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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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k a My Retirement & New Leadership at AoPS
rrusczyk   1573
N 4 hours ago by SmartGroot
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
1573 replies
rrusczyk
Mar 24, 2025
SmartGroot
4 hours ago
k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Practice AMC 10A
freddyfazbear   42
N 21 minutes ago by DhruvJha
Hey everyone!

I’m back with another practice test. Sorry this one took a while to pump out since I have been busy lately.

Post your score/distribution, favorite problems, and thoughts on the difficulty of the test down below. Hope you enjoy!


Practice AMC 10A

1. Find the sum of the infinite geometric series 1/2 + 7/36 + 49/648 + …
A - 18/11, B - 9/22, C - 9/11, D - 18/7, E - 9/14

2. What is the first digit after the decimal point in the square root of 420?
A - 1, B - 2, C - 3, D - 4, E - 5

3. Caden’s calculator is broken and two of the digits are swapped for some reason. When he entered in 9 + 10, he got 21. What is the sum of the two digits that got swapped?
A - 2, B - 3, C - 4, D - 5, E - 6

4. Two circles with radiuses 47 and 96 intersect at two points A and B. Let P be the point 82% of the way from A to B. A line is drawn through P that intersects both circles twice. Let the four intersection points, from left to right be W, X, Y, and Z. Find (PW/PX)*(PY/PZ).
A - 50/5863, B - 47/96, C - 1, D - 96/47, E - 5863/50

5. Two dice are rolled, and the two numbers shown are a and b. How many possible values of ab are there?
A - 17, B - 18, C - 19, D - 20, E - 21

6. What is the largest positive integer that cannot be expressed in the form 6a + 9b + 4c + 20d, where a, b, c, and d are positive integers?
A - 29, B - 38, C - 43, D - 76, E - 82

7. What is the absolute difference of the probabilities of getting at least 6/10 on a 10-question true or false test and at least 3/5 on a 5-question true or false test?
A - 0, B - 1/504, C - 1/252, D - 1/126, E - 1/63

8. How many arrangements of the letters in the word “ginger” are there such that the two vowels have an even number of letters (remember 0 is even) between them (including the original “ginger”)?
A - 72, B - 108, C - 144, D - 216, E - 432

9. After opening his final exam, Jason does not know how to solve a single question. So he decides to pull out his phone and search up the answers. Doing this, Jason has a success rate of anywhere from 94-100% for any given question he uses his phone on. However, if the teacher sees his phone at any point during the test, then Jason gets a 0.5 multiplier on his final test score, as well as he must finish the rest of the test questions without his phone. (Assume Jason uses his phone on every question he does until he finishes the test or gets caught.) Every question is a 5-choice multiple choice question. Jason has a 90% chance of not being caught with his phone. What is the expected value of Jason’s test score, rounded to the nearest tenth of a percent?
A - 89.9%, B - 90.0%, C - 90.1%, D - 90.2%, E - 90.3%

10. A criminal is caught by a police officer. Due to a lack of cooperation, the officer calls in a second officer so they can start the arrest smoothly. Officer 1 takes 26:18 to arrest a criminal, and officer 2 takes 13:09 to arrest a criminal. With these two police officers working together, how long should the arrest take?
A - 4:23, B - 5:26, C - 8:46, D - 17:32, E - 19:44

11. Suppose that on the coordinate grid, the x-axis represents economic freedom, and the y-axis represents social freedom, where -1 <= x, y <= 1 and a higher number for either coordinate represents more freedom along that particular axis. Accordingly, the points (0, 0), (1, 1), (-1, 1), (-1, -1), and (1, -1) represent democracy, anarchy, socialism, communism, and fascism, respectively. A country is classified as whichever point it is closest to. Suppose a theoretical new country is selected by picking a random point within the square bounded by anarchy, socialism, communism, and fascism as its vertices. What is the probability that it is fascist?
A - 1 - (1/4)pi, B - 1/5, C - (1/16)pi, D - 1/4, E - 1/8

12. Statistics show that people in Memphis who eat at KFC n days a week have a (1/10)(n+2) chance of liking kool-aid, and the number of people who eat at KFC n days a week is directly proportional to 8 - n (Note that n can only be an integer from 0 to 7, inclusive). A random person in Memphis is selected. Find the probability that they like kool-aid.
A - 13/30, B - 17/30, C - 19/30, D - 23/30, E - 29/30

