is none
there is none
NewbieGamer
2017-03-09 21:10:13
none
none
DemonPlat4
2017-03-09 21:10:13
NONE
NONE
Reef334
2017-03-09 21:10:13
none of them
none of them
cooljoseph
2017-03-09 21:10:13
there isn't one
there isn't one
GeronimoStilton
2017-03-09 21:10:13
There aren't any more!
There aren't any more!
MountainHeight
2017-03-09 21:10:13
no more
no more
amackenzie1
2017-03-09 21:10:13
it seems that from now on they all contain cubes
it seems that from now on they all contain cubes
Funnybunny5246
2017-03-09 21:10:13
none
none
copeland
2017-03-09 21:10:16
Hm. This is a harder question. Since 125 is in $(64,128]$, all the intervals from $n=64$ to $n=124$ contain $5^3=125.$ That last interval is $(124,248]$.
Hm. This is a harder question. Since 125 is in $(64,128]$, all the intervals from $n=64$ to $n=124$ contain $5^3=125.$ That last interval is $(124,248]$.
copeland
2017-03-09 21:10:19
Alright, I think it's time for a picture.
Alright, I think it's time for a picture.
copeland
2017-03-09 21:10:20
Here are the cubes along the top. Below them I've graphed all of the intervals $(n,2n]$. An interval contains a cube if it crosses one of the vertical lines. Those are the red ones. The problem suggests that there is a first $n$ such that every line below the line $(n,2n]$ crosses one of the vertical lines (so will be red).
Here are the cubes along the top. Below them I've graphed all of the intervals $(n,2n]$. An interval contains a cube if it crosses one of the vertical lines. Those are the red ones. The problem suggests that there is a first $n$ such that every line below the line $(n,2n]$ crosses one of the vertical lines (so will be red).
copeland
2017-03-09 21:10:22
copeland
2017-03-09 21:10:31
Why can't we fit an interval in between 125 and 216?
Why can't we fit an interval in between 125 and 216?
a1b2
2017-03-09 21:11:02
The lines are longer than the intervals
The lines are longer than the intervals
varun555
2017-03-09 21:11:02
There is no interval short enough
There is no interval short enough
tdeng
2017-03-09 21:11:08
2*125=250
2*125=250
mishai
2017-03-09 21:11:08
216/125<2
216/125<2
curry3030
2017-03-09 21:11:08
125*2 = 250
125*2 = 250
samuel
2017-03-09 21:11:08
because 216/125<2
because 216/125<2
GeronimoStilton
2017-03-09 21:11:08
Because $2(125-1) > 216$
Because $2(125-1) > 216$
EpicCarrotMaster
2017-03-09 21:11:08
216-125 < 125 * 2
216-125 < 125 * 2
liuh008
2017-03-09 21:11:14
What a pretty picture
What a pretty picture
copeland
2017-03-09 21:11:15
Why thanks!
Why thanks!
copeland
2017-03-09 21:11:16
Since $\dfrac{216}{125}<2$, there can't be a pair $n$ and $2n$ that are both between these bars.
Since $\dfrac{216}{125}<2$, there can't be a pair $n$ and $2n$ that are both between these bars.
copeland
2017-03-09 21:11:17
What about between 216 and 343?
What about between 216 and 343?
Z_Math404
2017-03-09 21:11:31
did deven make that
did deven make that
copeland
2017-03-09 21:11:32
I made that. You can tell because it's so pro.
I made that. You can tell because it's so pro.
DemonPlat4
2017-03-09 21:11:42
216 < 125$\cdot$2
216 < 125$\cdot$2
sxu
2017-03-09 21:11:42
also <2
also <2
mishai
2017-03-09 21:11:42
343/216<216/125<2
343/216<216/125<2
vishwathganesan
2017-03-09 21:11:42
343/216 is less that 2 too
343/216 is less that 2 too
copeland
2017-03-09 21:11:44
Since $\dfrac{343}{216}<2$, there isn't a good interval in here either.
Since $\dfrac{343}{216}<2$, there isn't a good interval in here either.
copeland
2017-03-09 21:11:45
What about further to the right?
What about further to the right?
IsaacZ123
2017-03-09 21:12:27
the further we go, the less the ratios become
the further we go, the less the ratios become
IYN
2017-03-09 21:12:27
The fractions get smaller and smaller
The fractions get smaller and smaller
vishwathganesan
2017-03-09 21:12:27
(n+1)^3/n^3 tends to one
(n+1)^3/n^3 tends to one
brainiac1
2017-03-09 21:12:27
the quotient (n+1/n)^3 only decreases as n increases, there will never be another place where a cube is more than twice the previous cube after 64
the quotient (n+1/n)^3 only decreases as n increases, there will never be another place where a cube is more than twice the previous cube after 64
islander7
2017-03-09 21:12:27
ratios get smaller
ratios get smaller
mishai
2017-03-09 21:12:27
(n+2)^3/(n+1)^3<(n+1)^3/n^3<2
(n+2)^3/(n+1)^3<(n+1)^3/n^3<2
Archimedes15
2017-03-09 21:12:27
It is the same issue... it will always be less than 2.
It is the same issue... it will always be less than 2.
QuestForKnowledge
2017-03-09 21:12:27
(x/(x-1))^3 decreases
(x/(x-1))^3 decreases
copeland
2017-03-09 21:12:28
We're looking at \[\dfrac{(k+1)^3}{k^3}=\frac{1+3k^2+3k+1}{k^3}=1+\frac3k+\frac3{k^2}+\frac1{k^3}.\] When $k$ gets bigger, this value gets smaller, so the ratio is always going to be less than 2.
We're looking at \[\dfrac{(k+1)^3}{k^3}=\frac{1+3k^2+3k+1}{k^3}=1+\frac3k+\frac3{k^2}+\frac1{k^3}.\] When $k$ gets bigger, this value gets smaller, so the ratio is always going to be less than 2.
copeland
2017-03-09 21:12:31
Just so we are clear what happened, what is $Q(2)?$
Just so we are clear what happened, what is $Q(2)?$
mishai
2017-03-09 21:13:06
32
32
vishwathganesan
2017-03-09 21:13:06
32
32
legolego
2017-03-09 21:13:06
32
32
a1b2
2017-03-09 21:13:06
$Q(2)=32$
$Q(2)=32$
Z_Math404
2017-03-09 21:13:06
32
32
Makorn
2017-03-09 21:13:06
It would be $32$
It would be $32$
DemonPlat4
2017-03-09 21:13:06
Q(2) = 32
Q(2) = 32
yrnsmurf
2017-03-09 21:13:06
32
32
IsaacZ123
2017-03-09 21:13:06
32
32
IYN
2017-03-09 21:13:06
32
32
copeland
2017-03-09 21:13:07
The last interval that doesn't contain a cube is $(31,62],$ so the final set of red blocks begins at $(32,64]$. That makes
The last interval that doesn't contain a cube is $(31,62],$ so the final set of red blocks begins at $(32,64]$. That makes
copeland
2017-03-09 21:13:10
$Q(2)=32$.
$Q(2)=32$.
copeland
2017-03-09 21:13:12
Now on to $Q(3)$. Hopefully this is faster.
Now on to $Q(3)$. Hopefully this is faster.
copeland
2017-03-09 21:13:14
There ought to be an inequality between $Q(2)$ and $Q(3)$. What is it?
There ought to be an inequality between $Q(2)$ and $Q(3)$. What is it?
yrnsmurf
2017-03-09 21:14:05
Q(3)=<Q(2)
Q(3)=<Q(2)
DemonPlat4
2017-03-09 21:14:05
Q(2)$\geq$Q(3)
Q(2)$\geq$Q(3)
a1b2
2017-03-09 21:14:05
$Q(x)>=Q(x+1)$
$Q(x)>=Q(x+1)$
vishwathganesan
2017-03-09 21:14:05
oops I meant Q(n) >= Q(n + 1)
oops I meant Q(n) >= Q(n + 1)
copeland
2017-03-09 21:14:11
If there's a cube in $(n,2n]$ then there's definitely a cube in $(n,3n]$ so $Q(3)\leq Q(2)$. In fact, $Q$ is going to be a decreasing function. Let's compute $Q(3)$ by hand again to see what we can see.
If there's a cube in $(n,2n]$ then there's definitely a cube in $(n,3n]$ so $Q(3)\leq Q(2)$. In fact, $Q$ is going to be a decreasing function. Let's compute $Q(3)$ by hand again to see what we can see.
copeland
2017-03-09 21:14:19
First of all for $m=2$ we found the last interval that doesn't contain a cube lies between 27 and 64. Is there an interval of the form $(n,3n]$ between 27 and 64?
First of all for $m=2$ we found the last interval that doesn't contain a cube lies between 27 and 64. Is there an interval of the form $(n,3n]$ between 27 and 64?
thetank
2017-03-09 21:14:48
no
no
bomb427006
2017-03-09 21:14:48
no
no
Doink
2017-03-09 21:14:48
no
no
First
2017-03-09 21:14:48
No
No
Mrkiller
2017-03-09 21:14:48
no
no
copeland
2017-03-09 21:14:50
Why not?
