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4,201
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_23
| 2
|
In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad \textbf{(B) } 77 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 79 \qquad \textbf{(E) } 80$
|
We will refer to the Diagram section. In this solution, all angle measures are in degrees.
We rotate $\triangle PUM$ by $180^\circ$ about $M$ to obtain $\triangle AU'M.$ Let $H$ be the intersection of $\overline{PA}$ and $\overline{GU'},$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(375); pair P, A, T, U, G, M, N, U1, H; P = origin; A = (10,0); U = intersectionpoint(Circle(P,1),P--P+2*dir(36)); G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56)); T = extension(P,U,A,G); M = midpoint(P--A); N = midpoint(U--G); U1 = rotate(180,M)*U; H = intersectionpoint(P--A,G--U1); fill(U--P--M--cycle^^M--U1--A--cycle,yellow); dot("$P$",P,1.5*SW,linewidth(4)); dot("$A$",A,1.5*SE,linewidth(4)); dot("$U$",U,1.5*(0,1),linewidth(4)); dot("$G$",G,1.5*NE,linewidth(4)); dot("$T$",T,1.5*(0,1),linewidth(4)); dot("$M$",M,1.5*S,linewidth(4)); dot("$N$",N,1.5*(0,1),linewidth(4)); dot("$U'$",U1,1.5*S,linewidth(4)); dot("$H$",H,1.5*NW,linewidth(4)); draw(P--A--T--cycle^^U--G^^M--N^^U--U1--A); draw(G--U1,dashed); label("$1$",midpoint(G--A),1.5*dir(30)); label("$1$",midpoint(A--U1),1.5*dir(-30)); label("$1$",midpoint(U--P),1.5*dir(150)); label("$36^\circ$",P,5*dir(18),fontsize(8)); label("$56^\circ$",A,2.5*dir(180-56/2),fontsize(8)); label("$36^\circ$",A,2.5*dir(180+25),fontsize(8)); Label L = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(P-(0,1.5)--A-(0,1.5), L=L, arrow=Arrows(),bar=Bars(15)); add(pathticks(U--N, 2, .5, 4, 8, red)); add(pathticks(N--G, 2, .5, 4, 8, red)); add(pathticks(U--M, 1, .5, 0, 8, red)); add(pathticks(M--U1, 1, .5, 0, 8, red)); [/asy] Note that $\triangle GU'A$ is an isosceles triangle with $GA=U'A=1,$ so $\angle AGU'=\angle AU'G=\frac{180-\angle GAU'}{2}=44.$ In $\triangle GHA,$ it follows that $\angle GHA=180-\angle GAH-\angle AGH=80.$
Since $\frac{UM}{UU'}=\frac{UN}{UG}=\frac12,$ we conclude that $\triangle UMN\sim\triangle UU'G$ by SAS, from which $\angle UMN=\angle UU'G$ and $\angle UNM=\angle UGU'.$ By the Converse of the Corresponding Angles Postulate, we deduce that $\overline{MN}\parallel\overline{U'G}.$
Finally, we have $\angle NMA=\angle GHA=\boxed{80}$ by the Corresponding Angles Postulate.
| 80
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4,202
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_23
| 3
|
In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad \textbf{(B) } 77 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 79 \qquad \textbf{(E) } 80$
|
Link $PN$ , extend $PN$ to $Q$ so that $QN=PN$ . Then link $QG$ and $QA$
$\because M,N$ are the midpoints of $PA$ and $PQ,$ respectively
$\therefore MN$ is the midsegment of $\bigtriangleup PAQ$
$\therefore \angle QAP=\angle NMP$
Notice that $\bigtriangleup PUN\cong \bigtriangleup QGN$
As a result, $QG=AG=UP=1$ $\angle AQG=\angle QAG$ $\angle GQN=\angle NPU$
Also, $\angle GQN+\angle QPA=\angle QPU+\angle QPA=\angle UPA=36^{\circ}$
As a result, $2\angle QAG=180^{\circ}-56^{\circ}-36^{\circ}=88^{\circ}$
Therefore, $\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\circ}$
Since we are asked for the acute angle between the two lines, the answer to this problem is $\boxed{80}$
| 80
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4,203
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_23
| 4
|
In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad \textbf{(B) } 77 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 79 \qquad \textbf{(E) } 80$
|
Let the mid-point of $\overline{AT}$ be $B$ and the mid-point of $\overline{GT}$ be $C$ .
Since $BC=CG-BG$ and $CG=AB-\frac{1}{2}$ , we can conclude that $BC=\frac{1}{2}$ .
Similarly, we can conclude that $BM-CN=\frac{1}{2}$ . Construct $\overline{ND}\parallel\overline{BC}$ and intersects $\overline{BM}$ at $D$ , which gives $MD=DN=\frac{1}{2}$ .
Since $\angle{ABD}=\angle{BDN}$ $MD=DN$ , we can find the value of $\angle{DMN}$ , which is equal to $\frac{1}{2}\angle T=44^{\circ}$ . Since $\overline{BM}\parallel\overline{PT}$ , which means $\angle{DMN}+\angle{MNP}+\angle{P}=180^{\circ}$ , we can infer that $\angle{MNP}=100^{\circ}$ .
As we are required to give the acute angle formed, the final answer would be $80^{\circ}$ , which is $\boxed{80}$
| 80
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4,204
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_23
| 5
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In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad \textbf{(B) } 77 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 79 \qquad \textbf{(E) } 80$
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Let the bisector of $\angle ATP$ intersect $PA$ at $X.$ We have $\angle ATX = \angle PTX = 44^{\circ},$ so $\angle TXA = 80^{\circ}.$ We claim that $MN$ is parallel to this angle bisector, meaning that the acute angle formed by $MN$ and $PA$ is $80^{\circ},$ meaning that the answer is $\boxed{80}$
| 80
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4,205
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_23
| 6
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In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad \textbf{(B) } 77 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 79 \qquad \textbf{(E) } 80$
|
Note that $X$ , the midpoint of major arc $PA$ on $(PAT)$ is the Miquel Point of $PUAG$ (Because $PU = AG$ ). Then, since $1 = \frac{UN}{NG} = \frac{PM}{MA}$ , this spiral similarity carries $M$ to $N$ . Thus, we have $\triangle XMN \sim \triangle XAG$ , so $\angle XMN = \angle XAG$
But, we have $\angle XAG = \angle PAG = \angle PAX = 56 - \frac{180 - \angle PXA}{2} =56 - \frac{180 - \angle T}{2} = 56 - \frac{\angle A + \angle P}{2} = 56 - \frac{56+36}{2} = 56 - 46 = 10$ ; thus $\angle XMN = 10$
Then, as $X$ is the midpoint of the major arc, it lies on the perpendicular bisector of $PA$ , so $\angle XMA = 90$ . Since we want the acute angle, we have $\angle NMA = \angle XMA - \angle XMN = 90 - 10 = 80$ , so the answer is $\boxed{80}$
| 80
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4,206
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_23
| 7
|
In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad \textbf{(B) } 77 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 79 \qquad \textbf{(E) } 80$
|
The argument of the average of any two unit vectors is average of the arguments of the two vectors. Thereby, the acute angle formed is \[\frac{36^\circ{} + 180^\circ{} - 56^\circ{}}{2} = \boxed{80}.\]
| 80
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4,207
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_25
| 1
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For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$
|
By geometric series, we have \begin{alignat*}{8} A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\ B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\right)&&=b\cdot\frac{10^n-1}{9}, \\ C_n&=c\bigl(\phantom{ }\underbrace{111\cdots1}_{2n\text{ digits}}\phantom{ }\bigr)&&=c\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}. \end{alignat*} By substitution, we rewrite the given equation $C_n - B_n = A_n^2$ as \[c\cdot\frac{10^{2n}-1}{9} - b\cdot\frac{10^n-1}{9} = a^2\cdot\left(\frac{10^n-1}{9}\right)^2.\] Since $n > 0,$ it follows that $10^n > 1.$ We divide both sides by $\frac{10^n-1}{9}$ and then rearrange: \begin{align*} c\left(10^n+1\right) - b &= a^2\cdot\frac{10^n-1}{9} \\ 9c\left(10^n+1\right) - 9b &= a^2\left(10^n-1\right) \\ \left(9c-a^2\right)10^n &= 9b-9c-a^2. &&(\bigstar) \end{align*} Let $y=10^n.$ Note that $(\bigstar)$ is a linear equation with $y,$ and $y$ is a one-to-one function of $n.$ Since $(\bigstar)$ has at least two solutions of $n,$ it has at least two solutions of $y.$ We conclude that $(\bigstar)$ must be an identity, so we get the following system of equations: \begin{align*} 9c-a^2&=0, \\ 9b-9c-a^2&=0. \end{align*} The first equation implies that $c=\frac{a^2}{9}.$ Substituting this into the second equation gives $b=\frac{2a^2}{9}.$
To maximize $a + b + c = a + \frac{a^2}{3},$ we need to maximize $a.$ Clearly, $a$ must be divisible by $3.$ The possibilities for $(a,b,c)$ are $(9,18,9),(6,8,4),$ or $(3,2,1),$ but $(9,18,9)$ is invalid. Therefore, the greatest possible value of $a + b + c$ is $6+8+4=\boxed{18}.$
| 18
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4,208
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_25
| 2
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For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$
|
Immediately start trying $n = 1$ and $n = 2$ . These give the system of equations $11c - b = a^2$ and $1111c - 11b = (11a)^2$ (which simplifies to $101c - b = 11a^2$ ). These imply that $a^2 = 9c$ , so the possible $(a, c)$ pairs are $(9, 9)$ $(6, 4)$ , and $(3, 1)$ . The first puts $b$ out of range but the second makes $b = 8$ . We now know the answer is at least $6 + 8 + 4 = 18$
We now only need to know whether $a + b + c = 20$ might work for any larger $n$ . We will always get equations like $100001c - b = 11111a^2$ where the $c$ coefficient is very close to being nine times the $a$ coefficient. Since the $b$ term will be quite insignificant, we know that once again $a^2$ must equal $9c$ , and thus $a = 9, c = 9$ is our only hope to reach $20$ . Substituting and dividing through by $9$ , we will have something like $100001 - \frac{b}{9} = 99999$ . No matter what $n$ really was, $b$ is out of range (and certainly isn't $2$ as we would have needed).
The answer then is $\boxed{18}$
| 18
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4,209
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_25
| 3
|
For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$
|
The given equation can be written as \[c \cdot (\phantom{ } \overbrace{1111 \ldots 1111}^{2n\text{ digits}}\phantom{ }) - b \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ })^2.\] Divide by $\overbrace{11 \ldots 11}^{n\text{ digits}}$ on both sides: \[c \cdot (\phantom{ } \overbrace{1000 \ldots 0001}^{n+1\text{ digits}}\phantom{ }) - b = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }).\] Next, split the first term to make it easier to deal with: \begin{align*} 2c + c \cdot (\phantom{ }\overbrace{99 \ldots 99}^{n\text{ digits}}\phantom{ }) - b &= a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) \\ 2c - b &= (a^2 - 9c) \cdot (\phantom{ }\overbrace{11 \ldots 11}^{n\text{ digits}}\phantom{ }). \end{align*} Because $2c - b$ and $a^2 - 9c$ are constants and because there must be at least two distinct values of $n$ that satisfy, $2c - b = a^2 - 9c = 0.$ Thus, we have \begin{align*} 2c&=b, \\ a^2&=9c. \end{align*} Knowing that $a,b,$ and $c$ are single digit positive integers and that $9c$ must be a perfect square, the values of $(a,b,c)$ that satisfy both equations are $(3,2,1)$ and $(6,8,4).$ Finally, $6 + 8 + 4 = \boxed{18}.$
| 18
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4,210
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_25
| 4
|
For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$
|
By PaperMath’s sum , the answer is $6+8+4=\boxed{18}$
| 18
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4,211
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_1
| 1
|
Kate bakes a $20$ -inch by $18$ -inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?
