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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
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   23
N a few seconds ago by vincentwant
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 $m+n$.
23 replies
megahertz13
Oct 13, 2024
vincentwant
a few seconds ago
How to be wide awake during math competitions.
orangebear   13
N 5 minutes ago by studymoremath
As three years of taking math competitions, I have noticed that some of the competitions that I did really bad in " I mean really bad in" where the cases where I was sleepy, and not fully awake "examples like the AMC 10 B 2024, AMC 10 A 2023, AMC 8 2023, AMC 8 2024, Mathcounts chapter 2024, etc." and the ones where I was awake and, alert like "AMC 8 2022, National Math-con 2024 "not mathcounts", Mathkangaroo 2021, Mathcounts state, etc" I did medium - too surprising myself. Do any of you people have any tips on how to be alert during math competitions?
13 replies
orangebear
Today at 1:51 AM
studymoremath
5 minutes ago
9 How do AMC-10 scores translate to MATHCOUNTS States
pingpongmerrily   49
N 6 minutes ago by bwu_2022
How do AMC scores translate to MathCounts States, given that the problems are pretty similar?

Use your own individual experience/guesses to determine what you think..

NOTE: THIS IS FOR STATES!!!
49 replies
pingpongmerrily
Nov 14, 2024
bwu_2022
6 minutes ago
9 AIME Qualification AMC 12A
studymoremath   1
N 21 minutes ago by studymoremath
Now with the AMC's over, I think everyone has more of an idea of cutoff ranges and difficulty levels this year. Are my chances high enough that I should start studying for AIME?
1 reply
studymoremath
Yesterday at 6:43 PM
studymoremath
21 minutes ago
No more topics!
rotated equilateral
StressedPineapple   13
N Nov 14, 2024 by buddy2007
Source: 2024 AMC 12B #19
Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta < 60^{\circ}$, to form $\triangle DEF$. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan\theta$?

IMAGE

$\textbf{(A)}~\displaystyle\frac{3}{4}\qquad\textbf{(B)}~\displaystyle\frac{5\sqrt{3}}{11}\qquad\textbf{(C)}~\displaystyle\frac{4}{5}\qquad\textbf{(D)}~\displaystyle\frac{11}{13}\qquad\textbf{(E)}~\displaystyle\frac{7\sqrt{3}}{13}$
13 replies
StressedPineapple
Nov 13, 2024
buddy2007
Nov 14, 2024
rotated equilateral
G H J
Source: 2024 AMC 12B #19
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StressedPineapple
4 posts
#1
Y by
Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta < 60^{\circ}$, to form $\triangle DEF$. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan\theta$?

[asy]
defaultpen(fontsize(13)); size(200);
pair O=(0,0),A=dir(225),B=dir(-15),C=dir(105),D=rotate(38.21,O)*A,E=rotate(38.21,O)*B,F=rotate(38.21,O)*C;
draw(A--B--C--A,gray+0.4);draw(D--E--F--D,gray+0.4); draw(A--D--B--E--C--F--A,black+0.9); dot(O); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F));
[/asy]

$\textbf{(A)}~\displaystyle\frac{3}{4}\qquad\textbf{(B)}~\displaystyle\frac{5\sqrt{3}}{11}\qquad\textbf{(C)}~\displaystyle\frac{4}{5}\qquad\textbf{(D)}~\displaystyle\frac{11}{13}\qquad\textbf{(E)}~\displaystyle\frac{7\sqrt{3}}{13}$
This post has been edited 2 times. Last edited by StressedPineapple, Yesterday at 6:18 PM
Reason: diagram
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shiamk
100 posts
#2
Y by
I got B 5sqrt3/11
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vsamc
3769 posts
#3
Y by
You get like (sin theta + sin(120-theta)) * ((14/sqrt3)^2/2) * 3 = 91sqrt3
which becomes sin(theta+30) = 13/14

so tan(theta+30) = 13/3sqrt3, and then just do tangent subtraction
This post has been edited 1 time. Last edited by vsamc, Nov 13, 2024, 5:34 PM
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Jack_w
59 posts
#4
Y by
Drop an altitude from one of the three obtuse triangles on the side; it has length $2\sqrt3$. Then you can find the length of the segment from the vertex to the foot and consequently find $\tan\theta$.
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ninjaforce
71 posts
#5 • 3 Y
Y by AtharvNaphade, eibc, Leo.Euler
[asy]
size(250);
import geometry;

// Define the side length of the equilateral triangle
real sideLength = 14;
real radius = sideLength / (2 * sin(60 * pi / 180)); // Radius of the circumscribed circle

// Define the rotation angle theta where tan(theta) = 5sqrt(3)/11
real theta = atan(5 * sqrt(3) / 11);