Why not?
curry3030
2017-03-09 21:15:02
no, 27*3 = 81
no, 27*3 = 81
Makorn
2017-03-09 21:15:02
$27\cdot3=81>64$
$27\cdot3=81>64$
tdeng
2017-03-09 21:15:02
27*3=81>64
27*3=81>64
vishwathganesan
2017-03-09 21:15:02
64/27 < 3
64/27 < 3
varun555
2017-03-09 21:15:02
64/27<3
64/27<3
copeland
2017-03-09 21:15:06
Since $\dfrac{64}{27}<3$, there aren't any intervals inside here.
Since $\dfrac{64}{27}<3$, there aren't any intervals inside here.
copeland
2017-03-09 21:15:07
Oh good! $Q(3)<Q(2)$.
Oh good! $Q(3)<Q(2)$.
copeland
2017-03-09 21:15:09
Are there any intervals $(n,3n]$ that fit between 8 and 27?
Are there any intervals $(n,3n]$ that fit between 8 and 27?
DemonPlat4
2017-03-09 21:15:36
yes (8,24]
yes (8,24]
a1b2
2017-03-09 21:15:36
$(8,24]$
$(8,24]$
Makorn
2017-03-09 21:15:36
Yes, $(8,24]$
Yes, $(8,24]$
GeronimoStilton
2017-03-09 21:15:36
$(8,24]$
$(8,24]$
vishwathganesan
2017-03-09 21:15:36
(8, 24]
(8, 24]
cooljoseph
2017-03-09 21:15:36
Yes, 8,24
Yes, 8,24
IsaacZ123
2017-03-09 21:15:36
(8,24] should be the last one
(8,24] should be the last one
checkmatetang
2017-03-09 21:15:36
(8,24]
(8,24]
IYN
2017-03-09 21:15:36
(8, 24]
(8, 24]
amackenzie1
2017-03-09 21:15:36
8, 24
8, 24
First
2017-03-09 21:15:36
Yes,8, 24
Yes,8, 24
copeland
2017-03-09 21:15:38
Yes, but just barely! $(8,24]$ fits, but $(9,27]$ does not!
Yes, but just barely! $(8,24]$ fits, but $(9,27]$ does not!
copeland
2017-03-09 21:15:39
Do we know what $Q(3)$ is yet?
Do we know what $Q(3)$ is yet?
GeronimoStilton
2017-03-09 21:16:09
$Q(3) = 9$
$Q(3) = 9$
Makorn
2017-03-09 21:16:09
$Q(3)=9$
$Q(3)=9$
a1b2
2017-03-09 21:16:09
$Q(3)=9$
$Q(3)=9$
stronto
2017-03-09 21:16:09
9
9
samuel
2017-03-09 21:16:09
yes 9
yes 9
islander7
2017-03-09 21:16:09
9
9
Mrkiller
2017-03-09 21:16:09
9?
9?
cooljoseph
2017-03-09 21:16:09
yes: 9
yes: 9
ninjataco
2017-03-09 21:16:11
9
9
copeland
2017-03-09 21:16:13
Yes, we know that after $n=8$ the intervals all contain cubes. The interval $(8,24]$ does not contain a cube, so the intervals all turn red at $(9,27]$. That makes
Yes, we know that after $n=8$ the intervals all contain cubes. The interval $(8,24]$ does not contain a cube, so the intervals all turn red at $(9,27]$. That makes
copeland
2017-03-09 21:16:14
$Q(3)=9.$
$Q(3)=9.$
copeland
2017-03-09 21:16:15
copeland
2017-03-09 21:16:20
Oh, wow. That graph tells us even more, right? Can you tell me what $Q(4)$ is going to be right away?
Oh, wow. That graph tells us even more, right? Can you tell me what $Q(4)$ is going to be right away?
mishai
2017-03-09 21:16:52
2
2
GeronimoStilton
2017-03-09 21:16:52
$2$.
$2$.
ninjataco
2017-03-09 21:16:52
2
2
NewbieGamer
2017-03-09 21:16:52
2
2
tdeng
2017-03-09 21:16:52
2
2
cooljoseph
2017-03-09 21:16:52
2
2
brainiac1
2017-03-09 21:16:52
2
2
SomethingNeutral
2017-03-09 21:16:52
actually 2
actually 2
yrnsmurf
2017-03-09 21:16:52
2
2
Ericaops
2017-03-09 21:16:52
2
2
stronto
2017-03-09 21:16:52
2
2
copeland
2017-03-09 21:16:56
When we draw the graph for $m=4$, it's going to be just like this graph but with longer bars. For example, instead of $(5,15]$ we'll have $(5,20]$. The new bars will always be red when the old bars were.
When we draw the graph for $m=4$, it's going to be just like this graph but with longer bars. For example, instead of $(5,15]$ we'll have $(5,20]$. The new bars will always be red when the old bars were.
copeland
2017-03-09 21:16:58
However, the bar at 8 is going to be $(8,32]$, which contains 27 so it will turn red.
However, the bar at 8 is going to be $(8,32]$, which contains 27 so it will turn red.
copeland
2017-03-09 21:17:02
What about the bar at 2?
What about the bar at 2?
QuestForKnowledge
2017-03-09 21:17:30
red
red
DemonPlat4
2017-03-09 21:17:30
it reaches 8
it reaches 8
SomethingNeutral
2017-03-09 21:17:30
2,8 contains 8
2,8 contains 8
stronto
2017-03-09 21:17:30
itll be red just barely
itll be red just barely
IsaacZ123
2017-03-09 21:17:30
(2,8] contains a cube
(2,8] contains a cube
owm
2017-03-09 21:17:30
it will also be red
it will also be red
copeland
2017-03-09 21:17:32
The bar at 2 will be $(2,8]$, which also contains a cube so will also turn red.
The bar at 2 will be $(2,8]$, which also contains a cube so will also turn red.
copeland
2017-03-09 21:17:36
However, the bar at 1 will be $(1,4]$ which does not contain a cube.
However, the bar at 1 will be $(1,4]$ which does not contain a cube.
copeland
2017-03-09 21:17:39
That makes $Q(4)=2$.
That makes $Q(4)=2$.
copeland
2017-03-09 21:17:40
Every bar after the first bar is red.
Every bar after the first bar is red.
copeland
2017-03-09 21:17:41
copeland
2017-03-09 21:17:44
For what other values of $m$ is $Q(m)=2$?
For what other values of $m$ is $Q(m)=2$?
vishwathganesan
2017-03-09 21:18:07
Q(4) = Q(5) = Q(6) = Q(7) = 2
Q(4) = Q(5) = Q(6) = Q(7) = 2
Makorn
2017-03-09 21:18:07
$Q(4)=Q(5)=Q(6)=Q(7)=2$
$Q(4)=Q(5)=Q(6)=Q(7)=2$
SomethingNeutral
2017-03-09 21:18:07
5,6,7
5,6,7
GeronimoStilton
2017-03-09 21:18:07
$m = 5,6,7$
$m = 5,6,7$
vishwathganesan
2017-03-09 21:18:07
5,6,7
5,6,7
legolego
2017-03-09 21:18:07
5, 6, 7
5, 6, 7
Kiola
2017-03-09 21:18:07
5 6 7
5 6 7
MountainHeight
2017-03-09 21:18:07
5-7
5-7
DemonPlat4
2017-03-09 21:18:07
Q(5) = Q(6) = Q(7) = 2
Q(5) = Q(6) = Q(7) = 2
IsaacZ123
2017-03-09 21:18:07
5,6,7
5,6,7
IYN
2017-03-09 21:18:07
5, 6, 7
5, 6, 7
liuh008
2017-03-09 21:18:07
m <= 7
m <= 7
copeland
2017-03-09 21:18:10
Since $(1,5]$, $(1,6]$, and $(1,7]$ do not contain cubes, $Q(5)=Q(6)=Q(7)=2.$
Since $(1,5]$, $(1,6]$, and $(1,7]$ do not contain cubes, $Q(5)=Q(6)=Q(7)=2.$
copeland
2017-03-09 21:18:10
Then what?
Then what?
pianoman24
2017-03-09 21:18:32
Wait as $m$ goes up, it looks like $Q(m)$ will eventually always be $1$
Wait as $m$ goes up, it looks like $Q(m)$ will eventually always be $1$
IsaacZ123
2017-03-09 21:18:32
and then the rest are just 1
and then the rest are just 1
Doink
2017-03-09 21:18:32
it's all 1 from there
it's all 1 from there
liuh008
2017-03-09 21:18:32
The rest are all 1.
The rest are all 1.
lego101
2017-03-09 21:18:32
then 1?
then 1?
owm
2017-03-09 21:18:32
Q(8)=1
Q(8)=1
vishwathganesan
2017-03-09 21:18:32
Q(n) for n >= 8 equals 1
Q(n) for n >= 8 equals 1
tdeng
2017-03-09 21:18:32
AFter that it's always 1
AFter that it's always 1
yrnsmurf
2017-03-09 21:18:32
so Q(x)=1 if x>7
so Q(x)=1 if x>7
mishai
2017-03-09 21:18:32
Q(8)=1
Q(8)=1
Reef334
2017-03-09 21:18:32
they all equal 1
they all equal 1
Mrkiller
2017-03-09 21:18:32
then they are all 1!!!
then they are all 1!!!
copeland
2017-03-09 21:18:34
After 7, every bar is red. Therefore $Q(m)=1$ for $m>7$.
