https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Welp...&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T20:14:50ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_15&diff=1475462014 AIME II Problems/Problem 152021-02-20T02:59:57Z<p>Welp...: /* Solution 3 (induction) */</p>
<hr />
<div>==Problem==<br />
For any integer <math>k\geq 1</math>, let <math>p(k)</math> be the smallest prime which does not divide <math>k.</math> Define the integer function <math>X(k)</math> to be the product of all primes less than <math>p(k)</math> if <math>p(k)>2</math>, and <math>X(k)=1</math> if <math>p(k)=2.</math> Let <math>\{x_n\}</math> be the sequence defined by <math>x_0=1</math>, and <math>x_{n+1}X(x_n)=x_np(x_n)</math> for <math>n\geq 0.</math> Find the smallest positive integer <math>t</math> such that <math>x_t=2090.</math><br />
<br />
==Solution==<br />
Note that for any <math>x_i</math>, for any prime <math>p</math>, <math>p^2 \nmid x_i</math>. This provides motivation to translate <math>x_i</math> into a binary sequence <math>y_i</math>. <br />
<br />
Let the prime factorization of <math>x_i</math> be written as <math>p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots</math>, where <math>p_i</math> is the <math>i</math>th prime number. Then, for every <math>p_{a_k}</math> in the prime factorization of <math>x_i</math>, place a <math>1</math> in the <math>a_k</math>th digit of <math>y_i</math>. This will result in the conversion <math>x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots</math>.<br />
<br />
Multiplication for the sequence <math>x_i</math> will translate to addition for the sequence <math>y_i</math>. Thus, we see that <math>x_{n+1}X(x_n) = x_np(x_n)</math> translates into <math>y_{n+1} = y_n+1</math>. Since <math>x_0=1</math>, and <math>y_0=0</math>, <math>x_i</math> corresponds to <math>y_i</math>, which is <math>i</math> in binary. Since <math>x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090</math>, <math>t = 10010101_2</math> = <math>\boxed{149}</math>.<br />
<br />
==Solution 2 (Painful Bash)==<br />
We go through the terms and look for a pattern. We find that<br />
<br />
<math>x_0 = 1</math> <math>x_8 = 7</math><br />
<br />
<math>x_1 = 2</math> <math>x_9 = 14</math><br />
<br />
<math>x_2 = 3</math> <math>x_{10} = 21</math><br />
<br />
<math>x_3 = 6</math> <math>x_{11} = 42</math><br />
<br />
<math>x_4 = 5</math> <math>x_{12} = 35</math><br />
<br />
<math>x_5 = 10</math> <math>x_{13} = 70</math><br />
<br />
<math>x_6 = 15</math> <math>x_{14} = 105</math><br />
<br />
<math>x_7 = 30</math> <math>x_{15} = 210</math><br />
<br />
Commit to the bash. Eventually, you will receive that <math>x_{149} = 2090</math>, so <math>\boxed{149}</math> is the answer. Trust me, this is worth the 10 index points.<br />
<br />
<math>\textbf{-RootThreeOverTwo}</math><br />
<br />
==Solution 3 (induction)==<br />
<br />
Let <math>P_k</math> denote the <math>k</math>th prime.<br />
Lemma: <math>x_{n+2^{k-1}} = P_k \cdot x_{n}</math> for all <math>0 \leq n \leq 2^{k-1} - 1.</math><br />
<br />
<math>\mathbf{\mathrm{Proof:}}</math><br />
<br />
We can prove this using induction. Assume that <math>x_{2^{k-1}-1} = \prod_{j=1}^{k-1} P_j. </math> Then, using the given recursion <math>x_{k+1} = \frac{x_np(x_n)}{X(x_n)}</math>, we would “start fresh” for <math>x_{2^{k-1}} = P_k.</math> It is then easy to see that then <math>\frac{x_n}{P_k}</math> just cycles through the previous <math>x_{2^{k-1}}</math> terms of <math>\{ x_n \}, </math> since the recursion process is the same and <math>P_k</math> being a factor of <math>x_n</math> is not affected until <math>n = 2 \cdot {2^{k-1}} = 2^k, </math> when given our assumption <math>x_{2^k - 1} = \prod_{j=1}^{k} P_j</math> and <math>n = 2^k</math> is now the least <math>n</math> such that <cmath>P_{k+1} = p(x_{2^{n-1}}), </cmath> in which <math>P_a = p(x_n)</math> where <math>a > k</math> is the only way that the aforementioned cycle would be affected. Specifically, by cancellation according to our recursion, <math>x_{2^k} = P_{k+1}, </math> and the values of <math>x_n</math> just starts cycling through the previous <math>x_{2^k}</math> terms again until <math>x_{2^{k+1}}</math> when a new prime shows up in the prime factorization of <math>x_n, </math> when it starts cycling again, and so on. Using our base cases of <math>x_0</math> and <math>x_1, </math> our induction is complete. <br />
Now, it is easy to see that <math>2090 = 2 \cdot 5 \cdot 11 \cdot 19 = P_1 \cdot P_3 \cdot P_5 \cdot P_8,</math> and by Lemma #1, the least positive integer <math>n</math> such that <math>19 | x_n</math> is <math>2^7. </math> By repeatedly applying our obtained recursion from Lemma #1, it is easy to see that our answer is just <math>2^7 + 2^4 + 2^2 + 2^0, </math> or <math>10010101_2 = \boxed{149}.</math><br />
<br />
-fidgetboss_4000<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=II|num-b=14|after=Last Problem}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_15&diff=1468642010 AIME II Problems/Problem 152021-02-15T18:40:10Z<p>Welp...: /* Solution 1 */</p>
<hr />
<div>== Problem 15 ==<br />
<br />
In triangle <math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>AB</math> with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m-n</math>.<br />
<br />
== Solution 1 ==<br />
<br />
Let <math>Y = MN \cap AQ</math>. <math>\frac {BQ}{QC} = \frac {NY}{MY}</math> since <math>\triangle AMN \sim \triangle ACB</math>. Since quadrilateral <math>AMPN</math> is cyclic, <math>\triangle MYA \sim \triangle PYN</math> and <math>\triangle MYP \sim \triangle AYN</math>, yielding <math>\frac {YM}{YA} = \frac {MP}{AN}</math> and <math>\frac {YA}{YN} = \frac {AM}{PN}</math>. Multiplying these together yields *<math>\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)</math>.<br />
<br />
<math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>. <br />
<br />
Now we claim that <math>\triangle PMD \sim \triangle PNE</math>. To prove this, we can use cyclic quadrilaterals.<br />
<br />
From <math>AMPN</math>, <math>\angle PNY \cong \angle PAM</math> and <math>\angle ANM \cong \angle APM</math>. So, <math>m\angle PNA = m\angle PNY + m\angle ANM = m\angle PAM + m\angle APM = 180-m\angle PMA</math> and <math>\angle PNA \cong \angle PMD</math>.<br />
<br />
From <math>ADPE</math>, <math>\angle PDE \cong \angle PAE</math> and <math>\angle EDA \cong \angle EPA</math>. Thus, <math>m\angle MDP = m\angle PDE + m\angle EDA = m\angle PAE + m\angle EPA = 180-m\angle PEA</math> and <math>\angle PDM \cong \angle PEN</math>.<br />
<br />
Thus, from AA similarity, <math>\triangle PMD \sim \triangle PNE</math>.<br />
<br />
Therefore, <math>\frac {PN}{PM} = \frac {NE}{MD}</math>, which can easily be computed by the angle bisector theorem to be <math>\frac {145}{117}</math>. It follows that *<math>\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>.<br />
<br />
*These two ratios are the same thing and can also be derived from the Ratio Lemma.<br />
Ratio Lemma :<math>\frac{BD}{DC} = \frac{AB}{AC} \cdot \frac{\sin \angle BAD}{\sin \angle CAD}</math>, for any cevian AD of a triangle ABC.<br />
For the sine ratios use Law of Sines on triangles APM and APN, <cmath>\frac{PM}{\sin \angle PAM}=\frac{AP}{\sin \angle AMP}=\frac{AP}{\sin \angle ANP}=\frac{PN}{\sin \angle PAN}</cmath>. The information needed to use the Ratio Lemma can be found from the similar triangle section above.<br />
<br />
Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero<br />
<br />
== Extension ==<br />
The work done in this problem leads to a nice extension of this problem:<br />
<br />
Given a <math>\triangle ABC</math> and points <math>A_1</math>, <math>A_2</math>, <math>B_1</math>, <math>B_2</math>, <math>C_1</math>, <math>C_2</math>, such that <math>A_1</math>, <math>A_2</math> <math>\in BC</math>, <math>B_1</math>, <math>B_2</math> <math>\in AC</math>, and <math>C_1</math>, <math>C_2</math> <math>\in AB</math>, then let <math>\omega_1</math> be the circumcircle of <math>\triangle AB_1C_1</math> and <math>\omega_2</math> be the circumcircle of <math>\triangle AB_2C_2</math>. Let <math>A'</math> be the intersection point of <math>\omega_1</math> and <math>\omega_2</math> distinct from <math>A</math>. Define <math>B'</math> and <math>C'</math> similarly. Then <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> concur. <br />
<br />
This can be proven using Ceva's theorem and the work done in this problem, which effectively allows us to compute the ratio that line <math>AA'</math> divides the opposite side <math>BC</math> into and similarly for the other two sides.<br />
<br />
== Solution 2 ==<br />
This problem can be solved with barycentric coordinates. Let triangle <math>ABC</math> be the reference triangle with <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, and <math>C=(0,0,1)</math>. Thus, <math>N=(1:1:0)</math> and <math>M=(1:0:1)</math>. Using the Angle Bisector Theorem, we can deduce that <math>D=(14:0:15)</math> and <math>(14:13:0)</math>. Plugging the coordinates for triangles <math>ANM</math> and <math>AED</math> into the circle formula, we deduce that the equation for triangle <math>ANM</math> is <math>-a^2yz-b^2zx-c^2xy+(\frac{c^2}{2}y+\frac{b^2}{2}z)(x+y+z)=0</math> and the equation for triangle <math>AED</math> is <math>-a^2yz-b^2zx-c^2xy+(\frac{14c^2}{27}y+\frac{14b^2}{29}z)(x+y+z)=0</math>. Solving the system of equations, we get that <math>\frac{c^2y}{54}=\frac{b^2z}{58}</math>. This equation determines the radical axis of circles <math>ANM</math> and <math>AED</math>, on which points <math>P</math> and <math>Q</math> lie. Thus, solving for <math>\frac{z}{y}</math> gets the desired ratio of lengths, and <math>\frac{z}{y}=\frac{58c^2}{54b^2}</math> and plugging in the lengths <math>b=13</math> and <math>c=15</math> gets <math>\frac{725}{507}</math>. From this we get the desired answer of <math>725-507=\boxed{218}</math>.<br />
-wertguk<br />
<br />
==See Also==<br />
{{AIME box|year=2010|n=II|num-b=14|after=Last Problem}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_12&diff=1186922008 AIME I Problems/Problem 122020-03-03T05:11:25Z<p>Welp...: </p>
<hr />
<div>== Problem ==<br />
On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let <math>M</math> be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when <math>M</math> is divided by <math>10</math>.<br />
<br />
== Solution 1 ==<br />
Let <math>n</math> be the number of car lengths that separates each car. Then their speed is at most <math>15n</math>. Let a ''unit'' be the distance between the cars (front to front). Then the length of each unit is <math>4(n + 1)</math>. To maximize, in a unit, the CAR comes first, THEN the empty space. So at time zero, the car is right at the eye.<br />
<br />
Hence, we count the number of units that pass the eye in an hour: <math>\frac {15,000n\frac{\text{meters}}{\text{hour}}}{4(n + 1)\frac{\text{meters}}{\text{unit}}} = \frac {15,000n}{4(n + 1)}\frac{\text{units}}{\text{hour}}</math>. We wish to maximize this.<br />
<br />
Observe that as <math>n</math> gets larger, the <math> + 1</math> gets less and less significant, so we take the limit as <math>n</math> approaches infinity<br />
<center><math>\lim_{n\rightarrow \infty}\frac {15,000n}{4(n + 1)} = \lim_{n\rightarrow \infty}\frac {15,000}{4} = 3750</math></center><br />
Now, as the speeds are clearly finite, we can never actually reach <math>3750</math> full UNITs. However, we only need to find the number of CARS. We can increase their speed so that the camera stops (one hour goes by) after the car part of the <math>3750</math>th unit has passed, but not all of the space behind it. Hence, <math>3750</math> cars is possible, and the answer is <math>\boxed {375}</math>.<br />
<br />
== Solution 2 ==<br />
Disclaimer: This is for the people who may not understand calculus, and is also how I did it.<br />
First, we assume several things. First, we assume the speeds must be multiples of 15 to maximize cars, because any less will be a waste. Second, we assume each car goes at the same speed. Third, we start with one car in front of the photoelectric eye.<br />
We first set the speed of the cars as <math>15k</math>. Then, the distance between them is <math>\frac{4}{1000} \times k~km</math>. Therefore, it takes the car closest to the eye not on the eye <math>\frac{\frac{k}{250}}{15k}</math> hours to get to the eye. There is one hour, so the amount of cars that can pass is <math>\frac{1}{\frac{\frac{k}{250}}{15k}}</math>, or <math>3750</math> cars. When divided by ten, you get the quotient of <math>\boxed{375}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=I|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_12&diff=1186912008 AIME I Problems/Problem 122020-03-03T05:11:03Z<p>Welp...: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let <math>M</math> be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when <math>M</math> is divided by <math>10</math>.<br />
<br />
== Solution 1 ==<br />
Let <math>n</math> be the number of car lengths that separates each car. Then their speed is at most <math>15n</math>. Let a ''unit'' be the distance between the cars (front to front). Then the length of each unit is <math>4(n + 1)</math>. To maximize, in a unit, the CAR comes first, THEN the empty space. So at time zero, the car is right at the eye.<br />
<br />
Hence, we count the number of units that pass the eye in an hour: <math>\frac {15,000n\frac{\text{meters}}{\text{hour}}}{4(n + 1)\frac{\text{meters}}{\text{unit}}} = \frac {15,000n}{4(n + 1)}\frac{\text{units}}{\text{hour}}</math>. We wish to maximize this.<br />
<br />
Observe that as <math>n</math> gets larger, the <math> + 1</math> gets less and less significant, so we take the limit as <math>n</math> approaches infinity<br />
<center><math>\lim_{n\rightarrow \infty}\frac {15,000n}{4(n + 1)} = \lim_{n\rightarrow \infty}\frac {15,000}{4} = \frac {15,000}{4} = 3750</math></center><br />
Now, as the speeds are clearly finite, we can never actually reach <math>3750</math> full UNITs. However, we only need to find the number of CARS. We can increase their speed so that the camera stops (one hour goes by) after the car part of the <math>3750</math>th unit has passed, but not all of the space behind it. Hence, <math>3750</math> cars is possible, and the answer is <math>\boxed {375}</math>.<br />
<br />
== Solution 2 ==<br />
Disclaimer: This is for the people who may not understand calculus, and is also how I did it.<br />
First, we assume several things. First, we assume the speeds must be multiples of 15 to maximize cars, because any less will be a waste. Second, we assume each car goes at the same speed. Third, we start with one car in front of the photoelectric eye.<br />
We first set the speed of the cars as <math>15k</math>. Then, the distance between them is <math>\frac{4}{1000} \times k~km</math>. Therefore, it takes the car closest to the eye not on the eye <math>\frac{\frac{k}{250}}{15k}</math> hours to get to the eye. There is one hour, so the amount of cars that can pass is <math>\frac{1}{\frac{\frac{k}{250}}{15k}}</math>, or <math>3750</math> cars. When divided by ten, you get the quotient of <math>\boxed{375}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=I|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=Incircle&diff=106013Incircle2019-05-30T05:11:10Z<p>Welp...: /* Formulas */</p>
<hr />
<div>{{stub}}<br />
<br />
<br />
An '''incircle''' of a [[convex]] [[polygon]] is a [[circle]] which is inside the figure and [[tangent line | tangent]] to each side. Every [[triangle]] and [[regular polygon]] has a unique incircle, but in general polygons with 4 or more sides (such as non-[[square (geometry) | square]] [[rectangle]]s) do not have an incircle. A quadrilateral that does have an incircle is called a [[Tangential Quadrilateral]]. For a triangle, the center of the incircle is the [[Incenter]].<br />
<br />
==Formulas==<br />
*The radius of an incircle of a triangle (the inradius) with sides <math>a,b,c</math> and area <math>A</math> is <math>\frac{2A}{a+b+c}</math><br />
*The radius of an incircle of a right triangle (the inradius) with legs <math>a,b</math> and hypotenuse <math>c</math> is <math>r=\frac{ab}{a+b+c}=\frac{a+b-c}{2}</math>.<br />
