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026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(A--E); draw(E--B); draw(C--F); draw(F--D); label("$E$",E,NW); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); draw(rightanglemark(A,E,B)); draw(rightanglemark(C,F,D)); [/asy]
Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$ , respectively. Also, let $\theta = \angle BOC$ . From the Law of Cosines on $\triangle BOC$ , we have $\cos \theta = \frac{3}{4}$
Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$ , we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$ . In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\triangle OBC \cong \triangle OCD$ , so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$ , so $\angle DCF = \theta$
Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$ , so $FC = 150$ . Similarly, $BE = 150$ , and $AD = 150 + 200 + 150 = \boxed{500}$ | null | 500 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); [/asy]
Let $s = 200$ . Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$ . Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$ , hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$
Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.$ This gives us: \[\boxed{500.}\] | E | 500. |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}$ . Similarly, the cosine is $\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}$ .
Since there are three sides, and since $\sin\theta=\sin\left(180-\theta\right)$ ,we seek to find $2r\sin 3\theta$ .
First, $\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}$ and $\cos 2\theta=\frac{3}{4}$ by Pythagorean. \[\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}\] \[2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{500}\] | E | 500 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\sqrt{OB^2-BM^2}=\sqrt7$ and $ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}$ . Thus the height of the trapezoid is $\sqrt7-\sqrt{8-x^2}$ , so the area is $\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})$ . By Brahmagupta's formula , the area is $\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Setting these two equal, we get $(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Dividing both sides by $x+1$ and then squaring, we get $7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)$ . Expanding the right hand side and canceling the $x^2$ terms gives us $15-2(\sqrt7)(\sqrt{8-x^2})=2x+3$ . Rearranging and dividing by two, we get $(\sqrt7)(\sqrt{8-x^2})=6-x$ . Squaring both sides, we get $56-7x^2=x^2-12x+36$ . Rearranging, we get $8x^2-12x-20=0$ . Dividing by 4 we get $2x^2-3x-5=0$ . Factoring we get, $(2x-5)(x+1)=0$ , and since $x$ cannot be negative, we get $x=2.5$ . Since $DA=2x$ $DA=5$ . Scaling up by 100, we get $\boxed{500}$ | E | 500 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations, L is used to write alpha= statement real RADIUS; pair A, B, C, D, E, F, O, L; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(A,D,O,B); F=extension(A,D,O,C); L=midpoint(C--D); O=(0,0); //Path Definitions path quad = A -- B -- C -- D -- cycle; //Initial Diagram draw(circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,NW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,NE); label("$E$",E,SW); label("$F$",F,SE); label("$O$",O,SE); dot(O,linewidth(5)); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0)); draw(anglemark(C,O,B)); label("$\theta$",O,3N); draw(anglemark(E,F,O)); label("$\alpha$",F,3SW); draw(anglemark(D,F,C)); label("$\alpha$",F,3NE); draw(anglemark(F,C,D)); label("$\alpha$",C,3SSE); draw(anglemark(C,D,F)); label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW); [/asy] Label the points as shown, and let $\angle{EOF} = \theta$ . Since $\overline{OB} = \overline{OC}$ , and $\triangle{OFE} \sim \triangle{OCB}$ , we get that $\angle{EFO} = 90-\frac{\theta}{2}$ . We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$ . Also, since $\triangle{OCB} \cong \triangle{ODC}$ $\angle{OCD} = \alpha$ . This means that $\angle{CDF} = 180-2\alpha = \theta$ , which leads to $\triangle{OCB} \sim \triangle{DCF}$ .
Since we know that $\overline{CD} = 200$ $\overline{DF} = 200$ , and by similar reasoning $\overline{AE} = 200$ .
Finally, again using similar triangles, we get that $\overline{CF} = 100\sqrt{2}$ , which means that $\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}$ . We can again apply similar triangles (or use Power of a Point) to get $\overline{EF} = 100$ , and finally $\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{500}$ - ColtsFan10 | E | 500 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | We first scale down by a factor of $200\sqrt{2}$ . Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ , so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$ , and we let $C$ correspond to the complex number $z$ . Then, $A$ corresponds to $z^3$ and $B$ corresponds to $z^2$ . We are given that $\lvert z \rvert = 1$ and $\lvert z-1 \rvert = 1/\sqrt{2}$ , and we wish to find $\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert$ . Let $z=a+bi$ , where $a$ and $b$ are real numbers. Then, $a^2+b^2=1$ and $a^2-2a+1+b^2=1/2$ ; solving for $a$ and $b$ yields $a=3/4$ and $b=\sqrt{7}/4$ . Thus, $AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}$ . Scaling back up gives us a final answer of $\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{500}$ | E | 500 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | Let angle $C$ be $2a$ . This way $BD$ will be $400sin(a)$ . Now we can trig bash. As the circumradius of triangle $BCD$ is $200\sqrt{2}$ , we can use the formula \[R=\frac{abc}{4A}\] and \[A=\frac{absin(C)}{2}\] and plug in all the values we got to get \[200\sqrt{2}=\frac{200^2 \cdot 400sin(a)}{4 \cdot (\frac{200^2 sin(2a)}{2})}\] . This boils down to \[\sqrt{2}=\frac{sin(a)}{sin{2a}}\] . This expression can further be simplified by the trig identity \[sin(2a)=2sin(a)cos(a)\] . This leads to the final simplified form \[2\sqrt{2}=\frac{1}{cos(a)}\] . Solving this expression gives us \[cos(a)=\frac{\sqrt{2}}{4}\] . However, as we want $sin(a)$ , we use the identity $sin^2+cos^2=1$ , and substitute to get that $sin(a)=\frac{\sqrt{14}}{4}$ , and therefore BD is $100\sqrt{14}$
Then, as $ABCD$ is a cyclic quadrilateral, we can use Ptolemy’s Theorem (with $AD=x$ ) to get \[14 \cdot 100^2=200x+200^2\] . Finally, we solve to get $\boxed{500}$ | E | 500 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); [/asy] Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$ . Hence, $[ABCD]$ is an isosceles trapezoid.
Let $\angle CDA=\alpha$ . Notice that the length of the altitude from $C$ to $AD$ is $200sin(\alpha)$ . Furthermore, the length of the altitude from $O$ to $BC$ is $100\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\sqrt{7}-200sin\alpha$ . Let $F$ the feet of the altitude from $O$ to $AD$ . Then, $FD=(200+400cos(\alpha))/2=100+200cos(\alpha)$ , because $AOD$ is isosceles.
Therefore, by the Pythagorean theorem, $(100+200cos(\alpha))^2+(100\sqrt{7}-200sin(\alpha))^2=80000$ . Simplifying, we have $1+cos(\alpha)=sin(\alpha) \cdot sqrt{7} \implies cos^2(\alpha)+2cos(\alpha)+1=sin^2(\alpha) \cdot 7 = 7-7cos^2(\alpha) \implies 8cos^2(\alpha)+2cos(\alpha) - 6 =0$ . Solving this quadratic, we have $cos(\alpha)=\frac{3}{4}, -1$ , but $0<\alpha<180 \implies cos(\alpha)=3/4$ . Therefore, $AD=200cos(\alpha)+200cos(\alpha)+200=\boxed{500}$ | null | 500 |
a2726b2bf367e4476f5ad377f5a830ac | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_25 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | We prime factorize $72,600,$ and $900$ . The prime factorizations are $2^3\times 3^2$ $2^3\times 3\times 5^2$ and $2^2\times 3^2\times 5^2$ , respectively. Let $x=2^a\times 3^b\times 5^c$ $y=2^d\times 3^e\times 5^f$ and $z=2^g\times 3^h\times 5^i$ . We know that \[\max(a,d)=3\] \[\max(b,e)=2\] \[\max(a,g)=3\] \[\max(b,h)=1\] \[\max(c,i)=2\] \[\max(d,g)=2\] \[\max(e,h)=2\] and $c=f=0$ since $\text{lcm}(x,y)$ isn't a multiple of 5. Since $\max(d,g)=2$ we know that $a=3$ . We also know that since $\max(b,h)=1$ that $e=2$ . So now some equations have become useless to us...let's take them out. \[\max(b,h)=1\] \[\max(d,g)=2\] are the only two important ones left. We do casework on each now. If $\max(b,h)=1$ then $(b,h)=(1,0),(0,1)$ or $(1,1)$ . Similarly if $\max(d,g)=2$ then $(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)$ . Thus our answer is $5\times 3=\boxed{15}.$ | A | 15 |
a2726b2bf367e4476f5ad377f5a830ac | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_25 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | It is well known that if the $\text{lcm}(a,b)=c$ and $c$ can be written as $p_1^ap_2^bp_3^c\dots$ , then the highest power of all prime numbers $p_1,p_2,p_3\dots$ must divide into either $a$ and/or $b$ . Or else a lower $c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots$ is the $\text{lcm}$
Start from $x$ $\text{lcm}(x,y)=72$ so $8\mid x$ or $9\mid x$ or both. But $9\nmid x$ because $\text{lcm}(x,z)=600$ and $9\nmid 600$ .
So $x=8,24$
$y$ can be $9,18,36$ in both cases of $x$ but NOT $72$ because $\text{lcm}(y,z)=900$ and $72\nmid 900$
So there are six sets of $x,y$ and we will list all possible values of $z$ based on those.
