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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
Past PUMaC results
mathkiddus   1
N 22 minutes ago by lpieleanu
Does anybody know where we can find old PUMaC results/placing cutoffs
1 reply
mathkiddus
2 hours ago
lpieleanu
22 minutes ago
study for AIME to qual for USJMO
Logicus14   0
33 minutes ago
hi

so this is my first year doing amc10 (9th grade) and I got a decent enough score (121.5) on amc10b and after some research I found out that I need to get AT LEAST 10 problems right to qualify for USJMO

did anyone get 10+ problems right? If you did how did you study? any general tips?


ty
0 replies
Logicus14
33 minutes ago
0 replies
Problem 2
evt917   47
N an hour ago by MC_ADe
Source: 2024 AMC 12B #2 / AMC 10B #2
What is $10! - 7! \cdot 6!$?

$
\textbf{(A) }-120 \qquad
\textbf{(B) }0 \qquad
\textbf{(C) }120 \qquad
\textbf{(D) }600 \qquad
\textbf{(E) }720 \qquad
$
47 replies
evt917
Nov 13, 2024
MC_ADe
an hour ago
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   21
N an hour ago by vsarg
Orz to the Legends (Music, AMC version - MegaMath Channel original)

Get pumped up for the AMCs!

[youtube]https://www.youtube.com/watch?v=KmiNI00uo-s&ab_channel=MegaMathChannel[/youtube]

Join our discord! https://discord.gg/hh7vntTb2E

Math Problem (AMC 10/12): Let $M$ be the smallest positive integer satisfying the property that $M^6$ is a multiple of both $2024^2$ and $2025^3$. How many positive divisors does $M$ have?

James has a geometric series with first term $1$ and common ratio $0<r<1$. Andy multiplies the first term by $\frac{7}{5}$, and squares the common ratio. If the sum of the first series equals the sum of the second series, the common sum can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
21 replies
megahertz13
Oct 13, 2024
vsarg
an hour ago
No more topics!
sqrtsqrt
zhoujef000   21
N Nov 14, 2024 by Andrew2019
Source: 2024 AMC 10B #13
Positive integers $x$ and $y$ satisfy the equation $\sqrt{x}+\sqrt{y}=\sqrt{1183}.$ What is the minimum possible value of $x+y?$

$\textbf{(A) }585 \qquad\textbf{(B) }595\qquad\textbf{(C) }623\qquad\textbf{(D) }700\qquad\textbf{(E) }791$
21 replies
zhoujef000
Nov 13, 2024
Andrew2019
Nov 14, 2024
sqrtsqrt
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 AMC 10B #13
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zhoujef000
215 posts
#1
Y by
Positive integers $x$ and $y$ satisfy the equation $\sqrt{x}+\sqrt{y}=\sqrt{1183}.$ What is the minimum possible value of $x+y?$

$\textbf{(A) }585 \qquad\textbf{(B) }595\qquad\textbf{(C) }623\qquad\textbf{(D) }700\qquad\textbf{(E) }791$
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alexanderhamilton124
242 posts
#2
Y by
Answer is B) i think

use AM-Gm to eliminate A)
This post has been edited 1 time. Last edited by alexanderhamilton124, Nov 13, 2024, 5:32 PM
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nsking_1209
82 posts
#3
Y by
B confirmed
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Alex-131
5144 posts
#4
Y by
B, outline sqrt(1183) = 13sqrt(7). wlog sqrt(x)= 7sqrt7 and sqrt(y) = 6sqrt7, solve
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pingpongmerrily
2424 posts
#5
Y by
i didn't even solve it just guessed with am-gm
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MathRook7817
234 posts
#6
Y by
yeah, just 6sqrt7 and 7sqrt7
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bachkieu
84 posts
#7
Y by
pretty sure i multiplied 85*7 wrong
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Tem8
154 posts
#8
Y by
$$x+y=(\sqrt{x}+\sqrt{y})^2 - 2\sqrt{xy}=1183-2\sqrt{xy}.$$By $AM \ge GM$, \[\frac{\sqrt{x}+\sqrt{y}}{2} \ge \sqrt{\sqrt{xy}}.\]Squaring both sides, we find that
\[\frac{1183}{4} \ge \sqrt{xy},\]so \[-2\sqrt{xy} \ge -\frac{1183}{2}\].
Thus, \[x+y\ge1183-\frac{1183}{2}=591.5,\]so option A is eliminated automatically.
Now, let's check whether $x+y=595$ is possible for integer $x,y$.

