https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Meljel&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T18:14:29ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=142329User talk:Meljel2021-01-16T23:18:09Z<p>Meljel: /* :) */</p>
<hr />
<div>== hi. ==<br />
<br />
if you're from hysb then bye<br />
<br />
== :) ==<br />
<asy>unitsize(0.3cm);filldraw((10,0)--(10,2)--(12,2)--(12,0)--cycle,red);filldraw((12,0)--(12,2)--(14,2)--(14,0)--cycle,red);filldraw((14,0)--(14,2)--(16,2)--(16,0)--cycle,red);filldraw((11,2)--(11,4)--(13,4)--(13,2)--cycle,red);filldraw((13,2)--(13,4)--(15,4)--(15,2)--cycle,red);filldraw((12,4)--(12,6)--(14,6)--(14,4)--cycle,red);filldraw((0,0)--(0,2)--(2,2)--(2,0)--cycle,red);filldraw((2,0)--(2,2)--(4,2)--(4,0)--cycle,black);filldraw((4,0)--(4,2)--(6,2)--(6,0)--cycle,red);filldraw((1,2)--(1,4)--(3,4)--(3,2)--cycle,yellow);filldraw((3,2)--(3,4)--(5,4)--(5,2)--cycle,yellow);filldraw((2,4)--(2,6)--(4,6)--(4,4)--cycle,yellow);filldraw((20,0)--(20,2)--(22,2)--(22,0)--cycle,yellow);filldraw((22,0)--(22,2)--(24,2)--(24,0)--cycle,red);filldraw((24,0)--(24,2)--(26,2)--(26,0)--cycle,black);filldraw((21,2)--(21,4)--(23,4)--(23,2)--cycle,black);filldraw((23,2)--(23,4)--(25,4)--(25,2)--cycle,yellow);filldraw((22,4)--(22,6)--(24,6)--(24,4)--cycle,red);filldraw((30,0)--(30,2)--(32,2)--(32,0)--cycle,red);filldraw((32,0)--(32,2)--(34,2)--(34,0)--cycle,red);filldraw((34,0)--(34,2)--(36,2)--(36,0)--cycle,yellow);filldraw((31,2)--(31,4)--(33,4)--(33,2)--cycle,red);filldraw((33,2)--(33,4)--(35,4)--(35,2)--cycle,black);filldraw((32,4)--(32,6)--(34,6)--(34,4)--cycle, yellow);</asy><br />
<br />
<math>1~~~~1+1=2~~~~~yes</math></div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=142328User talk:Meljel2021-01-16T23:12:58Z<p>Meljel: /* hi. */</p>
<hr />
<div>== hi. ==<br />
<br />
if you're from hysb then bye<br />
<br />
== :) ==</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=142327User talk:Meljel2021-01-16T23:05:02Z<p>Meljel: /* hihiihi */</p>
<hr />
<div>== hi. ==<br />
<br />
if ur from hysb then bye<br />
<br />
== :) ==</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=142326User talk:Meljel2021-01-16T23:04:33Z<p>Meljel: /* It's me meljel */</p>
<hr />
<div>== hi. ==<br />
<br />
if ur from hysb then bye<br />
<br />
== hihiihi ==<br />
<br />
<cmath>a^2-b^2=(a+b)(a-b)</cmath><br />
<cmath>(\cos \theta + i\sin \theta)^n = \cos n\theta + ii\sin n\theta</cmath></div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_19&diff=1066002006 AMC 10B Problems/Problem 192019-06-18T18:44:39Z<p>Meljel: /* Solution 3 (Using Answer Choices) */</p>
<hr />
<div>== Problem ==<br />
A circle of radius <math>2</math> is centered at <math>O</math>. Square <math>OABC</math> has side length <math>1</math>. Sides <math>AB</math> and <math>CB</math> are extended past <math>B</math> to meet the circle at <math>D</math> and <math>E</math>, respectively. What is the area of the shaded region in the figure, which is bounded by <math>BD</math>, <math>BE</math>, and the minor arc connecting <math>D</math> and <math>E</math>?<br />
<br />
<!-- [[Image:2006amc10b19.gif]] --><br />
<asy>defaultpen(linewidth(0.8));<br />
pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B;<br />
fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray);<br />
clip(B--Arc(O, 2, 30, 60)--cycle);<br />
draw(Circle(origin, 2));<br />
draw((-2,0)--(2,0)^^(0,-2)--(0,2));<br />
draw(A--D^^C--E);<br />
label("$A$", A, dir(point--A));<br />
label("$C$", C, dir(point--C));<br />
label("$O$", O, dir(point--O));<br />
label("$D$", D, dir(point--D));<br />
label("$E$", E, dir(point--E));<br />
label("$B$", B, SW);</asy><br />
<math> \mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3})<br />
\qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} </math><br />
<br />
== Solution 1 ==<br />
The shaded area is equivalent to the area of sector <math>DOE</math>, minus the area of triangle <math>DOE</math> plus the area of triangle <math>DBE</math>. <br />
<br />
Using the [[Pythagorean Theorem]], <math>(DA)^2=(CE)^2=2^2-1^2=3</math> so <math>DA=CE=\sqrt{3}</math>.<br />
<br />
Clearly, <math>DOA</math> and <math>EOC</math> are <math>30-60-90</math> triangles with <math>\angle EOC = \angle DOA = 60^\circ </math>. Since <math>OABC</math> is a square, <math> \angle COA = 90^\circ </math>.<br />
<br />
<math>\angle DOE</math> can be found by doing some subtraction of angles. <br />
<br />
<math> \angle COA - \angle DOA = \angle EOA </math><br />
<br />
<math> 90^\circ - 60^\circ = \angle EOA = 30^\circ </math><br />
<br />
<math> \angle DOA - \angle EOA = \angle DOE </math><br />
<br />
<math> 60^\circ - 30^\circ = \angle DOE = 30^\circ </math><br />
<br />
So, the area of sector <math>DOE</math> is <math> \frac{30}{360} \cdot \pi \cdot 2^2 = \frac{\pi}{3} </math>.<br />
<br />
The area of triangle <math>DOE</math> is <math> \frac{1}{2}\cdot 2 \cdot 2 \cdot \sin 30^\circ = 1 </math>.<br />
<br />
Since <math>AB=CB=1</math> , <math>DB=ED=(\sqrt{3}-1)</math>. So, the area of triangle <math>DBE</math> is <math>\frac{1}{2} \cdot (\sqrt{3}-1)^2 = 2-\sqrt{3}</math>. Therefore, the shaded area is <math> (\frac{\pi}{3}) - (1) + (2-\sqrt{3}) = \frac{\pi}{3}+1-\sqrt{3} \Longrightarrow \boxed{\mathrm{(A)}} </math><br />
<br />
==Solution 2==<br />