$\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360$
|
The area of the pan is $20\cdot18=360$ . Since the area of each piece is $2\cdot2=4$ , there are $\frac{360}{4} = \boxed{90}$ pieces.
| 90
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4,212
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_1
| 2
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Kate bakes a $20$ -inch by $18$ -inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?
$\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360$
|
By dividing each of the dimensions by $2$ , we get a $10\times9$ grid that makes $\boxed{90}$ pieces.
| 90
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4,213
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_3
| 1
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A line with slope $2$ intersects a line with slope $6$ at the point $(40,30)$ . What is the distance between the $x$ -intercepts of these two lines?
$\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50$
|
Using point slope form, we get the equations $y-30 = 6(x-40)$ and $y-30 = 2(x-40)$ . Simplifying, we get $6x-y=210$ and $2x-y=50$ . Letting $y=0$ in both equations and solving for $x$ gives the $x$ -intercepts: $x=35$ and $x=25$ , respectively. Thus the distance between them is $35-25=\boxed{10}$
| 10
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4,214
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_3
| 2
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A line with slope $2$ intersects a line with slope $6$ at the point $(40,30)$ . What is the distance between the $x$ -intercepts of these two lines?
$\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50$
|
In order for the line with slope $2$ to travel "up" $30$ units (from $y=0$ ), it must have traveled $30/2=15$ units to the right. Thus, the $x$ -intercept is at $x=40-15=25$ . As for the line with slope $6$ , in order for it to travel "up" $30$ units it must have traveled $30/6=5$ units to the right. Thus its $x$ -intercept is at $x=40-5=35$ . Then the distance between them is $35-25=\boxed{10}$
| 10
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4,215
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_4
| 1
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A circle has a chord of length $10$ , and the distance from the center of the circle to the chord is $5$ . What is the area of the circle?
$\textbf{(A) }25\pi \qquad \textbf{(B) }50\pi \qquad \textbf{(C) }75\pi \qquad \textbf{(D) }100\pi \qquad \textbf{(E) }125\pi \qquad$
|
Let $O$ be the center of the circle, $\overline{AB}$ be the chord, and $M$ be the midpoint of $\overline{AB},$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, M; O = (0,0); A = (-5,5); B = (5,5); M = midpoint(A--B); draw(Circle(O,5sqrt(2))); dot("$O$", O, 1.5*S, linewidth(4.5)); dot("$A$", A, 1.5*NW, linewidth(4.5)); dot("$B$", B, 1.5*NE, linewidth(4.5)); dot("$M$", M, 1.5*N, linewidth(4.5)); draw(A--B^^M--O^^A--O^^M--O^^B--O); label("$5$", midpoint(A--M), 1.5*N); label("$5$", midpoint(B--M), 1.5*N); label("$5$", midpoint(O--M), 1.5*E); label("$r$", midpoint(O--A), 1.5*SW); label("$r$", midpoint(O--B), 1.5*SE); [/asy] Note that $\overline{OM}\perp\overline{AB}.$ Since $OM=AM=BM=5,$ we conclude that $\triangle OMA$ and $\triangle OMB$ are congruent isosceles right triangles. It follows that $r=5\sqrt2,$ so the area of $\odot O$ is $\pi r^2=\boxed{50}$
| 50
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4,216
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_5
| 2
|
How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A) } 128 \qquad \textbf{(B) } 192 \qquad \textbf{(C) } 224 \qquad \textbf{(D) } 240 \qquad \textbf{(E) } 256$
|
We can construct our subset by choosing which primes are included and which composites are included. There are $2^4-1$ ways to select the primes (total subsets minus the empty set) and $2^4$ ways to select the composites. Thus, there are $15\cdot16=\boxed{240}$ ways to choose a subset of the eight numbers.
| 240
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4,217
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_6
| 1
|
Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?
$\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \frac{4Q}{DS} \qquad \textbf{(D) } \frac{DQ}{4S} \qquad \textbf{(E) } \frac{DS}{4Q}$
|
Each can of soda costs $\frac QS$ quarters, or $\frac{Q}{4S}$ dollars. Therefore, $D$ dollars can purchase $\frac{D}{\left(\tfrac{Q}{4S}\right)}=\boxed{4}$ cans of soda.
| 4
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4,218
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_6
| 2
|
Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?
$\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \frac{4Q}{DS} \qquad \textbf{(D) } \frac{DQ}{4S} \qquad \textbf{(E) } \frac{DS}{4Q}$
|
Note that $S$ is in the unit of $\text{can}.$ On the other hand, $Q$ and $D$ are both in the unit of $\text{cost}.$
The units of $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},$ and $\textbf{(E)}$ are $\frac{\text{cost}^2}{\text{can}},\text{can},\frac{1}{\text{can}},\frac{\text{cost}^2}{\text{can}},$ and $\text{can},$ respectively. Since the answer is in the unit of $\text{can},$ we eliminate $\textbf{(A)},\textbf{(C)},$ and $\textbf{(D)}.$ Moreover, it is clear that $D$ dollars can purchase more than $S=\frac{DS}{4Q}$ cans of soda, so we eliminate $\textbf{(E)}.$
Finally, the answer is $\boxed{4}.$
| 4
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4,219
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_7
| 1
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What is the value of \[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\] $\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$
|
From the Change of Base Formula, we have \[\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \cdot \log 27}{\log 3 \cdot \log 5} = \frac{(2\log 5)\cdot(3\log 3)}{\log 3 \cdot \log 5} = \boxed{6}.\]
| 6
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4,220
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_7
| 2
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What is the value of \[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\] $\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$
|
Using the chain rule of logarithms $\log _{a} b \cdot \log _{b} c = \log _{a} c,$ we get \begin{align*} \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27 &= (\log _{3} 7 \cdot \log _{7} 11 \cdots \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdots \log _{21} 25) \\ &= \log _{3} 27 \cdot \log _{5} 25 \\ &= 3 \cdot 2 \\ &= \boxed{6}
| 6
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4,221
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_8
| 1
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Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$ . Point $C$ , not equal to $A$ or $B$ , lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
$\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75$
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For each $\triangle ABC,$ note that the length of one median is $OC=12.$ Let $G$ be the centroid of $\triangle ABC.$ It follows that $OG=\frac13 OC=4.$
As shown below, $\triangle ABC_1$ and $\triangle ABC_2$ are two shapes of $\triangle ABC$ with centroids $G_1$ and $G_2,$ respectively: [asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, C1, C2, G1, G2, M1, M2; O = (0,0); A = (-12,0); B = (12,0); C1 = (36/5,48/5); C2 = (-96/17,-180/17); G1 = O + 1/3 * C1; G2 = O + 1/3 * C2; M1 = (4,0); M2 = (-4,0); draw(Circle(O,12)); draw(Circle(O,4),red); dot("$O$", O, (3/5,-4/5), linewidth(4.5)); dot("$A$", A, W, linewidth(4.5)); dot("$B$", B, E, linewidth(4.5)); dot("$C_1$", C1, dir(C1), linewidth(4.5)); dot("$C_2$", C2, dir(C2), linewidth(4.5)); dot("$G_1$", G1, 1.5*E, linewidth(4.5)); dot("$G_2$", G2, 1.5*W, linewidth(4.5)); draw(A--B^^A--C1--B^^A--C2--B); draw(O--C1^^O--C2); dot(M1,red+linewidth(0.8),UnFill); dot(M2,red+linewidth(0.8),UnFill); [/asy] Therefore, point $G$ traces out a circle (missing two points) with the center $O$ and the radius $\overline{OG},$ as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is $\pi\cdot OG^2=16\pi\approx\boxed{50}.$
| 50
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4,222
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_8
| 2
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Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$ . Point $C$ , not equal to $A$ or $B$ , lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
$\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75$
|
We assign coordinates. Let $A = (-12,0)$ $B = (12,0)$ , and $C = (x,y)$ lie on the circle $x^2 +y^2 = 12^2$ . Then, the centroid of $\triangle ABC$ is $G = \left(\frac{-12 + 12 + x}{3}, \frac{0 + 0 + y}{3}\right) = \left(\frac x3,\frac y3\right)$ . Thus, $G$ traces out a circle with a radius $\frac13$ of the radius of the circle that point $C$ travels on. Thus, $G$ traces out a circle of radius $\frac{12}{3} = 4$ , which has area $16\pi\approx \boxed{50}$
| 50
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4,223
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_8
| 3
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Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$ . Point $C$ , not equal to $A$ or $B$ , lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
$\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75$
|
First we can draw a few conclusions from the given information. Firstly we can see clearly that the distance from the centroid to the center of the circle will remain the same no matter $C$ is on the circle. Also we can see that because the two legs of every triangles will always originate on the diameter, using inscribed angle rules, we know that $\angle C = \frac{180^\circ}{2} = 90^\circ$ . Now we know that all triangles $ABC$ will be a right triangle. We also know that the closed curve will simply be a circle with radius equal to the centroid of each triangle. We can now pick any arbitrary triangle, calculate its centroid, and the plug it in to the area formula. Using a $45^\circ$ $45^\circ$ $90^\circ$ triangle in conjunction with the properties of a centroid, we can quickly see that the length of the centroid is $4$ now we can plug it in to the area formula where we get $16\pi\approx\boxed{50}$
| 50
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4,224
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9
| 1
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What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$
|
Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{j=1}i + \sum^{100}_{i=1} \sum^{100}_{j=1}j \\ &= \sum^{100}_{i=1} (100i) + 100 \sum^{100}_{j=1}j \\ &= 100 \sum^{100}_{i=1}i + 100 \sum^{100}_{j=1}j \\ &= 100\cdot5050 + 100\cdot5050 \\ &= \boxed{1010000} ~MRENTHUSIASM
| 0
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4,225
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9
| 2
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What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$
|
Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \left(\sum^{100}_{j=1} i + \sum^{100}_{j=1} j\right) \\ &= \sum^{100}_{i=1} (100i+5050) \\ &= 100\sum^{100}_{i=1} i + \sum^{100}_{i=1} 5050 \\ &= 100\cdot5050+5050\cdot100 \\ &= \boxed{1010000} ~Vfire ~MRENTHUSIASM
| 0
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4,226
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9
| 3
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What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$
|
Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ Since the nested summation is symmetric with respect to $i$ and $j,$ it follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{i=1} (2i) \\ &= 2\sum^{100}_{i=1} \sum^{100}_{i=1} i \\ &= 2\sum^{100}_{i=1} 5050 \\ &= 2\cdot(5050\cdot100) \\ &= \boxed{1010000} ~Vfire ~MRENTHUSIASM
| 0
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4,227
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9
| 4
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What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$
|
The sum contains $100\cdot100=10000$ terms, and the average value of both $i$ and $j$ is $\frac{101}{2}.$ Therefore, the sum becomes \[10000\cdot\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{1010000}.\] ~Rejas ~MRENTHUSIASM
| 0
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4,228
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9
| 5
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What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$
|
We start by writing out the first few terms: \[\begin{array}{ccccccccc} (1+1) &+ &(1+2) &+ &(1+3) &+ &\cdots &+ &(1+100) \\ (2+1) &+ &(2+2) &+ &(2+3) &+ &\cdots &+ &(2+100) \\ (3+1) &+ &(3+2) &+ &(3+3) &+ &\cdots &+ &(3+100) \\ [-1ex] &&&&\vdots&&&& \\ (100+1) &+ &(100+2) &+ &(100+3) &+ &\cdots &+ &(100+100). \end{array}\] From the first terms in the parentheses, the sum $1+2+3+\cdots+100$ occurs $100$ times vertically.