// Define the center of the circle
pair O = (0,0);

// Draw the circle in black
draw(circle(O, radius), linewidth(0.8) + black);

// Define the vertices of the original triangle, with A in the top left
pair A = dir(150) * radius;   // Position A in the top-left
pair C = dir(270) * radius;   // C is at the bottom
pair B = dir(30) * radius;    // B is at the top-right

// Draw the original triangle in black
draw(A--B--C--cycle, linewidth(0.8) + black);

// Label the vertices of the original triangle in black
label("$A$", A, NW, black);
label("$B$", B, NE, black);
label("$C$", C, S, black);

// Rotate the triangle by angle -theta for a clockwise rotation
pair A1 = rotate(-degrees(theta)) * A;
pair B1 = rotate(-degrees(theta)) * B;
pair C1 = rotate(-degrees(theta)) * C;

// Draw the rotated triangle in black
draw(A1--B1--C1--cycle, linewidth(0.8) + black);

// Label the vertices of the rotated triangle in black
label("$A'$", A1, NW, black);
label("$B'$", B1, NE, black);
label("$C'$", C1, S, black);

// Connect the specified pairs of vertices
draw(A--C1, dashed + black);
draw(C--B1, dashed + black);
draw(B--A1, dashed + black);

// Connect each corresponding vertex between the original and rotated triangle
draw(A--A1, dashed + black);
draw(B--B1, dashed + black);
draw(C--C1, dashed + black);

// Add the center point O
dot(O, black);
label("$O$", O, NE, black);

// Project points A and A' down onto the horizontal line
pair A_proj = (A.x, 0);
pair A1_proj = (A1.x, 0);

// Draw only the segment OA_proj on the horizontal line
draw(O--A_proj, dashed + black);

// Project A and A' onto the horizontal line with dotted lines
draw(A--A_proj, dotted + black);
draw(A1--A1_proj, dotted + black);

// Mark the projections
dot(A_proj, black);
dot(A1_proj, black);

// Label the projections
label("$A_{proj}$", A_proj, S, black);
label("$A'_{proj}$", A1_proj, S, black);

// Add lines from O to A and O to A'
draw(O--A, dotted + black);
draw(O--A1, dotted + black);

// Find the intersection of line A'A'_{proj} and AB
pair X = intersectionpoint(A1--A1_proj, A--B);

// Draw and label the intersection point X
dot(X, black);
label("$X$", X, S, black);
[/asy]

Note that \( \theta = \angle A'OA'_{\text{proj}} - \angle AOA_{\text{proj}} \). Calculation of the tangents of those angles and then an application of the tangent subtraction formula yields \(\tan \theta = \frac{5\sqrt{3}}{11}. \)
This post has been edited 4 times. Last edited by ninjaforce, Nov 13, 2024, 6:27 PM
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mathprodigy2011
58 posts
#6
Y by
its just breaking the hexagon into 6 triangles and using sin area formula. It took me 5 mins ish tho :(
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Toinfinity
588 posts
#7 • 1 Y
Y by Mathandski
Let $AD=x$, $DB=y$, then the area of $ADB$ is $\frac{91\sqrt{3}-49\sqrt{3}}{3}=14\sqrt{3}$. By Sine area formula, $xy=56$, and by LoC (angle $ADB$ = 120 by cyclic quad properties), $x^2+xy+y^2=196$. Then, solving these equations gives $x=2\sqrt{7}, y=4\sqrt{7}$, so $\sin{\frac{\theta}{2}}=\frac{x}{OC}=\frac{2\sqrt{7}}{\frac{14}{\sqrt{3}}}$. Then, by trig bash, you get $\tan{\theta}=\frac{5\sqrt{3}}{11}$.
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djmathman
7928 posts
#8 • 4 Y
Y by ostriches88, happymathEZ, OronSH, clarkculus
Mine.