$25\mid z$ because $z$ must source all powers of $5$ $z\in\{25,50,75,100,150,300\}$ $z\ne\{200,225\}$ because of $\text{lcm}$ restrictions.
By different sourcing of powers of $2$ and $3$
\[(8,9):z=300\] \[(8,18):z=300\] \[(8,36):z=75,150,300\] \[(24,9):z=100,300\] \[(24,18):z=100,300\] \[(24,36):z=25,50,75,100,150,300\]
$z=100$ is "enabled" by $x$ sourcing the power of $3$ $z=75,150$ is uncovered by $y$ sourcing all powers of $2$ . And $z=25,50$ is uncovered by $x$ and $y$ both at full power capacity.
Counting the cases, $1+1+3+2+2+6=\boxed{15}.$ | A | 15 |
a2726b2bf367e4476f5ad377f5a830ac | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_25 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | As said in previous solutions, start by factoring $72, 600,$ and $900$ . The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\] To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows such that $x,y, \text{and } z$ are the vertices and the LCMs are on the edges.
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^33^25^0$",X--Y,2W); label("$2^33^15^2$",X--Z,2E); label("$2^23^25^2$",Y--Z,2S); [/asy]
Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and } 5$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^3$",X--Y,2W); label("$2^3$",X--Z,2E); label("$2^2$",Y--Z,2S); [/asy]
Analyzing for powers of $2$ , it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$ . Turning towards the vertices $y$ and $z$ , we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$3^2$",X--Y,2W); label("$3^1$",X--Z,2E); label("$3^2$",Y--Z,2S); [/asy]
Using the same logic as we did for powers of $2$ , it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \text{ and } z$ , we can use the same logic to find the only ordered pairs $(x,z)$ that will work are $\{(1,0)(0,1)(1,1)\}$
The final and last case is the powers of $5$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$5^0$",X--Y,2W); label("$5^2$",X--Z,2E); label("$5^2$",Y--Z,2S); [/asy]
This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \text{ and } y$ cannot have a factor of $5$ in their prime factorization.
Multiplying all the possible arrangements for prime factors $2,3, \text{ and } 5$ , we get the answer: \[5\cdot3\cdot1=\boxed{15}\] | A | 15 |
973bfb866e09015c9c772a9441eae04a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_1 | What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \tfrac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | Factorizing the numerator, $\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}$ then becomes $\frac{\frac{5}{2}}{a^{2}}$ which is equal to $\frac{5}{2}\cdot 2^2$ which is $\boxed{10}$ | D | 10 |
973bfb866e09015c9c772a9441eae04a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_1 | What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \tfrac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | Substituting $\frac{1}{2}$ for $a$ in $\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}$ gives us $\boxed{10}$ . ~peelybonehead | D | 10 |
5db511e260a007bf05a2c7a03cea0a2a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_3 | Let $x=-2016$ . What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | Substituting carefully, $\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)$
becomes $|4032-2016|+2016=2016+2016=4032$ which is $\boxed{4032}$ | D | 4032 |
5db511e260a007bf05a2c7a03cea0a2a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_3 | Let $x=-2016$ . What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | Solution by e_power_pi_times_i
Substitute $-y = x = -2016$ into the equation. Now, it is $\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y$ . Since $y = 2016$ , it is a positive number, so $|y| = y$ . Now the equation is $\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y$ . This further simplifies to $2y-y+y = 2y$ , so the answer is $\boxed{4032}$ | D | 4032 |
b5aaa1fdf58908caf63dfef406617b97 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_5 | The mean age of Amanda's $4$ cousins is $8$ , and their median age is $5$ . What is the sum of the ages of Amanda's youngest and oldest cousins?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25$ | The sum of the ages of the cousins is $4$ times the mean, or $32$ .
There are an even number of cousins, so there is no single median, so $5$ must be the mean of the two in the middle.
Therefore the sum of the ages of the two in the middle is $10$ . Subtracting $10$ from $32$ produces $\textbf{(D)}\ \boxed{22}$ | null | 22 |
796a9d89d90463f4d4ad24da7833f322 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_6 | Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$ . What is the smallest possible value for the sum of the digits of $S$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$ | Let the two three-digit numbers she added be $a$ and $b$ with $a+b=S$ and $a<b$ . The hundreds digits of these numbers must be at least $1$ and $2$ , so $a\ge 100$ and $b\ge 200$
Say $a=100+p$ and $b=200+q$ ; then we just need $p+q=100$ with $p$ and $q$ having different digits which aren't $1$ or $2$ .There are many solutions, but $p=3$ and $q=97$ give $103+297=400$ which proves that $\boxed{4}$ is attainable. | B | 4 |
796a9d89d90463f4d4ad24da7833f322 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_6 | Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$ . What is the smallest possible value for the sum of the digits of $S$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$ | For this problem, to find the $3$ -digit integer with the smallest sum of digits, one should make the units and tens digit add to $0$ . To do that, we need to make sure the digits are all distinct. For the units digit, we can have a variety of digits that work. $7$ works best for the top number which makes the bottom digit $3$ . The tens digits need to add to $9$ because of the $1$ that needs to be carried from the addition of the units digits. We see that $5$ and $4$ work the best as we can't use $6$ and $3$ . Finally, we use $2$ and $1$ for our hundreds place digits.
Adding the numbers $257$ and $143$ , we get $400$ which means our answer is $\boxed{4}$ | B | 4 |
feb36a6f7fb03dfc794a098df9f9bbf9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_7 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | We can set up a system of equations where $x$ and $y$ are the two acute angles. WLOG, assume that $x$ $<$ $y$ in order for the complement of $x$ to be greater than the complement of $y$ . Therefore, $5x$ $=$ $4y$ and $90$ $-$ $x$ $=$ $2$ $(90$ $-$ $y)$ . Solving for $y$ in the first equation and substituting into the second equation yields \[\begin{split} 90 - x & = 2 (90 - 1.25x) \\ 1.5x & = 90 \\ x & = 60 \end{split}\]
Substituting this $x$ value back into the first equation yields $y$ $=$ $75$ , leaving $x$ $+$ $y$ equal to $\boxed{135}$ | C | 135 |
feb36a6f7fb03dfc794a098df9f9bbf9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_7 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | We let the measures be $5x$ and $4x$ giving us the ratio of $5:4$ . We know $90-4x>90-5x$ since this inequality gives $x>0$ , which is true since the measures of angles are never negative. We also know the bigger complement is twice the smaller, so
$90-4x=2(90-5x)$
$90-4x=180-10x$
$6x=90$
$x=15$
Therefore, the angles are $75$ and $60$ , which sum to $\boxed{135}$ | C | 135 |
089944d922099390ae23a937f1e00157 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_8 | What is the tens digit of $2015^{2016}-2017?$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8$ | Notice that, for $n\ge 2$ $2015^n\equiv 15^n$ is congruent to $25\pmod{100}$ when $n$ is even and $75\pmod{100}$ when $n$ is odd. (Check for yourself). Since $2016$ is even, $2015^{2016} \equiv 25\pmod{100}$ and $2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}$
So the answer is $\textbf{(A)}\ \boxed{0}$ | null | 0 |
089944d922099390ae23a937f1e00157 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_8 | What is the tens digit of $2015^{2016}-2017?$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8$ | Since we only need the tens digits, we only need to care about the multiplication of tens and ones. (If you want to use mathematical terms then we only need to look at the exponents in $mod 100$ .) We will use the " $\equiv$ " sign to denote congruence in modulus, basically taking the last two digits and ignoring everything else.
\[\begin{split} 15^{1}\equiv15 \\ 15^{2}\equiv25 \end{split}\] From here we only need to multiply $15\cdot25$ and we can ignore the hundreds digits. \[\begin{split} 15^{3}\equiv75 \\ 15^{4}\equiv25 \\ 15^{5}\equiv75 \end{split}\] Notice that for every $x\neq1$ $15^{x}\equiv25$ if $x$ is even, and $15^{x}\equiv75$ if $x$ is odd. Since $2015^{2016}$ has an even exponent, we conclude that the last two digits will be $25$ , and subtracting $25-17=0 \Longrightarrow \boxed{0}$ .