13. A southern plantation has a length of 60 meters and a width of 50 meters. On the plantation, there is 1 kg of cotton per square meter waiting to be picked. The master of the plantation initially calls over 25 cotton pickers, each picking cotton at a rate of 5 kg per hour starting at 9 AM. However, he wants all of the cotton to be picked by 9 PM, and realizes that he needs to speed up the process. At 12 PM, the master then encourages his pickers to work faster by whipping them, in which they then all speed up to 6 kg per hour. At 1 PM, the master calls in 15 more pickers which pick at 5 kg per hour. Unfortunately, at 3 PM, the clouds drift away and the hot sun starts beating down, which slows every picker down by 2 kg per hour. At 4 PM, the clouds return, and all pickers return to picking at 5 kg per hour. At 5 PM, the master calls in 30 more pickers, which again pick at 5 kg per hour. At 6 PM, he calls in 30 more pickers. At 7 PM, he whips all the pickers again, speeding them up to 6 kg per hour. But at 8 PM, n pickers suddenly crash out and stop working due to fatigue, and the rest all slow back down to 5 kg per hour because they are tired. The master does not have any more pickers, so if too many of them drop out, he is screwed and will have to go overtime. Find the maximum value of n such that all of the cotton can still be picked on time, done no later than 9 PM.
A - 51, B - 52, C - 53, D - 54, E - 55

14. Find the number of positive integers n less than 69 such that the average of all the squares from 1^2 to n^2, inclusive, is an integer.
A - 11, B - 12, C - 23, D - 24, E - 48

15. Find the number of ordered pairs (a, b) of integers such that (a - b)^2 = 625 - 2ab.
A - 6, B - 10, C - 12, D - 16, E - 20

16. What is the 420th digit after the decimal point in the decimal expansion of 1/13?
A - 4, B - 5, C - 6, D - 7, E - 8

17. Two congruent towers stand near each other. Both take the shape of a right rectangular prism. A plane that cuts both towers into two pieces passes through the vertical axes of symmetry of both towers and does not cross the floor or roof of either tower. Let the point that the plane crosses the axis of symmetry of the first tower be A, and the point that the plane crosses the axis of symmetry of the second tower be B. A is 81% of the way from the floor to the roof of the first tower, and B is 69% of the way from the floor to the roof of the second tower. What percent of the total mass of both towers combined is above the plane?
A - 19%, B - 25%, C - 50%, D - 75%, E - 81%

18. What is the greatest number of positive integer factors an integer from 1 to 100 can have?
A - 10, B - 12, C - 14, D - 15, E - 16

19. On an analog clock, the minute hand makes one full revolution every hour, and the hour hand makes one full revolution every 12 hours. Both hands move at a constant rate. During which of the following time periods does the minute hand pass the hour hand?
A - 7:35 - 7:36, B - 7:36 - 7:37, C - 7:37 - 7:38, D - 7:38 - 7:39, E - 7:39 - 7:40

20. Find the smallest positive integer that is a leg in three different Pythagorean triples.
A - 12, B - 14, C - 15, D - 20, E - 21

21. How many axes of symmetry does the graph of (x^2)(y^2) = 69 have?
A - 2, B - 3, C - 4, D - 5, E - 6

22. Real numbers a, b, and c are chosen uniformly and at random from 0 to 3. Find the probability that a + b + c is less than 2.
A - 4/81, B - 8/81, C - 4/27, D - 8/27, E - 2/3

23. Let f(n) be the sum of the positive integer divisors of n. Find the sum of the digits of the smallest odd positive integer n such that f(n) is greater than 2n.
A - 15, B - 18, C - 21, D - 24, E - 27