After 7, every bar is red. Therefore $Q(m)=1$ for $m>7$.
copeland
2017-03-09 21:18:35
So what is our final sum?
So what is our final sum?
vishwathganesan
2017-03-09 21:19:15
so total is 2059
so total is 2059
Z_Math404
2017-03-09 21:19:15
2059
2059
vishwathganesan
2017-03-09 21:19:15
so answer is 059
so answer is 059
mishai
2017-03-09 21:19:15
1*2010+4*2+9+32=2059
1*2010+4*2+9+32=2059
legolego
2017-03-09 21:19:15
59
59
DemonPlat4
2017-03-09 21:19:15
32+9+4*2+2010*1=2059 -> 059
32+9+4*2+2010*1=2059 -> 059
owm
2017-03-09 21:19:15
32+9+2+2+2+2+1+1....
32+9+2+2+2+2+1+1....
IsaacZ123
2017-03-09 21:19:15
59
59
IYN
2017-03-09 21:19:15
32+8+2+2+2+2+1*(2017-7)
32+8+2+2+2+2+1*(2017-7)
IsaacZ123
2017-03-09 21:19:15
2059
2059
Makorn
2017-03-09 21:19:15
$32+9+2+2+2+2+2+2010(1)=2059$
$32+9+2+2+2+2+2+2010(1)=2059$
copeland
2017-03-09 21:19:17
The final sum is
\[32+9+4\cdot2+1\cdot2010\equiv32+9+8+10\equiv\boxed{059}.\]
The final sum is
\[32+9+4\cdot2+1\cdot2010\equiv32+9+8+10\equiv\boxed{059}.\]
copeland
2017-03-09 21:19:24
Great work.
Great work.
copeland
2017-03-09 21:19:28
I'm really feeling it now.
I'm really feeling it now.
NewbieGamer
2017-03-09 21:19:41
14!
14!
W.Sun
2017-03-09 21:19:41
Yay! 2 More!
Yay! 2 More!
mishai
2017-03-09 21:19:41
yes my favorite problem
yes my favorite problem
lego101
2017-03-09 21:19:41
Number fourteen…I need some caffeine…
Number fourteen…I need some caffeine…
GeronimoStilton
2017-03-09 21:19:41
Problem 14!
Problem 14!
copeland
2017-03-09 21:19:44
Favorite?
Favorite?
copeland
2017-03-09 21:19:46
14. Let $a>1$ and $x>1$ satisfy\[\log_a\left(\log_a(\log_a 2)+\log_a24-128\right)=128\]and \[\log_a(\log_ax)=256.\] Find the remainder when $x$ is divided by 1000.
14. Let $a>1$ and $x>1$ satisfy\[\log_a\left(\log_a(\log_a 2)+\log_a24-128\right)=128\]and \[\log_a(\log_ax)=256.\] Find the remainder when $x$ is divided by 1000.
copeland
2017-03-09 21:19:48
There's a rule of thumb about AMC and AIME problems that when you see a logarithm, that means the problem is really scary but way less hard than it looks.
There's a rule of thumb about AMC and AIME problems that when you see a logarithm, that means the problem is really scary but way less hard than it looks.
copeland
2017-03-09 21:19:53
This problem proves that rule doesn't always hold.
This problem proves that rule doesn't always hold.
copeland
2017-03-09 21:20:00
There are lots of different avenues to take with this problem. Most of them fail. There are a couple of nice first steps, though. Any ideas?
There are lots of different avenues to take with this problem. Most of them fail. There are a couple of nice first steps, though. Any ideas?
Doink
2017-03-09 21:20:36
Take equation 1 to the power of a
Take equation 1 to the power of a
potatosensei
2017-03-09 21:20:36
undo the logs
undo the logs
stronto
2017-03-09 21:20:36
remove the first nested log
remove the first nested log
amackenzie1
2017-03-09 21:20:36
exp?
exp?
nukelauncher
2017-03-09 21:20:36
convert to exponential form
convert to exponential form
ppu
2017-03-09 21:20:36
rewrite in exponential form?
rewrite in exponential form?
pianoman24
2017-03-09 21:20:36
Write in exponent form instead of logarithm form
Write in exponent form instead of logarithm form
AwesomeAaron
2017-03-09 21:20:36
reverse the logs
reverse the logs
claserken
2017-03-09 21:20:36
Turn logs into exponentials?
Turn logs into exponentials?
copeland
2017-03-09 21:20:37
Let's exponentiate both sides with respect to base $a$:
Let's exponentiate both sides with respect to base $a$:
copeland
2017-03-09 21:20:38
\[\log_a(\log_a 2)+\log_a24-128=a^{128}\]
\[\log_a(\log_a 2)+\log_a24-128=a^{128}\]
copeland
2017-03-09 21:20:41
That looks nicer. Now we can push the 128 to the other side since I'm not really in the mood for negatives (or fractions) yet.
That looks nicer. Now we can push the 128 to the other side since I'm not really in the mood for negatives (or fractions) yet.
copeland
2017-03-09 21:20:43
\[\log_a(\log_a 2)+\log_a24=a^{128}+128\]
\[\log_a(\log_a 2)+\log_a24=a^{128}+128\]
copeland
2017-03-09 21:20:45
And now what?
And now what?
Z_Math404
2017-03-09 21:21:06
do it again
do it again
sxu
2017-03-09 21:21:06
a again
a again
ucap
2017-03-09 21:21:06
more exponents
more exponents
Makorn
2017-03-09 21:21:06
Combine the two logs
Combine the two logs
amackenzie1
2017-03-09 21:21:06
exponentiate again
exponentiate again
brainiac1
2017-03-09 21:21:06
sum the two logs
sum the two logs
nosaj
2017-03-09 21:21:06
combine the loggs
combine the loggs
DemonPlat4
2017-03-09 21:21:06
combine the logarithms
combine the logarithms
stronto
2017-03-09 21:21:06
combine logs on left side
combine logs on left side
cooljoseph
2017-03-09 21:21:06
exponentiate again
exponentiate again
duck_master
2017-03-09 21:21:06
raise it to a again!
raise it to a again!
legolego
2017-03-09 21:21:06
combine logs
combine logs
copeland
2017-03-09 21:21:08
We have $\log u+\log v =\log uv$, so
We have $\log u+\log v =\log uv$, so
copeland
2017-03-09 21:21:09
\[\log_a(24\log_a2)=a^{128}+128.\]
\[\log_a(24\log_a2)=a^{128}+128.\]
copeland
2017-03-09 21:21:10
Let's ditch those logs by exponentiating:
Let's ditch those logs by exponentiating:
copeland
2017-03-09 21:21:13
\[24\log_a2=a^{a^{128}+128}.\]
\[24\log_a2=a^{a^{128}+128}.\]
copeland
2017-03-09 21:21:14
And what's $a^{24\log_a2}$ equal?
And what's $a^{24\log_a2}$ equal?
mishai
2017-03-09 21:21:35
2^24
2^24
IsaacZ123
2017-03-09 21:21:35
2^24
2^24
yrnsmurf
2017-03-09 21:21:35
2^24
2^24
Z_Math404
2017-03-09 21:21:35
2^24
2^24
monkeybanana
2017-03-09 21:21:35
2^24
2^24
cooleybz2013
2017-03-09 21:21:35
2^24
2^24
tdeng
2017-03-09 21:21:35
2^24
2^24
copeland
2017-03-09 21:21:37
Next we have
Next we have
copeland
2017-03-09 21:21:38
\[2^{24}=a^{a^{a^{128}+128}}.\]
\[2^{24}=a^{a^{a^{128}+128}}.\]
copeland
2017-03-09 21:21:42
Those aren't the same base. Can we fix that?
Those aren't the same base. Can we fix that?
mishai
2017-03-09 21:22:08
a=2^n
a=2^n
math101010
2017-03-09 21:22:08
a=2^b
a=2^b
stronto
2017-03-09 21:22:08
let a=2^k
let a=2^k
yrnsmurf
2017-03-09 21:22:08
set a=2^b
set a=2^b
brainiac1
2017-03-09 21:22:08
assume a = 2^b
assume a = 2^b
yrnsmurf
2017-03-09 21:22:08
take the log base 2
take the log base 2
copeland
2017-03-09 21:22:10
Let's write $a=2^b$. Then we have
Let's write $a=2^b$. Then we have
copeland
2017-03-09 21:22:12
\[2^{24}=(2^b)^{(2^b)^{(2^b)^{128}+128}}=2^{b2^{b2^{128b}+128b}},\]so
\[2^{24}=(2^b)^{(2^b)^{(2^b)^{128}+128}}=2^{b2^{b2^{128b}+128b}},\]so
copeland
2017-03-09 21:22:16
\[24=b2^{b2^{128b}+128b}.\]
\[24=b2^{b2^{128b}+128b}.\]
copeland
2017-03-09 21:22:25
And do you see the magical simplification on the right side?
And do you see the magical simplification on the right side?
samuel
2017-03-09 21:23:16
a little
a little
potatosensei
2017-03-09 21:23:16
not really lol
not really lol
vishwathganesan
2017-03-09 21:23:16
no not really
no not really
copeland
2017-03-09 21:23:21
Alright, great.