~JH. L | A | 0 |
359798c6a9b6e3dca1b7bc061d971493 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_9 | All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$ , with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64$ . What is the length of $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$ | Let the point where the height of the triangle intersects with the base be $D$ . Now we can guess what $x$ is and find $y$ . If $x$ is $3$ , then $y$ is $9$ . The cords of $B$ and $C$ would be $(-3,9)$ and $(3,9)$ , respectively. The distance between $B$ and $C$ is $6$ , meaning the area would be $\frac{6 \cdot 9}{2}=27$ , not $64$ . Now we let $x=4$ $y$ would be $16$ . The cords of $B$ and $C$ would be $(-4,16)$ and $(4,16)$ , respectively. $BC$ would be $8$ , and the height would be $16$ . The area would then be $\frac{8 \cdot 16}{2}$ which is $64$ , so $BC$ is $\boxed{8}$ | C | 8 |
786cfe97132cfd9a4a54f8abf975f396 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_10 | A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$ | We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
$\left(\frac{3}{5}\right)^2=\frac{12}{x}$
We can then solve the equation to get $x=\frac{100}{3}$ which is closest to $\boxed{33.3}$ | D | 33.3 |
786cfe97132cfd9a4a54f8abf975f396 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_10 | A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$ | Also recall that the area of an equilateral triangle is $\frac{a^2\sqrt3}{4}$ so we can give a ratio as follows:
$\frac{\frac{9\sqrt3}{4}}{12}$ $=$ $\frac{\frac{25\sqrt3}{4}}{x}$
Cross multiplying and simplifying, we get $12 \cdot \frac{25}{9}$
Which is $33.\overline{3}$ $\approx$ $\boxed{33.3}$ | D | 33.3 |
786cfe97132cfd9a4a54f8abf975f396 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_10 | A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$ | Note that the ratio of the two triangle's weights is equal to the ratio of their areas, as the height is the same. The ratio of their areas is equal to the square of the ratio of their sides. So if $x$ denotes the weight of the second triangle, we have \[\frac{x}{12}=\frac{5^2}{3^2}=\frac{25}{9}\] Solving gives us $x \approx 33.33$ so the answer is $\boxed{33.3}$ | D | 33.3 |
6486448cc2ce29bdf9b1dce2282215bb | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_11 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$ | If the dimensions are $4a\times 4b$ , then one side will have $a+1$ posts (including corners) and the other $b+1$ (including corners).
The total number of posts is $2(a+b)=20$
This diagram represents the number of posts around the garden. [asy]size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen); for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7)); draw(arc((0,0),.5,-90,-270)--arc((4,0),.5,90,-90)--cycle,gray+dotted); draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted); draw((0,-1)--(5,-1),Arrows);draw((6,0)--(6,7),Arrows); label("$a+1$",(0,-1)--(5,-1),S);label("$b+1$",(6,0)--(6,7),E); label("$a$",(1,1));label("$b$",(4,5)); [/asy]
We solve the system $\begin{cases}b+1=2(a+1)\\a+b=10\end{cases}$ to get $a=3,\ b=7$ . Then the area is $4a\cdot 4b=336$ which is $\boxed{336}$ | B | 336 |
6486448cc2ce29bdf9b1dce2282215bb | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_11 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$ | To do this problem, we have to draw a rectangle. We also have to keep track of the fence posts. Putting a post on each corner leaves us with only $16$ posts. Now there are twice as many posts on the longer side then the shorter side. From this, we can see that we can put $8$ posts on the longer side and $4$ posts on the shorter side. On the shorter side, we have $3$ spaces between the $4$ posts. On the longer side, we have $7$ spaces between the $8$ fence posts. There are $4$ yards between each post. Therefore, the answer is $12\times28=\boxed{336}$ | B | 336 |
0bdeb97ca2aed9b3870a3969e433a764 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_12 | Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?
$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$ | The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is $\frac{\tbinom32}{\tbinom52}=\frac3{10}$ , so the answer is $1-0.3$ which is $\boxed{0.7}$ | D | 0.7 |
0bdeb97ca2aed9b3870a3969e433a764 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_12 | Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?
$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$ | There are $2$ cases to get an even number. Case 1: $\text{Even} \times \text{Even}$ and Case 2: $\text{Odd} \times \text{Even}$ . Thus, to get an $\text{Even} \times \text{Even}$ , you get $\frac {\binom {2}{2}}{\binom {5}{2}}= \frac {1}{10}$ . And to get $\text{Odd} \times \text{Even}$ , you get $\frac {\binom {3}{1}}{\binom {5}{2}}= \frac {6}{10}$ $\frac {1}{10}+\frac {6}{10}=\frac {7}{10}$ which is $0.7$ and the answer is $\boxed{0.7}$ | D | 0.7 |
0bdeb97ca2aed9b3870a3969e433a764 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_12 | Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?
$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$ | Note that we have three cases to get an even number: even $\times$ even, odd $\times$ even and even $\times$ odd.
The probability of case 1 is $\dfrac{2}{5}\cdot\dfrac{1}{4}$ , the probability of case 2 is $\dfrac{2}{5}\cdot\dfrac{3}{4}$ and the probability of case 3 is $\dfrac{3}{5}\cdot\dfrac12$
Adding these up we get $\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{3}{10} = \boxed{0.7}.$ | D | 0.7 |
ca61fb544748ea47765de7124a17a205 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_13 | At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000$ babies were in sets of quadruplets?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160$ | We can set up a system of equations where $a$ is the sets of twins, $b$ is the sets of triplets, and $c$ is the sets of quadruplets. \[\begin{split} 2a + 3b + 4c & = 1000 \\ b & = 4c \\ a & = 3b \end{split}\]
Solving for $c$ and $a$ in the second and third equations and substituting into the first equation yields \[\begin{split} 2 (3b) + 3b + 4 (0.25b) & = 1000 \\ 6b + 3b + b & = 1000 \\ b & = 100 \end{split}\]
Since we are trying to find the number of babies and NOT the number of sets of quadruplets, the solution is not $c$ , but rather $4c$ . Therefore, we strategically use the second initial equation to realize that $b$ $=$ $4c$ , leaving us with the number of babies born as quadruplets equal to $\boxed{100}$ | D | 100 |
ca61fb544748ea47765de7124a17a205 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_13 | At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000$ babies were in sets of quadruplets?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160$ | Say there are $12x$ sets of twins, $4x$ sets of triplets, and $x$ sets of quadruplets. That's $12x\cdot2=24x$ twins, $4x\cdot3=12x$ triplets, and $x\cdot4=4x$ quadruplets. A tenth of the babies are quadruplets and that's $\frac{1}{10}(1000)=\boxed{100}$ | D | 100 |
ab9458bad984170648c7e26cca586e02 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_14 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$ | The region is a right triangle which contains the following lattice points: $(0,0); (1,0)\rightarrow(1,3); (2,0)\rightarrow(2,6); (3,0)\rightarrow(3,9); (4,0)\rightarrow(4,12); (5,0)\rightarrow(5,15)$
[asy]size(10cm); for(int i=0;i<6;++i)for(int j=0;j<=3*i;++j)dot((i,j)); draw((0,-1)--(0,16),EndArrow);draw((-1,0)--(6,0),EndArrow); draw((-.5,-pi/2)--(5.2,5.2*pi),gray);draw((-1,-.1)--(6,-.1)^^(5.1,-1)--(5.1,16),gray); [/asy]
Squares $1\times 1$ :
Suppose that the top-right corner is $(x,y)$ , with $2\le x\le 5$ . Then to include all other corners, we need $1\le y\le 3(x-1)$ .
This produces $3+6+9+12=30$ squares.
Squares $2\times 2$ :
Here $3\le x\le 5$ . To include all other corners, we need $2\le y\le 3(x-2)$ .
This produces $2+5+8=15$ squares.
Squares $3\times 3$ :
Similarly, this produces $5$ squares.
No other squares will fit in the region. Therefore the answer is $\boxed{50}$ | D | 50 |
ab9458bad984170648c7e26cca586e02 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_14 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$ | The vertical line is just to the right of $x = 5$ , the horizontal line is just under $y = 0$ , and the sloped line will always be above the $y$ value of $3x$ .
This means they will always miss being on a coordinate with integer coordinates so you just have to count the number of squares to the left, above, and under these lines. After counting the number of $1\cdot1$ $2\cdot2$ , and $3\cdot3$ squares and getting $30$ $15$ , and $5$ respectively, and we end up with $\boxed{50}$ | D | 50 |
ab9458bad984170648c7e26cca586e02 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_14 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$ | The endpoint lattice points are $(1,3), (2,6), (3,9), (4,12), (5,15).$ Now we split this problem into cases.
Case 1: Square has length $\bf1$
The $x$ coordinates must be $(1,2)$ or $(2,3)$ and so on to $(4,5).$ The idea is that you start at $y=1$ and add at the endpoint, namely $y=3.$ The number ends up being $3+6+9+12 = 30$ squares for this case.
Case 2: Square has length $\bf2$
The $x$ coordinates must be $(1,3)$ or $(2,4)$ or $(3,15)$ and so now it starts at $y=2.$ It ends up being $2+5+8 = 15.$
Case 3: Square has length $\bf3$
The $x$ coordinates must be $(1,4)$ or $(2,5)$ so there is $1+4 = 5$ squares for this case.