From the pythagorean theorem, we can see that <math>DA</math> is <math>\sqrt{3}</math>. Then, <math>DB = DA - BA = \sqrt{3} - 1</math>. The area of the shaded element is the area of sector <math>DOE</math> minus the areas of triangle <math>DBO</math> and triangle <math>EBO</math> combined. Below is an image to help.<br />
<br />
<br />
<!-- [[Image:2006amc10b19.gif]] --><br />
<asy>defaultpen(linewidth(0.8));<br />
pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B;<br />
fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray);<br />
clip(B--Arc(O, 2, 30, 60)--cycle);<br />
draw(Circle(origin, 2));<br />
draw((-2,0)--(2,0)^^(0,-2)--(0,2));<br />
draw(A--D^^C--E^^D--O^^E--O^^B--O);<br />
label("$A$", A, dir(point--A));<br />
label("$C$", C, dir(point--C));<br />
label("$O$", O, dir(point--O));<br />
label("$D$", D, dir(point--D));<br />
label("$E$", E, dir(point--E));<br />
label("$B$", (1.33,1.04), SW);</asy><br />
<br />
<br />
Using the Base Altitude formula, where <math>DB</math> and <math>BE</math> are the bases and <math>OA</math> and <math>CO</math> are the altitudes, respectively, <math>[DBO] = [EBO] = \frac{\sqrt{3}-1}{2}</math>. The area of sector <math>DOE</math> is <math>\frac{1}{12}</math> of circle <math>O</math>. The area of circle <math>O</math> is <math>4\pi</math>, and therefore we have the area of sector <math>DBE</math> to be <math>\frac{\pi}{3} + 1 - \sqrt{3} \Longrightarrow \boxed{A}</math><br />
<br />
==Solution 3 (Using Answer Choices)==<br />
Like the first solutions, you find that the area of sector <math>DOE</math> is <math>\frac{\pi}{3}</math>. We also know that the triangles will not be in terms of <math>{\pi}</math>. Looking at the answers, choices <math>\text{(A)}</math> and <math>\text{(E)}</math> both contain <math>\frac{\pi}{3}</math>. However, based on the diagram, we observe that the answer must be less than <math>\frac {\pi}{3}</math>. Only <math>\boxed{A}</math> consists of a value less than <math>\frac{\pi}{3}</math>.<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2006|ab=B|num-b=18|num-a=20}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
[[Category:Circle Problems]]<br />
{{MAA Notice}}</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems&diff=1065942006 AMC 10A Problems2019-06-18T18:28:21Z<p>Meljel: /* Problem 17 */</p>
<hr />
<div>==Problem 1==<br />
Sandwiches at Joe's Fast Food cost <math> \textdollar 3 </math> each and sodas cost <math> \textdollar 2 </math> each. How many dollars will it cost to purchase 5 sandwiches and 8 sodas?<br />
<br />
<math>\mathrm{(A)}\ 31\qquad\mathrm{(B)}\ 32\qquad\mathrm{(C)}\ 33\qquad\mathrm{(D)}\ 34\qquad\mathrm{(E)}\ 35</math><br />
<br />
[[2006 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>?<br />
<br />
<math>\mathrm{(A)}\ -h\qquad \mathrm{(B)}\ 0\qquad \mathrm{(C)}\ h\qquad \mathrm{(D)}\ 2h\qquad\mathrm{(E)}\ h^3</math><br />
<br />
[[2006 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The ratio of Mary's age to Alice's age is <math>3:5</math>. Alice is <math>30</math> years old. How many years old is Mary?<br />
<br />
<math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18\qquad\mathrm{(C)}\ 20\qquad\mathrm{(D)}\ 24\qquad\mathrm{(E)}\ 50</math><br />
<br />
[[2006 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?<br />
<br />
<math>\mathrm{(A)}\ 17\qquad\mathrm{(B)}\ 19\qquad\mathrm{(C)}\ 21\qquad\mathrm{(D)}\ 22\qquad\mathrm{(E)}\ 23</math><br />
<br />
[[2006 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Doug and Dave shared a pizza with 8 equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half of the pizza. The cost of a plain pizza was 8 dollars, and there was an additional cost of 2 dollars for putting anchovies on one half. Dave ate all of the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each then paid for what he had eaten. How many more dollars did Dave pay than Doug?<br />
<br />
<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 5</math><br />
<br />
[[2006 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
What non-zero real value for <math>x</math> satisfies <math>(7x)^{14}=(14x)^7</math>?<br />
<br />
<math>\mathrm{(A)}\ \frac17\qquad\mathrm{(B)}\ \frac27\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 14</math><br />
<br />
[[2006 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=4;<br />
draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);<br />
draw((6,4)--(6,0)--(12,0)--(12,-4));<br />
label("$A$",(0,4),NW);<br />
label("$B$",(18,4),NE);<br />
label("$C$",(18,-4),SE);<br />
label("$D$",(0,-4),SW);<br />
label("$y$",(3,4),S);<br />
label("$y$",(15,-4),N);<br />
label("$18$",(9,4),N);<br />
label("$18$",(9,-4),S);<br />
label("$8$",(0,0),W);<br />
label("$8$",(18,0),E);<br />
dot((0,4));<br />
dot((18,4));<br />
dot((18,-4));<br />
dot((0,-4));</asy><br />
<br />
The <math> 8 \times 18 </math> rectangle <math>ABCD</math> is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is <math>y</math>?<br />
<br />
<math>\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10</math><br />
<br />
[[2006 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
A parabola with equation <math>y=x^2+bx+c</math> passes through the points <math> (2,3) </math> and <math> (4,3) </math>. What is <math>c</math>?<br />
<br />
