id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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4,201 | 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 ang... | 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, ... | 80 |
4,202 | 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 ang... | 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... | 80 |
4,203 | 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 ang... | 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 $D$ , whi... | 80 |
4,204 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_23 | 5 | 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 ang... | 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 |
4,205 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_23 | 6 | 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 ang... | 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 PA... | 80 |
4,206 | 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 ang... | 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 |
4,207 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_25 | 1 | 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... | 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}\r... | 18 |
4,208 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_25 | 2 | 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... | 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 make... | 18 |
4,209 | 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 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}... | 18 |
4,210 | 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... | By PaperMath’s sum , the answer is $6+8+4=\boxed{18}$ | 18 |
4,211 | 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 |
4,212 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_1 | 2 | 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 |
4,213 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_3 | 1 | 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 |
4,214 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_3 | 2 | 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 $x$ -inte... | 10 |
4,215 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_4 | 1 | 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... | 50 |
4,216 | 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 |
4,217 | 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) } \... | 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 |
4,218 | 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) } \... | 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{c... | 4 |
4,219 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_7 | 1 | 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 |
4,220 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_7 | 2 | 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 \... | 6 |
4,221 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_8 | 1 | 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 ... | 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 */... | 50 |
4,222 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_8 | 2 | 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 ... | 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 ... | 50 |
4,223 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_8 | 3 | 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 ... | 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 inscri... | 50 |
4,224 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9 | 1 | 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... | 0 |
4,225 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9 | 2 | 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 + ... | 0 |
4,226 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9 | 3 | 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} \su... | 0 |
4,227 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9 | 4 | 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 |
4,228 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9 | 5 | 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{arra... | 0 |
4,229 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_9 | 6 | 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^... | 0 |
4,230 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_10 | 1 | 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?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$ | 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 |
4,231 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_10 | 2 | 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?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$ | 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 canno... | 225 |
4,232 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_12 | 1 | 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) }2... | 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... | 18 |
4,233 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_13 | 1 | 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 =... | 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 = (... | 200 |
4,234 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_13 | 3 | 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 =... | 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,... | 200 |
4,235 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_13 | 4 | 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 =... | 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 \[... | 200 |
4,236 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_14 | 1 | 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 i... | 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 ... | 11 |
4,237 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_14 | 2 | 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 i... | 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 $9$ solut... | 11 |
4,238 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_14 | 3 | 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 i... | 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 Zo... | 11 |
4,239 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_14 | 4 | 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 i... | 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 $1$... | 11 |
4,240 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15 | 1 | How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$ | 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 ... | 96 |
4,241 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15 | 2 | How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$ | 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... | 96 |
4,242 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15 | 3 | How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$ | 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 condi... | 96 |
4,243 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15 | 4 | How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$ | 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. Th... | 96 |
4,244 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15 | 5 | How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$ | 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, $\frac{9}{10}$ $\t... | 96 |
4,245 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15 | 6 | How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$ | 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 $2$ : _ _ $5$ :... | 96 |
4,246 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_15 | 7 | How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$ | 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 $3$ ... | 96 |
4,247 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 1 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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... | 7 |
4,248 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 2 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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-a... | 7 |
4,249 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 3 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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 + 15... | 7 |
4,250 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 4 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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}... | 7 |
4,251 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 5 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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 $\boxed{7}.$ | 7 |
4,252 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 6 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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... | 7 |
4,253 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 7 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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}{... | 7 |
4,254 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 8 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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 denom... | 7 |
4,255 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 9 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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 $16-9= \boxed{7}.$ | 7 |
4,256 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 11 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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... | 7 |
4,257 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | 12 | 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 $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | 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 $p=9.$ Th... | 7 |
4,258 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18 | 1 | 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 $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | 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... | 17 |
4,259 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18 | 2 | 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 $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | 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 ... | 17 |
4,260 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18 | 3 | 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 $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | 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 m... | 17 |
4,261 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18 | 4 | 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 $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | 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 o... | 17 |
4,262 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18 | 5 | 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 $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | 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{20... | 17 |
4,263 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18 | 6 | 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 $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | \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 ... | 17 |
4,264 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18 | 7 | 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 $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | 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... | 17 |
4,265 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_19 | 3 | 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 \t... | 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 $323 + 17 = \boxed{340}$ | 340 |
4,266 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_19 | 5 | 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 \t... | 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 $4$ di... | 340 |
4,267 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_21 | 1 | 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\tex... | 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 circumradi... | 72 |
4,268 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_22 | 1 | 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 $P(-1) = -9$
$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$ | 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 orde... | 220 |
4,269 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_22 | 2 | 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 $P(-1) = -9$
$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$ | 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 ... | 220 |
4,270 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_23 | 1 | 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 degre... | 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 o... | 120 |
4,271 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_23 | 2 | 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 degre... | 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 $xyz$ ... | 120 |
4,272 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | 1 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | 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 visual... | 199 |
4,273 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | 2 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | 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 {-... | 199 |
4,274 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | 3 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | 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 th... | 199 |
4,275 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | 4 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | 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 $... | 199 |
4,276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | 5 | 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 rea... | 199 |
4,277 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | 6 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ . How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | 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 contradi... | 199 |
4,278 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_24 | 7 | 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 |
4,279 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | 1 | 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 $\om... | 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 $m\angle{O_2P_2P_1} = 30^\circ$ . M... | 552 |
4,280 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | 2 | 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 $\om... | 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 ... | 552 |
4,281 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | 3 | 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 $\om... | 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 $\trian... | 552 |
4,282 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | 4 | 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 $\om... | [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); de... | 552 |
4,283 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | 5 | 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 $\om... | 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 l... | 552 |
4,284 | 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 $\om... | 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 \... | 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... | 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... | 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 S... | 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... | 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 pe... | 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... | 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... | 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... | 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... | 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... | 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 = ... | 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 m... | 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 ... | 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... | 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)}\... | 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 t... | 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 $5$ posit... | 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 t... | 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 $12.$ ... | 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 $2... | 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: \[... | 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,... | 28 |
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