24. Find the last three digits of 24^10.
A - 376, B - 576, C - 626, D - 876, E - 926

25. A basketball has a diameter of 9 inches, and the hoop has a diameter of 18 inches. Peter decides to pick up the basketball and make a throw. Given that Peter has a 1/4 chance of accidentally hitting the backboard and missing the shot, but if he doesn’t, he is guaranteed that the frontmost point of the basketball will be within 18 inches of the center of the hoop at the moment when a great circle of the basketball crosses the plane containing the rim. No part of the ball will extend behind the backboard at any point during the throw, and the rim is attached directly to the backboard. What is the probability that Peter makes a green FN?
A - 3/128, B - 3/64, C - 3/32, D - 3/16, E - 3/8
42 replies
freddyfazbear
Mar 24, 2025
DhruvJha
21 minutes ago
Cauchy-Schwarz 6
prtoi   2
N 29 minutes ago by sqing
Source: Handout by Samin Riasat
Let a, b, c > 0. Prove that
$\sum_{cyc}^{}\sqrt{\frac{2a}{b+c}}\le\sqrt{3(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})}$
2 replies
1 viewing
prtoi
Yesterday at 4:30 PM
sqing
29 minutes ago
Another AM-GM problem
prtoi   2
N 31 minutes ago by sqing
Source: Handout by Samin Riasat
Prove that:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\ge3+n$
2 replies
prtoi
Yesterday at 4:11 PM
sqing
31 minutes ago
Cauchy Schwarz 4
prtoi   3
N 33 minutes ago by sqing
Source: Zhautykov Olympiad 2008
Let a, b, c be positive real numbers such that abc = 1.
Show that
$\frac{1}{b(a+b)}+\frac{1}{b(a+b)}+\frac{1}{b(a+b)}\ge\frac{3}{2}$
3 replies
prtoi
Yesterday at 4:25 PM
sqing
33 minutes ago
projection vector manipulation
RenheMiResembleRice   1
N 39 minutes ago by RenheMiResembleRice
Source: Yanting Ji, Hanxue Dou
If $proj_{b}v=\left(3,11\right)$, find $proj_{b}\left(v+\left(-282,396\right)\right)$
1 reply
RenheMiResembleRice
2 hours ago
RenheMiResembleRice
39 minutes ago
Cauchy-Schwarz 2
prtoi   4
N 40 minutes ago by sqing
Source: Handout by Samin Riasat
if $a^2+b^2+c^2+d^2=4$, prove that:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge4$
4 replies
1 viewing
prtoi
Yesterday at 4:19 PM
sqing
40 minutes ago
Collinearity with orthocenter
liberator   179
N 44 minutes ago by bjump
Source: IMO 2013 Problem 4
Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand
179 replies
liberator
Jan 4, 2016
bjump
44 minutes ago
Inspired by old results
sqing   2
N an hour ago by sqing
Source: Own
Let $a,b,c$ be real numbers.Prove that
$$\frac{  (a-b)(b-c)(c-a)}{ (a^2+1)(b^2+1)(c^2+1)}\leq\frac{3\sqrt 3}{8}$$$$\frac{  (a-b)(b-c)(c-a)}{ (a^2+2)(b^2+1)(c^2+2)}\leq\frac{3}{8}\sqrt{\frac{3}{2}}$$$$\frac{  (a-b)(b-c)(c-a)}{ (a^2+3)(b^2+1)(c^2+3)}\leq\frac{3}{8} $$$$\frac{  (a-b)(b-c)(c-a)}{ (a^2+3)(b^2+2)(c^2+3)}\leq\frac{3}{16} $$
2 replies
sqing
Yesterday at 12:17 PM
sqing
an hour ago
Mathcounts state
happymoose666   12
N an hour ago by fxx11
Hi everyone,
I just have a question. I live in PA and I sadly didn't make it to nationals this year. Is PA a competitive state? I'm new into mathcounts and not sure
12 replies
happymoose666
Mar 24, 2025
fxx11
an hour ago
a_1 = 2025 implies a_k < 1/2025?
navi_09220114   6
N an hour ago by navi_09220114
Source: Own. Malaysian APMO CST 2025 P1
A sequence is defined as $a_1=2025$ and for all $n\ge 2$, $$a_n=\frac{a_{n-1}+1}{n}$$Determine the smallest $k$ such that $\displaystyle a_k<\frac{1}{2025}$.