From the second terms in the parentheses, the sum $1+2+3+\cdots+100$ occurs $100$ times horizontally.
Recall that the sum of the first $100$ positive integers is $1+2+3+\cdots+100=\frac{101\cdot100}{2}=5050.$ Therefore, the answer is \[2\cdot\left(5050\cdot100\right)=\boxed{1010000}.\] ~RandomPieKevin ~MRENTHUSIASM
| 0
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4,229
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9
| 6
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What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$
|
When we expand the nested summation as shown in Solution 5, note that:
Together, the nested summation becomes \begin{align*} \sum^{100}_{k=1}\left[(k+1)k\right] + \sum^{99}_{k=1}\left[(k+101)(100-k)\right] &= \sum^{100}_{k=1}\left[k^2+k\right] + \sum^{99}_{k=1}\left[-k^2-k+10100\right] \\ &= \sum^{100}_{k=1}k^2 + \sum^{100}_{k=1}k - \sum^{99}_{k=1}k^2 - \sum^{99}_{k=1}k + \sum^{99}_{k=1}10100 \\ &= \left(\sum^{100}_{k=1}k^2 - \sum^{99}_{k=1}k^2\right) + \left(\sum^{100}_{k=1}k - \sum^{99}_{k=1}k\right) + \sum^{99}_{k=1}10100 \\ &= 100^2+100+10100\cdot99 \\ &= \boxed{1010000} ~MRENTHUSIASM
| 0
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4,230
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_10
| 1
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A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$
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To minimize the number of distinct values, we want to maximize the number of times a number appears. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 = \boxed{225}.$
| 225
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4,231
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_10
| 2
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A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$
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As in Solution 1, we want to maximize the number of time each number appears to do so. We can set up an equation $10 + 9( x - 1 )\geq2018,$ where $x$ is the number of values. Notice how we can then rearrange the equation into $1 + 9 ( 1 )+9 ( x - 1 )\geq2018,$ which becomes $9 x\geq2017,$ or $x\geq224\frac19.$ We cannot have a fraction of a value so we must round up to $\boxed{225}.$
| 225
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4,232
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_12
| 1
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Side $\overline{AB}$ of $\triangle ABC$ has length $10$ . The bisector of angle $A$ meets $\overline{BC}$ at $D$ , and $CD = 3$ . The set of all possible values of $AC$ is an open interval $(m,n)$ . What is $m+n$
$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }20 \qquad$
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Let $AC=x.$ By Angle Bisector Theorem, we have $\frac{AB}{AC}=\frac{BD}{CD},$ from which $BD=CD\cdot\frac{AB}{AC}=\frac{30}{x}.$
Recall that $x>0.$ We apply the Triangle Inequality to $\triangle ABC:$
Taking the intersection of the solutions gives \[(m,n)=(0,\infty)\cap(0,15)\cap(3,\infty)=(3,15),\] so the answer is $m+n=\boxed{18}.$
| 18
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4,233
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_13
| 1
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Square $ABCD$ has side length $30$ . Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$ . The centroids of $\triangle{ABP}$ $\triangle{BCP}$ $\triangle{CDP}$ , and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
[asy] unitsize(120); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); draw(A--B--C--D--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N*1.5+E*0.5); dot(A); dot(B); dot(C); dot(D); [/asy]
$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$
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As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\overline{AB},\overline{BC},\overline{CD},\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},$ respectively. [asy] /* Made by MRENTHUSIASM */ unitsize(210); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); pair M1 = midpoint(A--B); pair M2 = midpoint(B--C); pair M3 = midpoint(C--D); pair M4 = midpoint(D--A); pair G1 = centroid(A,B,P); pair G2 = centroid(B,C,P); pair G3 = centroid(C,D,P); pair G4 = centroid(D,A,P); filldraw(M1--M2--P--cycle,red); filldraw(M2--M3--P--cycle,yellow); filldraw(M3--M4--P--cycle,green); filldraw(M4--M1--P--cycle,lightblue); draw(A--B--C--D--cycle); draw(M1--M2--M3--M4--cycle); draw(G1--G2--G3--G4--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N); label("$M_1$", M1, W); label("$M_2$", M2, S); label("$M_3$", M3, E); label("$M_4$", M4, N); label("$G_1$", G1, 1.5S); label("$G_2$", G2, 1.5E); label("$G_3$", G3, 1.5NE); label("$G_4$", G4, 1.5E); dot(A); dot(B); dot(C); dot(D); dot(M1); dot(M2); dot(M3); dot(M4); dot(G1); dot(G2); dot(G3); dot(G4); [/asy] By SAS, we conclude that $\triangle G_1G_2P\sim\triangle M_1M_2P, \triangle G_2G_3P\sim\triangle M_2M_3P, \triangle G_3G_4P\sim\triangle M_3M_4P,$ and $\triangle G_4G_1P\sim\triangle M_4M_1P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\frac23.$
Note that quadrilateral $M_1M_2M_3M_4$ is a square of side-length $15\sqrt2.$ It follows that:
Together, quadrilateral $G_1G_2G_3G_4$ is a square of side-length $10\sqrt2,$ so its area is $\left(10\sqrt2\right)^2=\boxed{200}.$
| 200
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4,234
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_13
| 3
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Square $ABCD$ has side length $30$ . Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$ . The centroids of $\triangle{ABP}$ $\triangle{BCP}$ $\triangle{CDP}$ , and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
[asy] unitsize(120); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); draw(A--B--C--D--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N*1.5+E*0.5); dot(A); dot(B); dot(C); dot(D); [/asy]
$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$
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This solution refers to the diagram in Solution 1.
We place the diagram in the coordinate plane: Let $A=(0,30),B=(0,0),C=(30,0),D=(30,30),$ and $P=(3x,3y).$
Recall that for any triangle in the coordinate plane, the coordinates of its centroid are the averages of the coordinates of its vertices. It follows that $G_1=(x,y+10),G_2=(x+10,y),G_3=(x+20,y+10),$ and $G_4=(x+10,y+20).$
Note that $G_1G_3=G_2G_4=20$ and $\overline{G_1G_3}\perp\overline{G_2G_4}.$ Therefore, the area of quadrilateral $G_1G_2G_3G_4$ is \[\frac12\cdot G_1G_3\cdot G_2G_4=\boxed{200}.\]
| 200
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4,235
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_13
| 4
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Square $ABCD$ has side length $30$ . Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$ . The centroids of $\triangle{ABP}$ $\triangle{BCP}$ $\triangle{CDP}$ , and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
[asy] unitsize(120); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); draw(A--B--C--D--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N*1.5+E*0.5); dot(A); dot(B); dot(C); dot(D); [/asy]
$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$
|
Let $X,Y,Z,W$ be the midpoints of sides $AB,BC,CD,DE$ , respectively.
Notice that a homothety centered at P with ratio $\frac{2}{3}$ will send $XYZW$ to $G_{1}G_{2}G_{3}G_{4}$ , so $G_{1}G_{2}G_{3}G_{4}$ is a square with area $\left(\frac{2}{3}\right)^2 [XYZW]$ , but $[XYZW]=\frac{1}{2}[ABCD]$ so our desired area is \[\frac{4}{9}\cdot\frac{1}{2}\cdot900=\boxed{200}\]
| 200
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4,236
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_14
| 1
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Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$
|
Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \[\frac{n+c}{n+1}=1+\frac{c-1}{n+1}\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. The prime factorization of $c-1$ is either $p^8$ or $p^2q^2.$ Since $c-1<100,$ the only possibility is $c-1=2^2\cdot3^2=36,$ from which $c=37.$ We conclude that Joey is $c+1=38$ years old today.
Suppose that Joey's age is a multiple of Zoe's age after $k$ years, in which Joey and Zoe will be $k+38$ and $k+1$ years old, respectively. We are given that \[\frac{k+38}{k+1}=1+\frac{37}{k+1}\] is an integer for some positive integer $k.$ It follows that $37$ is divisible by $k+1,$ so the only possibility is $k=36.$ We conclude that Joey will be $k+38=74$ years old then, from which the answer is $7+4=\boxed{11}.$
| 11
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4,237
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_14
| 2
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Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$
|
Let Joey's age be $j$ , Chloe's age be $c$ , and we know that Zoe's age is $1$
We know that there must be $9$ values $k\in\mathbb{Z}$ such that $c+k=a(1+k)$ where $a$ is an integer.
Therefore, $c-1+(1+k)=a(1+k)$ and $c-1=(1+k)(a-1)$ . Therefore, we know that, as there are $9$ solutions for $k$ , there must be $9$ solutions for $c-1$ . We know that this must be a perfect square. Testing perfect squares, we see that $c-1=36$ , so $c=37$ . Therefore, $j=38$ . Now, since $j-1=37$ , by similar logic, $37=(1+k)(a-1)$ , so $k=36$ and Joey will be $38+36=74$ and the sum of the digits is $\boxed{11}$
| 11
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4,238
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_14
| 3
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Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$
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Here's a different way of stating Solution 2:
If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has $9$ factors. Therefore, the difference between Chloe and Zoe's age is $36$ , so Chloe is $37$ , and Joey is $38$ . The common factor that will divide both of their ages is $37$ , so Joey will be $74$ . The answer is $7 + 4 = \boxed{11}$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_14
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Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$
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Similar approach to above, just explained less concisely and more in terms of the problem (less algebraic).