<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 7\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ 11</math><br />
<br />
[[2006 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
How many sets of two or more consecutive positive integers have a sum of 15?<br />
<br />
<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 5</math><br />
<br />
[[2006 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
For how many real values of <math>x</math> is <math>\sqrt{120-\sqrt{x}}</math> an integer?<br />
<br />
<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 6\qquad\mathrm{(C)}\ 9\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ 11\qquad</math><br />
<br />
[[2006 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Which of the following describes the graph of the equation <math>(x+y)^2=x^2+y^2</math>?<br />
<br />
<math>\mathrm{(A)}\ \text{the empty set}\qquad\mathrm{(B)}\ \text{one point}\qquad\mathrm{(C)}\ \text{two lines}\qquad\mathrm{(D)}\ \text{a circle}\qquad\mathrm{(E)}\ \text{the entire plane}</math><br />
<br />
[[2006 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<asy><br />
size(150); pathpen = linewidth(0.6); defaultpen(fontsize(10));<br />
D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle);<br />
D((16,-8)--(24,-8));<br />
label('Dog', (24, -8), SE);<br />
MP('I', (8,-8), (0,0));<br />
MP('8', (16,-4), W);<br />
MP('8', (16,-12),W);<br />
MP('8', (20,-8), N);<br />
label('Rope', (20,-8),S);<br />
D((0,-20)--(16,-20)--(16,-36)--(0,-36)--cycle);<br />
D((16,-24)--(24,-24));<br />
MP("II", (8,-28), (0,0));<br />
MP('4', (16,-22), W);<br />
MP('8', (20,-24), N);<br />
label("Dog",(24,-24),SE);<br />
label("Rope", (20,-24), S);<br />
</asy><br />
<br />
Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.<br />
<br />
Which of these arrangements give the dog the greater area to roam, and by how many square feet?<br />
<br />
<math>\mathrm{(A)}\ \text{I, by}\ 8\pi\qquad\mathrm{(B)}\ \text{I, by}\ 6\pi\qquad\mathrm{(C)}\ \text{II, by}\ 4\pi\qquad\mathrm{(D)} \text{II, by}\ 8\pi\qquad\mathrm{(E)}\ \text{II, by}\ 10\pi</math><br />
<br />
[[2006 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
A player pays <math>\textdollar 5 </math> to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.) <br />
<br />
<math>\mathrm{(A)}\ 12\qquad\mathrm{(B)}\ 30\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 60\qquad\mathrm{(E)}\ 100</math><br />
<br />
[[2006 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<asy><br />
size(7cm); pathpen = linewidth(0.7);<br />
D(CR((0,0),10));<br />
D(CR((0,0),9.5));<br />
D(CR((0,-18.5),9.5));<br />
D(CR((0,-18.5),9));<br />
MP("$\vdots$",(0,-31),(0,0));<br />
D(CR((0,-39),3));<br />
D(CR((0,-39),2.5));<br />
D(CR((0,-43.5),2.5));<br />
D(CR((0,-43.5),2));<br />
D(CR((0,-47),2));<br />
D(CR((0,-47),1.5));<br />
D(CR((0,-49.5),1.5));<br />
D(CR((0,-49.5),1.0));<br />
<br />
D((12,-10)--(12,10)); MP('20',(12,0),E);<br />
D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);</asy><br />
<br />
A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the other rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring? <br />
<br />
<math>\mathrm{(A)}\ 171\qquad\mathrm{(B)}\ 173\qquad\mathrm{(C)}\ 182\qquad\mathrm{(D)}\ 188\qquad\mathrm{(E)}\ 210</math><br />
<br />
[[2006 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other? <br />
<br />
<math>\mathrm{(A)}\ 29\qquad\mathrm{(B)}\ 42\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 47\qquad\mathrm{(E)}\ 50</math><br />
<br />
[[2006 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<!--[[Image:2006_AMC10A-16.png]]--><br />
<asy><br />
size(200); pathpen = linewidth(0.7); pointpen = black;<br />
real t=2^0.5;<br />
D((0,0)--(4*t,0)--(2*t,8)--cycle);<br />
D(CR((2*t,2),2));<br />
D(CR((2*t,5),1));<br />
D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N);<br />
D((2*t,2)--(2*t,4)); D((2*t,5)--(2*t,6));<br />
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W);</asy><br />
<br />
A circle of radius 1 is tangent to a circle of radius 2. The sides of <math>\triangle ABC</math> are tangent to the circles as shown, and the sides <math>\overline{AB}</math> and <math>\overline{AC}</math> are congruent. What is the area of <math>\triangle ABC</math>?<br />
<br />
<math>\mathrm{(A)}\ \frac{35}{2}\qquad\mathrm{(B)}\ 15\sqrt{2}\qquad\mathrm{(C)}\ \frac{64}{3}\qquad\mathrm{(D)}\ 16\sqrt{2}\qquad\mathrm{(E)}\ 24</math><br />
<br />
[[2006 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<center>[[Image:2006_AMC10A-17.png]]</center><br />
<br />
In rectangle <math>ADEH</math>, points <math>B</math> and <math>C</math> trisect <math>\overline{AD}</math>, and points <math>G</math> and <math>F</math> trisect <math>\overline{HE}</math>. In addition, <math>AH=AC=2</math>, and <math>AD = 3</math>. What is the area of quadrilateral <math>WXYZ</math> shown in the figure? <br />
<br />
<math>\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ \frac{\sqrt{2}}{2}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2}\qquad\mathrm{(D)}\ \frac{2\sqrt{2}}{2}\qquad\mathrm{(E)}\ \frac{2\sqrt{3}}{3}</math><br />
<br />
[[2006 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible? <br />
<br />
<math>\mathrm{(A)}\ 10^4\times26^2\qquad\mathrm{(B)}\ 10^3\times26^3\qquad\mathrm{(C)}\ 5\times10^4\times26^2\qquad\mathrm{(D)}\ 10^2\times26^4\qquad\mathrm{(E)}\ 5\times10^3\times26^3</math><br />