Alright, great.
liuh008
2017-03-09 21:23:24
b * 2^(128b)=k
b * 2^(128b)=k
copeland
2017-03-09 21:23:31
Wait, what's that about?
Wait, what's that about?
copeland
2017-03-09 21:23:42
It looks like there might be some symmetry on the right hand side.
It looks like there might be some symmetry on the right hand side.
copeland
2017-03-09 21:23:45
Almost.
Almost.
copeland
2017-03-09 21:23:56
There's all this $2^{128b}$ nonsense.
There's all this $2^{128b}$ nonsense.
copeland
2017-03-09 21:24:00
But it's not quite the same.
But it's not quite the same.
copeland
2017-03-09 21:24:26
Can we invent another $b2^{128b}$ somewhere?
Can we invent another $b2^{128b}$ somewhere?
a1b2
2017-03-09 21:24:32
split up the sum in the exponent
split up the sum in the exponent
copeland
2017-03-09 21:24:35
Then what happens?
Then what happens?
Makorn
2017-03-09 21:25:49
$24=b2^{b2^{128b}}\cdotb^{128}$
$24=b2^{b2^{128b}}\cdotb^{128}$
rt03
2017-03-09 21:25:49
the RHS is $b2^{128b} \cdot 2^{b2^{128b}}$
the RHS is $b2^{128b} \cdot 2^{b2^{128b}}$
tdeng
2017-03-09 21:25:49
$b\cdot2^{128b}\cdot2^{b2^{128b}}$
$b\cdot2^{128b}\cdot2^{b2^{128b}}$
duck_master
2017-03-09 21:25:49
$b2^{128b} 2^{b2^{128b}}$
$b2^{128b} 2^{b2^{128b}}$
copeland
2017-03-09 21:25:52
There is a magical step in every solution to this problem. Here the magical step is
There is a magical step in every solution to this problem. Here the magical step is
copeland
2017-03-09 21:25:53
\[2^{24}=b2^{b2^{128b}+128b}=(b2^{128b})2^{b2^{128b}}.\]
\[2^{24}=b2^{b2^{128b}+128b}=(b2^{128b})2^{b2^{128b}}.\]
copeland
2017-03-09 21:25:54
That right side is $t2^t$ for some $t$. Can we write 24 in the same way?
That right side is $t2^t$ for some $t$. Can we write 24 in the same way?
mishai
2017-03-09 21:26:16
3*2^3
3*2^3
liuh008
2017-03-09 21:26:16
t = 3
t = 3
brainiac1
2017-03-09 21:26:16
24 = 3*2^3
24 = 3*2^3
SomethingNeutral
2017-03-09 21:26:16
3*2^3
3*2^3
tdeng
2017-03-09 21:26:16
3*2^3
3*2^3
rt03
2017-03-09 21:26:16
3 \cdot 2^3
3 \cdot 2^3
copeland
2017-03-09 21:26:18
$24=3\cdot2^3!$
$24=3\cdot2^3!$
copeland
2017-03-09 21:26:19
Therefore
Therefore
copeland
2017-03-09 21:26:21
\[3\cdot2^3=(b2^{128b})2^{b2^{128b}}.\]
\[3\cdot2^3=(b2^{128b})2^{b2^{128b}}.\]
copeland
2017-03-09 21:26:25
So?
So?
stronto
2017-03-09 21:26:47
b*2^128b = 3
b*2^128b = 3
mishai
2017-03-09 21:26:47
b*2^(128b)=3
b*2^(128b)=3
DemonPlat4
2017-03-09 21:26:47
b2^128b=3
b2^128b=3
randomsolver
2017-03-09 21:26:47
3=b2^(128b)
3=b2^(128b)
SeanGee
2017-03-09 21:26:47
b2^128b = 3
b2^128b = 3
copeland
2017-03-09 21:26:49
Since $f(t)=t2^t$ is monotonic, we have to have $b2^{128b}=3.$
Since $f(t)=t2^t$ is monotonic, we have to have $b2^{128b}=3.$
copeland
2017-03-09 21:26:50
We're still not done. We need another substitution. What should we make $b?$
We're still not done. We need another substitution. What should we make $b?$
mishai
2017-03-09 21:27:45
b=3/2^k
b=3/2^k
liuh008
2017-03-09 21:27:45
3/2^c could work
3/2^c could work
rt03
2017-03-09 21:27:45
$3 \cdot 2^z$
$3 \cdot 2^z$
copeland
2017-03-09 21:27:48
$b=3\cdot2^c$ is nice, but let's set $b=3\cdot2^{-c}.$ I want to get $b$ over to the right side of the equation. This gives
$b=3\cdot2^c$ is nice, but let's set $b=3\cdot2^{-c}.$ I want to get $b$ over to the right side of the equation. This gives
copeland
2017-03-09 21:27:52
\[3\cdot 2^{-c}2^{128\cdot3\cdot 2^{-c}}=3,\]so\[2^{3\cdot2^{7-c}}=2^c\] and
\[3\cdot 2^{-c}2^{128\cdot3\cdot 2^{-c}}=3,\]so\[2^{3\cdot2^{7-c}}=2^c\] and
copeland
2017-03-09 21:28:05
\[3\cdot 2^{7-c}=c.\]
\[3\cdot 2^{7-c}=c.\]
copeland
2017-03-09 21:28:09
Do you see a $c$ that solves this equation?
Do you see a $c$ that solves this equation?
mishai
2017-03-09 21:28:39
c=6
c=6
Makorn
2017-03-09 21:28:39
$c=6$
$c=6$
EpicCarrotMaster
2017-03-09 21:28:39
6
6
cooleybz2013
2017-03-09 21:28:39
6
6
pianoman24
2017-03-09 21:28:39
6.
6.
yrnsmurf
2017-03-09 21:28:39
6
6
math.fever
2017-03-09 21:28:39
6
6
nukelauncher
2017-03-09 21:28:39
6
6
BIGBUBBLE
2017-03-09 21:28:39
6
6
QuestForKnowledge
2017-03-09 21:28:39
shamelessly gueass 6
shamelessly gueass 6
mathcrazymar
2017-03-09 21:28:39
6
6
fangmu
2017-03-09 21:28:39
6
6
Funnybunny5246
2017-03-09 21:28:39
6
6
NewbieGamer
2017-03-09 21:28:39
6
6
Pudentane
2017-03-09 21:28:39
6
6
MrMXS
2017-03-09 21:28:39
$6$
$6$
lego101
2017-03-09 21:28:39
6
6
copeland
2017-03-09 21:28:41
$c=6$ solves this equation. If $c=6,$ what is $a?$
$c=6$ solves this equation. If $c=6,$ what is $a?$
vishwathganesan
2017-03-09 21:29:26
2^3/64
2^3/64
liuh008
2017-03-09 21:29:26
2^(3/64)
2^(3/64)
rt03
2017-03-09 21:29:26
$2^{\frac{3}{64}}$
$2^{\frac{3}{64}}$
Makorn
2017-03-09 21:29:26
$a=2^{\frac{6}{64}}$?
$a=2^{\frac{6}{64}}$?
QuestForKnowledge
2017-03-09 21:29:26
2^(3/64)
2^(3/64)
copeland
2017-03-09 21:29:28
If $c=6$ then $b=3\cdot2^{-6},$ and $a=2^{3\cdot 2^{-6}}.$
If $c=6$ then $b=3\cdot2^{-6},$ and $a=2^{3\cdot 2^{-6}}.$
copeland
2017-03-09 21:29:56
\[a=8^b=8^{2^{-6}}\]
\[a=8^b=8^{2^{-6}}\]
copeland
2017-03-09 21:30:05
If you got this far, then congrats. You're still not done. We need to find the $x$ that solves $\log_a\log_a x=256.$
If you got this far, then congrats. You're still not done. We need to find the $x$ that solves $\log_a\log_a x=256.$
copeland
2017-03-09 21:30:09
So we need some logs. What do we do?
So we need some logs. What do we do?
IsaacZ123
2017-03-09 21:31:30
unlock the logarithms
unlock the logarithms
math.fever
2017-03-09 21:31:30
convert a to logs
convert a to logs
potatosensei
2017-03-09 21:31:30
undo the logs
undo the logs
DemonPlat4
2017-03-09 21:31:30
change logarithms into exponentation
change logarithms into exponentation
copeland
2017-03-09 21:31:50
We have two paths. We could turn one equation into an exponential equation or turn the other equation into a logs equation.
We have two paths. We could turn one equation into an exponential equation or turn the other equation into a logs equation.
copeland
2017-03-09 21:32:03
Let's simplify our equation for $a$ a tiny bit first.
Let's simplify our equation for $a$ a tiny bit first.
copeland
2017-03-09 21:32:10
Let's take the log of our equation:\[1=\log_a8^{2^{-6}}=2^{-6}\log_a8,\] and \[\log_a8=2^6.\]
Let's take the log of our equation:\[1=\log_a8^{2^{-6}}=2^{-6}\log_a8,\] and \[\log_a8=2^6.\]
copeland
2017-03-09 21:32:12
Can we get 256 in there?