*For any polygon with an incircle, <math>A=sr</math>, where <math>A</math> is the area, <math>s</math> is the semiperimeter, and <math>r</math> is the inradius.<br />
*The coordinates of the incenter (center of incircle) are <math>(\dfrac{aA_x+bB_x+cC_x}{a+b+c}, \dfrac{aA_y+bB_y+cC_y}{a+b+c})</math>, if the coordinates of each vertex are <math>A(A_x, A_y)</math>, <math>B(B_x, B_y)</math>, and <math>C(C_x, C_y)</math>, the side opposite of <math>A</math> has length <math>a</math>, the side opposite of <math>B</math> has length <math>b</math>, and the side opposite of <math>C</math> has length <math>c</math>.<br />
<br />
*The formula for the semiperimeter is <math>s=\frac{a+b+c}{2}</math>.<br />
<br />
*And area of the triangle by Heron is <math>A^2=s(s-a)(s-b)(s-c)</math>.<br />
<br />
==See also==<br />
*[[Circumradius]]<br />
*[[Inradius]]<br />
*[[Kimberling center]]<br />
<br />
[[Category:Geometry]]</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=Incircle&diff=106012Incircle2019-05-30T05:10:33Z<p>Welp...: /* Formulas */</p>
<hr />
<div>{{stub}}<br />
<br />
<br />
An '''incircle''' of a [[convex]] [[polygon]] is a [[circle]] which is inside the figure and [[tangent line | tangent]] to each side. Every [[triangle]] and [[regular polygon]] has a unique incircle, but in general polygons with 4 or more sides (such as non-[[square (geometry) | square]] [[rectangle]]s) do not have an incircle. A quadrilateral that does have an incircle is called a [[Tangential Quadrilateral]]. For a triangle, the center of the incircle is the [[Incenter]].<br />
<br />
==Formulas==<br />
*The radius of an incircle of a triangle (the inradius) with sides <math>a,b,c</math> and area <math>A</math> is <math>\frac{2A}{a+b+c}</math><br />
*The radius of an incircle of a right triangle (the inradius) with legs <math>a,b</math> and hypotenuse <math>c</math> is <math>r=\frac{ab}{a+b+c}=\frac{a+b-c}{2}</math>.<br />
*For any polygon with an incircle, <math>A=sr</math>, where <math>A</math> is the area, <math>s</math> is the semiperimeter, and <math>r</math> is the inradius.<br />
*The coordinates of the incenter (center of incircle) are <math>((aA_x+bB_x+cC_x)/(a+b+c), (aA_y+bB_y+cC_y)/(a+b+c))</math>, if the coordinates of each vertex are <math>A(A_x, A_y)</math>, <math>B(B_x, B_y)</math>, and <math>C(C_x, C_y)</math>, the side opposite of <math>A</math> has length <math>a</math>, the side opposite of <math>B</math> has length <math>b</math>, and the side opposite of <math>C</math> has length <math>c</math>.<br />
<br />
*The formula for the semiperimeter is <math>s=\frac{a+b+c}{2}</math>.<br />
<br />
*And area of the triangle by Heron is <math>A^2=s(s-a)(s-b)(s-c)</math>.<br />
<br />
==See also==<br />
*[[Circumradius]]<br />
*[[Inradius]]<br />
*[[Kimberling center]]<br />
<br />
[[Category:Geometry]]</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_11&diff=947742015 AIME I Problems/Problem 112018-05-28T05:08:29Z<p>Welp...: reordered stuff</p>
<hr />
<div>==Problem==<br />
Triangle <math>ABC</math> has positive integer side lengths with <math>AB=AC</math>. Let <math>I</math> be the intersection of the bisectors of <math>\angle B</math> and <math>\angle C</math>. Suppose <math>BI=8</math>. Find the smallest possible perimeter of <math>\triangle ABC</math>.<br />
<br />
==Solution 1==<br />
<br />
Let <math>D</math> be the midpoint of <math>\overline{BC}</math>. Then by SAS Congruence, <math>\triangle ABD \cong \triangle ACD</math>, so <math>\angle ADB = \angle ADC = 90^o</math>.<br />
<br />
Now let <math>BD=y</math>, <math>AB=x</math>, and <math>\angle IBD = \dfrac{\angle ABD}{2} = \theta</math>.<br />
<br />
Then <math>\mathrm{cos}{(\theta)} = \dfrac{y}{8}</math><br />
<br />
and <math>\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{y^2-32}{32}</math>.<br />
<br />
Cross-multiplying yields <math>32y = x(y^2-32)</math>.<br />
<br />
Since <math>x,y>0</math>, <math>y^2-32</math> must be positive, so <math>y > 5.5</math>.<br />
<br />
Additionally, since <math>\triangle IBD</math> has hypotenuse <math>\overline{IB}</math> of length <math>8</math>, <math>BD=y < 8</math>.<br />
<br />
Therefore, given that <math>BC=2y</math> is an integer, the only possible values for <math>y</math> are <math>6</math>, <math>6.5</math>, <math>7</math>, and <math>7.5</math>.<br />
<br />
However, only one of these values, <math>y=6</math>, yields an integral value for <math>AB=x</math>, so we conclude that <math>y=6</math> and <math>x=\dfrac{32(6)}{(6)^2-32}=48</math>.<br />
<br />
Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>.<br />
<br />
==Solution 2 (No Trig)==<br />
Let <math>AB=x</math> and the foot of the altitude from <math>A</math> to <math>BC</math> be point <math>E</math> and <math>BE=y</math>. Since ABC is isosceles, <math>I</math> is on <math>AE</math>. By Pythagorean Theorem, <math>AE=\sqrt{x^2-y^2}</math>. Let <math>IE=a</math> and <math>IA=b</math>. By Angle Bisector theorem, <math>\frac{y}{a}=\frac{x}{b}</math>. Also, <math>a+b=\sqrt{x^2-y^2}</math>. Solving for <math>a</math>, we get <math>a=\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}</math>. Then, using Pythagorean Theorem on <math>\triangle BEI</math> we have <math>y^2+\left(\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}\right)^2=8^2=64</math>. Simplifying, we have <math>y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64</math>. Factoring out the <math>y^2</math>, we have <math>y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\right)=64</math>. Adding 1 to the fraction and simplifying, we have <math>\frac{y^2x(x+y)}{(x+y)^2}=32</math>. Crossing out the <math>x+y</math>, and solving for <math>x</math> yields <math>32y = x(y^2-32)</math>. Then, we continue as Solution 1 does.<br />
<br />
==Solution 3==<br />
Let <math>AB=x</math>, call the midpoint of <math>BC</math> point <math>E</math>, call the point where the incircle meets <math>AB</math> point <math>D</math>, and let <math>BE=y</math>. We are looking for the minimum value of <math>2(x+y)</math>. <math>AE</math> is an altitude because the triangle is isosceles. By Pythagoras on <math>BEI</math>, the inradius is <math>\sqrt{64-y^2}</math> and by Pythagoras on <math>ABE</math>, <math>AE</math> is <math>\sqrt{x^2-y^2}</math>. By equal tangents, <math>BE=BD=y</math>, so <math>AD=x-y</math>. Since <math>ID</math> is an inradius, <math>ID=IE</math> and using pythagoras on <math>ADI</math> yields <math>AI=</math><math>\sqrt{x^2-2xy+64}</math>. <math>ADI</math> is similar to <math>AEB</math> by <math>AA</math>, so we can write <math>\frac{x-y}{\sqrt{x^2-2xy+64}}=\frac{\sqrt{x^2-y^2}}{x}</math>. Simplifying, <math>\frac{x}{\sqrt{x^2-2xy+64}}=\sqrt{\frac{x+y}{x-y}}</math>. Squaring, subtracting 1 from both sides, and multiplying everything out, we get <math>yx^2-2xy^2+64y=yx^2 -32x+32y-xy^2</math>, which turns into <math>32y=x(y^2-32)</math>. Finish as in Solution 1.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=10|num-a=12}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Intermediate Geometry Problems]]</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&diff=938952011 AIME I Problems/Problem 112018-04-12T03:29:17Z<p>Welp...: </p>
<hr />
<div>== Problem ==<br />
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by 1000. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by 1000.<br />
<br />
== Solution 1==<br />
Note that <math>x \equiv y \pmod{1000} \Leftrightarrow x \equiv y \pmod{125}</math> and <math>x \equiv y \pmod{8}</math>. So we must find the first two integers <math>i</math> and <math>j</math> such that <math>2^i \equiv 2^j \pmod{125}</math> and <math>2^i \equiv 2^j \pmod{8}</math> and <math>i \neq j</math>. Note that <math>i</math> and <math>j</math> will be greater than 2 since remainders of <math>1, 2, 4</math> will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that <math>2^{100}\equiv 1\pmod{125}</math> (see [[Euler's Totient Theorem|Euler's theorem]]) and <math>2^0,2^1,2^2,\ldots,2^{99}</math> are all distinct modulo 125 (proof below). Thus, <math>i = 103</math> and <math>j =3</math> are the first two integers such that <math>2^i \equiv 2^j \pmod{1000}</math>. All that is left is to find <math>S</math> in mod <math>1000</math>. After some computation:<br />
<cmath><br />
S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}.<br />
</cmath><br />
To show that <math>2^0, 2^1,\ldots, 2^{99}</math> are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of <math>2^{20}\equiv 1\pmod{125}</math> or <math>2^{50}\equiv 1\pmod{125}</math>. However, writing <math>2^{10}\equiv 25 - 1\pmod{125}</math>, we can easily verify that <math>2^{20}\equiv -49\pmod{125}</math> and <math>2^{50}\equiv -1\pmod{125}</math>, giving us the needed contradiction.<br />
<br />
== Solution 2 ==<br />
Notice that our sum of remainders looks like <cmath>S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots.</cmath> We want to find the remainder of <math>S</math> upon division by <math>1000.</math> Since <math>1000</math> decomposes into primes as <math>1000 = 2^3 \cdot 5^3</math>, we can check the remainders of <math>S</math> modulo <math>2^3</math> and modulo <math>5^3</math> separately. <br />
<br />
Checking <math>S</math> modulo <math>2^3</math> is easy, so lets start by computing the remainder of <math>S</math> upon division by <math>5^3.</math> To do this, let's figure out when our sequence finally repeats.<br />
Notice that since the remainder when dividing any term of <math>S</math> (after the third term) by <math>1000</math> will be a multiple of <math>2^3</math>, when this summation finally repeats, the first term to repeat will be ''not'' be <math>1</math> since <math>2^3 \nmid 1.</math> Instead, the first term to repeat will be <math>2^3</math>, and then the sequence will continue once again <math>2^4, 2^5, \cdots.</math> <br />
<br />
Now, to compute <math>S</math> modulo <math>125</math>, we want to find the least positive integer <math>d</math> such that <math>2^d \equiv 1 \pmod {125}</math> since then <math>d</math> will just be the number of terms of <math>S</math> (after the third term!) before the sequence repeats. In other words, our sequence will be of the form <math>S = 1 + 2 + 4 + \left(2^3 + 2^4 + \cdots + 2^{2 + d}\right)</math> and then we will have <math>2^{d + 3} \equiv 2^3 \pmod {125}</math>, and the sequence will repeat from there. Here, <math>d</math> simply represents the order of <math>2</math> modulo <math>125</math>, denoted by <math>d = \text{ord}_{125}(2).</math> To begin with, we'll use a well-known property of the order to get a bound on <math>d.</math><br />
<br />
Since <math>\gcd(2, 125) = 1</math> and <math>\phi(125) = 100</math>, we know by Euler's Theorem that <math>2^{100} \equiv 1 \pmod {125}.</math> However, we do not know that <math>100</math> is the ''least'' <math>d</math> satisfying <math>2^d \equiv 1 \pmod {125}.</math> Nonetheless, it is a well known property of the order that <math>\text{ord}_{125}(2) = d | \phi(125) = 100.</math> Therefore, we can conclude that <math>d</math> must be a positive divisor of <math>100.</math><br />
<br />
Now, this still leaves a lot of possibilities, so let's consider a smaller modulus for the moment, say <math>\mod 5.</math> Clearly, we must have that <math>2^d \equiv 1 \pmod 5.</math> Since <math>2^4 \equiv 1 \pmod 5</math> and powers of two will then cycle every four terms, we know that <math>2^d \equiv 1 \pmod 5 \iff 4 | d.</math> Combining this relation with <math>d | 100</math>, it follows that <math>d \in \{4, 20, 100\}.</math> <br />
<br />
Now, it is trivial to verify that <math>d \ne 4.</math> In addition, we know that <cmath>2^{20} = \left(2^{10}\right)^2 = \left(1024\right)^2 \equiv 24^2 = 576 \not\equiv 1 \pmod {125}.</cmath> Therefore, we conclude that <math>d \ne 20.</math> Hence, we must have <math>d = 100.</math> (Notice that we could have guessed this by Euler's, but we couldn't have been certain without investigating the order more thoroughly).<br />
<br />
Now, since we have found <math>d = 100</math>, we know that <cmath>S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots + 2^{102}.</cmath> There are two good ways to finish from here: <br />
<br />
The first way is to use a trick involving powers of <math>2.</math> Notice that <cmath>S = 1 + 2 + 4 + ... + 2^{102} = 2^{103} - 1.</cmath> Certainly <cmath>S = 2^{103} - 1 \equiv -1 \equiv 7 \pmod {8}.</cmath> In addition, since we already computed <math>\text{ord}_{125}(2) = d = 100</math>, we know that <cmath>S = 2^{103} - 1 = 2^{100} \cdot 2^3 - 1 \equiv 2^3 - 1 \equiv 7 \pmod {125}.</cmath> Therefore, since <math>S \equiv 7 \pmod{8}</math> and <math>S \equiv 7 \pmod{125}</math>, we conclude that <math>S \equiv \boxed{007} \pmod {1000}.</math><br />
<br />
The second way is not as slick, but works better in a general setting when there aren't any convenient tricks as in Method 1. Let us split the terms of <math>S</math> into groups: <cmath>R = 1 + 2 + 4; \quad T = 2^3 + 2^4 + \cdots + 2^{102}.</cmath> It is easy to see that <math>R</math> is congruent to <math>7</math> modulo <math>1000.</math> <br />
<br />
Now, for <math>T</math>, notice that there are <math>100</math> terms in the summation, each with a different remainder upon division by <math>125.</math> Since each of these remainders is certainly relatively prime to <math>125</math>, these <math>100</math> remainders correspond to the <math>100</math> positive integers less than <math>125</math> that are relatively prime to <math>125.</math> Therefore, <br />
<cmath>\begin{align*}T &\equiv 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 + 11 + \cdots + 124 \pmod{125} \\<br />
&= \left(1 + 2 + 3 + \cdots + 125\right) - \left(5 + 10 + 15 + \cdots 125\right) \\<br />
&= \frac{125 \cdot 126}{2} - 5\left(1 + 2 + 3 + \cdots 25\right) \\<br />
&= 125 \cdot 63 - 5\left(\frac{25 \cdot 26}{2}\right) \\<br />
&= 125\left(63 - 13\right) \\<br />
&\equiv 0 \pmod{125}.\end{align*}</cmath><br />
Then, since <math>T</math> is divisible by <math>125</math> and <math>8</math>, it follows that <math>T</math> is divisible by <math>1000.</math> Therefore, <cmath>S = R + T \equiv R \equiv \boxed{007} \pmod{1000}.</cmath><br />
<br />
== Solution 3 ==<br />
We know <math>1, 2,</math> and <math>4</math> are in <math>R</math>. Any other element in <math>R</math> must be a multiple of <math>8</math>. All multiples of <math>8</math> under <math>1000</math> sum up to a multiple of <math>1000</math>. So we can ignore them. We need to remove all multiples of <math>5</math>, or <math>40</math> in what we counted because all elements of <math>R</math> can only be divisible by <math>2</math>. But, their sum is also a multiple of <math>1000</math>. Likewise, the sum of all multiples of <math>8k</math> for some <math>k</math> will be a multiple of <math>1000</math>. Thus, our answer is <math>1+2+4=\boxed{007}</math>.<br />
<br />
Solution by TheUltimate123<br />
<br />
== Solution 4 ==<br />
<br />
Recognize that as you cycle through progressively higher powers of two, you will have to begin repeating in a pattern at some point, since there are only a finite number of possible 3-digit endings and clearly there is no way for a small sub-pattern to form. The previous statement is true by induction since if a pattern began occurring, then it would need to occur for all values before that as well which is clearly untrue unless it encompasses all possible powers of two. Thus we can start thinking about a final value for it to start repeating. It can't be <math>001</math> or <math>501</math>, and it can't be <math>002</math> or <math>502</math> either because the last two digits of any power of 2 greater than <math>2^1</math> are divisible by 4. However, it can be <math>004</math> or <math>504</math>. From the fact that <math>1+2+4+8 \dots 2^n=2^{n+1}-1</math>, we can safely assume that the sum of all possible endings mod 1000 will be <math>2 \times4 -1=\boxed{007}</math>.<br />
<br />
== Solution 5 (assumptions) ==<br />