Let $C+n$ denote Chloe's age, $J+n$ denote Joey's age, and $Z+n$ denote Zoe's age, where $n$ is the number of years from now. We are told that $C+n$ is a multiple of $Z+n$ exactly nine times. Because $Z+n$ is $1$ at $n=0$ and will increase until greater than $C-Z$ , it will hit every natural number less than $C-Z$ , including every factor of $C-Z$ . For $C+n$ to be an integral multiple of $Z+n$ , the difference $C-Z$ must also be a multiple of $Z$ , which happens if $Z$ is a factor of $C-Z$ . Therefore, $C-Z$ has nine factors. The smallest number that has nine positive factors is $2^23^2=36$ . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know $Z=1$ and $J=C+1$ . Thus, \begin{align*} C-Z&=36, \\ J-Z&=37. \end{align*} By our above logic, the next time $J-Z$ is a multiple of $Z+n$ will occur when $Z+n$ is a factor of $J-Z$ . Because $37$ is prime, the next time this happens is at $Z+n=37$ , when $J+n=74$ . The answer is $7+4=\boxed{11}$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15
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How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$
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Let $\underline{ABC}$ be one such odd positive $3$ -digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$
As $A\in\{1,2,4,5,6,7,8,9\}$ and $C\in\{1,5,7,9\},$ there are $8$ possibilities for $A$ and $4$ possibilities for $C.$ Note that each ordered pair $(A,C)$ determines the value of $B$ modulo $3,$ so $B$ can be any element in one of the sets $\{0,6,9\},\{1,4,7\},$ or $\{2,5,8\}.$ We conclude that there are always $3$ possibilities for $B.$
By the Multiplication Principle, the answer is $8\cdot4\cdot3=\boxed{96}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15
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How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$
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Let $\underline{ABC}$ be one such odd positive $3$ -digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$
As $A\in\{1,2,4,5,6,7,8,9\},B\in\{0,1,2,4,5,6,7,8,9\},$ and $C\in\{1,5,7,9\},$ note that:
We apply casework to $A+B+C\equiv0\pmod3:$ \[\begin{array}{c|c|c||l} & & & \\ [-2.5ex] \boldsymbol{A\operatorname{mod}3} & \boldsymbol{B\operatorname{mod}3} & \boldsymbol{C\operatorname{mod}3} & \multicolumn{1}{c}{\textbf{Count}} \\ [0.5ex] \hline & & & \\ [-2ex] 0 & 0 & 0 & 2\cdot3\cdot1=6 \\ 0 & 1 & 2 & 2\cdot3\cdot1=6 \\ 0 & 2 & 1 & 2\cdot3\cdot2=12 \\ 1 & 0 & 2 & 3\cdot3\cdot1=9 \\ 1 & 1 & 1 & 3\cdot3\cdot2=18 \\ 1 & 2 & 0 & 3\cdot3\cdot1=9 \\ 2 & 0 & 1 & 3\cdot3\cdot2=18 \\ 2 & 1 & 0 & 3\cdot3\cdot1=9 \\ 2 & 2 & 2 & 3\cdot3\cdot1=9 \end{array}\] Together, the answer is $6+6+12+9+18+9+18+9+9=\boxed{96}.$
| 96
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15
| 3
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How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$
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Analyze that the three-digit integers divisible by $3$ start from $102.$ In the $200$ 's, it starts from $201.$ In the $300$ 's, it starts from $300.$ We see that the units digits is $0, 1,$ and $2.$
Write out the $1$ - and $2$ -digit multiples of $3$ starting from $0, 1,$ and $2.$ Count up the ones that meet the conditions. Then, add up and multiply by $3,$ since there are three sets of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$ 's ll contain the digit $3.$
Together, the answer is $3(12+12+12)-12=\boxed{96}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15
| 4
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How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$
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Consider the number of $2$ -digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$ -digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$ -digit number. However, we have $7 \equiv 1\pmod{3},$ and thus we need to count any $2$ -digit number $\equiv 2\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \boxed{96}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15
| 5
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How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$
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We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$ -digit numbers. Exactly $\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\cdot\frac{1}{3}=150.$ Of these $150$ numbers, $\frac{4}{5}$ $\textbf{do not}$ have $3$ in their ones (units) digit, $\frac{9}{10}$ $\textbf{do not}$ have $3$ in their tens digit, and $\frac{8}{9}$ $\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$ -digit integers is \[900\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{5}\cdot\frac{9}{10}\cdot\frac{8}{9}=\boxed{96}.\]
| 96
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15
| 6
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How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$
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We will start with the numbers that could work. This numbers include _ _ $1$ , _ _ $5$ , _ _ $7$ , _ _ $9$ . Let's work case by case.
Case $1$ : _ _ $1$ : The two blanks could be any number that is $2$ mod $3$ that does not include $3$ . We have $24$ cases for this case (we could count every case).
Case $2$ : _ _ $5$ : The $2$ blanks could be any number that is $1$ mod $3$ that does not include $3$ . But we could see that this case has exactly the same solutions to case $1$ because we have a $1-1$ correspondence. We can do the exact same for case $3$
Cases $4$ : _ _ $9$ : We need the blanks to be a multiple of $3$ , but does not contain 3. We have $(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87, 90, 96, 99)$ which also contains $24$ numbers. Therefore, we have $24 \cdot 4$ which is equal to $\boxed{96}.$
| 96
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15
| 7
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How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$
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This problem is solvable by inclusion exclusion principle. There are $\frac{999-105}{6} + 1 = 150$ odd $3$ -digit numbers divisible by $3$ . We consider the number of $3$ -digit numbers divisible by $3$ that contain either $1, 2$ or $3$ digits of $3$
For $\underline{AB3}$ $AB$ is any $2$ -digit number divisible by $3$ , which gives us $\frac{99-12}{3} + 1 = 30$ . For $\underline{A3B}$ , for each odd $B$ , we have $3$ values of $A$ that give a valid case, thus we have $5(3) = 15$ cases. For $\underline{3AB}$ , we also have $15$ cases, but when $B=3, 9$ $A$ can equal $0$ , so we have $17$ cases.
For $\underline{A33}$ , we have $3$ cases. For $\underline{3A3}$ , we have $4$ cases. For $\underline{33A}$ , we have $2$ cases. Finally, there is just one case for $\underline{333}$
By inclusion exclusion principle, we get $150 - 30 - 15 - 17 + 3 + 4 + 2 - 1 = \boxed{96}$ numbers.
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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More generally, let $a,b,c,d,p,$ and $q$ be positive integers such that $bc-ad=1$ and \[\frac ab < \frac pq < \frac cd.\] From $\frac ab < \frac pq,$ we have $bp-aq>0,$ or \[bp-aq\geq1. \hspace{15mm} (1)\] From $\frac pq < \frac cd,$ we have $cq-dp>0,$ or \[cq-dp\geq1. \hspace{15mm} (2)\] Since $bc-ad=1,$ note that:
To minimize $q,$ we set $q=b+d,$ from which $p=a+c.$ Together, we can prove that \[\frac{a+c-1}{b+d}\leq\frac ab<\frac{a+c}{b+d}<\frac cd\leq\frac{a+c+1}{b+d}. \hspace{15mm} (\bigstar)\] For this problem, we have $a=5,b=9,c=4,$ and $d=7,$ so $p=a+c=9$ and $q=b+d=16.$ The answer is $q-p=\boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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Define variables $a,b,c,d,p,$ and $q$ as Solution 1 does. Moreover, this solution refers to inequalities $(1)$ and $(2)$ in Solution 1.
Note that \begin{align*} \frac{1}{bd}&=\frac{bc-ad}{bd} \\ &=\frac cd - \frac ab \\ &=\left(\frac cd - \frac pq\right)+\left(\frac pq - \frac ab\right) \\ &=\frac{cq-dp}{dq}+\frac{bp-aq}{bp} \\ &\geq\frac{1}{dq}+\frac{1}{bq}. \end{align*} Multiplying both sides of $\frac{1}{bd}\geq\frac{1}{dq}+\frac{1}{bq}$ by $bdq,$ we get $q\geq b+d.$
For this problem, we have $a=5,b=9,c=4,$ and $d=7,$ so $q\geq b+d=16.$ At $q=16,$ we have \[\frac{8}{16}<\frac59<\frac{9}{16}<\frac47<\frac{10}{16},\] from which $p=9.$ Therefore, the answer is $q-p=\boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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Inverting the given inequality we get \[\frac{7}{4} < \frac{q}{p} < \frac{9}{5},\] which simplifies to \[35p < 20q < 36p.\] We can now substitute $q = p + k.$ Note we need to find $k:$ \[35p < 20p + 20k < 36p,\] which simplifies to \[15p < 20k < 16p.\] Clearly $p>k.$ We will now substitute $p = k + x$ to get \[15k + 15x < 20k < 16k + 16x.\] The inequality $15k + 15x < 20k$ simplifies to $3x < k.$
The inequality $20k < 16k + 16x$ simplifies to $k < 4x.$
Combining the two inequalities, we get \[3x < k < 4x.\] Since $x$ and $k$ are integers, the smallest values of $x$ and $k$ that satisfy the above equation are $2$ and $7$ respectively. Substituting these back in, we arrive with an answer of $\boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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Because $q$ and $p$ are positive integers with $p<q,$ we can let $q=p+k$ where $k\in{\mathbb{Z}}.$ Now, the problem condition reduces to \[\frac{5}{9}<\frac{p}{p+k}<\frac{4}{7}.\]
Our first inequality is $\frac{5}{9}<\frac{p}{p+k},$ which gives us $5p+5k<9p,$ or $\frac{5}{4}k<p.$
Our second inequality is $\frac{p}{p+k}<\frac{4}{7},$ which gives us $7p<4p+4k,$ or $p<\frac{4}{3}k.$
Hence, we have $\frac{5}{4}k<p<\frac{4}{3}k,$ or $15k<12p<16k.$
It is clear that we are aiming to find the least positive integer value of $k$ such that there is at least one value of $p$ that satisfies the inequality.
Now, simple casework through the answer choices of the problem reveals that $q-p=p+k-p=k\ge{\boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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We subtract $\frac{1}{2}$ from both sides of the equation, so \[\frac{1}{18} < \frac{p}{q}-\frac{1}{2} < \frac{1}{14}.\] Then for $q$ to be as small as possible, $\frac{p}{q}-\frac{1}{2}$ has to be $\frac{1}{16},$ so $\frac{p}{q}$ is $\frac{9}{16}$ and $q-p$ is $\boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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Cross-multiply the inequality to get \[35q < 63p < 36q.\] Then, we have $0 < 63p-35q < q,$ or \[0 < 7(9p-5q) < q.\] Since $p$ and $q$ are integers, $9p-5q$ is an integer. To minimize $q,$ start from $9p-5q=1,$ which gives $p=\frac{5q+1}{9}.$ This limits $q$ to be greater than $7,$ so test values of $q$ starting from $q=8.$ However, $q=8$ to $q=14$ do not give integer values of $p.$
Once $q>14,$ it is possible for $9p-5q$ to be equal to $2,$ so $p$ could also be equal to $\frac{5q+2}{9}.$ The next value, $q=15,$ is not a solution, but $q=16$ gives $p=\frac{5\cdot 16 + 1}{9} = 9.$ Thus, the smallest possible value of $q$ is $16,$ and the answer is $16-9= \boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
| 7
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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Start with $\frac{5}{9}.$ Repeat the following process until you arrive at the answer: if the fraction is less than or equal to $\frac{5}{9},$ add $1$ to the numerator; otherwise, if it is greater than or equal to $\frac{4}{7},$ add one $1$ to the denominator. We have \[\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}.\]
Therefore, the answer is $16 - 9 = \boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
| 8
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill.
The interval can also be written as $0.5556<x<0.5714.$ This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range.