Can we get 256 in there?
vishwathganesan
2017-03-09 21:33:09
multiply by 4
multiply by 4
Makorn
2017-03-09 21:33:09
$4log_a8=256$
$4log_a8=256$
owm
2017-03-09 21:33:09
multiply by 4
multiply by 4
GeronimoStilton
2017-03-09 21:33:09
By multiplying both sides by $4$, yes.
By multiplying both sides by $4$, yes.
tdeng
2017-03-09 21:33:09
256/4=2^6
256/4=2^6
math.fever
2017-03-09 21:33:09
2^2log_a(8)=2^8
2^2log_a(8)=2^8
copeland
2017-03-09 21:33:11
Yeah, multiply by $2^2$:\[256=2^2\cdot2^6=4\log_a8=\log_a8^4=\log_a2^{12}.\]
Yeah, multiply by $2^2$:\[256=2^2\cdot2^6=4\log_a8=\log_a8^4=\log_a2^{12}.\]
copeland
2017-03-09 21:33:12
This is progress! We've gotten rid of one of the layers of log. Since $256=\log_a\log_a x$ we need to solve
This is progress! We've gotten rid of one of the layers of log. Since $256=\log_a\log_a x$ we need to solve
copeland
2017-03-09 21:33:13
\[\log_a x=2^{12}.\]
\[\log_a x=2^{12}.\]
copeland
2017-03-09 21:33:17
Congrats. Now we're here. We're still not done.
Congrats. Now we're here. We're still not done.
copeland
2017-03-09 21:33:21
I want to exponentiate:\[x=a^{2^{12}}.\] Now what?
I want to exponentiate:\[x=a^{2^{12}}.\] Now what?
QuestForKnowledge
2017-03-09 21:34:00
plug in that a
plug in that a
nukelauncher
2017-03-09 21:34:00
substitute the value of a
substitute the value of a
amackenzie1
2017-03-09 21:34:00
we know $a$!
we know $a$!
liuh008
2017-03-09 21:34:00
a = 2^(3/2^6)
a = 2^(3/2^6)
duck_master
2017-03-09 21:34:00
a=2^(3/64)
a=2^(3/64)
stronto
2017-03-09 21:34:00
a = 2^(3/64)
a = 2^(3/64)
copeland
2017-03-09 21:34:02
We do know that $a=8^{2^{-6}}$, so \[x=(8^{2^{-6}})^{2^{12}}=8^{2^6}=2^{3\cdot2^6}=2^{192}.\]
We do know that $a=8^{2^{-6}}$, so \[x=(8^{2^{-6}})^{2^{12}}=8^{2^6}=2^{3\cdot2^6}=2^{192}.\]
copeland
2017-03-09 21:34:09
Oh great! We're still not done, but we're close. We just need to compute $2^{192}$ modulo 1000.
Oh great! We're still not done, but we're close. We just need to compute $2^{192}$ modulo 1000.
copeland
2017-03-09 21:34:11
What do we need?
What do we need?
duck_master
2017-03-09 21:34:34
CRT
CRT
brainiac1
2017-03-09 21:34:34
CRT and some luck
CRT and some luck
GeronimoStilton
2017-03-09 21:34:34
CRT!
CRT!
liuh008
2017-03-09 21:34:34
mod 8 and mod 125
mod 8 and mod 125
stronto
2017-03-09 21:34:34
CRT!
CRT!
QuestForKnowledge
2017-03-09 21:34:34
take mod and reduce
take mod and reduce
S0larPh03nix
2017-03-09 21:34:34
crt and euler
crt and euler
Z_Math404
2017-03-09 21:34:34
chinese remainder theroesm
chinese remainder theroesm
Funnybunny5246
2017-03-09 21:34:34
CRT
CRT
nukelauncher
2017-03-09 21:34:34
crt
crt
copeland
2017-03-09 21:34:36
Oh, finally. I was worried this wasn't an AIME at all.
Oh, finally. I was worried this wasn't an AIME at all.
copeland
2017-03-09 21:34:43
We've made it this far without the Chinese Remainder Theorem. Now's the time. I'll get you started. $1000=8\cdot125$.
We've made it this far without the Chinese Remainder Theorem. Now's the time. I'll get you started. $1000=8\cdot125$.
copeland
2017-03-09 21:34:45
What is $2^{192}\pmod8?$
What is $2^{192}\pmod8?$
SomethingNeutral
2017-03-09 21:34:54
0
0
ninjataco
2017-03-09 21:34:54
0
0
Makorn
2017-03-09 21:34:54
$0$
$0$
ppu
2017-03-09 21:34:54
0
0
owm
2017-03-09 21:34:54
0
0
Z_Math404
2017-03-09 21:34:54
0
0
copeland
2017-03-09 21:34:56
It's zero. All day long.
It's zero. All day long.
copeland
2017-03-09 21:34:57
Now we need to know $2^{192}\pmod{125}.$ First, what is the order of an element modulo $125?$ That is, what is $\phi(125)?$
Now we need to know $2^{192}\pmod{125}.$ First, what is the order of an element modulo $125?$ That is, what is $\phi(125)?$
legolego
2017-03-09 21:35:15
100
100
yrnsmurf
2017-03-09 21:35:15
100
100
nosaj
2017-03-09 21:35:15
100
100
ppu
2017-03-09 21:35:15
100
100
varun555
2017-03-09 21:35:15
100
100
SnakeYu
2017-03-09 21:35:15
100
100
tdeng
2017-03-09 21:35:15
100
100
copeland
2017-03-09 21:36:02
$\phi(p^n)=p^n-p^{n-1}$, so $\phi(125)=125-25=100.$
$\phi(p^n)=p^n-p^{n-1}$, so $\phi(125)=125-25=100.$
copeland
2017-03-09 21:36:24
So what should we compute instead of $2^{192}\pmod{125}?$
So what should we compute instead of $2^{192}\pmod{125}?$
andsun19
2017-03-09 21:36:31
2^-8 MOD 125
2^-8 MOD 125
amackenzie1
2017-03-09 21:36:31
$2^{-8}$
$2^{-8}$
SnakeYu
2017-03-09 21:36:31
2^-8
2^-8
rt03
2017-03-09 21:36:31
$2^{-8}$
$2^{-8}$
cjquines0
2017-03-09 21:36:31
$2^{-8}$ seems easier
$2^{-8}$ seems easier
celestialphoenix3768
2017-03-09 21:36:31
2^-8 mod 125
2^-8 mod 125
copeland
2017-03-09 21:36:35
We can compute $2^{-8}\pmod{125}$, which (another freebie) is \[256^{-1}\equiv6^{-1}\pmod{125}.\]
We can compute $2^{-8}\pmod{125}$, which (another freebie) is \[256^{-1}\equiv6^{-1}\pmod{125}.\]
copeland
2017-03-09 21:36:37
Can you see the inverse of 6 modulo 125?
Can you see the inverse of 6 modulo 125?
copeland
2017-03-09 21:36:41
Can you think of any multiples of 6 that are one more than a multiple of 125?
Can you think of any multiples of 6 that are one more than a multiple of 125?
ppu
2017-03-09 21:37:01
126 XD
126 XD
rt03
2017-03-09 21:37:01
126
126
a1b2
2017-03-09 21:37:01
126
126
Makorn
2017-03-09 21:37:01
$126=6\cdot21$
$126=6\cdot21$
lego101
2017-03-09 21:37:01
126
126
amackenzie1
2017-03-09 21:37:01
126
126
liuh008
2017-03-09 21:37:01
126
126
brainiac1
2017-03-09 21:37:01
126
126
owm
2017-03-09 21:37:01
126
126
copeland
2017-03-09 21:37:04
Oh, 126 is a multiple of 6. So what is $6^{-1}\pmod{125}?$
Oh, 126 is a multiple of 6. So what is $6^{-1}\pmod{125}?$
S0larPh03nix
2017-03-09 21:37:18
21
21
celestialphoenix3768
2017-03-09 21:37:18
21
21
tdeng
2017-03-09 21:37:18
21
21
GeronimoStilton
2017-03-09 21:37:18
$21$
$21$
S0larPh03nix
2017-03-09 21:37:18
21
21
dhruv
2017-03-09 21:37:18
21
21
SomethingNeutral
2017-03-09 21:37:18
21
21
stronto
2017-03-09 21:37:18
21
21
MountainHeight
2017-03-09 21:37:18
21
21
rt03
2017-03-09 21:37:18
21
21
copeland
2017-03-09 21:37:20
Since $6\cdot21=126$ we know that $6^{-1}\equiv21\pmod{1000}.$
Since $6\cdot21=126$ we know that $6^{-1}\equiv21\pmod{1000}.$
copeland
2017-03-09 21:37:22
Great, but we're still not done. Now we need to find a solution to the congruence:
\begin{align*}
x&\equiv0\pmod8\\
x&\equiv21\pmod{125}.
\end{align*}
Great, but we're still not done. Now we need to find a solution to the congruence:
\begin{align*}
x&\equiv0\pmod8\\
x&\equiv21\pmod{125}.
\end{align*}
copeland
2017-03-09 21:37:23
Anyone got it?
Anyone got it?
celestialphoenix3768
2017-03-09 21:38:01
896
896
GeronimoStilton
2017-03-09 21:38:01
$896$!