<br />
[[2006 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression? <br />
<br />
<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 59\qquad\mathrm{(D)}\ 89\qquad\mathrm{(E)}\ 178</math><br />
<br />
[[2006 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Six distinct positive integers are randomly chosen between 1 and 2006, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of 5? <br />
<br />
<math>\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ \frac{3}{5}\qquad\mathrm{(C)}\ \frac{2}{3}\qquad\mathrm{(D)}\ \frac{4}{5}\qquad\mathrm{(E)}\ 1</math><br />
<br />
[[2006 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
How many four-digit positive integers have at least one digit that is a 2 or a 3? <br />
<br />
<math>\mathrm{(A)}\ 2439\qquad\mathrm{(B)}\ 4096\qquad\mathrm{(C)}\ 4903\qquad\mathrm{(D)}\ 4904\qquad\mathrm{(E)}\ 5416</math><br />
<br />
[[2006 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Two farmers agree that pigs are worth &#36;300 and that goats are worth &#36;210. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a &#36;390 debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way? <br />
<br />
<math>\mathrm{(A)}\ 5\qquad\mathrm{(B)}\ 10\qquad\mathrm{(C)}\ 30\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 210</math><br />
<br />
[[2006 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Circles with centers <math>A</math> and <math>B</math> have radii <math>3</math> and <math>8</math>, respectively. A common internal tangent intersects the circles at <math>C</math> and <math>D</math>, respectively. Lines <math>AB</math> and <math>CD</math> intersect at <math>E</math>, and <math>AE=5</math>. What is <math>CD</math>?<br />
<br />
[[Image:2006_AMC12A-16.png|center]]<br />
<br />
<math>\mathrm{(A)}\ 13\qquad\mathrm{(B)}\ \frac{44}{3}\qquad\mathrm{(C)}\ \sqrt{221}\qquad\mathrm{(D)}\ \sqrt{255}\qquad\mathrm{(E)}\ \frac{55}{3}</math><br />
<br />
[[2006 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron? <br />
<br />
<math>\mathrm{(A)}\ \frac{1}{8}\qquad\mathrm{(B)}\ \frac{1}{6}\qquad\mathrm{(C)}\ \frac{1}{4}\qquad\mathrm{(D)}\ \frac{1}{3}\qquad\mathrm{(E)}\ \frac{1}{2}</math> <br />
<br />
[[2006 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once? <br />
<br />
<math>\mathrm{(A)}\ \frac{1}{2187}\qquad\mathrm{(B)}\ \frac{1}{729}\qquad\mathrm{(C)}\ \frac{2}{243}\qquad\mathrm{(D)}\ \frac{1}{81}\qquad\mathrm{(E)}\ \frac{5}{243}</math> <br />
<br />
[[2006 AMC 10A Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|before=[[2005 AMC 10B Problems]]|after=[[2006 AMC 10B Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2006 AMC 10A]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=142 2006 AMC A Math Jam Transcript]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems&diff=1064642010 AMC 12B Problems2019-06-17T02:53:29Z<p>Meljel: /* Problem 3 */</p>
<hr />
<div>== Problem 1 ==<br />
Makarla attended two meetings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?<br />
<br />
<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35</math><br />
<br />
[[2010 AMC 12B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
A big <math>L</math> is formed as shown. What is its area?<br />
<br />
<center><asy><br />
unitsize(4mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
draw((0,0)--(5,0)--(5,2)--(2,2)--(2,8)--(0,8)--cycle);<br />
label("8",(0,4),W);<br />
label("5",(5/2,0),S);<br />
label("2",(5,1),E);<br />
label("2",(1,8),N);<br />
</asy></center><br />
<br />
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 30</math><br />
<br />
[[2010 AMC 12B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A ticket to a school play cost <math>x</math> dollars, where <math>x</math> is a whole number. A group of 9<sup>th</sup> graders buys tickets costing a total of &#36;<math>48</math>, and a group of 10<sup>th</sup> graders buys tickets costing a total of &#36;<math>64</math>. How many values for <math>x</math> are possible?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2010 AMC 12B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A month with <math>31</math> days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math><br />
<br />
[[2010 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Lucky Larry's teacher asked him to substitute numbers for <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> in the expression <math>a-(b-(c-(d+e)))</math> and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> were <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>, respectively. What number did Larry substitute for <math>e</math>?<br />
<br />
<math>\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5</math><br />
<br />
[[2010 AMC 12B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
At the beginning of the school year, <math>50\%</math> of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", and <math>50\%</math> answered "No." At the end of the school year, <math>70\%</math> answered "Yes" and <math>30\%</math> answered "No." Altogether, <math>x\%</math> of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80</math><br />
<br />