The denominators to be considered are $9,10,11,12,\ldots.$ We check $\frac{6}{10}, \frac{6}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}.$ At this point we know that we've got our fraction and our answer is $16-9=\boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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Graph the regions $y > \frac{5}{9}x$ and $y < \frac{4}{7}x.$ Note that the lattice point $(16,9)$ is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is $\frac{9}{16}$ and the answer is $16-9= \boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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In ascending order, we can use answer choices, values for $q-p,$ as a method of figuring out our answer through the means of substitution. Let the assumed difference be $7.$ Then, $p=q-7.$ We thus have two inequalities: $\frac{5}{9} < \frac{q-7}{q}$ and $\frac{q-7}{q} < \frac{4}{7}.$
Solving for $q$ in these equalities, we get $\frac{63}{4} < q < \frac{49}{3}.$ So, $q$ is between $15.75$ and $16.\overline{3},$ making it $16$ as $q$ is a positive integer (again, at this point, this is still an assumption). This would set $p=16-7=9.$
Since $\frac{5}{9} < \frac{9}{16} < \frac{4}{7},$ the minimum difference is $\boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17
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Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$
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Assume that the difference $\frac{p}{q} - \frac{5}{9}$ results in a fraction of the form $\frac{1}{9q}.$ Then, \[9p - 5q = 1.\] Also assume that the difference $\frac{4}{7} - \frac{p}{q}$ results in a fraction of the form $\frac{1}{7q}.$ Then, \[4q - 7p = 1.\] Solving the system of equations yields $q=16$ and $p=9.$ Therefore, the answer is $\boxed{7}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18
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A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
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For all integers $n \geq 7,$ note that \begin{align*} f(n)&=f(n-1)-f(n-2)+n \\ &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\ &=-f(n-3)+2n-1 \\ &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\ &=-f(n-4)+f(n-5)+n+2 \\ &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\ &=f(n-6)+6. \end{align*} It follows that \begin{align*} f(2018)&=f(2012)+6 \\ &=f(2006)+12 \\ &=f(2000)+18 \\ & \ \vdots \\ &=f(2)+2016 \\ &=\boxed{2017} ~MRENTHUSIASM
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18
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A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
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For all integers $n\geq3,$ we rearrange the given equation: \[f(n)-f(n-1)+f(n-2)=n. \hspace{28.25mm}(1)\] For all integers $n\geq4,$ it follows that \[f(n-1)-f(n-2)+f(n-3)=n-1. \hspace{15mm}(2)\] For all integers $n\geq4,$ we add $(1)$ and $(2):$ \[f(n)+f(n-3)=2n-1. \hspace{38.625mm}(3)\] For all integers $n\geq7,$ it follows that \[f(n-3)+f(n-6)=2n-7. \hspace{32mm}(4)\] For all integers $n\geq7,$ we subtract $(4)$ from $(3):$ \[f(n)-f(n-6)=6. \hspace{47.5mm}(5)\] From $(5),$ we have the following system of $336$ equations: \begin{align*} f(2018)-f(2012)&=6, \\ f(2012)-f(2006)&=6, \\ f(2006)-f(2000)&=6, \\ & \ \vdots \\ f(8)-f(2)&=6. \end{align*} We add these equations up to get \[f(2018)-f(2)=6\cdot336=2016,\] from which $f(2018)=f(2)+2016=\boxed{2017}.$
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A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
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Preamble: In this solution, we define the sequence $A$ to satisfy $a_n = f(n),$ where $a_n$ represents the $n$ th term of the sequence $A.$ This solution will show a few different perspectives. Even though it may not be as quick as some of the solutions above, I feel like it is an interesting concept, and may be more motivated.
To begin, we consider the sequence $B$ formed when we take the difference of consecutive terms between $A.$ Define $b_n = a_{n+1} - a_n.$ Notice that for $n \ge 4,$ we have
Notice that subtracting the second equation from the first, we see that $b_{n} = b_{n-1} - b_{n-2} + 1.$
If you didn’t notice that $B$ repeated directly in the solution above, you could also, possibly more naturally, take the finite differences of the sequence $b_n$ so that you could define $c_n = b_{n+1} - b_n.$ Using a similar method as above through reindexing and then subtracting, you could find that $c_n = c_{n-1} - c_{n-2}.$ The sum of any six consecutive terms of a sequence which satisfies such a recursion is $0,$ in which you have that $b_{n} = b_{n+6}.$ In the case in which finite differences didn’t reduce to such a special recursion, you could still find the first few terms of $C$ to see if there are any patterns, now that you have a much simpler sequence. Doing so in this case, it can also be seen by seeing that the sequence $C$ looks like \[\underbrace{2, 1, -1, -2, -1, 1,}_{\text{cycle period}} 2, 1, -1, -2, -1, 1, \ldots\] in which the same result follows.
Using the fact that $B$ repeats every six terms, this motivates us to look at the sequence $B$ more carefully. Doing so, we see that $B$ looks like \[\underbrace{2, 3, 2, 0, -1, 0,}_{\text{cycle period}} 2, 3, 2, 0, -1, 0, \ldots\] (If you tried pattern finding on sequence $B$ directly, you could also arrive at this result, although I figured defining a second sequence based on finite differences was more motivated.)
Now, there are two ways to finish.
Finish Method #1: Notice that any six consecutive terms of $B$ sum to $6,$ after which we see that $a_n = a_{n-6} + 6.$ Therefore, $a_{2018} = a_{2012} + 6 = \cdots = a_{2} + 2016 = \boxed{2017}.$
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A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
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Start out by listing some terms of the sequence. \begin{align*} f(1)&=1 \\ f(2)&=1 \\ f(3)&=3 \\ f(4)&=6 \\ f(5)&=8 \\ f(6)&=8 \\ f(7)&=7 \\ f(8)&=7 \\ f(9)&=9 \\ f(10)&=12 \\ f(11)&=14 \\ f(12)&=14 \\ f(13)&=13 \\ f(14)&=13 \\ f(15)&=15 \\ & \ \vdots \end{align*} Notice that $f(n)=n$ whenever $n$ is an odd multiple of $3$ , and the pattern of numbers that follow will always be $+2$ $+3$ $+2$ $+0$ $-1$ $+0$ .
The largest odd multiple of $3$ smaller than $2018$ is $2013$ , so we have \begin{align*} f(2013)&=2013 \\ f(2014)&=2016 \\ f(2015)&=2018 \\ f(2016)&=2018 \\ f(2017)&=2017 \\ f(2018)&=\boxed{2017}
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A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
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Writing out the first few values, we get \[1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19,\ldots.\] We see that every number $x$ where $x \equiv 1\pmod 6$ has $f(x)=x,f(x+1)=f(x)=x,$ and $f(x-1)=f(x-2)=x+1.$ The greatest number that's $1\pmod{6}$ and less than $2018$ is $2017,$ so we have $f(2017)=f(2018)=\boxed{2017}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18
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A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
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\begin{align*} f(n)&=f(n-1)-f(n-2)+n \\ f(n-1)&=f(n-2)-f(n-3)+n-1 \end{align*} Subtracting those two and rearranging gives \begin{align*} f(n)-2f(n-1)+2f(n-2)-f(n-3)&=1 \\ f(n-1)-2f(n-2)+2f(n-3)-f(n-4)&=1 \end{align*} Subtracting those two gives $f(n)-3f(n-1)+4f(n-2)-3f(n-3)+f(n-4)=0.$
The characteristic polynomial is $x^4-3x^3+4x^2-3x+1=0.$
$x=1$ is a root, so using synthetic division results in $(x-1)(x^3-2x^2+2x-1)=0.$
$x=1$ is a root, so using synthetic division results in $(x-1)^2(x^2-x+1)=0.$
$x^2-x+1=0$ has roots $x=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}.$
And \[f(n)=(An+D)\cdot1^n+B\cdot\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)^n+C\cdot\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^n.\] Plugging in $n=1$ $n=2$ $n=3$ , and $n=4$ results in a system of $4$ linear equations $\newline$ Solving them gives $A=1, \ B=\frac{1}{2}-\frac{i\sqrt{3}}{2}, \ C=\frac{1}{2}+\frac{i\sqrt{3}}{2}, \ D=1.$ Note that you can guess $A=1$ by answer choices.
So plugging in $n=2018$ results in \begin{align*} 2018+1+\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)^{2019}+\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{2019}&=2019+(\cos(-60^{\circ})+\sin(-60^{\circ}))^{2019})+(\cos(60^{\circ})+\sin(60^{\circ}))^{2019}) \\ &=2019+(\cos(-60^{\circ}\cdot2019)+\sin(-60^{\circ}\cdot2019))+(\cos(60^{\circ}\cdot2019)+sin(60^{\circ}\cdot2019)) \\ &=\boxed{2017} ~ryanbear
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A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$
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We utilize patterns to solve this equation: \begin{align*} f(3)&=3, \\ f(4)&=6, \\ f(5)&=8, \\ f(6)&=8, \\ f(7)&=7, \\ f(8)&=8. \end{align*} We realize that the pattern repeats itself. For every six terms, there will be four terms that we repeat, and two terms that we don't repeat. We will exclude the first two for now, because they don't follow this pattern.
First, we need to know whether or not $2016$ is part of the skip or repeat. We notice that $f(6),f(12), \ldots,f(6n)$ all satisfy $6+6(n-1)=n,$ and we know that $2016$ satisfies this, leaving $n=336.$ Therefore, we know that $2016$ is part of the repeat section. But what number does it repeat?
We know that the repeat period is $2,$ and it follows that pattern of $1,1,8,8,7,7.$ Again, since $f(6) = f(5)$ and so on for the repeat section, $f(2016)=f(2015),$ so we don't need to worry about which one, since it repeats with period $2.$ We see that the repeat pattern of $f(6),f(12),\ldots,f(6n)$ follows $8,14,20,$ it is an arithmetic sequence with common difference $6.$ Therefore, $2016$ is the $335$ th term of this, but including $1,$ it is $336\cdot6+1=\boxed{2017}.$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_19
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Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$ . At some moment Mary wrote $323$ as a divisor of $n$ . What is the smallest possible value of the next divisor written to the right of $323$
$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$
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The prime factorization of $323$ is $17 \cdot 19$ . Our answer must be a multiple of either $17$ or $19$ or both. Since $17 < 19$ , the next smallest divisor that is divisble by $17$ would be $323 + 17 = \boxed{340}$
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Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,\ldots,\dfrac{n}{2},n$ . At some moment Mary wrote $323$ as a divisor of $n$ . What is the smallest possible value of the next divisor written to the right of $323$
$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$
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Note that $323$ multiplied by any of the answer choices results in a $5$ or $6$ -digit $n$ . So, we need a choice that shares a factor(s) with $323$ , such that the factors we'll need to add to the prime factorization of $n$ (in result to adding the chosen divisor) won't cause our number to multiply to more than $4$ digits.
The prime factorization of $323$ is $17\cdot19$ , and since we know $n$ is even, our answer needs to be
We see $340$ achieves this and is the smallest to do so ( $646$ being the other). So, we get $\boxed{340}$
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_21
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In $\triangle{ABC}$ with side lengths $AB = 13$ $AC = 12$ , and $BC = 5$ , let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle{ABC}$ . What is the area of $\triangle{MOI}$
$\textbf{(A)}\ \frac52\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{13}{4}\qquad\textbf{(E)}\ \frac72$
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In this solution, let the brackets denote areas.
We place the diagram in the coordinate plane: Let $A=(12,0),B=(0,5),$ and $C=(0,0).$
Since $\triangle ABC$ is a right triangle with $\angle ACB=90^\circ,$ its circumcenter is the midpoint of $\overline{AB},$ from which $O=\left(6,\frac52\right).$ Note that the circumradius of $\triangle ABC$ is $\frac{13}{2}.$
Let $s$ denote the semiperimeter of $\triangle ABC.$ The inradius of $\triangle ABC$ is $\frac{[ABC]}{s}=\frac{30}{15}=2,$ from which $I=(2,2).$
Since $\odot M$ is also tangent to both coordinate axes, its center is at $M=(a,a)$ and its radius is $a$ for some positive number $a.$ Let $P$ be the point of tangency of $\odot O$ and $\odot M.$ As $\overline{OP}$ and $\overline{MP}$ are both perpendicular to the common tangent line at $P,$ we conclude that $O,M,$ and $P$ are collinear. It follows that $OM=OP-MP,$ or \[\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.\] Solving this equation, we have $a=4,$ from which $M=(4,4).$
Finally, we apply the Shoelace Theorem to $\triangle MOI:$ \[[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{72}.\] Remark
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_22
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Consider polynomials $P(x)$ of degree at most $3$ , each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ . How many such polynomials satisfy $P(-1) = -9$
$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$
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Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that \[P(-1)=-a+b-c+d=-9.\] Let $a'=9-a$ and $c'=9-c.$ Note that both of $a'$ and $c'$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a',b,c',d)$ have one-to-one correspondence.
We rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or \[a'+b+c'+d=9.\] By the stars and bars argument, there are $\binom{9+4-1}{4-1}=\boxed{220}$ ordered quadruples $(a',b,c',d).$
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Consider polynomials $P(x)$ of degree at most $3$ , each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ . How many such polynomials satisfy $P(-1) = -9$
$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$
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Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $P(-1)=-a+b-c+d=-9,$ which rearranges to \[b+d+9=a+c.\] Note that $b+d+9$ is an integer from $9$ through $27,$ and $a+c$ is an integer from $0$ through $18.$ Therefore, both of $b+d+9$ and $a+c$ are integers from $9$ through $18.$ We construct the following table: \[\begin{array}{c|c|c|c||c} & & & & \\ [-2.5ex] \boldsymbol{b+d} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(b,d)} & \boldsymbol{a+c} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(a,c)} & \boldsymbol{\#}\textbf{ of Ordered Quadruples }\boldsymbol{(a,b,c,d)} \\ [0.5ex] \hline & & & & \\ [-2ex] 0 & 1 & 9 & 10 & 1\cdot10=10 \\ 1 & 2 & 10 & 9 & \phantom{0}2\cdot9=18 \\ 2 & 3 & 11 & 8 & \phantom{0}3\cdot8=24 \\ 3 & 4 & 12 & 7 & \phantom{0}4\cdot7=28 \\ 4 & 5 & 13 & 6 & \phantom{0}5\cdot6=30 \\ 5 & 6 & 14 & 5 & \phantom{0}6\cdot5=30 \\ 6 & 7 & 15 & 4 & \phantom{0}7\cdot4=28 \\ 7 & 8 & 16 & 3 & \phantom{0}8\cdot3=24 \\ 8 & 9 & 17 & 2 & \phantom{0}9\cdot2=18 \\ 9 & 10 & 18 & 1 & 10\cdot1=10 \end{array}\] We sum up the counts in the last column to get the answer $2\cdot(10+18+24+28+30)=\boxed{220}.$
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Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$
$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$
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This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
Without the loss of generality, let $AC=BC=1.$ As shown below, we place Earth in the $xyz$ -plane with $C=(0,0,0)$ such that the positive $x$ -axis runs through $A,$ the positive $y$ -axis runs through $0^\circ$ latitude and $160^\circ \text{ W}$ longitude, and the positive $z$ -axis runs through the North Pole. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^B--C--D--cycle); label("$A$",A,5*dir((2.5,-3,0))); label("$B$",B,3*dir(B)); label("$C$",C,1.5*(1,0,-1)); label("$D$",D,3*(-1/2,-1/2,0)); draw((-1.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-1.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-1.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); label("$x$",(1.5,0,0),2*dir((1.5,0,0))); label("$y$",(0,1.5,0),3*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); [/asy] It follows that $A=(1,0,0)$ and $D=(-t,t,0)$ for some positive number $t.$ Since $\triangle BCD$ is an isosceles right triangle, we have $B=\left(-t,t,\sqrt{2}t\right).$ By the Distance Formula, we get $(-t)^2+t^2+\left(\sqrt{2}t\right)^2=1,$ from which $t=\frac12.$
As $\vec{A} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $\vec{B} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix},$ we obtain \[\cos\angle ACB=\frac{\vec{A}\bullet\vec{B}}{\left\lVert\vec{A}\right\rVert\left\lVert\vec{B}\right\rVert}=-\frac12\] by the dot product, so $\angle ACB=\boxed{120}$ degrees.
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Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$
$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$
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This solution refers to the diagram in Solution 2.
In spherical coordinates $(\rho,\theta,\phi),$ note that $\rho,\theta,$ and $\phi$ represent the radial distance, the polar angle, and the azimuthal angle, respectively.
Without the loss of generality, let $AC=BC=1.$ As shown in Solution 2, we place Earth in the $xyz$ -plane with origin $C$ such that the positive $x$ -axis runs through $A,$ the positive $y$ -axis runs through $0^\circ$ latitude and $160^\circ \text{ W}$ longitude, and the positive $z$ -axis runs through the North Pole.
In spherical coordinates, we have $A=(1,90^\circ,0^\circ)$ and $B=(1,45^\circ,135^\circ).$ Now, we express $\vec{A}$ and $\vec{B}$ in Cartesian coordinates: \[\vec{A} = \begin{pmatrix}\sin90^\circ \cos0^\circ \\ \sin90^\circ \sin0^\circ \\ \cos90^\circ \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} \text{ and } \vec{B} = \begin{pmatrix}\sin45^\circ \cos135^\circ \\ \sin45^\circ \sin135^\circ \\ \cos45^\circ \end{pmatrix} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix}.\] We continue with the last paragraph of Solution 2 to get the answer $\angle ACB=\boxed{120}$ degrees.
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Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
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This rewrites itself to $x^2=10,000\{x\}$ where $\lfloor x \rfloor + \{x\} = x$
Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$ , then $1$ to $2$ with a hole at $x=2$ etc.
Here is a graph of $y=x^2$ and $y=16\{x\}$ for visualization.
[asy] import graph; size(400); xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5})); yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18})); real y(real x) {return x^2;} draw(circle((-4,16), 0.1)); draw(circle((-3,16), 0.1)); draw(circle((-2,16), 0.1)); draw(circle((-1,16), 0.1)); draw(circle((0,16), 0.1)); draw(circle((1,16), 0.1)); draw(circle((2,16), 0.1)); draw(circle((3,16), 0.1)); draw(circle((4,16), 0.1)); draw((-5,0)--(-4,16), black); draw((-4,0)--(-3,16), black); draw((-3,0)--(-2,16), black); draw((-2,0)--(-1,16), black); draw((-1,0)--(-0,16), black); draw((0,0)--(1,16), black); draw((1,0)--(2,16), black); draw((2,0)--(3,16), black); draw((3,0)--(4,16), black); draw(graph(y,-4.2,4.2),green); [/asy]
Now notice that when $x=\pm 100$ the graph has a hole at $(\pm 100,10,000)$ which the equation $y=x^2$ passes through and then continues upwards. Thus our set of possible solutions is bounded by $(-100,100)$ . We can see that $y=x^2$ intersects each of the lines once and there are $99-(-99)+1=199$ lines for an answer of $\boxed{199}$
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Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
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Same as the first solution, $x^2=10,000\{x\}$
We can write $x$ as $\lfloor x \rfloor+\{x\}$ . Expanding everything, we get a quadratic in $\{x\}$ in terms of $\lfloor x \rfloor$ \[\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0\]
We use the quadratic formula to solve for $\{x\}$ \[\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ ( 2\lfloor x \rfloor - 10,000 )^2 - 4\lfloor x \rfloor^2 }}{2} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ 4\lfloor x \rfloor^2 -40,000 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 }}{2}\]
Since $0 \leq \{x\} < 1$ , we get an inequality which we can then solve. After simplifying a lot, we get that $\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0$
Solving over the integers, $-101 < \lfloor x \rfloor < 99$ , and since $\lfloor x \rfloor$ is an integer, there are $\boxed{199}$ solutions. Each value of $\lfloor x \rfloor$ should correspond to one value of $x$ , so we are done.
| 199
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4,274
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24
| 3
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Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
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Let $x = a+k$ where $a$ is the integer part of $x$ and $k$ is the fractional part of $x$ .
We can then rewrite the problem below:
$(a+k)^2 + 10000a = 10000(a+k)$
From here, we get
$(a+k)^2 + 10000a = 10000a + 10000k$
Solving for $a+k = x$
$(a+k)^2 = 10000k$
$x = a+k = \pm100\sqrt{k}$
Because $0 \leq k < 1$ , we know that $a+k$ cannot be less than or equal to $-100$ nor greater than or equal to $100$ . Therefore:
$-99 \leq x \leq 99$
There are $199$ elements in this range, so the answer is $\boxed{199}$
| 199
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4,275
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24
| 4
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Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
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Notice the given equation is equivalent to $(\lfloor x \rfloor+\{x\})^2=10,000\{x\}$
Now we know that $\{x\} < 1$ so plugging in $1$ for $\{x\}$ we can find the upper and lower bounds for the values.
$(\lfloor x \rfloor +1)^2 = 10,000(1)$
$(\lfloor x \rfloor +1) = \pm 100$
$\lfloor x \rfloor = 99, -101$
And just like $\textbf{Solution 2}$ , we see that $-101 < \lfloor x \rfloor < 99$ , and since $\lfloor x \rfloor$ is an integer, there are $\boxed{199}$ solutions. Each value of $\lfloor x \rfloor$ should correspond to one value of $x$ , so we are done.
| 199
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4,276
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24
| 5
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Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
|
Firstly, if $x$ is an integer, then $10,000\lfloor x \rfloor=10,000x$ , so $x$ must be $0$
If $0<x<1$ , then we know the following:
$0<x^2<1$
$10,000\lfloor x \rfloor =0$
$0<10,000x<10,000$
Therefore, $0<x^2+10,000\lfloor x \rfloor <1$ , which overlaps with $0<10,000x<10,000$ . This means that there is at least one real solution between $0$ and $1$ . Since $x^2+10,000\lfloor x \rfloor$ increases quadratically and $10,000x$ increases linearly, there is only one solution for this case.
Similarly, if $1<x<2$ , then we know the following:
$1<x^2<4$
$10,000\lfloor x \rfloor =10,000$
$<10,000<10,000x<20,000$
By following similar logic, we can find that there is one solution between $1$ ad $2$
We can also follow the same process to find that there are negative solutions for $x$ as well.