Proposed by Ivan Chan Kai Chin
6 replies
navi_09220114
Feb 27, 2025
navi_09220114
an hour ago
Factorials, Sum of Powers, Diophantine Equations,
John_Mgr   3
N an hour ago by John_Mgr
Source: Nepal NMO 2025 p4
Find all the pairs of positive integers $n$ and $x$ such that: \[1^n+2^n+3^n+\cdots +n^n=x!\]
$Petko$ $Lazarov, Bulgaria$
3 replies
1 viewing
John_Mgr
Mar 15, 2025
John_Mgr
an hour ago
Inspired by IMO 1984
sqing   5
N an hour ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that
$$a^2+b^2+ ab +24abc\leq\frac{81}{64}$$Equality holds when $a=b=\frac{3}{8},c=\frac{1}{4}.$
$$a^2+b^2+ ab +18abc\leq\frac{343}{324}$$Equality holds when $a=b=\frac{7}{18},c=\frac{2}{9}.$
5 replies
sqing
Yesterday at 3:01 AM
sqing
an hour ago
USA Canada math camp
Bread10   35
N an hour ago by abbominable_sn0wman
How difficult is it to get into USA Canada math camp? What should be expected from an accepted applicant in terms of the qualifying quiz, essays and other awards or math context?
35 replies
Bread10
Mar 2, 2025
abbominable_sn0wman
an hour ago
LMT Spring 2025 and Girls' LMT 2025
vrondoS   13
N an hour ago by ethan2011
The Lexington High School Math Team is proud to announce LMT Spring 2025 and our inaugural Girls’ LMT 2025! LMT is a competition for middle school students interested in math. Students can participate individually, or on teams of 4-6 members. This announcement contains information for BOTH competitions.

LMT Spring 2025 will take place from 8:30 AM-5:00 PM on Saturday, May 3rd, at Lexington High School, 251 Waltham St., Lexington, MA 02421.

The competition will include two individual rounds, a Team Round, and a Guts Round, with a break for lunch and mini-events. A detailed schedule is available at https://lhsmath.org/LMT/Schedule.

There is a $15 fee per participant, paid on the day of the competition. Pizza will be provided for lunch, at no additional cost.

Register for LMT at https://lhsmath.org/LMT/Registration/Home.

Girls’ LMT 2025 will be held ONLINE on MathDash from 11:00 AM-4:15 PM EST on Saturday, April 19th, 2025. Participation is open to middle school students who identify as female or non-binary. The competition will include an individual round and a team round with a break for lunch and mini-events. It is free to participate.

Register for GLMT at https://www.lhsmath.org/LMT/Girls_LMT.

More information is available on our website: https://lhsmath.org/LMT/Home. Email lmt.lhsmath@gmail.com with any questions.
13 replies
+2 w
vrondoS
2 hours ago
ethan2011
an hour ago
Bashable CP on Q20? :skull:
pingpongmerrily   25
N Nov 13, 2024 by pingpongmerrily
Source: 2024 AMC 10B Problem 20
Three different pairs of shoes are placed in a row so that no left shoe is next to a right shoe from a different pair. In how many ways can these six shoes be lined up?

$
\textbf{(A) }60\qquad
\textbf{(B) }72\qquad
\textbf{(C) }90\qquad
\textbf{(D) }108\qquad
\textbf{(E) }120\qquad
$
25 replies
pingpongmerrily
Nov 13, 2024
pingpongmerrily
Nov 13, 2024
Bashable CP on Q20? :skull:
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Source: 2024 AMC 10B Problem 20
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pingpongmerrily
3522 posts
#1
Y by
Three different pairs of shoes are placed in a row so that no left shoe is next to a right shoe from a different pair. In how many ways can these six shoes be lined up?

$
\textbf{(A) }60\qquad
\textbf{(B) }72\qquad
\textbf{(C) }90\qquad
\textbf{(D) }108\qquad
\textbf{(E) }120\qquad
$
This post has been edited 3 times. Last edited by LauraZed, Nov 13, 2024, 5:31 PM
Reason: adjusting answer choices format
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Yihang2009
164 posts
#2
Y by
The answer was 60
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pingpongmerrily
3522 posts
#3
Y by
yup, i just bashed out $(12+6+6+6)\cdot 2$
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BS2012
945 posts
#4
Y by
I did 6*(4+4+2) using WLOG on the order of the left shoes
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alexanderhamilton124
378 posts
#5
Y by
A) 60 just do a case bash, very misplaced IMO
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gladIasked
623 posts
#6
Y by
nah this was one of the harder problems on here imo
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Squidget
432 posts
#7
Y by
I didn’t even look at this question during the test :sob:
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Tem8
238 posts
#8
Y by
Note that a right shoe has at most one neighboring left shoe and a right shoe has at most one neighboring left shoe. Thus, all combinations of shoes are:
RRLLLR, LRRLLR, RLLRRL, RLLLRR, RRRLLL, LRRRLL, LLRRRL, LLLRRR.
In total there are $6+6+6+6+3\cdot2\cdot2+6+6+3\cdot2\cdot2 = \boxed{\textbf{(A) } 60}$ combinations.
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megarnie
5541 posts
#9
Y by
Right: 1 2 3
Left: 4 5 6

Where $n$ pairs with $n + 3$.