$896$!
liuh008
2017-03-09 21:38:01
896
896
celestialphoenix3768
2017-03-09 21:38:01
896
896
brainiac1
2017-03-09 21:38:01
896
896
rd123
2017-03-09 21:38:01
896
896
S0larPh03nix
2017-03-09 21:38:01
896
896
islander7
2017-03-09 21:38:01
896
896
a1b2
2017-03-09 21:38:01
$\boxed{896}$
$\boxed{896}$
Z_Math404
2017-03-09 21:38:01
896
896
copeland
2017-03-09 21:38:03
The value $21-125=-104$ is actually a multiple of 8, so that's the right residue class. The answer is $1000+(-104)=\boxed{896}.$
The value $21-125=-104$ is actually a multiple of 8, so that's the right residue class. The answer is $1000+(-104)=\boxed{896}.$
copeland
2017-03-09 21:38:08
Great work, team. Uno mas.
Great work, team. Uno mas.
copeland
2017-03-09 21:38:13
(That's Spanish.)
(That's Spanish.)
lego101
2017-03-09 21:38:33
Number fifteen. THE FINAL BATTLE. It's you vs. the math, and no one can intervene.
Number fifteen. THE FINAL BATTLE. It's you vs. the math, and no one can intervene.
QuestForKnowledge
2017-03-09 21:38:33
one more
one more
IsaacZ123
2017-03-09 21:38:33
oh my
oh my
Makorn
2017-03-09 21:38:33
I only do Esperanto; it's logically equivalent to math
I only do Esperanto; it's logically equivalent to math
pie314159265
2017-03-09 21:38:33
i thought it was chinese jkjk
i thought it was chinese jkjk
poremon
2017-03-09 21:38:33
Yo entiendo
Yo entiendo
copeland
2017-03-09 21:38:38
15. The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt3,$ $5,$ and $\sqrt{37},$ as shown, is $\dfrac{m\sqrt p}n,$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.
15. The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt3,$ $5,$ and $\sqrt{37},$ as shown, is $\dfrac{m\sqrt p}n,$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.
copeland
2017-03-09 21:38:40
copeland
2017-03-09 21:38:52
Let's pick some variables. There are lots of choices, but I have a personal favorite. What are the options that you see?
Let's pick some variables. There are lots of choices, but I have a personal favorite. What are the options that you see?
math101010
2017-03-09 21:39:23
s=side length
s=side length
pianoman24
2017-03-09 21:39:23
How about $s$ for side length?
How about $s$ for side length?
Regardy
2017-03-09 21:39:23
the area?
the area?
tdeng
2017-03-09 21:39:23
Sidelength of the equilaterel triangle
Sidelength of the equilaterel triangle
copeland
2017-03-09 21:39:48
We could use side length or area. Those are options, but I don't like side length because it might not give us a unique triangle.
We could use side length or area. Those are options, but I don't like side length because it might not give us a unique triangle.
QuestForKnowledge
2017-03-09 21:39:55
alpha and omega and beta and ...
alpha and omega and beta and ...
copeland
2017-03-09 21:39:59
Those are lots of letters.
Those are lots of letters.
copeland
2017-03-09 21:40:05
Where else could we put them?
Where else could we put them?
yrnsmurf
2017-03-09 21:40:18
the coordinates of the vertices of the triangle
the coordinates of the vertices of the triangle
QuestForKnowledge
2017-03-09 21:40:21
for angles maybe, thats there oringinal purpose
for angles maybe, thats there oringinal purpose
nosaj
2017-03-09 21:40:23
we should have a variable for the horizontal and vertical positions of the two vertices on the legs
we should have a variable for the horizontal and vertical positions of the two vertices on the legs
copeland
2017-03-09 21:40:29
We could label $\theta$ as the variable below and work entirely in terms of that. We could also work with $\phi=90^\circ-\theta$ just as easily. We can also work with either of the lengths, $a$, or $b$, of that small right triangle.
We could label $\theta$ as the variable below and work entirely in terms of that. We could also work with $\phi=90^\circ-\theta$ just as easily. We can also work with either of the lengths, $a$, or $b$, of that small right triangle.
copeland
2017-03-09 21:40:30
copeland
2017-03-09 21:40:34
My inclination is to work with both of the variables $a$ and $b$. There are two reasons for this.
My inclination is to work with both of the variables $a$ and $b$. There are two reasons for this.
copeland
2017-03-09 21:40:36
First, I like symmetry, and I expect that the symmetrical relationship between $a$ and $b$ will help us deal with the variables together.
First, I like symmetry, and I expect that the symmetrical relationship between $a$ and $b$ will help us deal with the variables together.
copeland
2017-03-09 21:40:40
Second, I have a hunch that if we use only $a$, then whatever ugly expression gives us $b$ is going to be carried around through a lot of computations trying to make things harder on us.
Second, I have a hunch that if we use only $a$, then whatever ugly expression gives us $b$ is going to be carried around through a lot of computations trying to make things harder on us.
lego101
2017-03-09 21:40:47
Why wouldn't side length give a unique triangle?
Why wouldn't side length give a unique triangle?
copeland
2017-03-09 21:40:49
Good question.
Good question.
copeland
2017-03-09 21:41:06
If there's a unique minimum, then moving either direction from it should give you the numbers just larger than it.
If there's a unique minimum, then moving either direction from it should give you the numbers just larger than it.
copeland
2017-03-09 21:41:15
Let's think in terms of $a$ and $b$. There is a relation between them that makes the equilateral fit inside the big triangle. What's it time for now?
Let's think in terms of $a$ and $b$. There is a relation between them that makes the equilateral fit inside the big triangle. What's it time for now?
GeronimoStilton
2017-03-09 21:41:42
Please not coordinate bash!
Please not coordinate bash!
amgmflannigan
2017-03-09 21:41:42
coordinates
coordinates
GeronimoStilton
2017-03-09 21:41:42
Complex bash!
Complex bash!
nosaj
2017-03-09 21:41:42
uhh coordinate geometry?
uhh coordinate geometry?
copeland
2017-03-09 21:41:55
It's time for some analytic techniques. Let's skip complex numbers on this, though that works nicely, too. Here are some coordinates:
It's time for some analytic techniques. Let's skip complex numbers on this, though that works nicely, too. Here are some coordinates:
copeland
2017-03-09 21:41:56
copeland
2017-03-09 21:41:58
Now, if we randomly pick $a$ and $b$, we won't have $(x,y)$ on that line.
Now, if we randomly pick $a$ and $b$, we won't have $(x,y)$ on that line.
copeland
2017-03-09 21:42:01
In terms of $a$ and $b$, let's figure out $(x,y)$. Synthetically, how would you get to $(x,y)?$
In terms of $a$ and $b$, let's figure out $(x,y)$. Synthetically, how would you get to $(x,y)?$
nosaj
2017-03-09 21:42:52
rotatate 60 degrees!
rotatate 60 degrees!
yrnsmurf
2017-03-09 21:42:52
rotate (a,0) 60 degrees around (0,b)
rotate (a,0) 60 degrees around (0,b)
ninjataco
2017-03-09 21:42:52
rotate (a,0) around (0,b)
rotate (a,0) around (0,b)
cooljoseph
2017-03-09 21:42:52
rotating 60 degrees
rotating 60 degrees
IsaacZ123
2017-03-09 21:42:52
rotate the line with (0,b) and (a,0) by -60 degrees
rotate the line with (0,b) and (a,0) by -60 degrees
copeland
2017-03-09 21:43:01
That's one idea. Here's the other:
That's one idea. Here's the other:
owm
2017-03-09 21:43:07
Find the midpoint and slope of (0,b)-(a,0)
Find the midpoint and slope of (0,b)-(a,0)
liuh008
2017-03-09 21:43:07
Perpendicular bisector of (0,b) and (a,0)
Perpendicular bisector of (0,b) and (a,0)
mathfever
2017-03-09 21:43:07
perpendicular bisector of the line
perpendicular bisector of the line
Funnybunny5246
2017-03-09 21:43:07
Perpendicular bisector of ab
Perpendicular bisector of ab
copeland
2017-03-09 21:43:11
Let's go with the second.
Let's go with the second.
copeland
2017-03-09 21:43:20
One way to get there is to go to the midpoint of the segment from $(a,0)$ to $(0,b)$ and then walk along the altitude. The slope of the that base is $-\frac ba$. What's the slope of the altitude?
One way to get there is to go to the midpoint of the segment from $(a,0)$ to $(0,b)$ and then walk along the altitude. The slope of the that base is $-\frac ba$. What's the slope of the altitude?
raxu
2017-03-09 21:43:37
$\frac{a}{b}$
$\frac{a}{b}$
IsaacZ123
2017-03-09 21:43:37
a/b
a/b
math101010
2017-03-09 21:43:37
a/b
a/b
islander7
2017-03-09 21:43:37
a/b
a/b
nosaj
2017-03-09 21:43:37
a/b
a/b
copeland
2017-03-09 21:43:39
The slope of the altitude is $\dfrac ab$.
The slope of the altitude is $\dfrac ab$.
copeland
2017-03-09 21:43:45
If the edge length is $s$, what's the altitude?