We know that units digits repeat every <math>4</math>, and tens digits every <math>20</math>. Therefore, continuing the pattern, hundreds digits should repeat every <math>100</math>. But taking modulo <math>8</math>, we know that <math>1, 2, 4</math> will not be the same modulo <math>1000</math>. Hence, we start at <math>2^3</math> and move up <math>100</math>. That lands us at the first repeat, <math>2^{103}</math>. Just sum up to <math>2^{102}</math> to get a modulus of <math>\boxed{007}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_15&diff=938752010 AIME II Problems/Problem 152018-04-10T04:04:23Z<p>Welp...: /* Solution */</p>
<hr />
<div>== Problem 15 ==<br />
<br />
In triangle <math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>AB</math> with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m-n</math>.<br />
<br />
== Solution ==<br />
<br />
Let <math>Y = MN \cap AQ</math>. <math>\frac {BQ}{QC} = \frac {NY}{MY}</math> since <math>\triangle AMN \sim \triangle ACB</math>. Since quadrilateral <math>AMPN</math> is cyclic, <math>\triangle MYA \sim \triangle PYN</math> and <math>\triangle MYP \sim \triangle AYN</math>, yielding <math>\frac {YM}{YA} = \frac {MP}{AN}</math> and <math>\frac {YA}{YN} = \frac {AM}{PN}</math>. Multiplying these together yields <math>\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)</math>.<br />
<br />
<math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>. <br />
<br />
Now we claim that <math>\triangle PMD \sim \triangle PNE</math>. To prove this, we can use cyclic quadrilaterals.<br />
<br />
From <math>AMPN</math>, <math>\angle PNE \cong \angle PAM</math> and <math>\angle ANM \cong \angle APM</math>. So, <math>m\angle PNA = m\angle PNE + m\angle ANM = m\angle PAM + m\angle APM = 180-m\angle PMA</math> and <math>\angle PNA \cong \angle PMD</math>.<br />
<br />
From <math>ADPE</math>, <math>\angle PDE \cong \angle PAE</math> and <math>\angle EDA \cong \angle EPA</math>. Thus, <math>m\angle MDP = m\angle PDE + m\angle EDA = m\angle PAE + m\angle EPA = 180-m\angle PEA</math> and <math>\angle PDM \cong \angle PEN</math>.<br />
<br />
Thus, from AA similarity, <math>\triangle PMD \sim \triangle PNE</math>.<br />
<br />
Therefore, <math>\frac {PN}{PM} = \frac {NE}{MD}</math>, which can easily be computed by the angle bisector theorem to be <math>\frac {145}{117}</math>. It follows that <math>\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>.<br />
<br />
Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero<br />
<br />
== Extension ==<br />
The work done in this problem leads to a nice extension of this problem:<br />
<br />
Given a <math>\triangle ABC</math> and points <math>A_1</math>, <math>A_2</math>, <math>B_1</math>, <math>B_2</math>, <math>C_1</math>, <math>C_2</math>, such that <math>A_1</math>, <math>A_2</math> <math>\in BC</math>, <math>B_1</math>, <math>B_2</math> <math>\in AC</math>, and <math>C_1</math>, <math>C_2</math> <math>\in AB</math>, then let <math>\omega_1</math> be the circumcircle of <math>\triangle AB_1C_1</math> and <math>\omega_2</math> be the circumcircle of <math>\triangle AB_2C_2</math>. Let <math>A'</math> be the intersection point of <math>\omega_1</math> and <math>\omega_2</math> distinct from <math>A</math>. Define <math>B'</math> and <math>C'</math> similarly. Then <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> concur. <br />
<br />
This can be proven using Ceva's theorem and the work done in this problem, which effectively allows us to compute the ratio that line <math>AA'</math> divides the opposite side <math>BC</math> into and similarly for the other two sides.<br />
<br />
==See Also==<br />
{{AIME box|year=2010|n=II|num-b=14|after=Last Problem}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_I_2015_Problems/Problem_4&diff=93498Mock AIME I 2015 Problems/Problem 42018-03-25T23:33:48Z<p>Welp...: Created page with "==Problem 4== At the AoPS Carnival, there is a "Weighted Dice" game show. This game features two identical looking weighted 6 sided dice. For each integer <math>1\leq i\leq 6<..."</p>
<hr />
<div>==Problem 4==<br />
At the AoPS Carnival, there is a "Weighted Dice" game show. This game features two identical looking weighted 6 sided dice. For each integer <math>1\leq i\leq 6</math>, Die A has <math>\tfrac{i}{21}</math> probability of rolling the number <math>i</math>, while Die B has a <math>\tfrac{7-i}{21}</math> probability of rolling <math>i</math>. During one session, the host randomly chooses a die, rolls it twice, and announces that the sum of the numbers on the two rolls is <math>10</math>. Let <math>P</math> be the probability that the die chosen was Die A. When <math>P</math> is written as a fraction in lowest terms, find the sum of the numerator and denominator.<br />
<br />
==Solution==<br />
The probability that A lands at 10 is <cmath>\dfrac{4 \cdot 6 + 5 \cdot 5 + 6 \cdot 4}{441} = \dfrac{73}{441}.</cmath><br />
The probability that B lands at 10 is <cmath>\dfrac{1 \cdot 3 + 2 \cdot 2 + 3 \cdot 1}{441} = \dfrac{10}{441}.</cmath><br />
The probability that A was picked is thus <math>\dfrac{73}{73+10} = \dfrac{73}{83}</math>. The sum of numerator and denominator is thus <math>\boxed{156}</math>.</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_9&diff=932282018 AIME I Problems/Problem 92018-03-14T03:54:10Z<p>Welp...: </p>
<hr />
<div>Find the number of four-element subsets of <math>\{1,2,3,4,\dots, 20\}</math> with the property that two distinct elements of a subset have a sum of <math>16</math>, and two distinct elements of a subset have a sum of <math>24</math>. For example, <math>\{3,5,13,19\}</math> and <math>\{6,10,20,18\}</math> are two such subsets.<br />
<br />
==Solutions==<br />
<br />
==Solution 1==<br />
This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set <math>\{a, b, c, d\}</math>.<br />
<br />
Note that there are only two cases: 1 where <math>a + b = 16</math> and <math>c + d = 24</math> or 2 where <math>a + b = 16</math> and <math>a + c = 24</math>. Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you <math>a=d</math>, which cannot be true.<br />
<br />
Case 1.<br />
This is probably the simplest: just make a list of possible combinations for <math>\{a, b\}</math> and <math>\{c, d\}</math>. We get <math>\{1, 15\}\dots\{7, 9\}</math> for the first and <math>\{4, 20\}\dots\{11, 13\}</math> for the second. That appears to give us <math>7*8=56</math> solutions, right? NO. Because elements can't repeat, take out the supposed sets<br />
<cmath>\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 16, 4, 20\}, \{5, 11, 5, 19\},</cmath><cmath>\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}</cmath> That's ten cases gone. So <math>46</math> for Case 1.<br />
<br />
Case 2.<br />
We can look for solutions by listing possible <math>a</math> values and filling in the blanks. Start with <math>a=4</math>, as that is the minimum. We find <math>\{4, 12, 20, ?\}</math>, and likewise up to <math>a=15</math>. But we can't have <math>a=8</math> or <math>a=12</math> because <math>a=b</math> or <math>a=c</math>, respectively! Now, it would seem like there are <math>10</math> values for <math>a</math> and <math>17</math> unique values for each <math>?</math>, giving a total of <math>170</math>, but that is once again not true because there are some repeated values! We can subtract 1 from all pairs of sets that have two elements in common, because those can give us identical sets. There are 3 pairs about <math>a=8</math> and 3 pairs about <math>a=12</math>, meaning we lose <math>6</math>. That's <math>164</math> for Case 2.<br />
<br />
Total gives <math>\boxed{210}</math>.<br />
<br />
-expiLnCalc<br />
<br />
== Sol. 2 (C++) ==<br />
<br />
This code works:<br />
int num = 0;<br />
for(int i = 1; i <= 20; i++){<br />
for(int j = i+1; j <= 20; j++){<br />
for(int k = j+1; k <= 20; k++){<br />
for(int m = k+1; m <= 20; m++){<br />
if(i+j==16 || i + k == 16 || i + m == 16 || j + k == 16<br />
|| j + m == 16 || k + m == 16){<br />
if(i+j==24 || i+k==24 || i+m==24 || j+k==24 || j+m == 24 || k+m==24){<br />
num++; <br />
}<br />
}<br />
}<br />
}<br />
}<br />
}<br />
cout << num << endl;<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=92149AMC historical results2018-02-21T00:50:38Z<p>Welp...: /* 2018 */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
<br />
Stop trolling cutoffs<br />
https://artofproblemsolving.com/community/c5h1594477_a_treatise_on_cutoff_trolling<br />
<br />
===AMC 10A===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
===AMC 10B===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
===AMC 12A===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
===AMC 12B===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.56<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 111<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 111<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 93<br />
*DHR: 111<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100.5<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100.5<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91243AMC historical results2018-02-15T04:34:24Z<p>Welp...: /* AMC 10B */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*Distinguished Honor Roll: TBD<br />
<br />
===AMC 10B===<br />
*Mean Score: TBD<br />
*AIME Floor: TBD<br />
*Dist. Honor Roll: TBD<br />
<br />
===AMC 12A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*Distinguished Honor Roll floor: TBD<br />
<br />
===AMC 12B===<br />
This exam has not occurred yet.<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91242AMC historical results2018-02-15T04:34:09Z<p>Welp...: /* AMC 10B */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*Distinguished Honor Roll: TBD<br />
<br />
===AMC 10B===<br />
Mean Score: TBD<br />
AIME Floor: TBD<br />
Dist. Honor Roll: TBD<br />
<br />
===AMC 12A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*Distinguished Honor Roll floor: TBD<br />
<br />
===AMC 12B===<br />
This exam has not occurred yet.<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91241AMC historical results2018-02-15T04:33:59Z<p>Welp...: /* AMC 10B */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*Distinguished Honor Roll: TBD<br />
<br />
===AMC 10B===<br />
Mean Score: TBD\\<br />
AIME Floor: TBD\\<br />
Dist. Honor Roll: TBD<br />
<br />
===AMC 12A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*Distinguished Honor Roll floor: TBD<br />
<br />
===AMC 12B===<br />
This exam has not occurred yet.<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91240AMC historical results2018-02-15T04:33:47Z<p>Welp...: /* AMC 10B */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*Distinguished Honor Roll: TBD<br />
<br />
===AMC 10B===<br />
Mean Score: TBD<br />
AIME Floor: TBD<br />
Dist. Honor Roll: TBD<br />
<br />
===AMC 12A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*Distinguished Honor Roll floor: TBD<br />
<br />
===AMC 12B===<br />
This exam has not occurred yet.<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_9&diff=911472017 AIME I Problems/Problem 92018-02-13T02:28:43Z<p>Welp...: </p>
<hr />
<div>==Problem 9==<br />
Let <math>a_{10} = 10</math>, and for each integer <math>n >10</math> let <math>a_n = 100a_{n - 1} + n</math>. Find the least <math>n > 10</math> such that <math>a_n</math> is a multiple of <math>99</math>.<br />
<br />
==Solution 1==<br />
Writing out the recursive statement for <math>a_n, a_{n-1}, \dots, a_{10}</math> and summing them gives <cmath>a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10</cmath><br />
Which simplifies to <cmath>a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)</cmath><br />
Therefore, <math>a_n</math> is divisible by 99 if and only if <math>\frac{1}{2}(n+10)(n-9)</math> is divisible by 99, so <math>(n+10)(n-9)</math> needs to be divisible by 9 and 11. Assume that <math>n+10</math> is a multiple of 11. Writing out a few terms, <math>n=12, 23, 34, 45</math>, we see that <math>n=45</math> is the smallest <math>n</math> that works in this case. Next, assume that <math>n-9</math> is a multiple of 11. Writing out a few terms, <math>n=20, 31, 42, 53</math>, we see that <math>n=53</math> is the smallest <math>n</math> that works in this case. The smallest <math>n</math> is <math>\boxed{045}</math>.<br />
<br />
Note that we can also construct the solution using CRT by assuming either <math>11</math> divides <math>n+10</math> and <math>9</math> divides <math>n-9</math>, or <math>9</math> divides <math>n+10</math> and <math>11</math> divides <math>n-9</math>, and taking the smaller solution.<br />
<br />
==Solution 2==<br />
<cmath>a_n \equiv a_{n-1} + n \pmod {99} </cmath><br />
By looking at the first few terms, we can see that <br />
<cmath>a_n \equiv 10+11+12+ \dots + n \pmod {99} </cmath><br />
This implies<br />
<cmath>a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} </cmath><br />
Since <math>a_n \equiv 0 \pmod {99}</math>, we can rewrite the equivalence, and simplify <br />
<cmath>0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} </cmath><br />
<cmath>0 \equiv n(n+1) - 90 \pmod {99} </cmath><br />
<cmath>0 \equiv 4n^2+4n+36 \pmod {99} </cmath><br />
<cmath>0 \equiv (2n+1)^2+35 \pmod {99} </cmath><br />
<cmath>64 \equiv (2n+1)^2 \pmod {99} </cmath><br />
The only squares that are congruent to <math>64 \pmod {99}</math> are <math>(\pm 8)^2</math> and <math>(\pm 19)^2</math>, so <br />
<cmath>2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}</cmath><br />
<math>2n+1 \equiv -8 \pmod {99}</math> yields <math>n=45</math> as the smallest integer solution.<br />
<br />
<math>2n+1 \equiv 8 \pmod {99}</math> yields <math>n=53</math> as the smallest integer solution.<br />
<br />
<math>2n+1 \equiv -19 \pmod {99}</math> yields <math>n=89</math> as the smallest integer solution.<br />
<br />
<math>2n+1 \equiv 19 \pmod {99}</math> yields <math>n=9</math> as the smallest integer solution. However, <math>n</math> must be greater than <math>10</math>.<br />
<br />
The smallest positive integer solution greater than <math>10</math> is <math>n=\boxed{045}</math>.<br />
<br />
==Solution 3==<br />
<math>a_n=a_{n-1} + n \pmod{99}</math>. Using the steps of the previous solution we get up to <math>n^2+n \equiv 90 \pmod{99}</math>. This gives away the fact that <math>(n)(n+1) \equiv 0 \pmod{9} \implies n \equiv \{0, 8\} \pmod{9}</math> so either <math>n</math> or <math>n+1</math> must be a multiple of 9. <br />
<br />
Case 1 (<math>n|9</math>): Say <math>n=9x</math> and after simplification <math>x(9x+1) = 10 \pmod{90} \forall x \in \mathbb{Z}</math>. <br />
<br />
Case 2: (<math>n+1|9</math>): Say <math>n=9a-1</math> and after simplification <math>(9a-1)(a) = 10 \pmod{90} \forall a \in \mathbb{Z}</math>.<br />
<br />
As a result <math>a</math> must be a divisor of <math>10</math> and after doing some testing in both cases the smallest value that works is <math>x=5 \implies \boxed{045}</math>.<br />
<br />
~First<br />
<br />
==Solution 4 (not good, risky)==<br />
We just notice that <math>100 \equiv 1 \pmod{99}</math>, so we are just trying to find <math>10 + 11 + 12 + \cdots + n</math> modulo <math>99</math>, or <math>\dfrac{n(n+1)}{2} - 45</math> modulo <math>99</math>. Also, the sum to <math>44</math> is divisible by <math>99</math>, and is the first one that is. Thus, if we sum to <math>45</math> the <math>45</math> is cut off and thus is just a sum to <math>44</math>.<br />
<br />
Without checking whether there are other sums congruent to <math>45 \pmod{99}</math>, we can just write the answer to be <math>\boxed{045}</math>.<br />
<br />
==See also==<br />
{{AIME box|year=2017|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_1&diff=907012018 AMC 10A Problems/Problem 12018-02-09T01:01:25Z<p>Welp...: /* Solution */</p>
<hr />
<div>== Problem ==<br />
What is the value of<br />
<cmath>\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?</cmath><math>\textbf{(A) } \frac58 \qquad \textbf{(B) }\frac{11}7 \qquad \textbf{(C) } \frac85 \qquad \textbf{(D) } \frac{18}{11} \qquad \textbf{(E) } \frac{15}8 </math><br />
<br />
== Solution == <br />
<math>2+1=3</math>, so the reciprocal is <math>\frac{1}{3}</math>. Summing <math>1</math> and repeating leads to <math>\frac{3}{4}</math>. Rinse and repeat to get <math>\frac{4}{7}</math>. Adding <math>1</math> to that is <math>\frac{11}{7}</math> (B)<br />
<br />