Therefore, $30+15+5 = \boxed{50}$ | D | 50 |
7e8541822e50425b2a46e2e9a36027c5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_15 | All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity: [asy]size(4cm); for(int i=0;i<3;++i)for(int j=0;j<3;++j)filldraw(box((i,j),(i+1,j+1)),gray((i+j)%2*.2+.7));[/asy] But since there are only four even numbers in the set, the five darker squares must contain the odd numbers, which sum to $1+3+5+7+9=25.$ Therefore if the sum of the numbers in the corners is $18$ , the number in the center must be $\boxed{7}$ | C | 7 |
7e8541822e50425b2a46e2e9a36027c5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_15 | All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | Quick testing shows that \[3~2~1\] \[4~7~8\] \[5~6~9\] is a valid solution. $3+1+5+9 = 18$ , and the numbers follow the given condition. The center number is found to be $\boxed{7}$ .. — @adihaya talk ) 12:27, 21 February 2016 (EST) ~edited | C | 7 |
7e8541822e50425b2a46e2e9a36027c5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_15 | All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | First let the numbers be \[1 ~8~ 7\] \[2 ~ 9 ~6\] \[3 ~ 4~ 5\] with the numbers $1-8$ around the outsides and $9$ in the middle. We see that the sum of the four corner numbers is $16$ . If we switch $7$ and $9$ , then the corner numbers will add up to $18$ and the consecutive numbers will still be touching each other. The answer is $\boxed{7}$ . ~edited | C | 7 |
7e8541822e50425b2a46e2e9a36027c5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_15 | All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | Testing out the box with the center square taking on the value of 5 and 6, we find that they either do not satisfy the first or the second condition. Testing 7, we find that a valid configuration is \[1 ~8~ 9\] \[2 ~ 7 ~6\] \[3 ~ 4~ 5\]
$\boxed{7}$ | C | 7 |
401f08d599db85464c4422e600481e68 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | The sum of an infinite geometric series is of the form: \[\begin{split} S & = \frac{a_1}{1-r} \end{split}\] where $a_1$ is the first term and $r$ is the ratio whose absolute value is less than 1.
We know that the second term is the first term multiplied by the ratio.
In other words: \[\begin{split} a_1 \cdot r & = 1 \\ a_1 & = \frac{1}{r} \end{split}\]
Thus, the sum is the following: \[\begin{split} S & = \frac{\frac{1}{r}}{1-r} \\\\ S & =\frac{1}{r-r^2} \end{split}\]
Since we want the minimum value of this expression, we want the maximum value for the denominator, $-r^2+r$ .
The maximum x-value of a quadratic with leading coefficient $-a$ is $\frac{-b}{2a}$ \[\begin{split} r & = \frac{-(1)}{2(-1)} \\\\ r & = \frac{1}{2} \end{split}\]
Plugging $r$ $=$ $\frac{1}{2}$ into the quadratic yields: \[\begin{split} S & = \frac{1}{\frac{1}{2} -\left(\frac{1}{2}\right)^2} \\\\ S & = \frac{1}{\frac{1}{4}} \end{split}\]
Therefore, the minimum sum of our infinite geometric sequence is $\boxed{4}$ .
(Solution by akaashp11) | E | 4 |
401f08d599db85464c4422e600481e68 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | After observation we realize that in order to minimize our sum $\frac{a}{1-r}$ with $a$ being the reciprocal of r. The common ratio $r$ has to be in the form of $\frac{1}{x}$ with $x$ being an integer as anything more than $1$ divided by $x$ would give a larger sum than a ratio in the form of $\frac{1}{x}$
The first term has to be $x$ , so then in order to minimize the sum, we have minimize $x$
The smallest possible value for $x$ such that it is an integer that's greater than $1$ is $2$ . So our first term is $2$ and our common ratio is $\frac{1}{2}$ . Thus the sum is $\frac{2}{\frac {1}{2}}$ or $\boxed{4}$ .
Solution 2 by No_One | E | 4 |
401f08d599db85464c4422e600481e68 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | We can see that if $a$ is the first term, and $r$ is the common ratio between each of the terms, then we can get \[S=\frac{a}{1-r} \implies S-Sr=a\] Also, we know that the second term can be expressed as $a\cdot r$ notice if we multiply $S-Sr=a$ by $r$ , we would get \[r(S-Sr)=ar \implies Sr-Sr^2=1 \implies Sr^2-Sr+1=0\] This quadratic has real solutions if the discriminant is greater than or equal to zero, or \[S^2-4\cdot S \cdot 1 \ge 0\] This yields that $S\le 0$ or $S\ge 4$ .
However, since we know that $S$ has to be positive, we can safely conclude that the minimum possible value of $S$ is $\boxed{4}$ | E | 4 |
401f08d599db85464c4422e600481e68 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Let the first term of the geometric series $x$ . Since it must be decreasing, we have $x>1$ and the third term is $\frac{1}{x}$ . Realize that by AM-GM inequality $x+\frac{1}{x} \ge 2$ with equality if $x = 1$ . However, we established that $x>1$ so that means $x+\frac{1}{x} > 2$ . So the sum of the first three terms of the sequence $x + \frac{1}{x} + 1$ is greater than $3$ , and the geometric series keeps continuing infinitely. This means the sum continues increasing. The only answer choice greater than $3$ is $\boxed{4}$ . ~skyscraper | E | 4 |
401f08d599db85464c4422e600481e68 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Let the first term be $k.$ The sum of the series is $\frac{k}{1- \frac{1}{k}} =\frac{k^2}{k-1}.$ Rewrite this as $\frac{k^2-2k+1}{k-1} +\frac{2k-1}{k-1} = k-1+\frac{2k-2}{k-1} +\frac{1}{k-1} = (k-1) + \left(\frac{1}{k-1}\right) + 2.$ By AM-GM we know that $(k-1) + \left(\frac{1}{k-1}\right) \ge 2$ so the minimum is $2+2 = \boxed{4}.$ | E | 4 |
401f08d599db85464c4422e600481e68 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Set the first term is $a.$ , the common ratio should be $\frac{1}{a}.$
The sum to infinity of the series is $S=\frac{a}{1-\frac{1}{a}}=\frac{a^2}{a-1}.$
Since $S$ is positive, we have $a>1.$ Define the function $f(a)=\frac{a^2}{a-1}$ , the domain of this function is $a>1.$
Let $f^{'}(a)=\frac{2a^2-2a-a^2}{(a-1)^2}=\frac{a(a-2)}{(a-1)^2}=0.$ We solve that $a=2.$
It's easy to find that when $1<a<2, f^{'}(a)<0,$ when $a>2, f^{'}(a)>0.$ Thus $f(a)$ has a minimum value when $a=2.$ , which is $f(2)=4.$ Choose $\boxed{4}.$ | E | 4 |
404b5cf21fd5ca316bacf19511345972 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_17 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$ | Let us call the six faces of our cube $a,b,c,d,e,$ and $f$ (where $a$ is opposite $d$ $c$ is opposite $e$ , and $b$ is opposite $f$ .
Thus, for the eight vertices, we have the following products: $abc,abe,bcd,bde,acf,cdf,aef,$ and $def$ .
Let us find the sum of these products: \[abc+abe+bcd+bde+acf+cdf+aef+def\] We notice $b$ is a factor of the first four terms, and $f$ is a factor of the last four terms. \[b(ac+ae+cd+de)+f(ac+ae+cd+de)\] Now, we can factor even more:
\begin{align*} & (b+f)(ac+ae+cd+de) \\ = &(b+f)(a(c+e)+d(c+e)) \\ = &(b+f)(a+d)(c+e) \end{align*} We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is $(7+2)$ $(6+3)$ , and $(5+4)$ all factors. \begin{align*} & (7+2)(6+3)(5+4) \\ = & 9 \cdot 9 \cdot 9 \\ = & 729. \end{align*} Thus our answer is $\boxed{729}$ | D | 729 |
404b5cf21fd5ca316bacf19511345972 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_17 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$ | We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact that there are 3 odd numbers. This means that there must be one side that has an odd area! Any odd number added with even numbers is always odd. Given that both c) and e) are both even, d) is our only choice.
Thus our answer is $\boxed{729}$ | D | 729 |
404b5cf21fd5ca316bacf19511345972 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_17 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$ | We first find the factorization $(b+f)(a+d)(c+e)$ using the method in Solution 1. By using AM-GM, we get, $(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3$ . To maximize the factorization, we get the answer is $\left( \frac{27}{3} \right)^3 = \boxed{729}$ | D | 729 |
404b5cf21fd5ca316bacf19511345972 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_17 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$ | First, we intuitivly notice that multiplying large numbers together and smaller numbers together tends to produce larger sums, while multiplying large numbers with smaller numbers tends to produce smaller sums. From this, we guess that it is optimal to have $7, 6,$ and $5$ to be around one vertex to produce at least one large product of $210$ . We can immediately eliminate a) and b) as answer choices since they are very close to this product, as well as e) since it is exactly $8 \cdot 210$ and we know that every following product will be smaller than this. We also guess that that is is most likely optimal to create a pairing where a number and its complement (the number that's the difference of 9 and this number) are on opposite sides. Using these two guesses, we can construct a net of sides and brute force the solution (or estimate) leaving us with $\boxed{729}$ | D | 729 |
4e4698e524481f0f12fae08fbd93c713 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_18 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | Factor $345=3\cdot 5\cdot 23$
Suppose we take an odd number $k$ of consecutive integers, with the median as $m$ . Then $mk=345$ with $\tfrac12k<m$ .
Looking at the factors of $345$ , the possible values of $k$ are $3,5,15,23$ with medians as $115,69,23,15$ respectively.
Suppose instead we take an even number $2k$ of consecutive integers, with median being the average of $m$ and $m+1$ . Then $k(2m+1)=345$ with $k\le m$ .
Looking again at the factors of $345$ , the possible values of $k$ are $1,3,5$ with medians $(172,173),(57,58),(34,35)$ respectively.