If the edge length is $s$, what's the altitude?
swagger21
2017-03-09 21:44:24
s*sqrt3/2
s*sqrt3/2
math101010
2017-03-09 21:44:24
s*sqrt(3)/2
s*sqrt(3)/2
ThorJames
2017-03-09 21:44:24
s sqrt(3) / 2
s sqrt(3) / 2
raxu
2017-03-09 21:44:24
$\frac{sqrt{3}}{2}s$
$\frac{sqrt{3}}{2}s$
tdeng
2017-03-09 21:44:24
$\frac{s\sqrt{3}}{2}$
$\frac{s\sqrt{3}}{2}$
Funnybunny5246
2017-03-09 21:44:24
s sqrt(3)/2
s sqrt(3)/2
vishwathganesan
2017-03-09 21:44:24
s*sqrt3/2
s*sqrt3/2
GeronimoStilton
2017-03-09 21:44:24
$\frac{s\sqrt{3}}{2}$
$\frac{s\sqrt{3}}{2}$
MountainHeight
2017-03-09 21:44:24
s√3/2
s√3/2
MrMXS
2017-03-09 21:44:24
$\cfrac{s\sqrt{3}}{2}$
$\cfrac{s\sqrt{3}}{2}$
copeland
2017-03-09 21:44:27
The altitude is $s\dfrac{\sqrt3}2$.
The altitude is $s\dfrac{\sqrt3}2$.
copeland
2017-03-09 21:44:37
So if we start at $\left(\dfrac a2,\dfrac b2\right)$ and move $s\dfrac{\sqrt3}2$ in the $\dfrac ab$ direction where do we end up?
So if we start at $\left(\dfrac a2,\dfrac b2\right)$ and move $s\dfrac{\sqrt3}2$ in the $\dfrac ab$ direction where do we end up?
raxu
2017-03-09 21:45:21
$(x,y)=(\frac{a+\sqrt{3}b}{2},\frac{b+\sqrt{3}a}{2})$
$(x,y)=(\frac{a+\sqrt{3}b}{2},\frac{b+\sqrt{3}a}{2})$
IsaacZ123
2017-03-09 21:45:24
(a/2+a/(sqrt(a^2+b^2)*s*sqrt(3)/2,b/2+b/(sqrt(a^2+b^2)*s*sqrt(3)/2)
(a/2+a/(sqrt(a^2+b^2)*s*sqrt(3)/2,b/2+b/(sqrt(a^2+b^2)*s*sqrt(3)/2)
copeland
2017-03-09 21:45:26
The vector we want to travel along points toward $(b,a)$. The vector $(b,a)$ of course has length $s$, since $(a,-b)$ does. So the vector along the altitude is $\dfrac{\sqrt3}2(b,a)$ from the center of the base.
The vector we want to travel along points toward $(b,a)$. The vector $(b,a)$ of course has length $s$, since $(a,-b)$ does. So the vector along the altitude is $\dfrac{\sqrt3}2(b,a)$ from the center of the base.
copeland
2017-03-09 21:45:27
The third vertex is \[\left(\dfrac a2,\dfrac b2\right)+\left(\dfrac {b\sqrt3}2,\dfrac{a\sqrt3}2\right),\]which is
The third vertex is \[\left(\dfrac a2,\dfrac b2\right)+\left(\dfrac {b\sqrt3}2,\dfrac{a\sqrt3}2\right),\]which is
copeland
2017-03-09 21:45:29
\[\left(\dfrac{a+b\sqrt3}2,\dfrac{b+a\sqrt3}2\right).\]
\[\left(\dfrac{a+b\sqrt3}2,\dfrac{b+a\sqrt3}2\right).\]
copeland
2017-03-09 21:45:33
This is a nice expression since when we swap $a$ and $b,$ the coordinates swap (because the equilateral triangle flips over $y=x$).
This is a nice expression since when we swap $a$ and $b,$ the coordinates swap (because the equilateral triangle flips over $y=x$).
copeland
2017-03-09 21:45:34
What's the equation for the hypotenuse?
What's the equation for the hypotenuse?
raxu
2017-03-09 21:46:48
$y=5-\frac{5}{2\sqrt{3}}x$
$y=5-\frac{5}{2\sqrt{3}}x$
duck_master
2017-03-09 21:46:48
$\frac{x}{5}+\frac{y}{2\sqrt{3}}=1$
$\frac{x}{5}+\frac{y}{2\sqrt{3}}=1$
math101010
2017-03-09 21:46:48
y=(-2sqrt3/5)*x+5
y=(-2sqrt3/5)*x+5
Funnybunny5246
2017-03-09 21:46:48
ay+bx=ab
ay+bx=ab
vishwathganesan
2017-03-09 21:46:48
5y+2sqrt3x=something
5y+2sqrt3x=something
liuh008
2017-03-09 21:46:48
-5/(2sqrt{3})x + 5 = y
-5/(2sqrt{3})x + 5 = y
copeland
2017-03-09 21:46:51
If a line has intercepts $(p,0)$ and $(0,q)$, then its equation is $qx+py=pq$.
If a line has intercepts $(p,0)$ and $(0,q)$, then its equation is $qx+py=pq$.
copeland
2017-03-09 21:46:52
The equation for the hypotenuse is \[5x+2\sqrt3 y=10\sqrt3.\]
The equation for the hypotenuse is \[5x+2\sqrt3 y=10\sqrt3.\]
copeland
2017-03-09 21:47:00
We can combine these to get the relationship between $a$ and $b$. Awesome. I won't torture you with the work.
We can combine these to get the relationship between $a$ and $b$. Awesome. I won't torture you with the work.
copeland
2017-03-09 21:47:04
Combining $(x,y)=\left(\dfrac{a+b\sqrt3}2,\dfrac{b+a\sqrt3}2\right)$ with $5x+2\sqrt3 y=10\sqrt3$ gives
\[
5\left(\dfrac{a+b\sqrt3}2\right)+2\sqrt3\left(\dfrac{b+a\sqrt3}2\right)=10\sqrt3.
\]Multiplying by 2 and expanding gives\[
11a+7\sqrt3 b=20\sqrt3.
\]
Combining $(x,y)=\left(\dfrac{a+b\sqrt3}2,\dfrac{b+a\sqrt3}2\right)$ with $5x+2\sqrt3 y=10\sqrt3$ gives
\[
5\left(\dfrac{a+b\sqrt3}2\right)+2\sqrt3\left(\dfrac{b+a\sqrt3}2\right)=10\sqrt3.
\]Multiplying by 2 and expanding gives\[
11a+7\sqrt3 b=20\sqrt3.
\]
copeland
2017-03-09 21:47:14
What are we trying to minimize?
What are we trying to minimize?
swagger21
2017-03-09 21:47:43
a^2 + b^2
a^2 + b^2
math101010
2017-03-09 21:47:43
a^2+b^2
a^2+b^2
pie314159265
2017-03-09 21:47:43
a^2+b^2
a^2+b^2
ninjataco
2017-03-09 21:47:43
a^2 + b^2
a^2 + b^2
raxu
2017-03-09 21:47:43
$\frac{\sqrt{3}}{4}(a^2+b^2)$
$\frac{\sqrt{3}}{4}(a^2+b^2)$
cooljoseph
2017-03-09 21:47:43
$a^2+b^2$
$a^2+b^2$
treemath
2017-03-09 21:47:43
a^2+b^2
a^2+b^2
liuh008
2017-03-09 21:47:43
a^2 + b^2
a^2 + b^2
owm
2017-03-09 21:47:43
a^2+b^2
a^2+b^2
copeland
2017-03-09 21:47:48
We're trying to minimize the area of that gray triangle.
We're trying to minimize the area of that gray triangle.
copeland
2017-03-09 21:47:52
The area of an equilateral triangle with side $s$ is $\dfrac{s^2\sqrt3}4$. Since $s^2=a^2+b^2$. . .
The area of an equilateral triangle with side $s$ is $\dfrac{s^2\sqrt3}4$. Since $s^2=a^2+b^2$. . .
copeland
2017-03-09 21:47:53
We want to minimize\[\dfrac{\sqrt3}4(a^2+b^2)\] given that \[11a+7\sqrt3 b=20\sqrt3.\]
We want to minimize\[\dfrac{\sqrt3}4(a^2+b^2)\] given that \[11a+7\sqrt3 b=20\sqrt3.\]
copeland
2017-03-09 21:47:57
What tool do we need?
What tool do we need?
CornSaltButter
2017-03-09 21:48:31
Cauchy-schwarz inequality
Cauchy-schwarz inequality
randomsolver
2017-03-09 21:48:31
cauchy
cauchy
liuh008
2017-03-09 21:48:31
CauchyS?
CauchyS?
tdeng
2017-03-09 21:48:31
Cauchy Schwarz?
Cauchy Schwarz?
cooljoseph
2017-03-09 21:48:31
C-S?
C-S?
raxu
2017-03-09 21:48:31
Solve for $b$ in terms of $a$, or use Cauchy-Schwarz Inequality.
Solve for $b$ in terms of $a$, or use Cauchy-Schwarz Inequality.
copeland
2017-03-09 21:48:33
This is the perfect time for Cauchy-Schwartz. ( http://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality )
This is the perfect time for Cauchy-Schwartz. ( http://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality )
copeland
2017-03-09 21:48:34
We need two lists of numbers to mush together. What are our lists?
