Happy Thanksgiving! Please note that there are no classes November 25th-December 1st.

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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section on your LinkedIn profile.

Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

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
Nov 1, 2024
0 replies
A Riddle from Outer Space
vishwathganesan   5
N Yesterday at 4:22 AM by benryu
Here's an original riddle from three years ago, tweaked a little to be simpler:

On the distant planet of Vukovar, math is done in much the same way as it is here. They even use the same symbols as us, but all of their arithmetic seems to be wrong. Here's a list of equations which are all correct on Vukovar:

$2 + 3 = 6$
$402 + 158 = 671$
$2492 - 651 = 1630$
$7 \cdot 4 = 39$
$35 \cdot 62 = 2247$
$\frac{8029}{139} = 27$
$17 ^ 2 = 10841$

Given this information, what is the value of $(\frac{141}{25}) ^ 3 - (23 \cdot 48 + 74)$ on Vukovar?

Note: This riddle does not require advanced math. A basic knowledge of arithmetic should suffice. Calculators are allowed and could be helpful, but are not necessary.
5 replies
vishwathganesan
Mar 9, 2021
benryu
Yesterday at 4:22 AM
Brain Breaker
nellis25   12
N Yesterday at 4:06 AM by benryu
This multiple choice question has at least one correct answer. If you guessed at random, what is the probability your answer would be incorrect?

\begin{align*}
{\color{blue}{\textrm{A: }}} &\quad 0 &{\color{red}{\textrm{E: }}} &\quad 1/4    \\
{\color{red}{\textrm{B: }}} &\quad 1/4 &{\color{blue}{\textrm{F: }}} &\quad  1/8  \\
{\color{blue}{\textrm{C: }}} &\quad  1/2 &{\color{red}{\textrm{G: }}} &\quad 3/4  \\
{\color{red}{\textrm{D: }}} &\quad 3/4   &{\color{blue}{\textrm{H: }}} &\quad 1 
\end{align*}
[center]IMAGE[/center]
12 replies
nellis25
Jul 26, 2021
benryu
Yesterday at 4:06 AM
Missing Digits
nellis25   4
N Yesterday at 3:54 AM by benryu
The six-digit number 4A8A2A is divisible by the two-digit number 8A for some digit A in base 10. What is A?

[center]IMAGE[/center]
4 replies
nellis25
Aug 16, 2021
benryu
Yesterday at 3:54 AM
What's the Point
nellis25   12
N Yesterday at 3:51 AM by benryu
Grogg is trying to draw an equilateral triangle $\triangle ABC$ with a special point $P$ inside the triangle, so that $AP = 1, BP=2,$ and $CP=3.$ Can he do it? If so, what's the side-length of his triangle? If not, why not?
12 replies
nellis25
Sep 7, 2021
benryu
Yesterday at 3:51 AM
Matching Cards
nellis25   6
N Yesterday at 3:37 AM by benryu
Two shuffled standard 52-card decks are placed side-by-side.

What is the reciprocal of the probability that at least one card is in the same position in both decks, to nine decimal places?

[center]IMAGE[/center]
6 replies
nellis25
Jan 17, 2022
benryu
Yesterday at 3:37 AM
Tryptophan
nellis25   13
N Saturday at 10:49 PM by RAYRAY7512
After eating too much turkey, Alex, Winnie, and Lizzie got very sleepy. They all fell asleep at exactly 8:00pm and have the following sleep patterns:
[list]
[*] Lizzie sleeps for 5 hours at a time, wakes up, then immediately falls back asleep.
[*] Winnie sleeps for 4 hours at a time, wakes up, then immediately falls back asleep.
[*] Alex sleeps for 3 hours at a time, wakes up, then immediately falls back asleep.
[/list]
The little monsters will all get up if they all happen to wake up at the same time. What time will it be when the three little monsters get up?

[center]IMAGE[/center]
13 replies
nellis25
Nov 23, 2020
RAYRAY7512
Saturday at 10:49 PM
Pie Logic
nellis25   21
N Nov 22, 2024 by kabir_sahil
Six logicians have just finished their Thanksgiving repast, and are ready for dessert. Their host comes around and asks "Would all of you like some pumpkin pie?"

First logician: "I don't know."
Second logician: "I don't know."
Third logician: "I don't know."
Fourth logician: "I don't know."
Fifth logician: "I don't know."
Sixth logician: "No."

Who wants pumpkin pie?

[center]IMAGE[/center]
21 replies
nellis25
Nov 23, 2020
kabir_sahil
Nov 22, 2024
Like Twenty-four
nellis25   21
N Nov 20, 2024 by Yrock
Using exactly one multiplication, one addition, one division, and one subtraction, what is the smallest positive number you can create using the numbers 4, 5, 6, 7, and 8 once each? For instance, we can form the number ((8÷5) x 7) - (4+6)= 1.2 in this way.

[center]IMAGE[/center]
21 replies
nellis25
Dec 20, 2021
Yrock
Nov 20, 2024
Impostor(s)
nellis25   59
N Nov 14, 2024 by unicornz
In a popular video game, 10 players compete, and 2 are randomly selected to be impostors.

As the game begins, the green player, who is not an impostor, says "red orange sus" with no knowledge of who is or isn't an impostor. What is the probability that

[list]
[*] both red and orange are impostors?
[*] at least one of red and orange is an impostor?
[/list]

[center]IMAGE[/center]
59 replies
nellis25
Dec 8, 2020
unicornz
Nov 14, 2024
Boxes and Squares
nellis25   9
N Nov 14, 2024 by unicornz
Can you rearrange the boxes so that the sum of the numbers on any two neighboring boxes is a perfect square?