[[2010 AMC 12B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Shelby drives her scooter at a speed of <math>30</math> miles per hour if it is not raining, and <math>20</math> miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of <math>16</math> miles in <math>40</math> minutes. How many minutes did she drive in the rain?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30</math><br />
<br />
[[2010 AMC 12B Problems/Problem 7|Solution]]<br />
<br />
<br />
<br />
== Problem 8 ==<br />
Every high school in the city of Euclid sent a team of <math>3</math> students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed <math>37</math><sup>th</sup> and <math>64</math><sup>th</sup>, respectively. How many schools are in the city?<br />
<br />
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26</math><br />
<br />
[[2010 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Let <math>n</math> be the smallest positive integer such that <math>n</math> is divisible by <math>20</math>, <math>n^2</math> is a perfect cube, and <math>n^3</math> is a perfect square. What is the number of digits of <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math><br />
<br />
[[2010 AMC 12B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
The average of the numbers <math>1, 2, 3,\cdots, 98, 99,</math> and <math>x</math> is <math>100x</math>. What is <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
<br />
== Problem 11 ==<br />
A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability that it is divisible by <math>7</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
For what value of <math>x</math> does<br />
<br />
<cmath>\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?</cmath><br />
<br />
<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024</math><br />
<br />
[[2010 AMC 12B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
In <math>\triangle ABC</math>, <math>\cos(2A-B)+\sin(A+B)=2</math> and <math>AB=4</math>. What is <math>BC</math>?<br />
<br />
<math>\textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be positive integers with <math>a+b+c+d+e=2010</math> and let <math>M</math> be the largest of the sums <math>a+b</math>, <math>b+c</math>, <math>c+d</math> and <math>d+e</math>. What is the smallest possible value of <math>M</math>?<br />
<br />
<math>\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804</math><br />
<br />
[[2010 AMC 12B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
For how many ordered triples <math>(x,y,z)</math> of nonnegative integers less than <math>20</math> are there exactly two distinct elements in the set <math>\{i^x, (1+i)^y, z\}</math>, where <math>i=\sqrt{-1}</math>?<br />
<br />
<math>\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235</math><br />
<br />
[[2010 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
The entries in a <math>3 \times 3</math> array include all the digits from <math>1</math> through <math>9</math>, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br />
<br />
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60</math><br />
<br />
[[2010 AMC 12B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
A frog makes <math>3</math> jumps, each exactly <math>1</math> meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than <math>1</math> meter from its starting position?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than <math>100</math> points. What was the total number of points scored by the two teams in the first half?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math><br />
<br />
[[2010 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
A geometric sequence <math>(a_n)</math> has <math>a_1=\sin x</math>, <math>a_2=\cos x</math>, and <math>a_3= \tan x</math> for some real number <math>x</math>. For what value of <math>n</math> does <math>a_n=1+\cos x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8</math><br />
<br />
[[2010 AMC 12B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that<br />
<br />
<center><br />
<math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/><br />
<math>P(2) = P(4) = P(6) = P(8) = -a</math>.<br />
</center><br />
<br />
What is the smallest possible value of <math>a</math>?<br />
<br />
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math><br />
<br />
[[2010 AMC 12B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Let <math>ABCD</math> be a cyclic quadrilateral. The side lengths of <math>ABCD</math> are distinct integers less than <math>15</math> such that <math>BC\cdot CD=AB\cdot DA</math>. What is the largest possible value of <math>BD</math>?<br />
<br />
<math>\textbf{(A)}\ \sqrt{\dfrac{325}{2}} \qquad \textbf{(B)}\ \sqrt{185} \qquad \textbf{(C)}\ \sqrt{\dfrac{389}{2}} \qquad \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\dfrac{533}{2}}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Monic quadratic polynomials <math>P(x)</math> and <math>Q(x)</math> have the property that <math>P(Q(x))</math> has zeros at <math>x=-23, -21, -17,</math> and <math>-15</math>, and <math>Q(P(x))</math> has zeros at <math>x=-59,-57,-51</math> and <math>-49</math>. What is the sum of the minimum values of <math>P(x)</math> and <math>Q(x)</math>? <br />
<br />
<math>\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0</math><br />
<br />
[[2010 AMC 12B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
The set of real numbers <math>x</math> for which <br />
<br />
<cmath>\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1</cmath><br />