We need two lists of numbers to mush together. What are our lists?
IsaacZ123
2017-03-09 21:49:23
a^2+b^2 and 11a+7sqrt(3)b
a^2+b^2 and 11a+7sqrt(3)b
GeronimoStilton
2017-03-09 21:49:23
$a$ and $b$, $11$ and $7\sqrt{3}$
$a$ and $b$, $11$ and $7\sqrt{3}$
mishai
2017-03-09 21:49:23
(a^2+b^2)(121+147)>=(11a+7bsqrt3)^2
(a^2+b^2)(121+147)>=(11a+7bsqrt3)^2
raxu
2017-03-09 21:49:23
$11,7\sqrt{3};a,b$
$11,7\sqrt{3};a,b$
brainiac1
2017-03-09 21:49:23
(a^2+b^2)(121+147)
(a^2+b^2)(121+147)
swagger21
2017-03-09 21:49:23
(a^2 + b^2)(121 + 147) ≥ (11a + 7root3b)^2
(a^2 + b^2)(121 + 147) ≥ (11a + 7root3b)^2
lego101
2017-03-09 21:49:29
i knew there was going to be a cauchy schwarz problem on aime this year
i knew there was going to be a cauchy schwarz problem on aime this year
copeland
2017-03-09 21:49:37
Shocker, right?
Shocker, right?
copeland
2017-03-09 21:49:39
Certainly one of the lists is $a,b$. The other list is $7,\sqrt3$.
Certainly one of the lists is $a,b$. The other list is $7,\sqrt3$.
copeland
2017-03-09 21:49:47
Cauchy-Schwarz tells us that \[(11a+7\sqrt3b)^2\leq(a^2+b^2)(11^2+(7\sqrt3)^2).\]
Cauchy-Schwarz tells us that \[(11a+7\sqrt3b)^2\leq(a^2+b^2)(11^2+(7\sqrt3)^2).\]
copeland
2017-03-09 21:49:51
Expanding all that out and substituting what we know about the left side gives
\begin{align*}
a^2+b^2
\geq\quad&\frac{(11a+7\sqrt3b)^2}{(11^2+(7\sqrt3)^2)}\\
&=\frac{(20\sqrt3)^2}{(121+147)}\\
&=\frac{1200}{268}.\\
&=\frac{300}{67}.\\
\end{align*}
We even know that equality occurs when $(a,b)$ is proportional to $(11,7\sqrt3)$, so we are going to get a minimum from this.
Expanding all that out and substituting what we know about the left side gives
\begin{align*}
a^2+b^2
\geq\quad&\frac{(11a+7\sqrt3b)^2}{(11^2+(7\sqrt3)^2)}\\
&=\frac{(20\sqrt3)^2}{(121+147)}\\
&=\frac{1200}{268}.\\
&=\frac{300}{67}.\\
\end{align*}
We even know that equality occurs when $(a,b)$ is proportional to $(11,7\sqrt3)$, so we are going to get a minimum from this.
copeland
2017-03-09 21:50:01
That's nice. And what is the minimum area?
That's nice. And what is the minimum area?
GeronimoStilton
2017-03-09 21:50:22
You mean $11$ and $7\sqrt{3}$ is the second list. @copeland
You mean $11$ and $7\sqrt{3}$ is the second list. @copeland
brainiac1
2017-03-09 21:51:00
75sqrt3/67
75sqrt3/67
yrnsmurf
2017-03-09 21:51:00
75/67 sqrt 3
75/67 sqrt 3
IsaacZ123
2017-03-09 21:51:00
75sqrt(3)/67
75sqrt(3)/67
NewbieGamer
2017-03-09 21:51:00
75*rt3/67
75*rt3/67
Mudkipswims42
2017-03-09 21:51:00
$\dfrac{75\sqrt{3}}{67}$
$\dfrac{75\sqrt{3}}{67}$
amgmflannigan
2017-03-09 21:51:00
300/67*\sqrt{3}/4
300/67*\sqrt{3}/4
ninjataco
2017-03-09 21:51:00
300sqrt(3)/268
300sqrt(3)/268
liuh008
2017-03-09 21:51:00
75sqrt{3}
75sqrt{3}
GeronimoStilton
2017-03-09 21:51:00
$\frac{75\sqrt{3}}{67}$
$\frac{75\sqrt{3}}{67}$
mishai
2017-03-09 21:51:00
75sqrt3/67
75sqrt3/67
swagger21
2017-03-09 21:51:00
75root3/67
75root3/67
liuh008
2017-03-09 21:51:00
75sqrt{3]/67
75sqrt{3]/67
copeland
2017-03-09 21:51:02
The area is $\dfrac{s^2\sqrt3}4$ so the minimum area is $\dfrac{\sqrt3}4\cdot\dfrac{300}{67}=\dfrac{75\sqrt3}{67}.$
The area is $\dfrac{s^2\sqrt3}4$ so the minimum area is $\dfrac{\sqrt3}4\cdot\dfrac{300}{67}=\dfrac{75\sqrt3}{67}.$
copeland
2017-03-09 21:51:02
The final answer is. . .
The final answer is. . .
raxu
2017-03-09 21:51:33
$\frac{75\sqrt{3}}{67}$, giving us $\boxed{145}$.
$\frac{75\sqrt{3}}{67}$, giving us $\boxed{145}$.
mishai
2017-03-09 21:51:33
145
145
Metal_Bender19
2017-03-09 21:51:33
145
145
IsaacZ123
2017-03-09 21:51:33
145
145
raxu
2017-03-09 21:51:33
$\boxed{145}$
$\boxed{145}$
yrnsmurf
2017-03-09 21:51:33
145
145
Z_Math404
2017-03-09 21:51:33
145
145
GeronimoStilton
2017-03-09 21:51:33
$145$
$145$
MountainHeight
2017-03-09 21:51:33
145
145
Mudkipswims42
2017-03-09 21:51:33
145?
145?
ninjataco
2017-03-09 21:51:33
145
145
nature
2017-03-09 21:51:33
145
145
duck_master
2017-03-09 21:51:33
145
145
liuh008
2017-03-09 21:51:33
145
145
vishwathganesan
2017-03-09 21:51:33
oops 145
oops 145
randomsolver
2017-03-09 21:51:33
145
145
Jyzhang12
2017-03-09 21:51:33
145
145
curry3030
2017-03-09 21:51:33
145
145
amackenzie1
2017-03-09 21:51:33
145
145
illumination
2017-03-09 21:51:33
145
145
copeland
2017-03-09 21:51:36
\[75+3+67=\boxed{145}.\]
\[75+3+67=\boxed{145}.\]
copeland
2017-03-09 21:51:45
Guess what?
Guess what?
Z_Math404
2017-03-09 21:52:12
145 YAS WE DEFEATED ALL OF THEM WITH OUR MATH SWORDS!
145 YAS WE DEFEATED ALL OF THEM WITH OUR MATH SWORDS!
strategos21
2017-03-09 21:52:12
"and we are done"
"and we are done"
amackenzie1
2017-03-09 21:52:12
We did it!
We did it!
brainiac1
2017-03-09 21:52:12
and we're done
and we're done
awesomemaths
2017-03-09 21:52:12
partly spleeping now
partly spleeping now
IsaacZ123
2017-03-09 21:52:12
that was easier than 14
that was easier than 14
lego101
2017-03-09 21:52:12
We did it. We beat the problems. WE POWERED THROUGH
We did it. We beat the problems. WE POWERED THROUGH
amackenzie1
2017-03-09 21:52:12
Yay!
Yay!
vishwathganesan
2017-03-09 21:52:12
yay
yay
raxu
2017-03-09 21:52:12
We're done!
We're done!
IsaacZ123
2017-03-09 21:52:12
we're done
we're done
awesomemaths
2017-03-09 21:52:12
we are done
we are done
GeronimoStilton
2017-03-09 21:52:12
We finished the test in the required time limit!
We finished the test in the required time limit!
samuel
2017-03-09 21:52:12
We all got a 15 today!
We all got a 15 today!
lego101
2017-03-09 21:52:12
WE POWERED THROUGH
WE POWERED THROUGH
vishwathganesan
2017-03-09 21:52:12
we're not done though
we're not done though
xxu110
2017-03-09 21:52:12
what
what
cooleybz2013
2017-03-09 21:52:12
the end
the end
samuel
2017-03-09 21:52:12
What?
What?
copeland
2017-03-09 21:52:22
Yeah, I think 14 was epically hard.
Yeah, I think 14 was epically hard.
copeland
2017-03-09 21:52:34
That's all for the 2017 AIME A. Thanks for spending your evening with us.
That's all for the 2017 AIME A. Thanks for spending your evening with us.
copeland
2017-03-09 21:52:34
There will be a shindig just like this in a couple of weeks on March 24 when the inimitable Dave Patrick will work with you through the 2017 AIME II.
There will be a shindig just like this in a couple of weeks on March 24 when the inimitable Dave Patrick will work with you through the 2017 AIME II.
lego101
2017-03-09 21:52:58
Thank you so much!!!! This was amazing!
Thank you so much!!!! This was amazing!
copeland
2017-03-09 21:53:00
No, thank you guys.
No, thank you guys.
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