\begin{align*}
\sqrt{x}+\sqrt{y} &= \sqrt{1183}\\
x+y &= 595 \\
2\sqrt{xy} &= (\sqrt{x}+\sqrt{y})^2-(x+y)=588\\
xy &= 294^2 \\
(x-y)^2 &= (x+y)^2 - 4xy \\
&= 595^2 - 2^2 \cdot 294^2 \\
&= 595^2 - 588^2 \\
&= (595-588)(595+588) \\
&=7\cdot1183=91^2 \\
x-y&=91\\
x&=\frac{595+91}{2},y=\frac{595-91}{2}.
\end{align*}
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megarnie
5307 posts
#9
Y by
The answer is $\boxed{\textbf{(B)}\ 595}$, achieved by $x = 343$ and $y = 252$. Now we prove the bound.

Let $c = x + y$. Squaring the equation gives $2\sqrt{xy} = 1183 - c$, so $xy = \frac{(1183 - c)^2}{4}$. Now considering the quadratic with roots $x,y$, its discriminant, which is \[ (x+y)^2 - 4xy = c^2 - (1183 - c)^2 =1183 \cdot  (2c - 1183) = 7 \cdot (2c - 1183) \cdot 13^2 \]must be a perfect square.

This implies that $7 \mid 2c - 1183$, so $2c - 1183 \ge 7$ (as it must be odd), and therefore $c \ge 595$, as desired.

(from here, a way to find working $x,y$ is by doing $\frac{c \pm \sqrt{7 \cdot (2c-1183) \cdot 13^2}}{2} = \frac{c \pm 91}{2}$, giving $\{x,y\} = \{252,343\}$)
This post has been edited 1 time. Last edited by megarnie, Nov 13, 2024, 6:06 PM
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elizhang101412
1060 posts
#10
Y by
Tem8 wrote:
$$x+y=(\sqrt{x}+\sqrt{y})^2 - 2\sqrt{xy}=1183-2\sqrt{xy}.$$By $AM \ge GM$, \[\frac{\sqrt{x}+\sqrt{y}}{2} \ge \sqrt{\sqrt{xy}}.\]Squaring both sides, we find that
\[\frac{1183}{4} \ge \sqrt{xy},\]so \[-2\sqrt{xy} \ge -\frac{1183}{2}\].
Thus, \[x+y\ge1183-\frac{1183}{2}=591.5,\]so option A is eliminated automatically.
Now, let's check whether $x+y=595$ is possible for integer $x,y$.

\begin{align*}
\sqrt{x}+\sqrt{y} &= \sqrt{1183}\\
x+y &= 595 \\
2\sqrt{xy} &= (\sqrt{x}+\sqrt{y})^2-(x+y)=588\\
xy &= 294^2 \\
(x-y)^2 &= (x+y)^2 - 4xy \\
&= 595^2 - 2^2 \cdot 294^2 \\
&= 595^2 - 588^2 \\
&= (595-588)(595+588) \\
&=7\cdot1183=91^2 \\
x-y&=91\\
x&=\frac{595+91}{2},y=\frac{595-91}{2}.
\end{align*}

this was my sol !!
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lucaswujc
288 posts
#11
Y by
zhoujef000 wrote:
Positive integers $x$ and $y$ satisfy the equation $\sqrt{x}+\sqrt{y}=\sqrt{1183}.$ What is the minimum possible value of $x+y?$

$\textbf{(A) }585 \qquad\textbf{(B) }595\qquad\textbf{(C) }623\qquad\textbf{(D) }700\qquad\textbf{(E) }791$