[center]IMAGE[/center]
9 replies
nellis25
Dec 14, 2020
unicornz
Nov 14, 2024
Impostor(s)
nellis25   59
N Nov 14, 2024 by unicornz
In a popular video game, 10 players compete, and 2 are randomly selected to be impostors.

As the game begins, the green player, who is not an impostor, says "red orange sus" with no knowledge of who is or isn't an impostor. What is the probability that

[list]
[*] both red and orange are impostors?
[*] at least one of red and orange is an impostor?
[/list]

[center]IMAGE[/center]
59 replies
nellis25
Dec 8, 2020
unicornz
Nov 14, 2024
Impostor(s)
G H J
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nellis25
279 posts
#1 • 13 Y
Y by HudsonR, wm8611, UnknownMonkey, Mustafa372, aopspandy, studyinghardformathcounts, aidan0626, AlienGirl05, hh99754539, ilovepizza2020, Mango247, Mango247, Mango247
In a popular video game, 10 players compete, and 2 are randomly selected to be impostors.

As the game begins, the green player, who is not an impostor, says "red orange sus" with no knowledge of who is or isn't an impostor. What is the probability that
  • both red and orange are impostors?
  • at least one of red and orange is an impostor?
https://data.artofproblemsolving.com/images/keeplearning/Week%2036_Website_2_Imposters.png
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CreativeHedgehog
616 posts
#2 • 2 Y
Y by dolphinia, AlienGirl05
Solution
This post has been edited 1 time. Last edited by CreativeHedgehog, Dec 8, 2020, 11:35 PM
Reason: incorrect answer
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Anchit1234
1 post
#4
Y by
The probability of both of them being imposters is 1/36.

The probability of at least one of them being imposters is 2/9
This post has been edited 1 time. Last edited by Anchit1234, Dec 11, 2020, 3:16 AM
Reason: I misunderstood the question.
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solasky
1563 posts
#7 • 1 Y
Y by naomurakami
Click to reveal hidden text
Anchit1234 wrote:
The probability of both of them being imposters is 1/36.

The probability of at least one of them being imposters is 2/9

How did you get 2/9?
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wamofan
6825 posts
#8 • 8 Y
Y by sanyalr, zunaid_cnd, ChromeRaptor777, OronSH, studyinghardformathcounts, aidan0626, ilovepizza2020, xHypotenuse
The probability that red is impostor is 1 cuz red is always impostor :P :P :P
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sanyalr
15 posts
#9
Y by
Anchit1234 wrote:
The probability of both of them being imposters is 1/36.

The probability of at least one of them being imposters is 2/9

Nice to know when I'm playing Among Us as an innocent crewmate, and when there are only thee players left.
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Ansh2020
1412 posts
#10
Y by
solasky wrote:
Click to reveal hidden text
Anchit1234 wrote:
The probability of both of them being imposters is 1/36.

The probability of at least one of them being imposters is 2/9

How did you get 2/9?

The Formula for Probability = ('Number of outcomes we want')/('Number of possible outcomes')
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ethanhansummerfun
163 posts
#11 • 4 Y
Y by wm8611, zunaid_cnd, HamstPan38825, Skyman2011
god now aops knows among us
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wm8611
76 posts
#12
Y by
since green is not an imposter, we know that there is a 1/9 chance that red is the imposter. We also know there is a 1/9 chance that orange is the imposter. This is because there were 10 but since green is definitely not the imposter, we can take him out, resulting in the other 9 players. The possibility of getting both is 1/9 * 1/9 which gives 1/81. The probability of getting at least 2 of them has to be 1/9 because of the red and another 1/9 because of the orange. We get 2/9.
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hstrsoil
595 posts
#13
Y by
ethanhansummerfun wrote:
god now aops knows among us

imagine not. i got 2/9 btw.
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V0305
587 posts
#14 • 1 Y
Y by dolphinia
First, let's count the number of ways we can select two impostors from the 9 players. It equals nine choose two (I'm not sure how to get the notation, but don't make too much fuss by telling me how to; I'm okay with writing words instead of numbers). Nine choose two is (9*8)/(2*1). The 8 and 2 cancel, leaving 9*4 in the numerator and 1*1 in the denominator. 9*4 is 36, and 1*1 is just one. 36/1 is 36, so there are 36 ways to choose two impostors out of the nine players that could be impostors (the problem says the green player is not an impostor).
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goodskate
820 posts
#15
Y by
I like how nobody ever mentioned that this game is Among Us :P

EDIT: Oh wait @4above (post #11) mentioned it and @2above (post #13) quoted it

ANOTHER EDIT: Oh and post #9 also mentioned it
This post has been edited 5 times. Last edited by goodskate, Jan 8, 2021, 2:30 AM
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lprado
48 posts
#16 • 3 Y
Y by Mango247, Mango247, Mango247
I love among us
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ethanhansummerfun
163 posts
#17
Y by
*AMONG US REFERENCE*
V0305 wrote:
First, let's count the number of ways we can select two impostors from the 9 players. It equals nine choose two (I'm not sure how to get the notation, but don't make too much fuss by telling me how to; I'm okay with writing words instead of numbers). Nine choose two is (9*8)/(2*1). The 8 and 2 cancel, leaving 9*4 in the numerator and 1*1 in the denominator. 9*4 is 36, and 1*1 is just one. 36/1 is 36, so there are 36 ways to choose two impostors out of the nine players that could be impostors (the problem says the green player is not an impostor).
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I agree
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A_MatheMagician
2251 posts
#18
Y by
who doesn't know that this is among us just look at the image
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