<br />
is the union of intervals of the form <math>a<x\le b</math>. What is the sum of the lengths of these intervals?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}</math><br />
<br />
[[2010 AMC 12B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
For every integer <math>n\ge2</math>, let <math>\text{pow}(n)</math> be the largest power of the largest prime that divides <math>n</math>. For example <math>\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2</math>. What is the largest integer <math>m</math> such that <math>2010^m</math> divides<br />
<br />
<center><br />
<math>\prod_{n=2}^{5300}\text{pow}(n)</math>?<br />
</center><br />
<br />
<br />
<math>\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78</math><br />
<br />
[[2010 AMC 12B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
<br />
{{AMC12 box|year=2010|ab=B|before=[[2010 AMC 12A Problems]]|after=[[2011 AMC 12A Problems]]}}<br />
<br />
{{MAA Notice}}</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=100197User talk:Meljel2019-01-08T05:49:52Z<p>Meljel: /* It's me meljel lol */</p>
<hr />
<div>== It's me meljel lol ==<br />
<br />
<cmath> \begin{align*}<br />
ax^2 + bx + c &= 0\\<br />
a(x^2 + \frac bax + \frac ca) &= 0\\<br />
a({(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca) &= 0\\<br />
{(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca &= 0\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac ca\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac {4ac}{4a^2}\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2 - 4ac}{4a^2}\\<br />
x + \frac {b}{2a} &= \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}\\<br />
x + \frac {b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= - \frac {b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\\<br />
\end{align*}<br />
</cmath><br />
<br />
== hihiihi ==<br />
<br />
<cmath>a^2-b^2=(a+b)(a-b)</cmath><br />
<cmath>(\cos \theta + i\sin \theta)^n = \cos n\theta + ii\sin n\theta</cmath></div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=100196User talk:Meljel2019-01-08T05:48:03Z<p>Meljel: /* It's me meljel lol */ new section</p>
<hr />
<div>== It's me meljel lol ==<br />
<br />
<cmath> \begin{align*}<br />
ax^2 + bx + c &= 0\\<br />
a(x^2 + \frac bax + \frac ca) &= 0\\<br />
a({(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca) &= 0\\<br />
{(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca &= 0\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac ca\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac {4ac}{4a^2}\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2 - 4ac}{4a^2}\\<br />
x + \frac {b}{2a} &= \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}\\<br />
x + \frac {b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= - \frac {b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\\<br />
\end{align*}<br />
</cmath><br />
<br />
== It's me meljel lol ==<br />
<br />
hi</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=100195User talk:Meljel2019-01-08T05:47:51Z<p>Meljel: Blanked the page</p>
<hr />
<div></div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=100194User talk:Meljel2019-01-08T05:47:38Z<p>Meljel: /* It's me meljel lol */ new section</p>
<hr />
<div><cmath> \begin{align*}<br />
ax^2 + bx + c &= 0\\<br />
a(x^2 + \frac bax + \frac ca) &= 0\\<br />
a({(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca) &= 0\\<br />
{(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca &= 0\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac ca\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac {4ac}{4a^2}\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2 - 4ac}{4a^2}\\<br />
x + \frac {b}{2a} &= \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}\\<br />
x + \frac {b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= - \frac {b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\\<br />
\end{align*}<br />
</cmath><br />
<br />
== It's me meljel lol ==<br />
<br />
hi</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=83441User talk:Meljel2017-02-12T02:18:59Z<p>Meljel: This is Fun</p>
<hr />
<div><cmath> \begin{align*}<br />
ax^2 + bx + c &= 0\\<br />
a(x^2 + \frac bax + \frac ca) &= 0\\<br />
a({(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca) &= 0\\<br />
{(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca &= 0\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac ca\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac {4ac}{4a^2}\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2 - 4ac}{4a^2}\\<br />
x + \frac {b}{2a} &= \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}\\<br />
x + \frac {b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= - \frac {b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\\<br />
\end{align*}<br />
</cmath></div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_25&diff=833972017 AMC 10A Problems/Problem 252017-02-10T04:36:47Z<p>Meljel: /* Solution 3 (Shorter and Not Casework) */</p>
<hr />
<div>==Problem==<br />
How many integers between <math>100</math> and <math>999</math>, inclusive, have the property that some permutation of its digits is a multiple of <math>11</math> between <math>100</math> and <math>999?</math> For example, both <math>121</math> and <math>211</math> have this property.<br />