Thus the answer is $\boxed{7}$ | E | 7 |
4e4698e524481f0f12fae08fbd93c713 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_18 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | We need to find consecutive numbers (an arithmetic sequence that increases by $1$ ) that sums to $345$ . This calls for the sum of an arithmetic sequence given that the first term is $k$ , the last term is $g$ and with $n$ elements, which is: $\frac {n \cdot (k+g)}{2}$
We look for sequences of $n$ consecutive numbers starting at $k$ and ending at $k+n-1$ . We can now substitute $g$ with $k+n-1$ . Now we substitute our new value of $g$ into $\frac {n \cdot (k+g)}{2}$ to get that the sum is $\frac {n \cdot (k+k+n-1)}{2} = 345$
This simplifies to $\frac {n \cdot (2k+n-1)}{2} = 345$ . This gives a nice equation. We multiply out the 2 to get that $n \cdot (2k+n-1)=690$ . This leaves us with 2 integers that multiply to $690$ which leads us to think of factors of $690$ . We know the factors of $690$ are: $1,2,3,5,6,10,15,23,30,46,69,115,138,230,345,690$ . So through inspection (checking), we see that only $2,3,5,6,10,15$ and $23$ work. This gives us the answer of $\boxed{7}$ ways. | E | 7 |
4e4698e524481f0f12fae08fbd93c713 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_18 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | At the very end of Solution 2, where we find the factors of 690, instead of inspection, notice that all numbers will work until you get to $30$ , and that is because $\frac{345}{30}=11.5$ , which means $11$ and $12$ must be the middle 2 numbers; however, a sequence of length $30$ with middle numbers $11$ and $12$ that consists only of integers would go into the negatives, so any number from 30 onwards wouldn't work, and since $1$ is a trivial, non-counted solution, we get $\boxed{7}$ -ColtsFan10 | E | 7 |
4e4698e524481f0f12fae08fbd93c713 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_18 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | The median of the sequence $m$ is either an integer or a half integer. Let $m=\frac{i}{2}, i \in N$ , then $P=i\cdot n=2\cdot 3 \cdot 5 \cdot 23$
On the other hand we have two constraints:
1) $m \geq \frac{n+1}{2} \iff i>n$ because the integers in the sequence are all positive, and $n>1$
2) If $n$ is odd then $m$ is an integer, $i$ is even; if $n$ is even then $m$ is a half integer, $i$ is odd. Therefore, $n$ and $i$ have opposite parity.
Now $P$ has $16$ factors and it is not a perfect square. There are $8-1=7$ choices for $1 < n < \sqrt{P}$ . Also since $2|P, 4\nmid P$ , we know $n$ and $\frac{P}{n}$ must have opposite parity. Therefore the answer is $\boxed{7}$ | E | 7 |
beb7a3d764e8a85a7ede11f805cce7c1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_20 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | The center of dilation must lie on the line $A A'$ , which can be expressed as $y = \dfrac{4x}{3} - \dfrac{2}{3}$ . Note that the center of dilation must have an $x$ -coordinate less than $2$ ; if the $x$ -coordinate were otherwise, then the circle under the transformation would not have an increased $x$ -coordinate in the coordinate plane. Also, the ratio of dilation must be equal to $\dfrac{3}{2}$ , which is the ratio of the radii of the circles. Thus, we are looking for a point $(x,y)$ such that $\dfrac{3}{2} \left( 2 - x \right) = 5 - x$ (for the $x$ -coordinates), and $\dfrac{3}{2} \left( 2 - y \right) = 6 - y$ . We do not have to include absolute value symbols because we know that the center of dilation has a lower $x$ -coordinate, and hence a lower $y$ -coordinate, from our reasoning above. Solving the two equations, we get $x = -4$ and $y = - 6$ . This means that any point $(a,b)$ on the plane will dilate to the point $\left( \dfrac{3}{2} (a + 4) - 4, \dfrac{3}{2} (b + 6) - 6 \right)$ , which means that the point $(0,0)$ dilates to $\left( 6 - 4, 9 - 6 \right) = (2,3)$ . Thus, the origin moves $\sqrt{2^2 + 3^2} = \boxed{13}$ units. | null | 13 |
beb7a3d764e8a85a7ede11f805cce7c1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_20 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ /* by adihaya */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -7., xmax = 9., ymin = -7., ymax = 9.6; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,50.49019607843137253,1.); pen uuuuuu = rgb(0.666666666,0.26666666666666666,0.26666666666666666); pen qqzzff = rgb(0.,0.6,1.); pen ffwwqq = rgb(1.,0.4,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair O = (3.,0.), A = (2.,2.), B = (2.,1.), C = (4.203155585,5.592712848525), D = (5.,4.), F = (-3.999634206191805,-5.999512274922407), G = (-3.999634206191812,-5.9995122749224175); /* by adihaya */ draw((2.482656878,0.)---(4.482568783,0.48268779)--(2.,0.48272202065687797)--B--cycle, qqwuqq); draw((5.482722020656878,0.)--(7.4827220878,1.48277797)--(5.,0.48272687797)--(5.,0.)--cycle, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(shift(A) * scale(2., 2.)*unitcircle); draw(shift((5.,6.)) * scale(3.000060969351735, 3.000060969351735)*unitcircle); draw((xmin, 1.3333333333333333*xmin-0.6666666666666666)--(xmax, 1.3333333333333333*xmax-0.6666666666666666)); /* line */ draw((2.,ymin)--(2.,ymax)); /* line */ draw((5.,ymin)--(5.,ymax)); /* line */ draw((xmin, 1.6666869897839114*xmin + 0.6666260204321771)--(xmax, 1.6666869897839114*xmax + 0.6666260204321771)); /* line */ draw(O--F, qqzzff); draw(F--A, ffwwqq); /* dots and labels */ dot(O,blue); label("$O$", (0.08696973475182286,0.23426871275979863), NE * labelscalefactor,blue); dot(A,blue); label("$A$", (2.089474351594523,2.23677332960249), NE * labelscalefactor,blue); dot((5.,6.),blue); label("$A'$", (5.093231276858573,6.2190268290055695), NE * labelscalefactor,blue); dot(B,xdxdff); label("$B$", (2.089474351594523,0.23426871275979863), NE * labelscalefactor,xdxdff); label("$c$", (0.9971991060439592,3.2607813723061394), NE * labelscalefactor); dot(C,xdxdff); label("$C$", (3.2955282685566036,3.829674729363722), NE * labelscalefactor,xdxdff); label("$d$", (3.477574142815031,8.107752774436745), NE * labelscalefactor); label("$a$", (7.255026033677397,9.404829628528034), NE * labelscalefactor); label("$b$", (2.1804972887237364,9.404829628528034), NE * labelscalefactor); label("$e$", (4.615360856930201,9.404829628528034), NE * labelscalefactor); dot(D,linewidth(3.pt) + uuuuuu); /* Solution by adihaya */ label("$D$", (2.089474351594523,4.125499275033665), NE * labelscalefactor,uuuuuu); dot((5.,9.000060969351734),linewidth(3.pt) + uuuuuu); label("$E$", (5.093231276858573,9.131760817140394), NE * labelscalefactor,uuuuuu); label("$f$", (4.933941136882449,9.404829628528034), NE * labelscalefactor); dot(F,linewidth(3.pt) + uuuuuu); label("$\Large{(-4,-6)}$", (-3.73599362467515,-6.273871291978948), NE * labelscalefactor,uuuuuu); label("$\Large{2\sqrt{13}}$", (-2.916787190512227,-2.0868161840351394), NE * labelscalefactor,qqzzff); dot(G,linewidth(3.pt) + uuuuuu); label("$G$", (-3.9180394989335774,-5.864268074897489), NE * labelscalefactor,uuuuuu); label("$\Large{10}$", (0.2690156090102501,-0.6759606585323339), NE * labelscalefactor,ffwwqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(0.9090909090909091) * currentpicture; /* end of picture */[/asy] Using analytic geometry, we find that the center of dilation is at $(-4,-6)$ and the coefficient/factor is $1.5$ . Then, we see that the origin is $2\sqrt{13}$ from the center, and will be $1.5 \times 2\sqrt{13} = 3\sqrt{13}$ from it afterwards.
Thus, it will move $3\sqrt{13} - 2\sqrt{13} = \boxed{13}$ | null | 13 |
beb7a3d764e8a85a7ede11f805cce7c1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_20 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | Using the ratios of radii of the circles, $\frac{3}{2}$ , we find that the scale factor is $1.5$ . If the origin had not moved, this indicates that the center of the
circle would be $(3,3)$ , simply because of $(2 \cdot 1.5, 2 \cdot 1.5)$ . Since the center has moved from $(3,3)$ to $(5,6)$ , we apply the distance formula and get: $\sqrt{(6-3)^2 + (5-3)^2} = \boxed{13}$ | null | 13 |
beb7a3d764e8a85a7ede11f805cce7c1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_20 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | Before dilation, notice that the two axes are tangent to the circle with center $(2,2)$ . Using this, we can draw new axes tangent to the radius 3 circle with center $(5,6)$ , resulting in a "new origin" that is 3 units left and 3 units down from the center $(5,6)$ , or $(2,3)$ . Using the distance formula, the distance from $(0,0)$ and $(2,3)$ is $\boxed{13}$ .