-Welp...<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2018|ab=A|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_24&diff=906992018 AMC 10A Problems/Problem 242018-02-09T00:56:22Z<p>Welp...: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Triangle <math>ABC</math> with <math>AB=50</math> and <math>AC=10</math> has area <math>120</math>. Let <math>D</math> be the midpoint of <math>\overline{AB}</math>, and let <math>E</math> be the midpoint of <math>\overline{AC}</math>. The angle bisector of <math>\angle BAC</math> intersects <math>\overline{DE}</math> and <math>\overline{BC}</math> at <math>F</math> and <math>G</math>, respectively. What is the area of quadrilateral <math>FDBG</math>?<br />
<br />
<math><br />
\textbf{(A) }60 \qquad<br />
\textbf{(B) }65 \qquad<br />
\textbf{(C) }70 \qquad<br />
\textbf{(D) }75 \qquad<br />
\textbf{(E) }80 \qquad<br />
</math><br />
<br />
== Solution ==<br />
<br />
By angle bisector theorem, <math>BG=\frac{5a}{6}</math>. By similar triangles, <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>, where <math>h</math> is the length of the altitude to <math>BC</math>. Then <math>\frac{ah}{2}=120</math> and we wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}</math>.<br />
<br />
== (More basic) Solution 2 (bad) ==<br />
<br />
<math>\overline{DE}</math> is midway from <math>A</math> to <math>\overline{BC}</math>, and <math>DE = \frac{BC}{2}</math>. Therefore, <math>DE</math> is a quarter of the area, which is <math>30</math>. Using the angle bisector theorem in the same fashion as the previous problem, we get that one segment is <math>5</math> times the other. We want the larger piece, as described by the problem. Because the heights are identical, one area is <math>5</math> times the other, and <math>\frac{5}{6} \cdot 90 = \boxed{75}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_24&diff=906982018 AMC 10A Problems/Problem 242018-02-09T00:56:02Z<p>Welp...: added bad solution</p>
<hr />
<div>== Problem ==<br />
<br />
Triangle <math>ABC</math> with <math>AB=50</math> and <math>AC=10</math> has area <math>120</math>. Let <math>D</math> be the midpoint of <math>\overline{AB}</math>, and let <math>E</math> be the midpoint of <math>\overline{AC}</math>. The angle bisector of <math>\angle BAC</math> intersects <math>\overline{DE}</math> and <math>\overline{BC}</math> at <math>F</math> and <math>G</math>, respectively. What is the area of quadrilateral <math>FDBG</math>?<br />
<br />
<math><br />
\textbf{(A) }60 \qquad<br />
\textbf{(B) }65 \qquad<br />
\textbf{(C) }70 \qquad<br />
\textbf{(D) }75 \qquad<br />
\textbf{(E) }80 \qquad<br />
</math><br />
<br />
== Solution ==<br />
<br />
By angle bisector theorem, <math>BG=\frac{5a}{6}</math>. By similar triangles, <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>, where <math>h</math> is the length of the altitude to <math>BC</math>. Then <math>\frac{ah}{2}=120</math> and we wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}</math>. (trumpeter)<br />
<br />
== (More basic) Solution 2 (bad) ==<br />
<br />
<math>\overline{DE}</math> is midway from <math>A</math> to <math>\overline{BC}</math>, and <math>DE = \frac{BC}{2}</math>. Therefore, <math>DE</math> is a quarter of the area, which is <math>30</math>. Using the angle bisector theorem in the same fashion as the previous problem, we get that one segment is <math>5</math> times the other. We want the larger piece, as described by the problem. Because the heights are identical, one area is <math>5</math> times the other, and <math>\frac{5}{6} \cdot 90 = \boxed{75}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_23&diff=906952018 AMC 10A Problems/Problem 232018-02-09T00:48:25Z<p>Welp...: </p>
<hr />
<div>== Problem ==<br />
<br />
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square <math>S</math> so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from <math>S</math> to the hypotenuse is 2 units. What fraction of the field is planted?<br />
<br />
<asy><br />
draw((0,0)--(4,0)--(0,3)--(0,0));<br />
draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br />
fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br />
label("$4$", (2,0), N);<br />
label("$3$", (0,1.5), E);<br />
label("$2$", (.8,1), E);<br />
label("$S$", (0,0), NE);<br />
draw((0.3,0.3)--(1.4,1.9), dashed);<br />
</asy><br />
<br />
<math>\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75} </math><br />
<br />
==Solution==<br />
Let the square have side length <math>x</math>. Connect the upper-right vertex of square <math>S</math> with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is <math>6</math>.<br />
<br />
Square <math>S</math> has area <math>x^2</math>, and the two thin triangle regions have area <math>\dfrac{x(3-x)}{2}</math> and <math>\dfrac{x(4-x)}{2}</math>. The final triangular region with the hypotenuse as its base and height <math>2</math> has area <math>5</math>. Thus, we have <cmath>x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6</cmath><br />
<br />
Solving gives <math>x=\dfrac{2}{7}</math>. The area of <math>S</math> is <math>\dfrac{4}{49}</math> and the desired ratio is <math>\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}</math>.<br />
<br />
==Solution 2==<br />
Let the square have side length <math>s</math>. If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the big similar triangle is <math>\frac{5}{3}(2)=\frac{10}{3}</math>. Now, let's extend this big similar right triangle to the left until it hits the side of length 3. Now, the length is <math>\frac{10}{3}+s</math>, and using the ratios of the side lengths, the height is <math>\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}</math>. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so <cmath>\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}</cmath><br />
<br />
Now comes the easy part: finding the ratio of the areas: <math>\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}.}</math><br />
<br />
==Solution 3==<br />
We use coordinate geometry. Let the right angle be at <math>(0,0)</math> and the hypotenuse be the line <math>3x+4y = 12</math> for <math>0\le x\le 3</math>. Denote the position of <math>S</math> as <math>(s,s)</math>, and by the point to line distance formula, we know that <cmath>\frac{|3s+4s-12|}{5} = 2</cmath> <cmath>\Rightarrow |7s-12| = 10</cmath> Obviously <math>s<\frac{22}{7}</math>, so <math>s = \frac{2}{7}</math>, and from here the rest of the solution follows to get <math>\boxed{\frac{145}{147}}</math>.<br />
<br />
==Solution 4==<br />
Let the side length of the square be <math>x</math>. First off, let us make a similar triangle with the segment of length <math>2</math> and the top-right corner of <math>S</math>. Therefore, the longest side of the smaller triangle must be <math>2 \cdot \frac54 = \frac52</math>. We then do operations with that side in terms of <math>x</math>. We subtract <math>x</math> from the bottom, and <math>\frac{3x}{4}</math> from the top. That gives us the equation of <math>3-\frac{7x}{4} = \frac{5}{2}</math>. Solving, <cmath>12-7x = 10 \implies x = \frac{2}{7}.</cmath><br />
<br />
Thus, <math>x^2 = \frac{4}{49}</math>, so the fraction of the triangle (area <math>6</math>) covered by the square is <math>\frac{2}{147}</math>. The answer is then <math>\boxed{\dfrac{145}{147}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=A|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2018|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=90669AMC historical results2018-02-08T23:01:00Z<p>Welp...: add 2018</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 56.99<br />
*AIME floor: 96<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: <br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: <br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100 (Top 5% 106.5)<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_25&diff=903832017 AMC 10A Problems/Problem 252018-02-08T00:51:19Z<p>Welp...: /* Solution 4: (Risky) */</p>
<hr />
<div>==Problem==<br />
How many integers between <math>100</math> and <math>999</math>, inclusive, have the property that some permutation of its digits is a multiple of <math>11</math> between <math>100</math> and <math>999?</math> For example, both <math>121</math> and <math>211</math> have this property.<br />
<br />
<math> \mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486</math><br />
<br />
==Solution 1==<br />
<br />
Let the three-digit number be <math>ACB</math>:<br />
<br />
If a number is divisible by <math>11</math>, then the difference between the sums of alternating digits is a multiple of <math>11</math>.<br />
<br />
There are two cases:<br />
<math>A+B=C</math> and <math>A+B=C+11</math><br />
<br />
We now proceed to break down the cases. Note: let <math>A \geq C</math> so that we avoid counting the same permutations and having to subtract them later.<br />
<br />
<br />
<math>\textbf{Case 1}</math>: <math>A+B=C</math>. <br />
<br />
<br />
<br />
<math>\textbf{Part 1}</math>: <math>B=0</math><br />
<math>A=C</math>, this case results in 110, 220, 330...990. There are two ways to arrange the digits in each of those numbers.<br />
<math>2 \cdot 9 = 18</math><br />
<br />
<math>\textbf{Part 2}</math>: <math>B>0</math><br />
<math>B=1, A+1=C</math>, this case results in 121, 231,... 891. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to <math>45</math> cases.<br />
<br />
<math>\textbf{Part 3}</math>: <math>B=2, A+2=C</math>, this case results in 242, 352,... 792. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to <math>33</math> cases.<br />
<br />
<math>\textbf{Part 4}</math>: <math>B=3, A+3=C</math>, this case results in 363, 473,...693. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to <math>21</math> cases.<br />
<br />
<math>\textbf{Part 5}</math>: <math>B=4, A+4=C</math>, this case results in 484 and 594. There are <math>6</math> ways to arrange the digits in 594 and 3 ways for 484. This leads to <math>9</math> cases.<br />
<br />
This case has <math>18+45+33+21+9=126</math> subcases.<br />
<br />
<br />
<br />
<math>\textbf{Case 2}</math>: <math>A+B=C+11</math>. <br />
<br />
<br />
<math>\textbf{Part 1}</math>: <math>C=0, A+B=11</math>, this cases results in 209, 308, ...506. There are <math>4</math> ways to arrange each of those cases. This leads to <math>16</math> cases.<br />
<br />
<math>\textbf{Part 2}</math>: <math>C=1, A+B=12</math>, this cases results in 319, 418, ...616. There are <math>6</math> ways to arrange each of those cases, except the last. This leads to <math>21</math> cases.<br />
<br />
<math>\textbf{Part 3}</math>: <math>C=2, A+B=13</math>, this cases results in 429, 528, ...627. There are <math>6</math> ways to arrange each of those cases. This leads to <math>18</math> cases.<br />
<br />
...<br />
If we continue this counting, we receive <math>16+21+18+15+12+9+6+3=100</math> subcases.<br />
<br />
<math>100+126=\boxed{\textbf{(A) } 226}</math><br />
<br />
==Solution 2==<br />
<br />
We note that we only have to consider multiples of 11 and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of 11 has:<br />
<br />
<math>\textbf{Case 1:}</math> All three digits are the same. <br />
By inspection, we find that there are no multiples of 11 here.<br />
<br />
<math>\textbf{Case 2:}</math> Two of the digits are the same, and the third is different.<br />
<br />
<math>\textbf{Case 2a:}</math><br />
There are 8 multiples of 11 without a zero that have this property:<br />
121, 242, 363, 484, 616, 737, 858, 979.<br />
Each contributes 3 valid permutations, so there are <math>8 \cdot 3 = 24</math> permutations in this subcase.<br />
<br />
<math>\textbf{Case 2b:}</math><br />
There are 9 multiples of 11 with a zero that have this property:<br />
110, 220, 330, 440, 550, 660, 770, 880, 990.<br />
Each one contributes 2 valid permutations (the first digit can't be zero), so there are <math>9 \cdot 2 = 18</math> permutations in this subcase.<br />
<br />
<math>\textbf{Case 3:}</math> All the digits are different.<br />
Since there are <math>\frac{990-110}{11}+1 = 81</math> multiples of 11 between 100 and 999, there are <math>81-8-9 = 64</math> multiples of 11 remaining in this case. However, 8 of them contain a zero, namely 209, 308, 407, 506, 605, 704, 803, and 902. Each of those multiples of 11 contributes <math>2 \cdot 2=4</math> valid permutations, but we overcounted by a factor of 2; every permutation of 209, for example, is also a permutation of 902. Therefore, there are <math>8 \cdot 4 / 2 = 16</math>. Therefore, there are <math>64-8=56</math> remaining multiples of 11 without a 0 in this case. Each one contributes <math>3! = 6</math> valid permutations, but once again, we overcounted by a factor of 2 (note that if a number ABC is a multiple of 11, then so is CBA). Therefore, there are <math>56 \cdot 6 / 2 = 168</math> valid permutations in this subcase.<br />
<br />
Adding up all the permutations from all the cases, we have <math>24+18+16+168 = \boxed{\textbf{(A) } 226}</math>.<br />
<br />
==Solution 3 ==<br />
<br />
We can overcount and then subtract.<br />
We know there are <math>81</math> multiples of <math>11</math>.<br />
<br />
We can multiply by <math>6</math> for each permutation of these multiples. (Yet some multiples don't have 6)<br />
<br />
Now divide by <math>2</math>, because if a number <math>abc</math> with digits <math>a</math>, <math>b</math>, and <math>c</math> is a multiple of 11, then <math>cba</math> is also a multiple of 11 so we have counted the same permutations twice. <br />
<br />
Basically, each multiple of 11 has its own 3 permutations (say <math>abc</math> has <math>abc</math> <math>acb</math> and <math>bac</math> whereas <math>cba</math> has <math>cba</math> <math>cab</math> and <math>bca</math>). We know that each multiple of 11 has at least 3 permutations because it cannot have 3 repeating digits.<br />
<br />
Hence we have <math>243</math> permutations without subtracting for overcounting.<br />
Now note that we overcounted cases in which we have 0's at the start of each number. So, in theory, we could just answer <math>A</math> and move on.<br />
<br />
If we want to solve it, then we continue.<br />
<br />
We overcounted cases where the middle digit of the number is 0 and the last digit is 0.<br />
<br />
Note that we assigned each multiple of 11 3 permutations.<br />
<br />
The last digit is <math>0</math> gives <math>9</math> possibilities where we overcounted by <math>1</math> permutation for each of <math>110, 220, ... , 990</math>.<br />
<br />
The middle digit is 0 gives 8 possibilities where we overcount by 1.<br />
<math>605, 704, 803, 902</math> and <math>506, 407, 308, 209</math><br />
<br />
Subtracting <math>17</math> gives <math>\boxed{\textbf{(A) } 226}</math>.<br />
<br />
Now, we may ask if there is further overlap (I.e if two of <math>abc</math> and <math>bac</math> and <math>acb</math> were multiples of <math>11</math>) Thankfully, using divisibility rules, this can never happen as taking the divisibility rule mod 11 and adding we get that <math>2a</math>, <math>2b</math>, or <math>2c</math> is congruent to <math>0\ (mod\ 11)</math>. Since <math>a, b, c</math> are digits, this can never happen as none of them can equal 11 and they can't equal 0 as they are the leading digit of a 3 digit number in each of the cases.<br />
<br />
==Solution 4: (Risky)==<br />
<br />
Notice that (D) and (E) are the odd ones out, as both are in the 400s range, and the majority of the answers (3/5), are in the 200s range, so we can omit them. We now look at (A), (B), and (C). We quickly omit (C), as it is divisible by 10, and therefore the odd one out. We now have narrowed down to (A) and (B). At this point, you might flip a coin or just guess (A), but we can use another strategic elimination (as follows). According to a list of the probabilities of each answer choice, on average, for any AMC question, (A) is generally more likely than (B), so it pays to guess <math>\boxed{\text{(A)}}</math> at this point.<br />
<br />
Or, if you have already done some work you know that there is an even number of numbers with odd value, which means that an odd value would not work.<br />
So, (A) is better.<br />
<br />
You should probably use another strategy if you have more time.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=88896Gmaas2017-12-14T00:18:45Z<p>Welp...: /* Known Facts About Gmaas */</p>