There are not an infinite amount of solutions, so at one point there will be no solutions when $n<x<n+1$ for some integer $n$ . For there to be no solutions in a given range means that the range of $10,000\lfloor x \rfloor + x^2$ does not intersect the range of $10,000x$ $x^2$ will always be positive, and $10,000\lfloor x \rfloor$ is less than $10,000$ less than $10,000x$ , so when $x^2 >= 10,000$ , the equation will have no solutions. This means that there are $99$ positive solutions, $99$ negative solutions, and $0$ for a total of $\boxed{199}$ solutions.
| 199
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4,277
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24
| 6
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Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
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General solution to this type of equation $f(x, \lfloor x \rfloor) = 0$
$x^2 - 10000x + 10000 \lfloor x \rfloor =0$
$x=5000 \pm 100 \sqrt{2500- \lfloor x \rfloor}$ $\lfloor x \rfloor \le 2500$
$\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$
If $x= 5000 + 100 \sqrt{2500 - \lfloor x \rfloor}$ $x \ge 5000$ , it contradicts $x < \lfloor x \rfloor + 1 \le 2501$
So $x= 5000 - 100 \sqrt{2500 - \lfloor x \rfloor}$
Let $k = \lfloor x \rfloor$ $x= 5000 - 100 \sqrt{2500 - k}$
$k \le 5000 - 100 \sqrt{2500 - k} < k + 1$
$0 \le 5000 - k - 100 \sqrt{2500 - k} < 1$
$0 \le 2500 - k - 100 \sqrt{2500 - k} + 2500 < 1$
$0 \le (\sqrt{2500 - k} - 50)^2 < 1$
$-1 < \sqrt{2500 - k} - 50 < 1$
$49 < \sqrt{2500 - k} < 51$
$-101 < k < 99$
So the number of $k$ 's values is $99-(-101)-1=199$ . Because $x=5000-100\sqrt{2500-k}$ , for each value of $k$ , there is a value for $x$ . The answer is $\boxed{199}$
| 199
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4,278
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24
| 7
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Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
|
Subtracting $10000\lfloor x\rfloor$ from both sides gives $x^2=10000(x-\lfloor x\rfloor)=10000\{x\}$ . Dividing both sides by $10000$ gives $\left(\frac{x}{100}\right)^2=\{x\}<1$ $\left(\frac{x}{100}\right)^2<1$ when $-100<x<100$ so the answer is $\boxed{199}$
| 199
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4,279
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25
| 1
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Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$
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Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively and draw $O_1O_2$ $O_1P_1$ , and $O_2P_2$ . Note that $\angle{O_1P_1P_2}$ and $\angle{O_2P_2P_3}$ are both right. Furthermore, since $\triangle{P_1P_2P_3}$ is equilateral, $m\angle{P_1P_2P_3} = 60^\circ$ and $m\angle{O_2P_2P_1} = 30^\circ$ . Mark $M$ as the base of the altitude from $O_2$ to $P_1P_2.$ Since $\triangle P_2O_2M$ is a 30-60-90 triangle, $O_2M = 2$ and $P_2M = 2\sqrt{3}$ . Also, since $O_1O_2 = 8$ and $O_1P_1 = 4$ , we can find $P_1M = \sqrt{8^2 - (4 + 2)^2} = 2\sqrt{7}$ . Thus, $P_1P_2 = P_1M + P_2M = 2\sqrt{3} + 2\sqrt{7}$ . This makes \[\left[P_1P_2P_3\right] = \frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.\] So, our answer is $252 + 300 = \boxed{552}$
| 552
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4,280
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25
| 2
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Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$
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Let $O_i$ be the center of circle $\omega_i$ for $i=1,2,3$ , and let $K$ be the intersection of lines $O_1P_1$ and $O_2P_2$ . Because $\angle P_1P_2P_3 = 60^\circ$ , it follows that $\triangle P_2KP_1$ is a $30-60-90$ triangle. Let $x=P_1K$ ; then $P_2K = 2x$ and $P_1P_2 = x \sqrt 3$ . The Law of Cosines in $\triangle O_1KO_2$ gives \[8^2 = (x+4)^2 + (2x-4)^2 - 2(x+4)(2x-4)\cos 60^\circ,\] which simplifies to $3x^2 - 12x - 16 = 0$ . The positive solution is $x = 2 + \tfrac23\sqrt{21}$ . Then $P_1P_2 = x\sqrt 3 = 2\sqrt 3 + 2\sqrt 7$ , and so the area of $\triangle P_1P_2P_3$ is \[\frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.\] The requested sum is $300 + 252 = \boxed{552}$
| 552
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4,281
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25
| 3
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Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$
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Let $O_i$ be the center of circle $\omega_i$ for $i=1,2,3$ . Let $X$ be the centroid of $\triangle{O_1O_2O_3}$ , which also happens to be the centroid of $\triangle{P_1P_2P_3}$ . Because $m\angle{O_1P_1P_2} = 90^\circ$ and $m\angle{O_1P_1X} = 30^\circ$ $m\angle{O_1P_1X} = 60^\circ$ $O_2M$ is $2/3$ the height of $\triangle{P_1P_2P_3}$ , thus $O_2M$ is $8*\sqrt{3}/3$
Applying cosine law on $\triangle{O_1P_1X}$ , one finds that $P_1X = 2 + 2*\sqrt{21}/3$ . Multiplying by $3/2$ to solve for the height of $\triangle{P_1P_2P_3}$ , one gets $3 + \sqrt{21}$ . Simply multiplying by $2/\sqrt{3}$ and then calculating the equilateral triangle's area, one would get the final result of $\sqrt{300} + \sqrt{252}$
This makes the answer $252 + 300 = 552$ $\boxed{552}$
| 552
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4,282
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25
| 4
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Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$
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[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*30); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); label("$\Gamma$", A, W*15); draw(Circle(A,2),red); pair X=foot(A,P1,P3); dot(X,blue); draw(A--X,blue); label("$2\sqrt{7}$", X--P3); label("$2\sqrt{3}$",X--P1); label("$X$",X,dir(-80),blue); [/asy]
First, note that because the $\angle P_1=\angle P_2=\angle P_3=\pi/3$ , the arcs inside the shaded equilateral triangle are each $2\pi/3$ . Also, the distances between the centers of any two of the $3$ given circles are each $8$ .
Draw the circle $\Gamma$ concentric with $\omega_1$ with radius $2$ . Because the arc of $\omega_1$ inside said triangle is $2\pi/3$ $\Gamma$ touches $P_1P_3$ , say at a point $X$ . Thus, $P_1P_3$ is a common tangent of $\omega_3$ and $\Gamma$ , and it can be seen from inspection of the given diagram that the line is an common internal tangent.
The length of the common internal tangent segment $XP_3$ of $\Gamma$ and $\omega_3$ is then $\sqrt{8^2-(2+4)^2}=2\sqrt{7}$ , and it is easily seen that $XP_1=4\sin \pi/3=2\sqrt{3}$ .
Because $P_1P_3=2(\sqrt{3}+\sqrt{7})$ , the area of the shaded equilateral triangle is $\sqrt{3}(\sqrt{3}+\sqrt{7})^2=10\sqrt{3}+6\sqrt{7}$ . We get $\sqrt{300}+\sqrt{252}\Rightarrow\boxed{552}.$
| 552
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4,283
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25
| 5
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Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$
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This seems like a coordinate bashable problem. To do this, we notice that it is easier to graph the equilateral triangle first, then the circles, rather than the other way around. Let's forget the lengths of the radius for a moment and focus instead on the ratio of the circles' radius to $\triangle P_1P_2P_3$ 's side length.
WLOG let $P_1P_2 = 2$ . Place triangle $P_1P_2P_3$ on the coordinate plane with $P_1(0,\sqrt3)$ $P_2(1,0)$ , and $P_3(-1,0)$ . Let $r$ be the radius of the circles.
Now, we find the coordinates of the centers of circles $\omega_1$ and $\omega_2$ . Since $\omega_2$ is tangent to the x-axis at $(1,0)$ , the center of $\omega_2$ is $(1,r)$ . Draw a right triangle with legs parallel to the x and y axes, and with hypotenuse as the segment from the center of $\omega_1$ to $P_1$ . Since the slope of $\overline{P_1P_2}$ is $-\sqrt3$ , the slope of the hypotenuse is $\frac{1}{\sqrt{3}}$ , so the right triangle is $30-60-90$ . It's easy to see that the center of $\omega_1$ is $\left(-\frac{r\sqrt{3}}{2}, \sqrt{3} - \frac{r}{2}\right)$
Since $\omega_1$ and $\omega_2$ are tangent, the distance between the centers is $2r$ , so we have \[\sqrt{\left(1 + \frac{r\sqrt{3}}{2}\right)^2 + \left(\frac{3r}{2} - \sqrt{3}\right)^2} = 2r\] \[\left(1 + \frac{r\sqrt{3}}{2}\right)^2 + \left(\frac{3r}{2} - \sqrt{3}\right)^2 = 4r^2\] \[1 + r\sqrt{3} + \frac{3r^2}{4} + \frac{9r^2}{4} - 3r\sqrt{3} + 3 = 4r^2\] \[4 - 2r\sqrt{3} + 3r^2 = 4r^2\] \[r^2 + 2r\sqrt{3} - 4 = 0\] By the Quadratic Formula, $r = \frac{-2\sqrt{3} \pm \sqrt{12 + 16}}{2} = -\sqrt{3}\pm\sqrt{7}$ .
We take the positive value to get the radius of the circle is $\sqrt{7} - \sqrt{3}$
Therefore, the ratio of the radius to the side length of the equilateral triangle is $\sqrt{7} - \sqrt{3} : 2 = 4 : \frac{8}{\sqrt{7} - \sqrt{3}} = 4 : 2(\sqrt{7} + \sqrt{3})$
The side length of the equilateral triangle is $2(\sqrt{7} + \sqrt{3})$ , so its area is $(\sqrt{7} + \sqrt{3})^2\sqrt{3} = 10\sqrt{3} + 6\sqrt{7} = \sqrt{300} + \sqrt{252} \implies 300 + 252 = \boxed{552}$
| 552
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4,284
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25
| 6
|
Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ . See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ . What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$
|
Let $O_1$ $O_2$ , and $O_3$ be the centers of $\omega_1$ $\omega_2$ , and $\omega_3$
Let $P_1P_2 = a$ $\angle O_2O_1P_1= \alpha$
\[[O_1P_1O_2P_2O_3P_3] = [P_1P_2P_3] + 3 \cdot [O_1P_1P_3] = [O_1O_2O_3] + 3 \cdot [O_1P_1O_2]\]
\[[P_1P_2P_3] + 3 \cdot [O_1P_1P_3] = \frac12 \cdot a \cdot \frac{ \sqrt{3} }{2} a + 3 \cdot \frac12 \cdot 4 \cdot a \cdot \sin 30^{\circ} = \frac{ \sqrt{3} }{4} a^2 + 3a\]
\[[O_1O_2O_3] + 3 \cdot [O_1P_1O_2] = \frac12 \cdot 8 \cdot \frac{ \sqrt{3} }{2} \cdot 8 + 3 \cdot \frac12 \cdot 4 \cdot 8 \cdot \sin \alpha = 16 \sqrt{3} + 48 \sin \alpha\]
\[\frac{ \sqrt{3} }{4} a^2 + 3a = 16 \sqrt{3} + 48 \sin \alpha\]
\[\sin \alpha = \frac{ \sqrt{3} }{192} a^2 + \frac{a}{16} - \frac{ \sqrt{3} }{3}\]
By the Law of Cosine from $\triangle O_2P_1P_2$ $O_2P_1^2 = a^2 + 4^2 - 2 \cdot a \cdot 4 \cdot \cos 30^{\circ} = a^2 + 16 - 4a \sqrt{3}$
By the Law of Cosine from $\triangle O_1O_2P_1$ $O_2P_1^2 = 4^2 + 8 ^ 2 - 2 \cdot 4 \cdot 8 \cdot \cos \alpha = 80 - 64 \cos \alpha$
\[a^2 + 16 - 4a \sqrt{3} = 80 - 64 \cos \alpha\]
\[\cos \alpha = -(\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)\]
\[\because \sin^2 \alpha + \cos^2 \alpha = 1\]
\[\therefore (\frac{ \sqrt{3} }{192} a^2 + \frac{a}{16} - \frac{ \sqrt{3} }{3})^2 + (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 1\]
\[[ \frac{1}{\sqrt{3}} (\frac{ a^2 }{64} + \frac{ \sqrt{3} }{16} a - 1)]^2 + (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 1\]
\[(\frac{ a^2 }{64} + \frac{ \sqrt{3} }{16} a - 1)^2 + 3 \cdot (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 3\]
\[(a^2 + 4 \sqrt{3} a - 64)^2 + 3 \cdot (a^2 - 4 \sqrt{3} a - 64)^2 = 3 \cdot 64^2\]
\[(a^2 - 64)^2 + 2 (a^2 - 64) \cdot 4 \sqrt{3} a + (4 \sqrt{3} a)^2 + 3(a^2 - 64)^2 - 6 (a^2 - 64) \cdot 4 \sqrt{3} a + 3(4 \sqrt{3} a)^2 = 3 \cdot 64^2\]
\[4 (a^2 - 64)^2 - 4 (a^2 - 64) \cdot 4 \sqrt{3} a + 192 a^2 - 3 \cdot 64^2 = 0\]
\[4 (a^2 - 64)(a^2 - 64 - 4 \sqrt{3} a) + 192(a^2 - 64) = 0\]
\[4 (a^2 - 64)(a^2 - 64 - 4 \sqrt{3} a + 48) = 0\]
\[(a^2 - 64)(a^2 - 4 \sqrt{3} a - 16) = 0\]
\[a^2 -64 = 0 \quad or \quad a^2 - 4 \sqrt{3} a - 16 = 0\]
If $a^2 =64$ $a= 8$ $P_1O_2^2 = 8^2 + 4^2 - 2 \cdot 8 \cdot 4 \cdot \cos 30^{\circ} = 80 - 32 \sqrt{3} < 16$ $P_1O_2 < 4$ , meaning that $P_1$ is inside the circle.