~Mightyeagle | null | 13 |
beb7a3d764e8a85a7ede11f805cce7c1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_20 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | Because the dilation changes the circle of radius 2 to a circle of radius 3, the scale factor must be $\frac{3}{2}$ . The center of the circle with radius 3 is 3 units to the right and 4 units above the center of the circle with radius 2, so the center of dilation must be 6 units to the left and 8 units below the center of the radius 2 circle. So, the center of dilation is the point $(-4, -6)$ . The distance from that point to the origin is $2\sqrt{13}$ , and halving that to get the distance the origin moves gives $\boxed{13}$ . ~Meeshbot | null | 13 |
6eda98457d9930d25e2c94eac2538c0a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_21 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | Another way to solve this problem is using cases.
Though this may seem tedious, we only have to do one case since the area enclosed is symmetrical.
The equation for this figure is $x^2+y^2=|x|+|y|$ To make this as easy as possible,
we can make both $x$ and $y$ positive. Simplifying the equation for $x$ and $y$ being positive,
we get the equation $x^{2} +y^{2} -x-y = 0.$
Using the complete the square method, we get $\left(x-\frac{1}{2}\right)^{2} + \left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$
Therefore, the origin of this section of the shape is at $\left(\frac{1}{2}, \frac{1}{2}\right).$
Using the equation we can also see that the radius has a length of $\frac{\sqrt{2}}{2}$
With this shape we see that this shape can be cut into a right triangle and a semicircle.
The length of the hypotenuse of the triangle is $\sqrt{2}$ so using special right triangles, we see that
the area of the triangle is $\frac{1}{2}$ . The semicircle has the area of $\frac{1}{4}\pi$
But this is only $1$ case. There are $4$ cases in total so we have to multiply $\frac{1}{2}+\frac{1}{4}\pi$ by $4$
After multiplying, our answer is: \[\boxed{2}.\] | B | 2 |
6eda98457d9930d25e2c94eac2538c0a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_21 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | We solve with cases. The cases are:
Case 1: $x\geq0, y\geq0.$ Case 2: $x\geq0, y<0.$ Case 3: $x<0, y\geq0.$ Case 4: $x<0, y<0.$
We can quickly realize that the whole figure is symmetrical; so when you figure out the first case, you get the first part is $\left(x-\dfrac12\right)^2+\left(x-\dfrac12\right)^2$ you can figure out the whole figure: (scaled 8x). [asy] size(400); import TrigMacros; rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true); draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237)); [/asy] The way we figure out the area is by splitting it the following way: [asy] size(400); import TrigMacros; rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true); draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real g(real x) { return x+8; } draw(graph(g,0,-8), red+linewidth(1.5)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real h(real x) { return -x-8; } draw(graph(h,-8,0), red+linewidth(1.5)); real z(real x) { return x-8; } draw(graph(z,0,8), red+linewidth(1.5)); pair A,B,C,D; A = (8,0); B = (0,8); C = (-8,0); D = (0,-8); fill(A--B--C--D--cycle, red); [/asy]
We know each of the red lines is a diameter of the circle which is $\sqrt2$ . So the area of the red is 2. We know that the area of each semicircle is $\dfrac14 \pi$ so the area of the semicircles combines is $\pi$ . Thus we get $\boxed{2}.$ | B | 2 |
9e90187e86e07d26f777eda4fd3275dd | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_22 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$ | There are $10 \cdot 2+1=21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
[asy]size(7cm);label("A",(5,5));label("C",(10,0));label("B",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label("A",(20,5));label("C",(25,0));label("B",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); [/asy] But we know that every team beat exactly $10$ other teams, so for each possible $A$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $B$ and $C$ as $A$ beat exactly 10 teams and we are choosing 2 of them. Therefore there are $21\cdot\tbinom{10}2=945$ forks, and all the rest must be cycles.
Thus the answer is $1330-945=385$ which is $\boxed{385}$ | A | 385 |
9e90187e86e07d26f777eda4fd3275dd | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_22 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$ | Since there are $21$ teams and for each set of three teams there is a cycle, there are a total of $\tbinom{21}3=1330$ cycles of three teams. Because about $1/4$ of the cycles $\{A, B, C\}$ satisfy the conditions of the problems, our answer is close to $1/4 \cdot 1330=332.5$ . Looking at the answer choices, we find that $332.5$ is closer to $385$ than any other answer choices, and that the next closest is $665$ which is twice of $332.5$ , so our answer is $385$ which is $\boxed{385}$ | A | 385 |
9e90187e86e07d26f777eda4fd3275dd | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_22 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$ | Let's arrange all the teams in a circle and assume that each team won against the first 10 teams clockwise of themselves, and lost against the first 10 teams counter-clockwise of themselves.
Consider a working set of $3$ teams: moving clockwise, we know that the first team beat the team clockwise of itself, that team beat the next team clockwise of itself, and the final team beat the first team, which would be clockwise of itself. However, we also must remember that if the first team beat the second team, moving clockwise, the second team cannot be more than $10$ teams away from the first team; the same applies to the second and third team, and the third and first team.
Let's say, WLOG, that the first team, team $A$ , is at position $0$ out of $20$ on the circle.
If team $A$ is to beat team $B$ , since we are assuming each team beats the 10 members clockwise of themselves, there are $10$ places on the circle that team $B$ could be: positions $1$ through $10$ . Also, if team $C$ is to beat team $A$ , team $C$ must be located from positions $11$ to $20$
If Team $B$ is in position $n$ $C$ must be located from position $n+1$ to position $n+10$ , since $B$ beats $C$ . We also just found that $C$ 's position must be between $11$ and $20$ inclusive. So, when $B$ is in position $1$ $C$ can only be in position $11$ . When $B$ is in position $2$ $C$ can be in position $11$ or $12$ . In general, when $B$ is in position $n$ , there are $n$ choices for where $C$ can be. So, there are $1+2+3+ \dots + 10 = 55$ ways to place $B$ and $C$ . There are $21$ players to choose as player $A$ , and each working set will be counted $3$ times, so our final answer is $55 \cdot 21 \div 3 = \boxed{385,}$ | A | 385, |
9e90187e86e07d26f777eda4fd3275dd | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_22 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$ | This is a Graph Theory problem with directed graph. There are $21$ teams in total. WLOG, pick team $A$ , there are $10$ teams that lost to $A$ and $10$ teams that won over $A$ . Call the group of teams that lost to $A$ group $L$ , and the group of teams that won over $A$ group $W$
[asy] size(7cm); label("A",(20,5)); label("Group W",(27,0)); label("Group L",(13,0)); draw((19,4)--(16,1),EndArrow); draw((16,0)--(24,0),EndArrow); draw((24,1)--(21,4),EndArrow); [/asy]
Any team from group $L$ that won a team from group $W$ will form a cycle with $A$ . Now we need to count how many teams in group $L$ won over a team from group $W$ . The total number of wins in group $L$ is $10 \cdot 10 =100$ . There are $\tbinom{10}2=45$ wins among the teams inside group $L$ . So group $L$ has $100-45=55$ wins over group $W$
\[\frac{21 \cdot 55}{3}=\boxed{385,}\] | A | 385, |
f1a260dbd4b111e7bd531e8062cfe938 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_24 | How many four-digit integers $abcd$ , with $a \neq 0$ , have the property that the three two-digit integers $ab<bc<cd$ form an increasing arithmetic sequence? One such number is $4692$ , where $a=4$ $b=6$ $c=9$ , and $d=2$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20$ | The numbers are $10a+b, 10b+c,$ and $10c+d$ . Note that only $d$ can be zero, the numbers $ab$ $bc$ , and $cd$ cannot start with a zero, and $a\le b\le c$
To form the sequence, we need $(10c+d)-(10b+c)=(10b+c)-(10a+b)$ . This can be rearranged as $10(c-2b+a)=2c-b-d$ . Notice that since the left-hand side is a multiple of $10$ , the right-hand side can only be $0$ or $10$ . (A value of $-10$ would contradict $a\le b\le c$ .) Therefore we have two cases: $a+c-2b=1$ and $a+c-2b=0$
Case 1
If $c=9$ , then $b+d=8,\ 2b-a=8$ , so $5\le b\le 8$ . This gives $2593, 4692, 6791, 8890$ .
If $c=8$ , then $b+d=6,\ 2b-a=7$ , so $4\le b\le 6$ . This gives $1482, 3581, 5680$ .
If $c=7$ , then $b+d=4,\ 2b-a=6$ , so $b=4$ , giving $2470$ .
There is no solution for $c=6$ .
Added together, this gives us $8$ answers for Case 1.
Case 2
This means that the digits themselves are in an arithmetic sequence. This gives us $9$ answers, \[1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789.\] Adding the two cases together, we find the answer to be $8+9=$ $\boxed{17}$ | D | 17 |
f1a260dbd4b111e7bd531e8062cfe938 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_24 | How many four-digit integers $abcd$ , with $a \neq 0$ , have the property that the three two-digit integers $ab<bc<cd$ form an increasing arithmetic sequence? One such number is $4692$ , where $a=4$ $b=6$ $c=9$ , and $d=2$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20$ | Looking at the answer options, all the numbers are pretty small so it is easy to make a list.