<hr />
<div>=== Known Facts About Gmaas ===<br />
- sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br />
<br />
-Gmaas plays roblox mobile edition and likes minecraft, candy crush, and club penguin island<br />
<br />
-gmaas is roy moore's horse in the shape of a cat<br />
<br />
-Gmaas is a known roblox/club penguin player and is a legend at it. He has over 289547987693 robux and <math>190348 in CP.<br />
<br />
-This is all hypothetical EDIT: This is all factual <br />
<br />
- He is capable of salmon powers, according to PunSpark (ask him)<br />
<br />
The Gmaas told Richard Rusczyk to make AoPS<br />
<br />
-The Gmaas is everything. Yes you are part of the Gmaas-Dw789<br />
<br />
-The Gmaas knows every dimension up to 9999999999999999999th dimension<br />
<br />
-Certain theories provide evidence that he IS darth plagueis the wise<br />
<br />
-Gmaas has met George Washington<br />
<br />
-Gmaas is "TIRED OF PEOPLE ADDING TO HIS PAGE!!" (Maas 45).<br />
<br />
-Gmaas has multiple accounts; some of them are pifinity, Lord_Baltimore, Spacehead1AU, and Electro3.0<br />
<br />
-Owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br />
<br />
- Is Grayson Maas, confirmed.<br />
<br />
- Made this page<br />
<br />
- King of the first men, the anduls<br />
<br />
-is a well known professor at MEOWston Academy<br />
<br />
- rightful heir to the iron throne<br />
<br />
- has the power of catnip<br />
<br />
- Gmass can destroy the universe with a thought<br />
<br />
-Gmass is a Tuna addict<br />
<br />
- won the reward of being cutest and (fattest) cat ever<br />
<br />
-Last sighting 1571-stretch-algebra-a 12/6/17<br />
<br />
- owner of sseraj not pet<br />
<br />
- embodiment of life and universe and beyond <br />
<br />
- Watches memes<br />
<br />
-After Death became the GOD OF HYPERDEATH and obtained over 9000 souls <br />
<br />
-Gmass's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br />
<br />
-Gmass is a certified Slytherin and probably the cutest cat ever.<br />
<br />
-Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br />
<br />
-Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br />
<br />
- Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br />
<br />
-Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br />
<br />
-Gmaas is a supreme overlord who must be given a miencraf DIMOND<br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br />
<br />
- Gmaas is 5space's favorite animal. [https://www.youtube.com/watch?v=hSlb1ezRqfA&vl=en(Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
-Gmaas is my favorite pokemon<br />
<br />
- Gmaas is cool<br />
<br />
- He is often overfed (with probability </math>\frac{3972}{7891}<math>), or malnourished (with probability </math>\frac{3919}{7891}<math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS</cmath> supercars, excluding the Purrari and the 138838383 Teslas. <br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
- He is distant relative of Mathcat1234<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
<br />
~Alligator Swamp A 1072 <br />
<br />
~Alligator Swamp B 1073<br />
<br />
~Introduction to Algebra A (1170)<br />
<br />
~Introduction to Algebra B (1529)<br />
<br />
~Welcome to Panda Town Gate 1076<br />
<br />
~Welcome to Gmaas Town Gate 1221<br />
<br />
~Welcome to Gmaas Town Gate 1125<br />
<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
<br />
~The other side of the ice in Antarctica<br />
<br />
~Feisty Alligator Swamp 1115<br />
<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
<br />
~Introduction to Counting and Probability 1142 <br />
<br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
<br />
~Intermediate Counting and Probability 1137<br />
<br />
~Intermediate Counting and Probability 1207<br />
<br />
~Posting student surveys<br />
<br />
~USF Castle Walls 1203<br />
<br />
~Dark Lord's Hut 1210<br />
<br />
~AMC 10 Problem Series 1200<br />
<br />
~Intermediate Number Theory 1138<br />
<br />
~Intermediate Number Theory 1476<br />
<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
<br />
~Algebra B 1112<br />
<br />
~Intermediate Algebra 1561 7:17 PM 12/11/17<br />
<br />
<br />
<br />
<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (Especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
EDIT: Nobody has yet seen him atop a tribal base yet.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: Gmaas rarely disguises himself as a penguin.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br />
<br />
-Aha!! An impostor!! <br />
http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br />
(look at the acronym).<br />
<br />
- Gmaas might have been viewing (with a </math>\frac{99999}{100000}<math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- EDIT: Gmaas is a he.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mew.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the </math>\tan(90)<math> ritual. Draw a pentagram and write the numerical value of </math>\tan(90)<math> in the middle, and he will be summoned.<br />
<br />
- EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br />
https://artofproblemsolving.com/community/c287916h1291232<br />
<br />
- EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- EDIT: That has never happened and thus it does not contain the singularity of a black hole. <br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
-Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond.<br />
EDIT: that is a lie. he IS THE Illuminati. april fools(it's actually july, i do april fools year round)<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks.<br />
<br />
- Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br />
<br />
-The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF or Cobra, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br />
<br />
- EDIT: The above fact is somewhat irrelevant.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has </math>57843504<math> regenerations and has used </math>3<math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
-Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br />
<br />
- Gmaas loves to eat turnips. At </math>\frac{13}{32}<math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
-Gmaas has a secret hidden garden full of turnips under sseraj's house.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance in the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything on sight? Nobody knows. (Except him.)<br />
<br />
-Gmaas lives on Gallifrey and in Gotham City (he has sleepovers with Batman).<br />
<br />
-Gmaas is an excellent driver.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
-Donald Trump once sang "All Hail the Chief" to Gmaas, 3 days after being sworn in as US President.<br />
<br />
- Gmaas likess to talk with rrusczyk from time to time.<br />
<br />
- gmaas can shoot fire from his paws.<br />
<br />
- Gmaas is the reason why the USF has the longest thread on AoPS.<br />
<br />
- He (or she) is an avid watcher of the popular T.V. show "Bernie Sanders and the Gauntlet of DOOM"<br />
<br />
- sseraj, in 1521 Introduction to Number Theory, posted an image of gmaas after saying "Who wants to see 5space?" at around 5:16 PM Mountain Time, noting gmaas was "also 5space"<br />
<br />
-EDIT: he also did it in Introduction to Algebra A once<br />
<br />
- Gmaas is now my HD background on my Mac.<br />
<br />
- In 1521 Into to Number Theory, sseraj posted an image of a 5space gmaas fusion. (First sighting) <br />
<br />
- Also confirmed that gmaas doesn't like ketchup because it was the only food left the photo.<br />
<br />
- In 1447 Intro to Geometry, sseraj posted a picture of gmaas with a rubik's cube suggesting that gmaas's has an average solve time of </math>-GMAAS<math> seconds.<br />
<br />
-Gmass beat Superman in a fight with ease<br />
<br />
-Gmass was an admin of Roblox<br />
<br />
-Gmass traveled around the world,paying so much </math>MONEY<math> just to eat :D<br />
<br />
-Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br />
Summary:<br />
<br />
-When Gmaas subtracts </math>0.\overline{99}<math> from </math>1<math>, the difference is greater than </math>0$.<br />
<br />
-Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br />
<br />
-Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br />
<br />
-The results of the revival is top secret, and nobody knows what happened.<br />
<br />
-sseraj, in 1496 Prealgebra 2, said that gmaas is Santacat.<br />
<br />
-sseraj likes to post a picture of gmaas in every class he passes by.<br />
<br />
-sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Ender is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmass is now wandering space in search for a home.<br />
<br />
-Gmaas is the lord of the pokemans<br />
<br />
-Gmass can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will<br />
<br />
-Picture of Gmass http://i.imgur.com/PP9xi.png<br />
<br />
-Known by Mike Miller<br />
<br />
-Gmaas got mad at sseraj once, so he locked him in his own freezer<br />
<br />
-Gmaas is an obviously omnipotent cat.<br />
<br />
-ehawk11 met him<br />
<br />
-sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br />
<br />
-sseraj has posted pictures of gmass in intro to algebra, before class started, with the title, "caption contest" anyone who posted a caption mysteriously vanished in the middle of the night. <br />
<br />
- gmass has once slept on your bed and made it wet<br />
<br />
-It is rumored that rrusczyk is actually gmaas in disguise<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- [s]Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".[/s]Sorry, this was a rick roll troll.<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath and themoocow are also writing a book about Gmaas<br />
<br />
-Ornx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br />
<br />
- gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br />
<br />
-oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br />
<br />
-Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br />
<br />
-No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br />
<br />
-In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br />
https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br />
https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br />
https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br />
https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br />
<br />
<br />
- Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br />
<br />
-<math>Another</math> sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br />
<br />
-Gmaas has been sighted several times on the Global Announcements forum</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=88720Gmaas2017-11-30T00:37:20Z<p>Welp...: /* Known Facts About Gmaas */</p>
<hr />
<div>=== Known Facts About Gmaas ===<br />
-This is all hypothetical<br />
<br />
-Certain theories provide evidence that he IS darth plagueis the wise<br />
<br />
-Gmaas has met George Washington <br />
<br />
-Owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br />
<br />
- Is Grayson Maas, confirmed.<br />
<br />
- Made this page<br />
<br />
- King of the first men, the anduls<br />
<br />
-is a well known professor at MEOWston Academy<br />
<br />
- rightful heir to the iron throne<br />
<br />
- has the power of catnip<br />
<br />
- Gmass can destroy the universe with a thought<br />
<br />
-Gmass is a Tuna addict<br />
<br />
- won the reward of being cutest and (fattest) cat ever<br />
<br />
-Last sighting 1582-algebra-a 10/24/17<br />
<br />
- owner of sseraj not pet<br />
<br />
- embodiment of life and universe and beyond <br />
<br />
- Watches memes<br />
<br />
-After Death became the GOD OF HYPERDEATH and obtained over 9000 souls <br />
<br />
-Gmass's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br />
<br />
-Gmass is a certified Slytherin and probably the cutest cat ever.<br />
<br />
-Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br />
<br />
-Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br />
<br />
- Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br />
<br />
-Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br />
<br />
-Gmaas is a supreme overlord who must be given a miencraf DIMOND<br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br />
<br />
- Gmaas is 5space's favorite animal. [https://www.youtube.com/watch?v=hSlb1ezRqfA&vl=en(Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
-Gmaas is my favorite pokemon<br />
<br />
- Gmaas is cool<br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAS</cmath> supercars, excluding the Purrari and the 138838383 Teslas. <br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
- He is distant relative of Mathcat1234<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
<br />
~Alligator Swamp A 1072 <br />
<br />
~Alligator Swamp B 1073<br />
<br />
~Introduction to Algebra A (1170)<br />
<br />
~Introduction to Algebra B (1529)<br />
<br />
~Welcome to Panda Town Gate 1076<br />
<br />
~Welcome to Gmaas Town Gate 1221<br />
<br />
~Welcome to Gmaas Town Gate 1125<br />
<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
<br />
~The other side of the ice in Antarctica<br />
<br />
~Feisty Alligator Swamp 1115<br />
<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
<br />
~Introduction to Counting and Probability 1142 <br />
<br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
<br />
~Intermediate Counting and Probability 1137<br />
<br />
~Intermediate Counting and Probability 1207<br />
<br />
~Posting student surveys<br />
<br />
~USF Castle Walls 1203<br />
<br />
~Dark Lord's Hut 1210<br />
<br />
~AMC 10 Problem Series 1200<br />
<br />
~Intermediate Number Theory 1138<br />
<br />
~Intermediate Number Theory 1476<br />
<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
<br />
~Algebra B 1112<br />
<br />
<br />
<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (Especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
EDIT: Nobody has yet seen him atop a tribal base yet.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: The above disguises are rare for a Gmaas, except the cat.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br />
<br />
-Aha!! An impostor!! <br />
http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br />
(look at the acronym).<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99999}{100000}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mew.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
--This above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br />
https://artofproblemsolving.com/community/c287916h1291232<br />
--The above 'proof' is non-conclusive. math101010 had only put an approximation.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
-Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond.<br />
EDIT: that is a lie. he IS THE Illuminati. april fools(it's actually july, i do april fools year round)<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michal Stevens (also known as Vsaace XD), or his page name, Purrshanks.<br />
<br />
- Gmaas rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br />
EDIT: the reason why Dacammel left the USF is that gmass entrusted his TV to him, and not wanting to be discovered by LF or Cobra, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has <math>57843504</math> regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
-Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part<br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
-Gmaas has a secret hidden garden full of turnips under sseraj's house.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance in the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything on sight? Nobody knows. (Except him.)<br />
<br />
-Gmaas lives on Gallifrey and in Gotham City (he has sleepovers with Batman).<br />
<br />
-Gmaas is an excellent driver.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
-Donald Trump once sang "All Hail the Chief" to Gmaas, 3 days after being sworn in as US President.<br />
<br />
- Gmaas like to talk with rrusczyk from time to time.<br />
<br />
- gmaas can shoot fire from his paws.<br />
<br />