However, $P_1$ is not inside the circle.
\[So, \quad a^2 - 4 \sqrt{3} a - 16 = 0\]
\[a = \frac{ 4 \sqrt{3} + \sqrt{ (4 \sqrt{3})^2 - 4(-16) } }{2} = 2 \sqrt{3} + 2 \sqrt{7}\]
\[[P_1P_2P_3] = a^2 \cdot \frac{ \sqrt{3} }{4} = (2 \sqrt{3} + 2 \sqrt{7})^2 \cdot \frac{ \sqrt{3} }{4} = \sqrt{300} + \sqrt{252}\]
Therefore, the answer is $300 + 252 = \boxed{552}$
| 552
|
4,285
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_1
| 1
|
Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, 3-popsicle boxes for $$2$ , and 5-popsicle boxes for $$3$ . What is the greatest number of popsicles that Pablo can buy with $$8$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$
|
We can take two 5-popsicle boxes and one 3-popsicle box with $$8$ . Note that it is optimal since one popsicle is at the rate of $$1$ per popsicle, three popsicles at $$\frac{2}{3}$ per popsicle and finally, five popsicles at $$\frac{3}{5}$ per popsicle, hence we want as many $$3$ sets as possible. It is clear that the above is the optimal method. $\boxed{13}$
| 13
|
4,286
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_2
| 1
|
The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$
|
Let $x, y$ be our two numbers. Then $x+y = 4xy$ . Thus,
$\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = 4$
$\boxed{4}$
| 4
|
4,287
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_2
| 2
|
The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$
|
We can let $x=y.$ Then, we have that $2x=4x^2$ making $x=\tfrac{1}{2}.$ The answer is $\dfrac{2}{x}=4=\boxed{4}.$ Solasky talk
| 4
|
4,288
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_4
| 1
|
Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%$
|
Let $j$ represent how far Jerry walked, and $s$ represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Since Silvia walked the diagonal, she walked the hypotenuse of a $45$ $45$ $90$ triangle with leg length $1$ . Thus, $s = \sqrt{2} = 1.414...$ We can then take $\frac{j-s}{j} \approx \frac{2 - 1.4}{2} = 0.3 = 30\%$ $\boxed{30}$
| 30
|
4,289
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5
| 1
|
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$
|
All of the handshakes will involve at least one person from the $10$ who knows no one. Label these ten people $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ $I$ $J$
Person $A$ from the group of $10$ will initiate a handshake with everyone else ( $29$ people). Person $B$ initiates $28$ handshakes plus the one already counted from person $A$ . Person $C$ initiates $27$ new handshakes plus the two we already counted. This continues until person $J$ initiates $20$ handshakes plus the nine we already counted from $A$ ... $I$
$29+28+27+26+25+24+23+22+21+20 = \boxed{245}$
| 245
|
4,290
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5
| 2
|
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$
|
Let the group of people who all know each other be $A$ , and let the group of people who know no one be $B$ . Handshakes occur between each pair $(a,b)$ such that $a\in A$ and $b\in B$ , and between each pair of members in $B$ . Thus, the answer is
$|A||B|+{|B|\choose 2} = 20\cdot 10+{10\choose 2} = 200+45 = \boxed{245}$
| 245
|
4,291
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5
| 3
|
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$
|
The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are ${30\choose 2}$ and ${20\choose 2}$ , respectively. Thus, the total amount of handshakes is ${30\choose 2} - {20\choose 2} = 435 - 190= \boxed{245}$
| 245
|
4,292
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5
| 4
|
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$
|
Each of the $10$ people who do not know anybody will shake hands with all $20$ of the people who do know each other. This means there will be at least $20 * 10 = 200$ handshakes. In addition, those $10$ people will also shake hands with each other, giving us another $9+8+7+6+5+4+3+2+1 = 45$ handshakes. Therefore, there is a total of $200+45 = \boxed{245}$ handshakes.
| 245
|
4,293
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5
| 5
|
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$
|
Every one of the $20$ people who know each other will shake hands with each of the $10$ people who know no one, so there are $20\cdot 10 = 200$ handshakes here. Each of the $10$ people will also shake hands with each other, so there will be ${10 \choose 2}$ $=45$ handshakes for this case. In total, there are $200+45 = \boxed{245}$ handshakes.
| 245
|
4,294
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_6
| 1
|
Joy has $30$ thin rods, one each of every integer length from $1 \text{ cm}$ through $30 \text{ cm}$ . She places the rods with lengths $3 \text{ cm}$ $7 \text{ cm}$ , and $15 \text{cm}$ on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 17 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 19 \qquad\textbf{(E)}\ 20$
|
The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than $15 - (3 + 7) = 5$ and shorter than $15 + 3 + 7 = 25$ . This means Joy can use the $19$ possible integer rod lengths that fall into $[6, 24]$ . However, she has already used the rods of length $7$ cm and $15$ cm so the answer is $19 - 2 = 17$ $\boxed{17}$
| 17
|
4,295
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_7
| 1
|
Define a function on the positive integers recursively by $f(1) = 2$ $f(n) = f(n-1) + 1$ if $n$ is even, and $f(n) = f(n-2) + 2$ if $n$ is odd and greater than $1$ . What is $f(2017)$
$\textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036$
|
This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the base case, $f(1)=2$ . We also know that when $n$ is odd, $f(n)=f(n-2)+2$ . Thus we know that $f(2017)=f(2015)+2$ . Thus we know that n will always be odd in the recursion of $f(2017)$ , and we add $2$ each recursive cycle, which there are $1008$ of. Thus the answer is $1008*2+2=2018$ , which is answer $\boxed{2018}$ .
Note that when you write out a few numbers, you find that $f(n)=n+1$ for any $n$ , so $f(2017)=2018$
| 18
|
4,296
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_11
| 1
|
Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of $2017$ . She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 63\qquad\textbf{(C)}\ 117\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 163$
|
We know that the sum of the interior angles of the polygon is a multiple of $180$ . Note that $\left\lceil\frac{2017}{180}\right\rceil = 12$ and $180\cdot 12 = 2160$ , so the angle Claire forgot is $\equiv 2160-2017=143\mod 180$ . Since the polygon is convex, the angle is $\leq 180$ , so the answer is $\boxed{143}$
| 143
|
4,297
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_12
| 1
|
There are $10$ horses, named Horse 1, Horse 2, $\ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S = 2520$ . Let $T>0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$
|
We know that Horse $k$ will be at the starting point after $n$ minutes if $k|n$ . Thus, we are looking for the smallest $n$ such that at least $5$ of the numbers $\{1,2,\cdots,10\}$ divide $n$ . Thus, $n$ has at least $5$ positive integer divisors.
We quickly see that $12$ is the smallest number with at least $5$ positive integer divisors and that $1,2,3,4,6$ are each numbers of horses. Thus, our answer is $1+2=\boxed{3}$
| 3
|
4,298
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_12
| 2
|
There are $10$ horses, named Horse 1, Horse 2, $\ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S = 2520$ . Let $T>0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$
|
In order for at least $5$ horses to finish simultaneously, the current time needs to have at least $5$ divisors. Thus the number must have a form of either $p^4$ or $p^2*q$ , which have $5$ and $6$ factors respectively. The smallest number of the first form is $16$ , and the smallest number of the second form is $12.$ Thus, our answer is $1+2=\boxed{3}$
| 3
|
4,299
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_13
| 1
|
Driving at a constant speed, Sharon usually takes $180$ minutes to drive from her house to her mother's house. One day Sharon begins the drive at her usual speed, but after driving $\frac{1}{3}$ of the way, she hits a bad snowstorm and reduces her speed by $20$ miles per hour. This time the trip takes her a total of $276$ minutes. How many miles is the drive from Sharon's house to her mother's house?
$\textbf{(A)}\ 132 \qquad\textbf{(B)}\ 135 \qquad\textbf{(C)}\ 138 \qquad\textbf{(D)}\ 141 \qquad\textbf{(E)}\ 144$
|
Let total distance be $x$ . Her speed in miles per minute is $\tfrac{x}{180}$ . Then, the distance that she drove before hitting the snowstorm is $\tfrac{x}{3}$ . Her speed in snowstorm is reduced $20$ miles per hour, or $\tfrac{1}{3}$ miles per minute. Knowing it took her $276$ minutes in total, we create equation: \[\text{Time before Storm}\, + \, \text{Time after Storm} = \text{Total Time} \Longrightarrow\] \[\frac{\text{Distance before Storm}}{\text{Speed before Storm}} + \frac{\text{Distance in Storm}}{\text{Speed in Storm}} = \text{Total Time} \Longrightarrow \frac{\tfrac{x}{3}}{\tfrac{x}{180}} + \frac{\tfrac{2x}{3}}{\tfrac{x}{180} - \tfrac{1}{3}} = 276\]
Solving equation, we get $x=135$ $\Longrightarrow \boxed{135}$
| 135
|
4,300
|
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_14
| 1
|
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 16 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 40$
|
Alice may sit in the center chair, in an end chair, or in a next-to-end chair. Suppose she sits in the center chair. The 2nd and 4th chairs (next to her) must be occupied by Derek and Eric, in either order, leaving the end chairs for Bob and Carla in either order; this yields $2! * 2! = 4$ ways to seat the group.
Next, suppose Alice sits in one of the end chairs. Then the chair beside her will be occupied by either Derek or Eric. The center chair must be occupied by Bob or Carla, leaving the last two people to fill the last two chairs in either order. $2$ ways to seat Alice times $2$ ways to fill the next chair times $2$ ways to fill the center chair times $2$ ways to fill the last two chairs yields $2 * 2 * 2 * 2 = 16$ ways to fill the chairs.
Finally, suppose Alice sits in the second or fourth chair. Then the chairs next to her must be occupied by Derek and Eric in either order, and the other two chairs must be occupied by Bob and Carla in either order. This yields $2 * 2 * 2 = 8$ ways to fill the chairs.
In total, there are $4 + 8 + 16$ ways to fill the chairs, so the answer is $\boxed{28}$
| 28
|
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