Now WLOG that $1$ is the second one that shows up. Now WLOG that the first three are either $213, 214,$ or $413$, and then multiply by $2 \cdot 6$ at the end.

Case 1: First three are $213$
Then the last three can be $654$ or $645$ because $5,4$ cannot interact with $3$.

Case 2: First three are $214$
Then the last three can be $563$ only (oops I counted $365$ in test so I put $72$).

Case 3: First three are $413$.
Then the last three can be $256$ or $652$.

Hence the answer is $(2 + 1 + 2) \cdot 12 = \boxed{\textbf{(A)}\ 60}$.
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EthanSpoon
637 posts
#10
Y by
Ya it is A

I spent 5 minutes trying to figure out which cases I overcounted
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EGMO
145 posts
#11
Y by
bad problem highkey
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EpicBird08
1740 posts
#12
Y by
Here is a video solution to this problem.
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cxrupptedpat
226 posts
#13
Y by
what i got 120
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tyrantfire4
1086 posts
#14
Y by
:skull: !!!!
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pingpongmerrily
3522 posts
#15
Y by
cxrupptedpat wrote:
what i got 120

how
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williamxiao
2506 posts
#16
Y by
I’m cry I can’t count my cases

For organized casework we can split it into cases based on the edges. Label left shoes as abc and right as 123

Both edges are left: You have 3x2 ways to choose the edges, WLOG a _ _ _ _ b. Notice that c must be in the second or fifth slot or it must be touching two different right shoes. We have 2 ways to choose where c goes, and the remaining right shoes have only 1 ordering (ex. ac312b) for 12 cases.

One left one right: 2 ways to choose which edge is left, 3 ways to choose which left shoes it is. WLOG leftmost is a. We split this into sub cases based on whether the last one is 1 or not 1.
Last is 1: The one to the right of a must also be a left shoes, which we can do in 2 ways. WLOG ab…1 W see that the remaining three must be c32, so this has 2 cases.
Last is 2/3: we have two ways to choose what the rightmost shoe is, WLOG 2. We have a….2.We do casework on where b is: if b is second, we have abc312. If it is third, we fail. If it is fourth, we fail. If it is 5th, we have a13cb2. This has 2x2 cases

In total there are 6(2+4)=36 cases for this

Both edges are right: follow same logic as case 1, 12 cases

Sum 12+36+12 to get 60
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MathRook7817
629 posts
#17
Y by
yeah, its just bash
i skipped this one :(
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SpottedHawk
21 posts
#18
Y by
I was gonna bash C(6,3) cases but then realized i cant do this and wasted so much time on this problem :((
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EaZ_Shadow
1122 posts
#19
Y by
same I got 60 and I finished it with only 10 min left :) there was like 7 cases
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lucaswujc
293 posts
#20
Y by
bashable fr it was 60
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pingpongmerrily
3522 posts
#21
Y by
is there a way to solve this without bashing?
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wuwang2002
1192 posts
#22
Y by
harder than p25
how did i get 72 :skull:
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Schintalpati
599 posts
#23
Y by
pingpongmerrily wrote:
is there a way to solve this without bashing?

Yes this is basically the exact same as other solutions, but I’m explaining it 2x as much as needed when solving it. So first, realizing AaBbCc are the three pairs. One possible combo is AabBCc this has 3! ways. aABbcC has 3! ways. Then ABCcba has 3!*2!*2!. Then abcCAB has 3!*2!*2! ways so 60 is answer.
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Mr.Sharkman
490 posts
#24
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The answer is 60. It is just 4 cases.
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akliu
1736 posts
#25
Y by
I caseworked here with respect to how many right shoes were in between some two left shoes. It boils down to three main cases:
- 2 shoes, LRRLLR (and reversed) -> 12
- 2 shoes, RLRRLL (and reversed) - 12
- 0 shoes, LLLRRR (and reversed) -> 24
- 0 shoes, RLLLRR (and reversed) -> 12
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pingpongmerrily
3522 posts
#26
Y by
i did something similar to @bove, I tried doing multiplication principle and it failed so i bashed with casework
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