<br />
<math> \mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486</math><br />
<br />
==Solution 1==<br />
<br />
Let the three-digit number be <math>ACB</math>:<br />
<br />
If a number is divisible by <math>11</math>, then the difference between the sums of alternating digits is a multiple of <math>11</math>.<br />
<br />
There are two cases:<br />
<math>A+B=C</math> and <math>A+B=C+11</math><br />
<br />
We now proceed to break down the cases.<br />
<br />
<br />
<math>\textbf{Case 1}</math>: <math>A+B=C</math>. <br />
<br />
<br />
<br />
<math>\textbf{Part 1}</math>: <math>B=0</math><br />
<math>A=C</math>, this case results in 110, 220, 330...990. There are two ways to arrange the digits in each of those numbers.<br />
<math>2 \cdot 9 = 18</math><br />
<br />
<math>\textbf{Part 2}</math>: <math>B>0</math><br />
<math>B=1, A+1=C</math>, this case results in 121, 231,... 891. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to <math>45</math> cases.<br />
<br />
<math>\textbf{Part 3}</math>: <math>B=2, A+2=C</math>, this case results in 242, 352,... 792. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to <math>33</math> cases.<br />
<br />
<math>\textbf{Part 4}</math>: <math>B=3, A+3=C</math>, this case results in 363, 473,...693. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to <math>21</math> cases.<br />
<br />
<math>\textbf{Part 5}</math>: <math>B=4, A+4=C</math>, this case results in 484 and 594. There are <math>6</math> ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to <math>9</math> cases.<br />
<br />
This case has <math>18+45+33+21+9=126</math> subcases.<br />
<br />
<br />
<br />
<math>\textbf{Case 2}</math>: <math>A+B=C+11</math>. <br />
<br />
<br />
<math>\textbf{Part 1}</math>: <math>C=0, A+B=11</math>, this cases results in 209, 308, ...506. There are <math>4</math> ways to arrange each of those cases. This leads to <math>16</math> cases.<br />
<br />
<math>\textbf{Part 2}</math>: <math>C=1, A+B=12</math>, this cases results in 319, 418, ...616. There are <math>6</math> ways to arrange each of those cases, except the last. This leads to <math>21</math> cases.<br />
<br />
<math>\textbf{Part 3}</math>: <math>C=2, A+B=13</math>, this cases results in 429, 528, ...617. There are <math>6</math> ways to arrange each of those cases. This leads to <math>18</math> cases.<br />
<br />
...<br />
If we continue this counting, we receive <math>16+21+18+15+12+9+6+3=100</math> subcases.<br />
<br />
<math>100+126=\boxed{\textbf{(A) } 226}</math><br />
<br />
~Mathguy1492<br />
<br />
==Solution 2==<br />
<br />
We note that we only have to consider multiples of 11 and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of 11 has:<br />
<br />
<math>\textbf{Case 1:}</math> All three digits are the same. <br />
By inspection, we find that there are no multiples of 11 here.<br />
<br />
<math>\textbf{Case 2:}</math> Two of the digits are the same, and the third is different.<br />
<br />
<math>\textbf{Case 2a:}</math><br />
There are 8 multiples of 11 without a zero that have this property:<br />
121, 242, 363, 484, 616, 737, 858, 979.<br />
Each contributes 3 valid permutations, so there are <math>8 \cdot 3 = 24</math> permutations in this subcase.<br />
<br />
<math>\textbf{Case 2b:}</math><br />
There are 9 multiples of 11 with a zero that have this property:<br />
110, 220, 330, 440, 550, 660, 770, 880, 990.<br />
Each one contributes 2 valid permutations (the first digit can't be zero), so there are <math>9 \cdot 2 = 18</math> permutations in this subcase.<br />
<br />
<math>\textbf{Case 3:}</math> All the digits are different.<br />
Since there are <math>\frac{990-110}{11}+1 = 81</math> multiples of 11 between 100 and 999, there are <math>81-8-9 = 64</math> multiples of 11 remaining in this case. However, 8 of them contain a zero, namely 209, 308, 407, 506, 605, 704, 803, and 902. Each of those multiples of 11 contributes <math>2 \cdot 2=4</math> valid permutations, but we overcounted by a factor of 2; every permutation of 209, for example, is also a permutation of 902. Therefore, there are <math>8 \cdot 4 / 2 = 16</math>. Therefore, there are <math>64-8=56</math> remaining multiples of 11 without a 0 in this case. Each one contributes <math>3! = 6</math> valid permutations, but once again, we overcounted by a factor of 2 (note that if a number ABC is a multiple of 11, then so is CBA). Therefore, there are <math>56 \cdot 6 / 2 = 168</math> valid permutations in this subcase.<br />
<br />
Adding up all the permutations from all the cases, we have <math>24+18+16+168 = \boxed{\textbf{(A) } 226}</math>.<br />
<br />
==Solution 3 (Shorter and Not Casework)==<br />
<br />
We can overcount and then subtract.<br />
We know there are <math>81</math> multiples of <math>11</math>.<br />
<br />
We can multiply by <math>6</math> for each permutation of these multiples. (Yet some multiples don't have 6)<br />
<br />
Now divide by <math>2</math>, because if a number <math>abc</math> with digits <math>a</math>, <math>b</math>, and <math>c</math> is a multiple of 11, then <math>cab</math> is also a multiple of 11 so we have counted the same permutations twice. <br />
<br />