$12|23|34 \rightarrow 1234$
$13|35|57 \rightarrow 1357$
$14|48|82 \rightarrow 1482$
$23|34|45 \rightarrow 2345$
$24|46|68 \rightarrow 2468$
$24|47|70 \rightarrow 2470$
$25|59|93 \rightarrow 2593$
$34|45|56 \rightarrow 3456$
$35|57|79 \rightarrow 3579$
$35|58|81 \rightarrow 3581$
$45|56|67 \rightarrow 4567$
$46|69|92 \rightarrow 4692$
$56|67|78 \rightarrow 5678$
$56|68|80 \rightarrow 5680$
$67|78|89 \rightarrow 6789$
$67|79|91 \rightarrow 6791$
$88|89|90 \rightarrow 8890$
Counting all the cases we get our answer of $17$ which is $\boxed{17}$ -srisainandan6 | D | 17 |
f321275c248ef22ebe1fb4fe91d12e7a | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_2 | A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$ | Let $a$ be the amount of triangular tiles and $b$ be the amount of square tiles.
Triangles have $3$ edges and squares have $4$ edges, so we have a system of equations.
We have $a + b$ tiles total, so $a + b = 25$
We have $3a + 4b$ edges total, so $3a + 4b = 84$
Multiplying the first equation by $3$ on both sides gives $3a + 3b = 3(25) = 75$
Second equation minus the first equation gives $b = 9$ , so the answer is $\boxed{9}$ | D | 9 |
f321275c248ef22ebe1fb4fe91d12e7a | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_2 | A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$ | If all of the tiles were triangles, there would be $75$ edges. This is not enough, so there needs to be some squares. Trading a triangle for a square results in one additional edge each time, so we must trade out $9$ triangles for squares. Answer: $\boxed{9}$ | D | 9 |
f321275c248ef22ebe1fb4fe91d12e7a | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_2 | A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$ | Let $x$ be the number of square tiles. A square has $4$ edges, so the total number of edges from the square tiles is $4x$ . There are $25$ total tiles, which means that there are $25-x$ triangle tiles. A triangle has $3$ edges, so the total number of edges from the triangle tiles is $3(25-x)$ . Together, the total number of edges is $4x+3(25-x)=84$ . Solving our equation, we get that $x=9$ which means that our answer is $\boxed{9}$ | D | 9 |
5d01f154fb6f2e28afe3d94d5824e123 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_3 | Ann made a $3$ -step staircase using $18$ toothpicks as shown in the figure. How many toothpicks does she need to add to complete a $5$ -step staircase?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45); for(int i=0;i<=2;i=i+1) { for(int j=0;j<=3-i;j=j+1) { filldraw(shift((i,j))*h,black); filldraw(shift((j,i))*v,black); } }[/asy] | We can see that a $1$ -step staircase requires $4$ toothpicks and a $2$ -step staircase requires $10$ toothpicks. Thus, to go from a $1$ -step to $2$ -step staircase, $6$ additional toothpicks are needed and to go from a $2$ -step to $3$ -step staircase, $8$ additional toothpicks are needed. Applying this pattern, to go from a $3$ -step to $4$ -step staircase, $10$ additional toothpicks are needed and to go from a $4$ -step to $5$ -step staircase, $12$ additional toothpicks are needed. Our answer is $10+12=\boxed{22}$ | D | 22 |
2bde84a7af0af40495ba32dd0ba29cfb | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_5 | Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$ | If the average of the first $14$ peoples' scores was $80$ , then the sum of all of their tests is $14 \cdot 80 = 1120$ . When Payton's score was added, the sum of all of the scores became $15 \cdot 81 = 1215$ . So, Payton's score must be $1215-1120 = \boxed{95}$ | E | 95 |
2bde84a7af0af40495ba32dd0ba29cfb | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_5 | Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$ | The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first $14$ students each scored $80$ . If Payton also scored an $80$ , the average would still be $80$ . In order to increase the overall average to $81$ , we need to add one more point to all of the scores, including Payton's. This means we need to add a total of $15$ more points, so Payton needs $80+15 = \boxed{95}$ | E | 95 |
dddef45504626f12448c3ddd3ba637a0 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_6 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | Let $a$ be the bigger number and $b$ be the smaller.
$a + b = 5(a - b)$
Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives
$\frac{a}{b} = \frac32$ , so the answer is $\boxed{32}$ | B | 32 |
dddef45504626f12448c3ddd3ba637a0 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_6 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{32}$ | B | 32 |
08de7d9b6246df073d3559a42f1d9878 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_7 | How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$ | $73-13 = 60$ , so the amount of terms in the sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$ is the same as in the sequence $0$ $3$ $6$ $\dotsc$ $57$ $60$
In this sequence, the terms are the multiples of $3$ going up to $60$ , and there are $20$ multiples of $3$ in $60$
However, the number 0 must also be included, adding another multiple. So, the answer is $\boxed{21}$ | B | 21 |
08de7d9b6246df073d3559a42f1d9878 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_7 | How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$ | Using the formula for arithmetic sequence's nth term, we see that $a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21$ $\boxed{21}$ | B | 21 |
08de7d9b6246df073d3559a42f1d9878 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_7 | How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$ | Minus each of the terms by $12$ to make the the sequence $1 , 4 , 7,..., 61$ $\frac{61-1}{3}=20, 20 + 1 = 21$
$\boxed{21}$ | B | 21 |
08de7d9b6246df073d3559a42f1d9878 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_7 | How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$ | Subtract each of the terms by $10$ to make the sequence $3 , 6 , 9,..., 60, 63$ . Then divide the each term in the sequence by $3$ to get $1, 2, 3,..., 20, 21$ . Now it is clear to see that there are $21$ terms in the sequence. $\boxed{21}$ | B | 21 |
cc9aabc25427c0c0552a8f98f73225b7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_8 | Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ $1$
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$ | This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.
The first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the number of years until Pete is twice as old as his cousin.
$20+x=2(8+x)$
$20+x=16+2x$
$x=4$
The answer is $\boxed{4}$ | B | 4 |
788f3688d814f058e2054afdc1c52b7c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_9 | Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?
$\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}$ | Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \[r_2=\frac{11r_1}{10}\] \[\pi r_1^2h_1=\pi r_2^2h_2\] Substituting the first equation into the second and dividing both sides by $\pi$ , we get \[r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.\] Therefore, $\boxed{21}$ | D | 21 |
9a89512467dd03e8d9791f33209dfa41 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_10 | How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | The first thing one would want to do is place a possible letter that works and then stem off of it. For example, if we start with an $a$ , we can only place a $c$ or $d$ next to it. Unfortunately, after that step, we can't do too much, since:
$acbd$ is not allowed because of the $cb$ , and $acdb$ is not allowed because of the $cd$
We get the same problem if we start with a $d$ , since a $b$ will have to end up in the middle, causing it to be adjacent to an $a$ or $c$
If we start with a $b$ , the next letter would have to be a $d$ , and since we can put an $a$ next to it and then a $c$ after that, this configuration works. The same approach applies if we start with a $c$
So the solution must be the two solutions that were allowed, one starting from a $b$ and the other with a $c$ , giving us:
\[1 + 1 = \boxed{2}.\] | C | 2 |
9a89512467dd03e8d9791f33209dfa41 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_10 | How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | Case 1: the first letter is A
Subcase 1: the second letter is C
The next letter must either be B or D, both of which do not satisfy the conditions.
Subcase 2: the second letter is D
The third letter is forced to be B and the fourth is forced to be C, but this doesn't work because B and C are next to each other.
Case 2: the first letter is B
The next letter is forced to be D, the third letter forced to be A and the last forced to be C. This works.
Case 3: the first letter is C
The next letter is forced to be A, the third letter is D and the last letter is B. This works.
Case 4: the first letter is D
Subcase 1: the second letter is A
The third letter has to be C and the fourth has to be B but this doesn't work because B and C are next to each other.
Subcase 2: the second letter is B
The third letter cannot be A or C, so this doesn't work.
Summing the cases, there are two that work: $BDAC$ and $CADB$ $\Longrightarrow \boxed{2}$ | C | 2 |
6f5052de043fc9a2a69e3fefdd2a331c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_12 | Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$ | Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$
$y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1$
$y^2 + \pi^2 = 2\pi y + 1$
$y^2 - 2\pi y + \pi^2 = 1$
$(y-\pi)^2 = 1$
$y-\pi = \pm 1$
$y = \pi + 1$
$y = \pi - 1$
There are only two solutions to the equation $(y-\pi)^2 = 1$ , so one of them is the value of $a$ and the other is $b$ . The order does not matter because of the absolute value sign.
$| (\pi + 1) - (\pi - 1) | = 2$
The answer is $\boxed{2}$ | C | 2 |
6f5052de043fc9a2a69e3fefdd2a331c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_12 | Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$ | This solution is very closely related to Solution #1 and just simplifies the problem earlier to make it easier.