- Gmaas is the reason why the USF has the longest thread<br />
<br />
- He (or she) is an avid watcher of the popular T.V. show "Bernie Sanders and the Gauntlet of DOOM"<br />
<br />
- sseraj, in 1521 Introduction to Number Theory, posted an image of gmaas after saying "Who wants to see 5space?" at around 5:16 PM Mountain Time, noting gmaas was "also 5space"<br />
<br />
- Gmaas is now my HD background on my mac.<br />
<br />
-EDIT: he also did it in Introduction to Algebra A once<br />
<br />
- In 1521 Into to Number Theory, sseraj posted an image of a 5space gmaas fusion. (First sighting) <br />
<br />
- Also confirmed that gmaas doesn't like ketchup because it was the only food left the photo.<br />
<br />
- In 1447 Intro to Geometry, sseraj posted a picture of gmaas with a rubik's cube suggesting that gmaas's has an average solve time of <math>-GMAAS</math> seconds.<br />
<br />
-Gmass beat Superman in a fight with ease<br />
<br />
-Gmass was an admin of Roblox<br />
<br />
-Gmass traveled around the world,paying so much <math>MONEY</math> just to eat :D<br />
<br />
-Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br />
Summary:<br />
<br />
-When Gmaas subtracts <math>0.\overline{99}</math> from <math>1</math>, the difference is greater than <math>0</math>.<br />
<br />
-Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br />
<br />
-Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br />
<br />
-The results of the revival is top secret, and nobody knows what happened.<br />
<br />
-sseraj, in 1496 Prealgebra 2, said that gmaas is Santacat.<br />
<br />
-sseraj likes to post a picture of gmaas in every class he passes by.<br />
<br />
-sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Ender is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmass is now wandering space in search for a home.<br />
<br />
-Gmaas is the lord of the pokemans<br />
<br />
-Gmass can communicate, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will<br />
<br />
-Picture of Gmass http://i.imgur.com/PP9xi.png<br />
<br />
-Known by Mike Miller<br />
<br />
-Gmaas got mad at sseraj once, so he locked him in his own freezer<br />
<br />
-Gmaas is an obviously omnipotent cat.<br />
<br />
-ehawk11 met him<br />
<br />
-sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br />
<br />
-sseraj has posted pictures of gmass in intro to algebra, before class started, with the title, "caption contest" anyone who posted a caption mysteriously vanished in the middle of the night. after conferring with nearby cats.<br />
<br />
- has once slept on your bed and made it wet<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- [s]Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".[/s]Sorry, this was a rick roll troll.<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath and themoocow are also writing a book about Gmaas<br />
<br />
-Ornx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br />
<br />
- gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br />
<br />
-oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br />
<br />
-Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br />
<br />
-No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br />
<br />
-In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br />
https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br />
https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br />
https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br />
https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br />
<br />
<br />
- Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br />
<br />
-<math>Another</math> sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br />
<br />
-Gmaas has been sighted several times on the Global Announcements forum</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=1972_IMO_Problems/Problem_1&diff=882411972 IMO Problems/Problem 12017-11-13T05:27:57Z<p>Welp...: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum. <br />
<br />
<br />
==Solution==<br />
<br />
Note that there are <math>2^{10}-2=1022</math> distinct subsets of our set of 10 two-digit numbers. Also note that the sum of the elements of any subset of our set of 10 two-digit numbers must be between 10 and <math>91+92+93+94+95+96+97+98+99</math>, which is less than <math>100+100+100+100+100+100+100+100+100=1000 < 1022</math>. There are even less attainable sums. The [[Pigeonhole Principle]] then implies that there are two distinct subsets whose members have the same sum. Let these sets be <math>A</math> and <math>B</math>. Note that <math>A- (A\cap B)</math> and <math>B- (A\cap B)</math> are two distinct sets whose members have the same sum. These two sets are subsets of our set of 10 distinct two-digit numbers, so this proves the claim. <math>\square</math><br />
<br />
==Solution (Cheap Way)==<br />
<br />
By definition an empty set is disjoint to any other set. Therefore, our subsets will be empty set and empty set.<br />
<br />
Solution by: <i>mathwiz0803</i><br />
<br />
{{alternate solutions}}<br />
<br />
==See Also==<br />
<br />
{{IMO box|year=1972|before=First Problem|num-a=2}}<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366655&sid=dfbd2a1b3c3bf405ed1933757f19bbec#p366655 Discussion on AoPS/MathLinks]<br />
<br />
[[Category:Olympiad Combinatorics Problems]]</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_21&diff=877472010 AMC 12B Problems/Problem 212017-10-09T23:09:10Z<p>Welp...: /* Solution 2 */</p>
<hr />
<div>== Problem 21 ==<br />
Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that<br />
<br />
<center><br />
<math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/><br />
<math>P(2) = P(4) = P(6) = P(8) = -a</math>.<br />
</center><br />
<br />
What is the smallest possible value of <math>a</math>?<br />
<br />
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math><br />
<br />
== Solution 1 ==<br />
There must be some polynomial <math>Q(x)</math> such that <math>P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)</math><br />
<br />
Then, plugging in values of <math>2,4,6,8,</math> we get <br />
<br />
<cmath>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</cmath><br />
<cmath>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a</cmath><br />
<cmath>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</cmath><br />
<cmath>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</cmath><br />
<cmath>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</cmath><br />
Thus, <math>a</math> must be a multiple of <math>\text{lcm}(15,9,15,105)=315</math>.<br />
<br />
Now we show that there exists <math>Q(x)</math> such that <math>a=315.</math> We have<br />
<cmath>Q(2)=42, Q(4)=-70, Q(6)=42, Q(8)=-6</cmath><br />
Thus, <math>Q(x)=R(x)(x-2)(x-6)+42</math> for some <math>R(x).</math> From here it is clear that <math>Q(x)</math> exists, since we can take <math>R(x)=-8x+60.</math><br />
<br />
Therefore, our answer is <math> \boxed{\textbf{(B)}\ 315.} </math><br />
<br />
== Solution 2 (Calculus)==<br />
The evenly-spaced data suggests using [[discrete derivative|discrete derivatives]] to tackle this problem. First, note that any polynomial of degree <math>n</math><br />
<br />
<center><math>P(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n</math></center><br />
<br />
can also be written as<br />
<br />
<center><math>P(x) = b_0 + b_1 (x-1) + b_2 (x-1)(x-2) + \ldots + b_n (x-1)(x-1) \cdots (x-n)</math>.</center><br />
<br />
Moreover, the coefficients <math>a_i</math> are integers for <math>i=0, 1, 2, \ldots n</math> iff the coefficients <math>b_i</math> are integers for <math>i=0, 1, 2, \ldots n</math>. This latter form is convenient for calculating discrete derivatives of <math>P(x)</math>.<br />
<br />
The discrete derivative of a function <math>f(x)</math> is the related function <math>\Delta f(x)</math> defined as<br />
<br />
<center><math>\Delta f(x) = f(x+1) - f(x)</math>.</center><br />
<br />
With this definition, it's easy to see that for any positive integer <math>k</math> we have<br />
<br />
<center><math>\Delta [(x-1)(x-2)\cdots(x-k)] = k(x-1)(x-2)\cdots(x-[k-1])</math>.</center><br />
<br />
This in turn allows us to use successive discrete derivatives evaluated at <math>x=1</math> to calculate all of the coefficients <math>b_i</math> using<br />
<br />
<center><math>P(1)=b_0</math>, <math>\Delta P(1) = b_1</math>, <math>\Delta^2 P(1) = 2 b_2</math>, <math>\ldots</math>, <math>\Delta^7 P(1) = 7! b_7</math>.</center><br />
<br />
We can also calculate the following table of discrete derivatives based on the data points given in the problem statement:<br />
<br />
<center><br />
<table frame='box' rules='all' cellpadding='3'><br />
<tr><th /><th colspan='8'><math>x</math></th></tr><br />
<br />
<tr><td align='right'></td><td align='center'><math>1</math></td><td align='center'><math>2</math></td><td align='center'><math>3</math></td><td align='center'><math>4</math></td><td align='center'><math>5</math></td><td align='center'><math>6</math></td><td align='center'><math>7</math></td><td align='center'><math>8</math></td></tr><br />
<br />
<tr><td align='right'><math>P(x)</math></td><td align='right'><math>a</math></td><td align='right'><math>-a</math></td><td align='right'><math>a</math></td><td align='right'><math>-a</math></td><td align='right'><math>a</math></td><td align='right'><math>-a</math></td><td align='right'><math>a</math></td><td align='right'><math>-a</math></td></tr><br />
<br />
<tr><td align='right'><math>\Delta P(x)</math></td><td align='right'><math>-2a</math></td><td align='right'><math>2a</math></td><td align='right'><math>-2a</math></td><td align='right'><math>2a</math></td><td align='right'><math>-2a</math></td><td align='right'><math>2a</math></td><td align='right'><math>-2a</math></td><td /></tr><br />
<br />
<tr><td align='right'><math>\Delta^2 P(x)</math></td><td align='right'><math>4a</math></td><td align='right'><math>-4a</math></td><td align='right'><math>4a</math></td><td align='right'><math>-4a</math></td><td align='right'><math>4a</math></td><td align='right'><math>-4a</math></td><td /><td /></tr><br />
<br />
<tr><td colspan='9' align='center'><math>\vdots</math></td></tr><br />
<br />
<tr><td align='right'><math>\Delta^7 P(x)</math></td><td align='right'><math>-2^7 a</math></td><td /><td /><td /><td /><td /><td /><td /></tr><br />
</table><br />
</center><br />
<br />
Thus we can read down the column for <math>x=1</math> to find that <math>k! b_k = (-2)^k a</math> for <math>k = 0, 1, \ldots, 7</math>. Interestingly, even if we choose <math>P(x)</math> to have degree greater than <math>7</math>, the <math>8</math> coefficients of lowest order always satisfy these conditions. Moreover, it's straightforward to show that the <math>P(x)</math> of degree <math>7</math> with <math>b_k</math> satisfying these conditions will fit the data given in the problem statement. Thus, to find minimal necessary and sufficient conditions on the value of <math>a</math>, we need only consider these <math>8</math> equations. As a result, <math>P(x)</math> with integer coefficients fitting the given data exists iff <math>k!</math> divides <math>2^k a</math> for <math>k = 0, 1, \ldots, 7</math>. In other words, it's necessary and sufficient that<br />
<br />
<center><br />
<math>0! | a</math>,<br />
<br />
<math>1! | 2a</math>,<br />
<br />
<math>2! | 2^2 a</math>,<br />
<br />
<math>3! | 2^3 a</math>,<br />
<br />
<math>4! | 2^4 a</math>,<br />
<br />
<math>5! | 2^5 a</math>,<br />
<br />
<math>6! | 2^6 a</math>, and<br />
<br />
<math>7! | 2^7 a</math>.<br />
</center><br />
<br />
The last condition holds iff <math>7 \cdot 3 \cdot 5 \cdot 3 = 315</math> divides evenly into <math>a</math>. Since such <math>a</math> will also satisfy the first <math>7</math> conditions, our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=20|num-a=22|ab=B}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_17&diff=877362010 AMC 12B Problems/Problem 172017-10-09T22:49:23Z<p>Welp...: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
The entries in a <math>3 \times 3</math> array include all the digits from <math>1</math> through <math>9</math>, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60</math><br />
<br />
== Solution 1==<br />
The first 4 numbers will form one of 3 tetris "shapes".<br />
<br />
First, let's look at the numbers that form a <math>2\times2</math> block, sometimes called tetris <math> O</math>:<br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & \\<br />
\hline 3 & 4 & \\<br />
\hline & & \\<br />
\hline \end{tabular}</math><br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 1 & 3 & \\<br />
\hline 2 & 4 & \\<br />
\hline & & \\<br />
\hline \end{tabular}</math><br />
<br />
Second, let's look at the numbers that form a vertical "L", sometimes called tetris <math> J</math>:<br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 1 & 4 & \\<br />
\hline 2 & & \\<br />
\hline 3 & & \\<br />
\hline \end{tabular}</math><br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 1 & 3 & \\<br />
\hline 2 & & \\<br />
\hline 4 & & \\<br />
\hline \end{tabular}</math><br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & \\<br />
\hline 3 & & \\<br />
\hline 4 & & \\<br />
\hline \end{tabular}</math><br />
<br />
Third, let's look at the numbers that form a horizontal "L", sometimes called tetris <math> L</math>:<br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & 3 \\<br />
\hline 4 & & \\<br />
\hline & & \\<br />
\hline \end{tabular}</math><br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & 4 \\<br />
\hline 3 & & \\<br />
\hline & & \\<br />
\hline \end{tabular}</math><br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 1 & 3 & 4 \\<br />
\hline 2 & & \\<br />
\hline & & \\<br />
\hline \end{tabular}</math><br />
<br />
Now, the numbers 6-9 will form similar shapes (rotated by 180 degrees, and anchored in the lower-right corner of the 3x3 grid).<br />
<br />
If you match up one tetris shape from the numbers 1-4 and one tetris shape from the numbers 6-9, there is only one place left for the number 5 to be placed.<br />
<br />
So what shapes will physically fit in the 3x3 grid, together?<br />
<br />
<math> \begin{array}{ccl} 1 - 4 \text{ shape} & 6 - 9 \text{ shape} & \text{number of pairings} \\<br />
O & J & 2\times 3 = 6 \\<br />
O & L & 2\times 3 = 6 \\<br />
J & O & 3\times 2 = 6 \\<br />
J & J & 3 \times 3 = 9 \\<br />
L & O & 3 \times 2 = 6 \\<br />
L & L & 3 \times 3 = 9 \\<br />
O & O & \qquad \text{They don't fit} \\<br />
J & L & \qquad \text{They don't fit} \\<br />
L & J & \qquad \text{They don't fit} \\<br />
\end{array}</math><br />
<br />
The answer is <math> 4\times 6 + 2\times 9 = \boxed{\text{(D) }42}</math>.<br />
<br />
== Solution 2==<br />
This solution is trivial by the hook length theorem. The hooks look like this:<br />
<br />
<math> \begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\<br />
\hline 4 & 3 & 2\\<br />
\hline 3 & 2 & 1\\<br />
\hline \end{tabular}</math><br />
<br />
So, the answer is <math>\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}</math> = <math>\boxed{\text{(D) }42}</math><br />
<br />
P.S. The hook length formula is a formula to calculate the number of standard Young tableaux of a Young diagram. Numberphile has an easy-to-understand video about it here: https://www.youtube.com/watch?v=vgZhrEs4tuk The full proof is quite complicated and is not given in the video, although the video hints at possible proofs.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=16|num-a=18|ab=B}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_16&diff=877352010 AMC 12B Problems/Problem 162017-10-09T22:44:38Z<p>Welp...: /* Solution */</p>
<hr />
<div>== Problem 16 ==<br />
Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}</math><br />
<br />
== Solution ==<br />
<br />
We group this into groups of <math>3</math>, because <math>3|2010</math>. <br />
<br />
If <math>3|a</math>, we are done. There is a probability of <math>\frac{1}{3}</math> that that happens.<br />
<br />