Let $O$ be the center of triangles $ABC$ and $DEF$, $P$ be the foot of the altitude from $D$ to $\overline{AB}$, and $M$ be the midpoint of $\overline{AB}$. Remark that the condition $\theta\leq 60^\circ$ implies that $P$ lies on segment $\overline{AM}$ (as opposed to segment $\overline{BM}$). Furthermore, observe that $O$ is the center of circle $\odot(ADB)$, so by the Inscribed Angle Theorem $\angle DOA = 2\angle DBA$. It suffices to compute $\tan\angle DBA$ and then use the Tangent Double Angle Formula to obtain $\tan\theta$.
[asy]
import olympiad;
size(200);
defaultpen(linewidth(0.6)+fontsize(11));
real t = 48;
pair A = dir(90), B = dir(210), C = dir(330), D = dir(90 + t), E = dir(210 + t), F = dir(330 + t), O = origin, P = foot(D,A,B), M = (A+B)/2;
draw(D--P--A^^rightanglemark(D,P,A,2));
draw(A--B--C--cycle^^D--E--F--cycle^^A--O--D--B);
draw(unitcircle,linetype("3 3"));
label("$A$",A,dir(O--A));
label("$B$",B,dir(O--B));
label("$C$",C,dir(O--C));
label("$D$",D,dir(O--D));
label("$E$",E,dir(O--E));
label("$F$",F,dir(O--F));
label("$O$",O,S);
label("$P$",P,SE);
[/asy]
The area of $\triangle ABC$ is $\tfrac{14^2\sqrt 3}4 = 49\sqrt 3$. Because hexagon $ADBECF$ consists of $\triangle ABC$ plus three copies of $\triangle ADB$, the area of $\triangle ADB$ is $\tfrac{91\sqrt 3 - 49\sqrt 3}3 = 14\sqrt 3$. This means $DP = 2 \sqrt 3$. Additionally, because $AB = 14$, $OM = \tfrac{7\sqrt 3}3$ and $OD = \tfrac{14\sqrt 3}3$. It follows from the Pythagorean Theorem that
\begin{align*}
MP &= \sqrt{OD^2 - (OM + DP)^2}\\
   &= \sqrt{\left(\frac{14 \sqrt 3}3\right)^2 - \left(\frac{13\sqrt 3}3\right)^2}\\
   &= \sqrt{\frac{14^2 - 13^2}3} = \sqrt{\frac{27}3} = 3.
\end{align*}Finally, $BP = 7 + 3 = 10$, so $\tan\angle DBA = \tfrac{2\sqrt 3}{10} = \tfrac{\sqrt 3}5$ and
\[
\tan \theta = \tan(2\angle DOB) = \frac{\frac{2\sqrt 3}5}{1 - \frac{3}{25}} = \frac{5\sqrt 3}{11}.
\]
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YaoAOPS
1317 posts
#9
Y by
djmathman wrote:
Mine.

Let $O$ be the center of triangles $ABC$ and $DEF$, $P$ be the foot of the altitude from $D$ to $\overline{AB}$, and $M$ be the midpoint of $\overline{AB}$. Remark that the condition $\theta\leq 60^\circ$ implies that $P$ lies on segment $\overline{AM}$ (as opposed to segment $\overline{BM}$). Furthermore, observe that $O$ is the center of circle $\odot(ADB)$, so by the Inscribed Angle Theorem $\angle DOA = 2\angle DBA$. It suffices to compute $\tan\angle DBA$ and then use the Tangent Double Angle Formula to obtain $\tan\theta$.
[asy]
import olympiad;
size(200);
defaultpen(linewidth(0.6)+fontsize(11));
real t = 48;
pair A = dir(90), B = dir(210), C = dir(330), D = dir(90 + t), E = dir(210 + t), F = dir(330 + t), O = origin, P = foot(D,A,B), M = (A+B)/2;
draw(D--P--A^^rightanglemark(D,P,A,2));
draw(A--B--C--cycle^^D--E--F--cycle^^A--O--D--B);
draw(unitcircle,linetype("3 3"));
label("$A$",A,dir(O--A));
label("$B$",B,dir(O--B));
label("$C$",C,dir(O--C));
label("$D$",D,dir(O--D));
label("$E$",E,dir(O--E));
label("$F$",F,dir(O--F));
label("$O$",O,S);
label("$P$",P,SE);
[/asy]
The area of $\triangle ABC$ is $\tfrac{14^2\sqrt 3}4 = 49\sqrt 3$. Because hexagon $ADBECF$ consists of $\triangle ABC$ plus three copies of $\triangle ADB$, the area of $\triangle ADB$ is $\tfrac{91\sqrt 3 - 49\sqrt 3}3 = 14\sqrt 3$. This means $DP = 2 \sqrt 3$. Additionally, because $AB = 14$, $OM = \tfrac{7\sqrt 3}3$ and $OD = \tfrac{14\sqrt 3}3$. It follows from the Pythagorean Theorem that
\begin{align*}
MP &= \sqrt{OD^2 - (OM + DP)^2}\\
   &= \sqrt{\left(\frac{14 \sqrt 3}3\right)^2 - \left(\frac{13\sqrt 3}3\right)^2}\\
   &= \sqrt{\frac{14^2 - 13^2}3} = \sqrt{\frac{27}3} = 3.
\end{align*}Finally, $BP = 7 + 3 = 10$, so $\tan\angle DBA = \tfrac{2\sqrt 3}{10} = \tfrac{\sqrt 3}5$ and
\[
\tan \theta = \tan(2\angle DOB) = \frac{\frac{2\sqrt 3}5}{1 - \frac{3}{25}} = \frac{5\sqrt 3}{11}.
\]