Factoring out $1183$ yield s$7\cdot13\cdot13$; $\sqrt{x}$ $\sqrt{y}$ must be in the form $a\sqrt{b} + c\sqrt{b}$, so we choose 7 to be B, then we know $(a+c)^2$ has to equal 169, and $a+c$ has to equal 13. Since x and y are both positive, integers, to minimize it we let $a = 6$ and $c = 7$ (Closest integers summing to 13), which will equal $7\sqrt{7}+6\sqrt{7} = 13\sqrt{7} = \sqrt{1183}$ so $x = 36 \cdot 7$ and $y = 49 \cdot 7$, summing them results in $\textbf{(B)}  595$
This post has been edited 6 times. Last edited by lucaswujc, Nov 13, 2024, 6:16 PM
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andrewcheng
482 posts
#12
Y by
B confirmed 1183=13^2*7
smallest x and y will clearly be closest to each other by concavity of the sqrt function
13+-1/2=6 and 7
(6^2+7^2)*7=595
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eg4334
391 posts
#13
Y by
$\sqrt{1183} = 13\sqrt{7}$, and from here it is clear that $6\sqrt{7}$ and $7\sqrt{7}$ is optimal.
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pieMax2713
4131 posts
#14
Y by
zhoujef000 wrote:
Positive integers $x$ and $y$ satisfy the equation $\sqrt{x}+\sqrt{y}=\sqrt{1183}.$ What is the minimum possible value of $x+y?$

$\textbf{(A) }585 \qquad\textbf{(B) }595\qquad\textbf{(C) }623\qquad\textbf{(D) }700\qquad\textbf{(E) }791$

i took the 12b and didnt get this but here's a silly calculus confirmation
This post has been edited 1 time. Last edited by pieMax2713, Nov 13, 2024, 7:26 PM
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lpieleanu
2248 posts
#15
Y by
Solution
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orangebear
357 posts
#16 • 1 Y
Y by Alex-131
When you are so depressed from problem 12 that you forgot, how to do these problems
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Mr.Sharkman
386 posts
#17
Y by
First, notice that $1183$ is divisible by $7.$ (this can be found from checking small primes to divide into 1183). So, $1183 = 7 \cdot 169 = 7 \cdot 13^{2}.$ So, if $x = 7a^{2},$ and $y = 7b^{2},$ then $a+b = 13.$ By AM-GM, the minimum is when $a=6, b=7,$ so we get $7 \cdot (6^{2}+7^{2}) = 595.$
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amaops1123
1644 posts
#18
Y by
NOOOOOOOOOOOO I GOT A
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carcool0
21 posts
#19 • 4 Y
Y by goldenuni678, Kempu33334, ranu540, Tem8
Why is everyone using am-gm. I factored it and got 13sqrt(7) and i just assumed that it would be 6sqrt(7) and 7sqrt(7).
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Raine999
15 posts
#20
Y by
zhoujef000 wrote:
Positive integers $x$ and $y$ satisfy the equation $\sqrt{x}+\sqrt{y}=\sqrt{1183}.$ What is the minimum possible value of $x+y?$

$\textbf{(A) }585 \qquad\textbf{(B) }595\qquad\textbf{(C) }623\qquad\textbf{(D) }700\qquad\textbf{(E) }791$

$1183 = 7\cdot13^2$

$\sqrt{1183} = 13\sqrt{7}$

Intuitively, $\sqrt{x} = 6\sqrt{7}, \sqrt{y} = 7\sqrt{7}$

Therefore, $x+y = 252 + 343 = \boxed{595}$
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MathPerson12321
3168 posts
#21
Y by
Alex-131 wrote:
B, outline sqrt(1183) = 13sqrt(7). wlog sqrt(x)= 7sqrt7 and sqrt(y) = 6sqrt7, solve

Yeah thats the best solution without AMGM cheese
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Andrew2019
2218 posts
#22
Y by
carcool0 wrote:
Why is everyone using am-gm. I factored it and got 13sqrt(7) and i just assumed that it would be 6sqrt(7) and 7sqrt(7).

use am-gm to show it cant be lower
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