Otherwise, we have <math>3|bc+b+1</math>, which means that <math>b(c+1) \equiv 2\pmod{3}</math>. So either <cmath>b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}</cmath> or <cmath>b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3</cmath> which will lead to the property being true. There are a <math>\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}</math> chance for each bundle of cases to be true. Thus, the total for the cases is <math>\frac{2}{9}</math>. But we have to multiply by <math>\frac{2}{3}</math> because this only happens with a <math>\frac{2}{3}</math> chance. So the total is actually <math>\frac{4}{27}</math>.<br />
<br />
The grand total is <cmath>\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}</cmath><br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_15&diff=877342010 AMC 12B Problems/Problem 152017-10-09T22:28:37Z<p>Welp...: /* Solution */</p>
<hr />
<div>== Problem 15 ==<br />
For how many ordered triples <math>(x,y,z)</math> of nonnegative integers less than <math>20</math> are there exactly two distinct elements in the set <math>\{i^x, (1+i)^y, z\}</math>, where <math>i=\sqrt{-1}</math>?<br />
<br />
<math>\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235</math><br />
<br />
== Solution ==<br />
We have either <math>i^{x}=(1+i)^{y}\neq z</math>, <math>i^{x}=z\neq(1+i)^{y}</math>, or <math>(1+i)^{y}=z\neq i^x</math>.<br />
<br />
For <math>i^{x}=(1+i)^{y}</math>, this only occurs at <math>1</math>. <math>(1+i)^{y}=1</math> has only one solution, namely, <math>y=0</math>. <math>i^{x}=1</math> has five solutions between zero and nineteen, <math>x=0, x=4, x=8, x=12</math>, and <math>x=16</math>. <math>z\neq 1</math> has nineteen integer solutions between zero and nineteen. So for <math>i^{x}=(1+i)^{y}\neq z</math>, we have <math>5\cdot 1\cdot 19=95</math> ordered triples.<br />
<br />
For <math>i^{x}=z\neq(1+i)^{y}</math>, again this only occurs at <math>1</math>. <math>(1+i)^{y}\neq 1</math> has nineteen solutions, <math>i^{x}=1</math> has five solutions, and <math>z=1</math> has one solution, so again we have <math>5\cdot 1\cdot 19=95</math> ordered triples.<br />
<br />
For <math>(1+i)^{y}=z\neq i^x</math>, this occurs at <math>1</math> and <math>16</math>. <math>(1+i)^{y}=1</math> and <math>z=1</math> both have one solution while <math>i^{x}\neq 1</math> has fifteen solutions. <math>(1+i)^{y}=16</math> and <math>z=16</math> both have one solution, namely, <math>y=8</math> and <math>z=16</math>, while <math>i^{x}\neq 16</math> has twenty solutions (<math>i^x</math> only cycles as <math>1, i, -1, -i</math>). So we have <math>15\cdot 1\cdot 1+20\cdot 1\cdot 1=35</math> ordered triples.<br />
<br />
In total we have <math>{95+95+35=\boxed{\text{(D) }225}}</math> ordered triples<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=14|num-a=16|ab=B}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=877332010 AMC 12B Problems/Problem 182017-10-09T22:20:22Z<p>Welp...: /* Solution 2 (Geometric) */</p>
<hr />
<div>== Problem ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
== Solution ==<br />
===Solution 1 (Complex Numbers)===<br />
We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set<br />
<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath><br />
has magnitude less than or equal to <math> 1</math>. Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
===Solution 2 (Geometric)===<br />
<br />
The first frog hop doesn't matter because no matter where the frog hops is lands on the border of the circle you want it to end in. The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps.<br />
<br />
<br />
No matter where we start, we will have the small circle tangent to a point on the big circle. This is just like how <math>\frac{1}{2}x^2</math> and <math>x^2</math> are tangent. The area ratio of the two circles is <cmath>\frac{\pi}{4\pi} = \boxed{\frac{1}{4} \text{(C)}}</cmath>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=877322010 AMC 12B Problems/Problem 182017-10-09T22:19:30Z<p>Welp...: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
== Solution ==<br />
===Solution 1 (Complex Numbers)===<br />
We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set<br />
<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath><br />
has magnitude less than or equal to <math> 1</math>. Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
===Solution 2 (Geometric)===<br />
<br />
The first frog hop doesn't matter because no matter where the frog hops is lands on the border of the circle you want it to end in. The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps.<br />
<br />
<br />
No matter where we start, we will have the small circle tangent to a point on the big circle. This is just like how <math>\frac{1}[2}x^2</math> and <math>x^2</math> are tangent. The area ratio of the two circles is <math></math>\frac{\pi}{4\pi} = \boxed{\frac{1}{4} \text{(C)}$.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_18&diff=877312010 AMC 12B Problems/Problem 182017-10-09T22:09:12Z<p>Welp...: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
== Solution ==<br />
===Solution 1 (Complex Numbers)===<br />
We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set<br />
<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath><br />
has magnitude less than or equal to <math> 1</math>. Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
===Solution 2===<br />
<br />
The first frog hop doesn't matter because no matter where the frog hops is lands on the border of the circle you want it to end in. The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps. The entirety of the circle you want the frog to land in is enclosed in this larger disk, so find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_22&diff=874502017 AMC 12B Problems/Problem 222017-09-10T19:15:05Z<p>Welp...: /* Solution */</p>
<hr />
<div>==Problem 22==<br />
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br />
<br />
<math>\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}</math><br />
<br />
==Solution==<br />
It amounts to filling in a <math>4 \times 4</math> matrix. Columns <math>C_1 - C_4</math> are the random draws each round; rows <math>R_A - R_D</math> are the coin changes of each player. Also, let <math>\%R_A</math> be the number of nonzero elements in <math>R_A</math>.<br />
<br />
WLOG, let <math>C_1 = \begin{pmatrix} 1\\-1\\0\\0\end{pmatrix}</math>. Parity demands that <math>\%R_A</math> and <math>\%R_B</math> must equal <math>2</math> or <math>4</math>. <br />
<br />
Case 1: <math>\%R_A = 4</math> and <math>\%R_B = 4</math>. There are <math>\binom{3}{2}=3</math> ways to place <math>2</math> <math>-1</math>'s in <math>R_A</math>, so there are <math>3</math> ways.<br />
<br />
Case 2: <math>\%R_A = 2</math> and <math>\%R_B=4</math>. There are <math>3</math> ways to place the <math>-1</math> in <math>R_A</math>, <math>2</math> ways to place the remaining <math>-1</math> in <math>R_B</math> (just don't put it under the <math>-1</math> on top of it!), and <math>2</math> ways for one of the other two players to draw the green ball. (We know it's green because Bernardo drew the red one.) We can just double to cover the case of <math>\%R_A = 4</math>, <math>\%R_B = 2</math> for a total of <math>24</math> ways.<br />
<br />
Case 3: <math>\%R_A=\%R_B=2</math>. There are three ways to place the <math>-1</math> in <math>R_A</math>. Now, there are two cases as to what happens next.<br />
<br />
Sub-case 3.1: The <math>1</math> in <math>R_B</math> goes directly under the <math>-1</math> in <math>R_A</math>. There's obviously <math>1</math> way for that to happen. Then, there are <math>2</math> ways to permute the two pairs of <math>1, -1</math> in <math>R_C</math> and <math>R_D</math>. (Either the <math>1</math> comes first in <math>R_C</math> or the <math>1</math> comes first in <math>R_D</math>.)<br />
<br />
Sub-case 3.2: The <math>1</math> in <math>R_B</math> doesn't go directly under the <math>-1</math> in <math>R_A</math>. There are <math>2</math> ways to place the <math>1</math>, and <math>2</math> ways to do the same permutation as in Sub-case 3.1. <br />
Hence, there are <math>3(2+2 \cdot 2)=18</math> ways for this case.<br />
<br />
There's a grand total of <math>45</math> ways for this to happen, along with <math>12^3</math> total cases. The probability we're asking for is thus <math>\frac{45}{(12^3)}=\frac{5}{192}.</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_22&diff=874492017 AMC 12B Problems/Problem 222017-09-10T19:00:38Z<p>Welp...: /* Solution */</p>
<hr />
<div>==Problem 22==<br />
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br />
<br />
<math>\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}</math><br />
<br />
==Solution==<br />
It amounts to filling in a <math>4 \times 4</math> matrix. Columns <math>C_1 - C_4</math> are the random draws each round; rows <math>R_A - R_D</math> are the coin changes of each player. Also, let <math>\%R_A</math> be the number of nonzero elements in <math>R_A</math>.<br />
<br />
WLOG, let <math>C_1 = \begin{pmatrix} 1\\-1\\0\\0\end{pmatrix}</math>. Parity demands that <math>\%R_A</math> and <math>\%R_B</math> must equal <math>2</math> or <math>4</math>. <br />
<br />
Case 1: <math>\%R_A = 4</math> and <math>\%R_B = 4</math>. There are <math>\binom{3}{2}=3</math> ways to place <math>2</math> <math>-1</math>'s in <math>R_A</math>, so there are <math>3</math> ways.<br />
<br />
Case 2: <math>\%R_A = 2</math> and <math>\%R_B=4</math>. There are <math>3</math> ways to place the <math>-1</math> in <math>R_A</math>, <math>2</math> ways to place the remaining <math>-1</math> in <math>R_B</math> (just don't put it under the <math>-1</math> on top of it!), and <math>2</math> ways for one of the other two players to draw the green ball. (We know it's green because Bernardo drew the red one.) We can just double to cover the case of <math>\%R_A = 4</math>, <math>\%R_B = 2</math> for a total of <math>24</math> ways.<br />
<br />
Case 3: <math>\%R_A=\%R_B=2</math>. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.<br />
<br />
In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_24&diff=872892017 AMC 10A Problems/Problem 242017-08-31T03:51:23Z<p>Welp...: /* Solution 5 */</p>
<hr />
<div>==Problem==<br />
For certain real numbers <math>a</math>, <math>b</math>, and <math>c</math>, the polynomial <cmath>g(x) = x^3 + ax^2 + x + 10</cmath>has three distinct roots, and each root of <math>g(x)</math> is also a root of the polynomial <cmath>f(x) = x^4 + x^3 + bx^2 + 100x + c.</cmath>What is <math>f(1)</math>?<br />
<br />
<math>\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005</math><br />
<br />
==Solution 1==<br />
<math>f(x)</math> must have four roots, three of which are roots of <math>g(x)</math>. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of <math>f(x)</math> and <math>g(x)</math> are the same, we know that<br />
<br />
<cmath>f(x)=g(x)(x-r)</cmath><br />
<br />
where <math>r\in\mathbb{C}</math> is the fourth root of <math>f(x)</math>. Substituting <math>g(x)</math> and expanding, we find that<br />
<br />
<cmath>\begin{align*}f(x)&=(x^3+ax^2+x+10)(x-r)\\<br />
&=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.\end{align*}</cmath><br />
<br />
Comparing coefficients with <math>f(x)</math>, we see that<br />
<br />
<cmath>\begin{align*}<br />
a-r&=1\\<br />
1-ar&=b\\<br />
10-r&=100\\<br />
-10r&=c.\\<br />
\end{align*}</cmath><br />
<br />
(Solution 1.1 picks up here.)<br />
<br />
Let's solve for <math>a,b,c,</math> and <math>r</math>. Since <math>10-r=100</math>, <math>r=-90</math>, so <math>c=(-10)(-90)=900</math>. Since <math>a-r=1</math>, <math>a=-89</math>, and <math>b=1-ar=-8009</math>. Thus, we know that<br />
<br />
<cmath>f(x)=x^4+x^3-8009x^2+100x+900.</cmath><br />
<br />
Taking <math>f(1)</math>, we find that<br />
<br />
<cmath>\begin{align*}<br />
f(1)&=1^4+1^3-8009(1)^2+100(1)+900\\<br />
&=1+1-8009+100+900\\<br />
&=\boxed{\bold{(C)}\, -7007}.\\<br />
\end{align*}</cmath><br />
<br />
==Solution 1.1==<br />
A faster ending to Solution 1 is as follows.<br />
We shall solve for only <math>a</math> and <math>r</math>. Since <math>10-r=100</math>, <math>r=-90</math>, and since <math>a-r=1</math>, <math>a=-89</math>. Then,<br />
<cmath>\begin{align*}<br />
f(1)&=(1+r)(x^3+ax^2+x+10)\\<br />
&=(91)(-77)\\<br />
&=\boxed{\bold{(C)}\, -7007}.\\<br />
\end{align*}</cmath><br />
Thanks for reading, Rowechen Zhong.<br />
==Solution 2==<br />
We notice that the constant term of <math>f(x)=c</math> and the constant term in <math>g(x)=10</math>. Because <math>f(x)</math> can be factored as <math>g(x) \cdot (x- r)</math> (where <math>r</math> is the unshared root of <math>f(x)</math>, we see that using the constant term, <math>-10 \cdot r = c</math> and therefore <math>r = -\frac{c}{10}</math>.<br />
Now we once again write <math>f(x)</math> out in factored form: <br />
<br />
<cmath>f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)(x+\frac{c}{10})</cmath>.<br />
<br />
We can expand the expression on the right-hand side to get:<br />
<br />
<cmath>f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c</cmath><br />
<br />
Now we have <math>f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c=x^4+x^3+bx^2+100x+c</math>.<br />
<br />
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations:<br />
<cmath>10+\frac{c}{10}=100 \Rightarrow c=900</cmath><br />
<cmath>a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89</cmath><br />
<br />
and finally,<br />
<br />
<cmath>1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009</cmath>.<br />
<br />
We know that <math>f(1)</math> is the sum of its coefficients, hence <math>1+1+b+100+c</math>. We substitute the values we obtained for <math>b</math> and <math>c</math> into this expression to get <math>f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{\textbf{(C)}\,-7007}</math>.<br />
<br />
==Solution 3==<br />
<br />
Let <math>r_1,r_2,</math> and <math>r_3</math> be the roots of <math>g(x)</math>. Let <math>r_4</math> be the additional root of <math>f(x)</math>. Then from Vieta's formulas on the quadratic term of <math>g(x)</math> and the cubic term of <math>f(x)</math>, we obtain the following:<br />
<br />
<cmath>\begin{align*}<br />
r_1+r_2+r_3&=-a \\ <br />
r_1+r_2+r_3+r_4&=-1<br />
\end{align*}</cmath><br />
<br />
Thus <math>r_4=a-1</math>.<br />
<br />
Now applying Vieta's formulas on the constant term of <math>g(x)</math>, the linear term of <math>g(x)</math>, and the linear term of <math>f(x)</math>, we obtain:<br />
<br />
<cmath>\begin{align*}<br />
r_1r_2r_3 & = -10\\<br />
r_1r_2+r_2r_3+r_3r_1 &= 1\\ <br />
r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 & = -100\\<br />
\end{align*}</cmath><br />
<br />
Substituting for <math>r_1r_2r_3</math> in the bottom equation and factoring the remainder of the expression, we obtain:<br />
<br />
<cmath>-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100</cmath><br />
<br />
It follows that <math>r_4=-90</math>. But <math>r_4=a-1</math> so <math>a=-89</math><br />
<br />
Now we can factor <math>f(x)</math> in terms of <math>g(x)</math> as<br />
<br />
<cmath>f(x)=(x-r_4)g(x)=(x+90)g(x)</cmath><br />
<br />
Then <math>f(1)=91g(1)</math> and<br />
<br />
<cmath>g(1)=1^3-89\cdot 1^2+1+10=-77</cmath><br />
<br />
Hence <math>f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}</math>.<br />
<br />
==Solution 4 (Slight guessing)==<br />
Let the roots of <math>g(x)</math> be <math>r_1</math>, <math>r_2</math>, and <math>r_3</math>. Let the roots of <math>f(x)</math> be <math>r_1</math>, <math>r_2</math>, <math>r_3</math>, and <math>r_4</math>. From Vieta's, we have:<br />
<cmath>\begin{align*}<br />
r_1+r_2+r_3=-a \\<br />
r_1+r_2+r_3+r_4=-1 \\<br />
r_4=a-1<br />
\end{align*}</cmath><br />
The fourth root is <math>a-1</math>. Since <math>r_1</math>, <math>r_2</math>, and <math>r_3</math> are common roots, we have:<br />
<cmath>\begin{align*}<br />
f(x)=g(x)(x-(a-1)) \\<br />
f(1)=g(1)(1-(a-1)) \\<br />
f(1)=(a+12)(2-a) \\<br />
f(1)=-(a+12)(a-2) \\<br />
\end{align*}</cmath><br />
Let <math>a-2=k</math>:<br />
<cmath>\begin{align*}<br />
f(1)=-k(k+14)<br />
\end{align*}</cmath><br />
Note that <math>-7007=-1001\cdot(7)=-(7\cdot(11)\cdot(13))\cdot(7)=-91\cdot(77)</math><br />
This gives us a pretty good guess of <math>\boxed{\textbf{(C)}\, -7007}</math>.<br />
<br />
==Solution 5==<br />