$y^2 + x^4 = 2x^2 y + 1$ can be written as $x^4-2x^2y+y^2=1$ . Recognizing that this is a binomial square, simplify this to $(x^2-y)^2=1$ . This gives us two equations:
$x^2-y=1$ and $x^2-y=-1$
One of these $y$ 's is $a$ and one is $b$ . Substituting $\sqrt{\pi}$ for $x$ , we get $a=\pi+1$ and $b=\pi-1$
So, $|a-b|=|(\pi+1)-(\pi-1)|=2$
The answer is $\boxed{2}$ | C | 2 |
6f5052de043fc9a2a69e3fefdd2a331c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_12 | Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$ | This solution is similar to Solution #1 but uses a different way to find $y$ at the end.
Just like Solution #1, we arrive at the conclusion that $y^2 - 2\pi y + \pi^2 = 1$
Simplifying we get:
$y^2 - 2\pi y + \pi^2 -1 = 0$
We now can factor this quadratic. We must find two terms that multiply to $\pi^2 -1$ and add to $2\pi$
These terms are $\pi+1$ and $\pi-1$
Subtracting one from the other, we get $2$
Thus, the answer is $\boxed{2}$ | C | 2 |
d9bcf51a17b7f28a416e73e8c3115acf | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_13 | Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | Let Claudia have $x$ 5-cent coins and $\left( 12 - x \right)$ 10-cent coins. It is easily observed that any multiple of $5$ between $5$ and $5x + 10(12 - x) = 120 - 5x$ inclusive can be obtained by a combination of coins. Thus, $24 - x = 17$ combinations can be made, so $x = 7$ . But the answer is not $7,$ because we are asked for the number of 10-cent coins, which is $12 - 7 = \boxed{5}$ | C | 5 |
d9bcf51a17b7f28a416e73e8c3115acf | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_13 | Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of $5.$ To have exactly $17$ different multiples of $5,$ we will need to make up to $85$ cents. If all twelve coins were 5-cent coins, we will have $60$ cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain $5$ cents, and as we need to gain $25$ cents, the answer is $\boxed{5}$ | C | 5 |
d9bcf51a17b7f28a416e73e8c3115acf | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_13 | Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | Notice that for every $d$ dimes, any multiple of $5$ less than or equal to $10d + 5(12-d)$ is a valid arrangement. Since there are $17$ in our case, we have $10d + 5(12-d)=17 \cdot 5 \Rightarrow d=5$ . Therefore, the answer is $\boxed{5}$ | C | 5 |
7523763b4c2f76068d95796670c33d23 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$ | The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is $\boxed{4}$ . Similarly, the arrow would be pointing downward at 6:00. It would already have completed three 180 degree turns. Therefore, two 180 degree turns would be completed at 4:00. | C | 4 |
7523763b4c2f76068d95796670c33d23 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$ | The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. $\boxed{4}$ | C | 4 |
7523763b4c2f76068d95796670c33d23 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$ | The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so, therefore, the disk must travel its circumference before the arrow goes up. Its circumference is $20\pi$ , so that is $20\pi$ traveled on a $60\pi$ arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. $\boxed{4}$ | C | 4 |
7523763b4c2f76068d95796670c33d23 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$ | Suppose that the small disk also had a clock face on it, and that both disks were toothed wheels, free to rotate around their centers. The part of the picture where they engage would look like this: [asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label("12",1.56*dir(90)); label("1",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label("6",.74*dir(-90)+(0,3)); label("5",.74*dir(-60)+(0,3)); label("4",.74*dir(-30)+(0,3)); [/asy] The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\circ$ clockwise (i.e. 2 hours). [asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label("$30^\circ$",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*"12",1.56*dir(120)); label(rotate(30)*"1",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label("$60^\circ$",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*"6",.74*dir(-150)+(0,3)); label(rotate(-60)*"5",.74*dir(-120)+(0,3)); label(rotate(-60)*"4",.74*dir(-90)+(0,3)); [/asy] However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\circ$ clockwise to see what happens when the small cog rolls around it. [asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label("12",1.56*dir(90)); label("1",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label("$90^\circ$",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*"6",.74*dir(-180)+c); label(rotate(-90)*"5",.74*dir(-150)+c); label(rotate(-90)*"4",.74*dir(-120)+c); [/asy] It turns out that, when the point of tangency moves $30^\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\boxed{4}$ | C | 4 |
7523763b4c2f76068d95796670c33d23 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$ | We can approach this problem with angle measures. As the circumference of the disk is $10\pi,$ and the clock $20 \pi,$ we have that 30 degrees, or the angular measure between hours, of the disk is only 15 degrees of the clock. This yields that every two hour ticks that the clock rotates, on the third one, the ticks will meet. However, the disk must rotate 360 degrees in order to come back to its original position, so the angular measure that the disk has covered relative to the clock is simply $12 \cdot 15 \cdot \frac{2}{3},$ or $120^\circ$ from the 12 starting point, so $\boxed{4}$ | C | 4 |
7523763b4c2f76068d95796670c33d23 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$ | If the big clock were a flat plane, then the smaller clock could travel $\dfrac{40\pi}{20\pi}=2$ full revolutions.
But we also need to account for rotation. If we mark a red dot on the bottom of the small clock/bottom of the arrow, and then drag it around the clock, the direction of the arrow would still change. After traveling around the big clock, the small clock would travel $1$ full rotation.
Considering these two movements, the small clock travels 3 full rotations around the big clock so the arrow is next pointing upwards at $\dfrac{12 \text{ o'clock}}{3}=\boxed{4}$ | C | 4 |
7523763b4c2f76068d95796670c33d23 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$ | The center of rotation is in the center of the smaller circle, but extends to the center of the larger circle. That means the circumference of the circle in relation to the arrow is $60 \pi$ . The other circle is $20 \pi$ and so that is $\frac{1}{3}$ . So $\frac{12}{3} = 4$ which is $\boxed{4}$ | C | 4 |
7523763b4c2f76068d95796670c33d23 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$ | Let the center of the disk be Planet X with orbit eccentricity $0$ and the center of the clock be the Sun. Note that the question would then be asking for the solar day of Planet X, rather than the sidereal day. Because planet X is rotating around its axis in the same direction as it is revolving around the Sun, the solar day can be calculated as $d_{solar}=\frac{1}{\frac{1}{y}+\frac{1}{d_{siderial}}}$ , where $y$ is the length of the year of Planet X and $d_{siderial}$ is one sidereal day. It is easy to see that $y=12$ and $d_{siderial}=6$ , therefore the answer is $\frac{1}{\frac{1}{12}+\frac{1}{6}}=\frac{1}{\frac{1}{4}}=\boxed{4}$ ~sigmapie | C | 4 |
2b17858e13e959fe937ae629d1acd45e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_15 | Consider the set of all fractions $\frac{x}{y}$ , where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$ , the value of the fraction is increased by $10\%$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}$ | You can create the equation $\frac{x+1}{y+1}=\frac{11x}{10y}$
Cross multiplying and combining like terms gives $xy + 11x - 10y = 0$
This can be factored into $(x - 10)(y + 11) = -110$
$x$ and $y$ must be positive, so $x > 0$ and $y > 0$ , so $x - 10> -10$ and $y + 11 > 11$
Using the factors of 110, we can get the factor pairs: $(-1, 110),$ $(-2, 55),$ and $(-5, 22).$
But we can't stop here because $x$ and $y$ must be relatively prime.
$(-1, 110)$ gives $x = 9$ and $y = 99$ $9$ and $99$ are not relatively prime, so this doesn't work.
$(-2, 55)$ gives $x = 8$ and $y = 44$ . This doesn't work.
$(-5, 22)$ gives $x = 5$ and $y = 11$ . This does work.
We found one valid solution so the answer is $\boxed{1}$ | B | 1 |
2b17858e13e959fe937ae629d1acd45e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_15 | Consider the set of all fractions $\frac{x}{y}$ , where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$ , the value of the fraction is increased by $10\%$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}$ | The condition required is $\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}$
Observe that $x+1 > \frac{11}{10}\cdot x$ so $x$ is at most $9.$
By multiplying by $\frac{y+1}{x}$ and simplifying we can rewrite the condition as $y=\frac{11x}{10-x}$ . Since $x$ and $y$ are integer, this only has solutions for $x\in\{5,8,9\}$ . However, only the first yields a $y$ that is relative prime to $x$
There is only one valid solution so the answer is $\boxed{1}$ | B | 1 |
b0505938806fc67ffa6f318f02a9b0a7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_18 | Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$ . Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$
$\textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21$ | Notice that $1000$ is $3E8$ when converted to hexadecimal ( $3 \cdot 16^2 + 14 \cdot 16^1 + 8 \cdot 16^0$ ). We will proceed by constructing numbers that consist of only numeric digits in hexadecimal.
The first digit could be $0,$ $1,$ $2,$ or $3,$ and the second two could be any digit $0 - 9$ , giving $4 \cdot 10 \cdot 10 = 400$ combinations. However, this includes $000,$ so this number must be diminished by $1.$ Therefore, there are $399$ valid $n$ corresponding to those $399$ positive integers less than $1000$ that consist of only numeric digits. (Notice that $399$ is the least hexadecimal number using only decimal digits before $3E8$ .) Therefore, our answer is $3 + 9 + 9 = \boxed{21}$ | E | 21 |
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