Strange question but how did you decide what $\theta$ / area to use for this problem, given that the solution is pretty agnostic to the choice of $\theta$?
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djmathman
7928 posts
#10
Y by
YaoAOPS wrote:
Strange question but how did you decide what $\theta$ / area to use for this problem, given that the solution is pretty agnostic to the choice of $\theta$?
If I remember correctly, I chose the numbers to satisfy a few constraints.
  • In my solution (which is, admittedly, a pretty unorthodox one), I compute $\tan \angle DPB$ by writing $BP = BM + MP$. I wanted $MP$ to be an integer, because otherwise $\tan\angle DPB$ would be some annoying mixed radical and the computation would spiral during the double angle step at the end.
  • The numbers needed to be fairly small. They also, ideally, should only involve integers and $\sqrt 3$ terms, because those naturally appear when working with equilateral triangles (no funny business here!).
  • That said, the numbers couldn't be too small, because I wanted to avoid silly degenerate cases like $MP = 0$. (That is, I wanted the configuration to be "general" enough so that people actually needed to work to find the answer. Otherwise, it would feel as if the problem was set up for no good reason.)
This post has been edited 3 times. Last edited by djmathman, Nov 13, 2024, 7:15 PM
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plang2008
271 posts
#11
Y by
In contest solution (using dj diagram):

First, it is easy to show that $AO^2 = \frac{196}{3}$. Now consider $[AOD] + [DOB]$. Using $\frac12 ab \sin C$, we have $\frac{98}{3} (\sin \theta + \sin(120^\circ - \theta)) = \frac{91\sqrt3}{3}$.

Let $\alpha = 60^\circ - \theta$. Use sum-to-product: this gives $\sin(60^\circ - \alpha) + \sin(60^\circ + \alpha) = 2\sin 60^\circ \cos \alpha = \sqrt3\cos \alpha$.

Now we have $\frac{98\sqrt3}{3}\cos \alpha = \frac{91\sqrt3}{3}$. Solving gives $\cos \alpha = \frac{13}{14}$. Then $\sin \alpha = \frac{\sqrt{27}}{14}$ so $\tan \alpha = \frac{\sqrt{27}}{13}$.

Now we have $\tan \theta = \tan(60^\circ - \alpha) = \frac{\sqrt3 - \tan \alpha}{1 + \sqrt3 \tan \alpha} = \boxed{\frac{5\sqrt3}{11}}$.
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apotosaurus
75 posts
#12
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Set $\theta=2\gamma$. Then $AO = \frac{14}{\sqrt 3}$ and $AD = \frac{28}{\sqrt 3}\sin(\gamma)$. Furthermore, $\angle DAB = 60-\gamma$ so the altitude from $D$ in $\triangle ADB$ is $\frac{28}{\sqrt 3}\sin(\gamma)\sin(60-\gamma)$. Then $[ADB]=\frac{196}{\sqrt 3}\sin(\gamma)\sin(60-\gamma)$. Since $3[ADB]+[ABC]=91\sqrt 3$, $[ADB]=\frac{42}{\sqrt 3}$, so $\sin(\gamma)\sin(60-\gamma)=\frac 3{14}$. Then $\cos(60-\theta)-\frac 12 = \frac 37$, so \[\cos(60-\theta) = \frac{13}{14}.\]This also gives $\sin(60-\theta)=\frac{3\sqrt 3}{14}$, so $\tan(60-\theta) = \frac{3\sqrt 3}{13}$. Finally, \[\tan(\theta) = \tan(60-(60-\theta)) = \frac{\sqrt 3 - \frac{3\sqrt 3}{13}}{1 + \frac 9{13}}=\frac{10\sqrt 3}{22} = \boxed{\frac{5\sqrt 3}{11}}.\]
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shihan
243 posts
#13
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[asy]
defaultpen(fontsize(13)); size(200);
pair O=(0,0),A=dir(225),B=dir(-15),C=dir(105),D=rotate(38.21,O)*A,E=rotate(38.21,O)*B,F=rotate(38.21,O)*C;
draw(A--B--C--A,gray+0.4);draw(D--E--F--D,gray+0.4); draw(A--D--B--E--C--F--A,black+0.9); dot(O); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F));
[/asy]
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buddy2007
2052 posts
#14
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just draw lines from the center to each of the vertices, use 0.5absin and you get that sin(x) + sin(120 - x) is 13sqrt3/14 and you check answer choices and you realize that B works
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