First off, let's get rid of the <math>x^4</math> term by finding <math>h(x)=f(x)-xg(x)</math>. This polynomial consists of the difference of two polynomials with <math>3</math> common factors, so it must also have these factors. The polynomial is <math>h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c</math>, and must be equal to <math>(1-a)g(x)</math>. Equating the coefficients, we get <math>3</math> equations. We will tackle the situation one equation at a time, starting the <math>x</math> terms. Looking at the coefficients, we get <math>\dfrac{90}{1-a} = 1</math>. <cmath>\therefore 90=1-a.</cmath> The solution to the previous is obviously <math>a=-89</math>. We can now find <math>b</math> and <math>c</math>. <math>\dfrac{b-1}{1-a} = a</math>, <cmath>\therefore b-1=a(1-a)=-89*90=-8010</cmath> and <math>b=-8009</math>. Finally <math>\dfrac{c}{1-a} = 10</math>, <cmath>\therefore c=10(1-a)=10*90=900</cmath> Solving the original problem, <math>f(1)=1 + 1 + b + 100 + 1 = 102+b+c=102+900-8009=\boxed{\textbf{(C)}\, -7007}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2017|ab=A|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_20&diff=871232009 AMC 12A Problems/Problem 202017-08-19T23:50:55Z<p>Welp...: /* Solution 3 (which won't work when justification is required) */</p>
<hr />
<div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #20]] and [[2009 AMC 10A Problems|2009 AMC 10A #23]]}}<br />
<br />
== Problem ==<br />
Convex quadrilateral <math>ABCD</math> has <math>AB = 9</math> and <math>CD = 12</math>. Diagonals <math>AC</math> and <math>BD</math> intersect at <math>E</math>, <math>AC = 14</math>, and <math>\triangle AED</math> and <math>\triangle BEC</math> have equal areas. What is <math>AE</math>?<br />
<br />
<math>\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6</math><br />
[[Category: Introductory Geometry Problems]]<br />
__TOC__<br />
<br />
== Solution 1 ==<br />
Let <math>[ABC]</math> denote the area of triangle <math>ABC</math>. <math>[AED] = [BEC]</math>, so <math>[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]</math>. Since triangles <math>ABD</math> and <math>ABC</math> share a base, they also have the same height and thus <math>\overline{AB}||\overline{CD}</math> and <math>\triangle{AEB}\sim\triangle{CED}</math> with a ratio of <math>3: 4</math>. <math>AE = \frac {3}{7}\times AC</math>, so <math>AE = \frac {3}{7}\times 14 = 6\ \boxed{\textbf{(E)}}</math>.<br />
<br />
<center><asy>pathpen = linewidth(0.7);pointpen = black;<br />
pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("12",(C+D)/2);<br />
fill(A--D--E--cycle,rgb(0.8,0.8,0.8));fill(B--C--E--cycle,rgb(0.8,0.8,0.8));D(A--B--C--D--cycle);D(A--C);D(B--D);D(E);<br />
</asy></center><br />
<br />
== Solution 2 ==<br />
Using the sine area formula on triangles <math>AED</math> and <math>BEC</math>, as <math>\angle AED = \angle BEC</math>, we see that <br />
<br />
<center><cmath>(AE)(ED) = (BE)(EC)\quad \Longrightarrow\quad \frac {AE}{EC} = \frac {BE}{ED}.</cmath></center><br />
<br />
Since <math>\angle AEB = \angle DEC</math>, triangles <math>AEB</math> and <math>DEC</math> are similar. Their ratio is <math>\frac {AB}{CD} = \frac {3}{4}</math>. Since <math>AE + EC = 14</math>, we must have <math>EC = 8</math>, so <math>AE = 6\ \textbf{(E)}</math>.<br />
<br />
==Solution 3 (which won't work when justification is required)==<br />
The easiest way for areas of triangles to be equal would be if they were congruent. A way for that to work would be if <math>ABCD</math> were simply an isosceles trapezoid! Since <math>AC = 14</math> and <math>AE:EC = 3:4</math> (look at the side lengths and you'll know why!), <math>\boxed{AE = 6}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2009|ab=A|num-b=19|num-a=21}}<br />
{{AMC10 box|year=2009|ab=A|num-b=22|num-a=24}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_20&diff=870302004 AMC 10A Problems/Problem 202017-08-14T22:16:57Z<p>Welp...: /* Solution 2 (Non-trig) */</p>
<hr />
<div>==Problem==<br />
Points <math>E</math> and <math>F</math> are located on square <math>ABCD</math> so that <math>\triangle BEF</math> is equilateral. What is the ratio of the area of <math>\triangle DEF</math> to that of <math>\triangle ABE</math>?<br />
<br />
<center>[[Image:AMC10_2004A_20.png]]</center><br />
<br />
<math> \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} </math><br />
<br />
==Solution==<br />
Since triangle <math>BEF</math> is equilateral, <math>EA=FC</math>, and <math>EAB</math> and <math>FCB</math> are <math>SAS</math> congruent. Thus, triangle <math>DEF</math> is an isosceles right triangle. So we let <math>DE=x</math>. Thus <math>EF=EB=FB=x\sqrt{2}</math>. If we go angle chasing, we find out that <math>\angle AEB=75^{\circ}</math>, thus <math>\angle ABE=15^{\circ}</math>. <math>\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}</math>. Thus <math>\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}</math>, or <math>AE=\frac{x(\sqrt{3}-1)}{2}</math>. Thus <math>AB=\frac{x(\sqrt{3}+1)}{2}</math>, and <math>[ABE]=\frac{x^2}{4}</math>, and <math>[DEF]=\frac{x^2}{2}</math>. Thus the ratio of the areas is <math>\boxed{\mathrm{(D)}\ 2}</math><br />
<br />
==Solution 2 (Non-trig) ==<br />
Without loss of generality, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math>. It suffices that <math>AE = 1 - x</math>. Then triangles <math>ABE</math> and <math>CBF</math> are congruent by HL, so <math>CF = AE</math> and <math>DE = DF</math>. We find that <math>BE = EF = x \sqrt{2}</math>, and so, by the Pythagorean Theorem, we have<br />
<math>(1 - x)^2 + 1 = 2x^2.</math> This yields <math>x^2 + 2x = 2</math>, so <math>x^2 = 2 - 2x</math>. Thus, the desired ratio of areas is<br />
<cmath>\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.</cmath><br />
<br />
==See also==<br />
*[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131332 AoPS topic]<br />
{{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Ratio Problems]]<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_20&diff=870292004 AMC 10A Problems/Problem 202017-08-14T22:16:28Z<p>Welp...: /* Solution 2 (Non-trig) */</p>
<hr />
<div>==Problem==<br />
Points <math>E</math> and <math>F</math> are located on square <math>ABCD</math> so that <math>\triangle BEF</math> is equilateral. What is the ratio of the area of <math>\triangle DEF</math> to that of <math>\triangle ABE</math>?<br />
<br />
<center>[[Image:AMC10_2004A_20.png]]</center><br />
<br />
<math> \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} </math><br />
<br />
==Solution==<br />
Since triangle <math>BEF</math> is equilateral, <math>EA=FC</math>, and <math>EAB</math> and <math>FCB</math> are <math>SAS</math> congruent. Thus, triangle <math>DEF</math> is an isosceles right triangle. So we let <math>DE=x</math>. Thus <math>EF=EB=FB=x\sqrt{2}</math>. If we go angle chasing, we find out that <math>\angle AEB=75^{\circ}</math>, thus <math>\angle ABE=15^{\circ}</math>. <math>\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}</math>. Thus <math>\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}</math>, or <math>AE=\frac{x(\sqrt{3}-1)}{2}</math>. Thus <math>AB=\frac{x(\sqrt{3}+1)}{2}</math>, and <math>[ABE]=\frac{x^2}{4}</math>, and <math>[DEF]=\frac{x^2}{2}</math>. Thus the ratio of the areas is <math>\boxed{\mathrm{(D)}\ 2}</math><br />
<br />
==Solution 2 (Non-trig) ==<br />
Without loss of generality, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math>. It suffices that <math>AE = 1 - x</math>. Then triangles <math>ABE</math> and <math>CBF</math> are congruent by HL, so <math>CF = AE</math> and <math>DE = DF</math>. We find that <math>BE = EF = x \sqrt{2}</math>, and so, by the Pythagorean Theorem, we have<br />
<math>(1 - x)^2 + 1 = 2x^2.</math> This yields <math>x^2 + 2x = 2</math>, so <math>x^2 = 2 - 2x</math>. Thus, the desired ratio of areas is<br />
<cmath>\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{D) }2}.</cmath><br />
<br />
==See also==<br />
*[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131332 AoPS topic]<br />
{{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Ratio Problems]]<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_2&diff=869432017 AIME I Problems/Problem 22017-08-08T04:32:19Z<p>Welp...: /* Solution */</p>
<hr />
<div>==Problem 2==<br />
When each of <math>702</math>, <math>787</math>, and <math>855</math> is divided by the positive integer <math>m</math>, the remainder is always the positive integer <math>r</math>. When each of <math>412</math>, <math>722</math>, and <math>815</math> is divided by the positive integer <math>n</math>, the remainder is always the positive integer <math>s \neq r</math>. Find <math>m+n+r+s</math>.<br />
<br />
==Solution==<br />
Let's work on both parts of the problem separately. First, <cmath>855 \equiv 787 \equiv 702 \equiv r \pmod{m}.</cmath> We take the difference of <math>855</math> and <math>787</math>, and also of <math>787</math> and <math>702</math>. We find that they are <math>85</math> and <math>68</math>, respectively. Since the greatest common divisor of the two differences is <math>17</math> (and the only one besides one), it's safe to assume that <math>m = 17</math>.<br />
<br />
Then, we divide <math>855</math> by <math>17</math>, and it's easy to see that <math>r = 5</math>. Dividing <math>787</math> and <math>702</math> by <math>17</math> also yields remainders of <math>5</math>, which means our work up to here is correct.<br />
<br />
Doing the same thing with <math>815</math>, <math>722</math>, and <math>412</math>, the differences between <math>815</math> and <math>722</math> and <math>722</math> and <math>412</math> are <math>310</math> and <math>93</math>, respectively. Since the only common divisor (besides <math>1</math>, of course) is <math>31</math>, <math>n = 31</math>. Dividing all <math>3</math> numbers by <math>31</math> yields a remainder of <math>9</math> for each, so <math>s = 9</math>. Thus, <math>m + n + r + s = 17 + 5 + 31 + 9 = \boxed{62}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2017|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_19&diff=835422016 AMC 10A Problems/Problem 192017-02-14T04:12:32Z<p>Welp...: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
<br />
In rectangle <math>ABCD,</math> <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ratio <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s,</math> and <math>t</math> is <math>1.</math> What is <math>r+s+t</math>?<br />
<br />
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math><br />
<br />
== Solution 1==<br />
<br />
<asy><br />
size(6cm);<br />
pair D=(0,0), C=(6,0), B=(6,3), A=(0,3);<br />
draw(A--B--C--D--cycle);<br />
draw(B--D);<br />
draw(A--(6,2));<br />
draw(A--(6,1));<br />
label("$A$", A, dir(135));<br />
label("$B$", B, dir(45));<br />
label("$C$", C, dir(-45));<br />
label("$D$", D, dir(-135));<br />
label("$Q$", extension(A,(6,1),B,D),dir(-90));<br />
label("$P$", extension(A,(6,2),B,D), dir(90));<br />
label("$F$", (6,1), dir(0));<br />
label("$E$", (6,2), dir(0));<br />
</asy><br />
<br />
Use similar triangles. Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. As <math>\triangle ADP</math> and <math>\triangle BEP</math> are similar, we see that <math>\frac{PD}{PB}=\frac{3}{1}</math>. Thus <math>PB=\frac{BD}{4}</math>. Therefore, <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math><br />
<br />
==Solution 2==<br />
<br />
Coordinate Bash:<br />
We can set coordinates for the points. <math>D=(0,0), C=(6,0), B=(6,3),</math> and <math>A=(0,3)</math>. The line <math>BD</math>'s equation is <math>y = \frac{1}{2}x</math>, line <math>AE</math>'s equation is <math>y = -\frac{1}{6}x + 3</math>, and line <math>AF</math>'s equation is <math>y = -\frac{1}{3}x + 3</math>. Adding the equations of lines <math>BD</math> and <math>AE</math>, we find that the coordinates of <math>P</math> is <math>(\frac{9}{2},\frac{9}{4})</math>. Furthermore we find that the coordinates <math>Q</math> is <math>(\frac{18}{5}, \frac{9}{5})</math>. Using the [[Pythagorean Theorem]], the length of <math>QD</math> is <math>\sqrt{(\frac{18}{5})^2+(\frac{9}{5})^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> = <math>\sqrt{(\frac{9}{2})^2+(\frac{9}{4})^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.</math> <math>PQ = DP - DQ = \frac{9\sqrt{5}}{5} - \frac{9\sqrt{5}}{4} = \frac{9\sqrt{5}}{20}.</math> The length of <math>DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}</math>. Then <math>BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.</math> The ratio <math>BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.</math> Then <math>r, s,</math> and <math>t</math> is <math>5, 3,</math> and <math>12</math>, respectively. The problem tells us to find <math>r + s + t</math>, so <math>5 + 3 + 12 = \boxed{\textbf{(E) }20}</math><br />
<br />
An alternate solution is to perform the same operations, but only solve for the x-coordinates. By similar triangles, the ratios will be the same.<br />
<br />
==Solution 3==<br />
<br />
Extend <math>AF</math> to meet <math>CD</math> at point <math>T</math>. Since <math>FC=1</math> and <math>BF=2</math>, <math>TC=3</math> by similar triangles <math>\triangle TFC</math> and <math>\triangle AFB</math>. It follows that <math>\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}</math>. Now, using similar triangles <math>\triangle BEP</math> and <math>\triangle DAP</math>, <math>\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}</math>. WLOG let <math>BP=1</math>. Solving for <math>PQ, QD</math> gives <math>PQ=\frac{3}{5}</math> and <math>QD=\frac{12}{5}</math>. So our desired ratio is <math>5:3:12</math> and <math>5+3+12=\boxed{\textbf{(E) } 20}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Welp...https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=80815AoPS Wiki:FAQ2016-10-28T01:16:36Z<p>Welp...: /* For my homework, there is suppose to be a green bar but it's orange, why? */</p>
<hr />
<div>{{shortcut|[[A:FAQ]]}}<br />
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This is a community created list of Frequently Asked Questions about Art of Problem Solving. If you have a request to edit or add a question on this page, please make it [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=416&t=414129 here].<br />
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[img]http://i.imgur.com/aBcDeFgH.jpg[/img]<br />
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== Contests ==<br />
==== Where can I find past contest questions and solutions? ====<br />
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==== How do I get problems onto the contest page? ====<br />
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==== What are the guidelines for posting problems to be added to the contests section? ====<br />
:Refer to the [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=144&t=195579 guidelines in this post].<br />
<br />
==== Why is the wiki missing many contest questions? ====<br />
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<br />
==== What if I find an error on a problem? ====<br />
Please post an accurate description of the problem in [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=426693 this thread]<br />
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== LaTeX and Asymptote ==<br />
==== What is LaTeX, and how do I use it? ====<br />
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==== What is the olympiad package? ====<br />
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==AoPS Acronyms==<br />
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Please see the [//artofproblemsolving.com/ftw/faq For the Win! FAQ] for many common questions.<br />
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<br />
==== For my homework, there is suppose to be a green bar but it's orange, why? ====<br />
<br />
:For the bar to turn green, the writing problem must be attempted. Once it is attempted the bar will turn green.<br />
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==== I need more time for my homework, what should I do? ====<br />
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:Send an email to extensions@aops.com with your username, class name and ID, if known, and reason for extension. Someone should get back to you within a couple days.</div>Welp...