Basically, each multiple of 11 has its own 3 permutations (say <math>abc</math> has <math>abc</math> <math>acb</math> and <math>bac</math> whereas <math>cab</math> has <math>cab</math> <math>cba</math> and <math>bca</math>). We know that each multiple of 11 has at least 3 permutations because it cannot have 3 repeating digits.<br />
<br />
Hence we have <math>243</math> permutations without subtracting for over count.<br />
Now note that we overcounted cases is which we have 0's at the start of each number. So in theory we could just answer <math>A</math> and move on.<br />
<br />
We can also solve it :/<br />
We overcount cases where the Middle digit of the number is 0 and the last digit is 0.<br />
<br />
Note that we assigned each multiple of 11 3 permutations.<br />
<br />
The last digit is <math>0</math> gives <math>9</math> possibilities where we overcount by <math>1</math> permutation for each of <math>110, 220, ... , 990</math>.<br />
<br />
The middle digit is 0 gives 8 possibilities where we overcount by 1.<br />
<math>605, 704, 803, 902</math> and <math>506, 407, 308, 209</math><br />
<br />
Subtracting <math>17</math> gives <math>\boxed{\textbf{(A) } 226}</math>.<br />
<br />
Now, we may ask if there is further overlap (I.e if two of <math>abc</math> and <math>bac</math> and <math>acb</math> were multiples of <math>11</math>) Thankfully, using divisibility rules, this can never happen as taking the divisibility rule mod 11 and adding we get that <math>2a</math>, <math>2b</math>, or <math>2c</math> is congruent to <math>0\ (mod\ 11)</math>. Since <math>a, b, c</math> are digits, this can never happen as none of them can equal 11 and they can't equal 0 as they are the leading digit of a 3 digit number in each of the cases<br />
<br />
~fuzz1<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Meljel&diff=72743User talk:Meljel2015-11-02T20:13:56Z<p>Meljel: Created page with "\begin{align*} ax^2 + bx + c &= 0\\ a(x^2 + \frac bax + \frac ca) &= 0\\ a({(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca) &= 0\\ {(x + \frac {b}{2a})}^2 - \frac{b^2}{4..."</p>
<hr />
<div>\begin{align*}<br />
ax^2 + bx + c &= 0\\<br />
a(x^2 + \frac bax + \frac ca) &= 0\\<br />
a({(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca) &= 0\\<br />
{(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca &= 0\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac ca\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac {4ac}{4a^2}\\<br />
{(x + \frac {b}{2a})}^2 &= \frac{b^2 - 4ac}{4a^2}\\<br />
x + \frac {b}{2a} &= \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}\\<br />
x + \frac {b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= - \frac {b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\<br />
x &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\\<br />
\end{align*}</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=Aops_font&diff=71463Aops font2015-08-06T00:17:38Z<p>Meljel: /* Full List */</p>
<hr />
<div>The Aops font is a font that can be used to make special symbols in the forum. You enclose some text inside [aops] and [/aops] and the text is rendered in the aops font.<br />
<br />
= Purpose=<br />
The [aops][/aops] tags allow you to render items from the custom font that AoPS uses -- you'll recognize these icons from the rest of the site.<br />
<br />
For example, to type the symbol for the edit post button <span class="aops-font">L</span>, you would type <br />
:[aops]L[/aops]<br />
<br />
=Use=<br />
== In the Forums==<br />
To use the font in the forums, simply put [aops] before your text and [/aops] after it. The text will be converted as described in the tables below.<br />
== In the Wiki==<br />
Because bbcode does not work in the AoPS wiki, you need to use the following HTML code:<br />
:<code> <nowiki><span class="aops-font">text here!</span></nowiki></code><br />
<br />
=Characters=<br />
==Commonly Used==<br />
<div style="font-size:large;"><br />
{| class="wikitable"<br />
! Symbol<br />
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| <span class="aops-font">k</span><br />
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|}<br />
==Full List==<br />
<div style="font-size:large;"><br />
{| class="wikitable"<br />
! Character<br />
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| <span class="aops-font">=</span> <br />
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</div><br />
[[Category:AoPS forums]]</div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=User:Meljel&diff=63312User:Meljel2014-09-11T04:05:40Z<p>Meljel: Created page with "Hello there! <math>\sqrt5=\binom{\frac{\frac56+\frac98}{Fractions!}}{More Fractions!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!}</math>"</p>
<hr />
<div>Hello there!<br />
<br />
<math>\sqrt5=\binom{\frac{\frac56+\frac98}{Fractions!}}{More Fractions!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!}</math></div>Meljelhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems&diff=627832010 AMC 8 Problems2014-08-06T19:30:52Z<p>Meljel: /* Problem 18 */</p>
<hr />
<div>==Problem 1==<br />
At Euclid Middle School the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are <math>11</math> students in Mrs. Germain's class, <math>8</math> students in Mr. Newton's class, and <math>9</math> students in Mrs. Young's class taking the AMC 8 this year. How many mathematics students at Euclid Middle School are taking the contest? <br />
<br />
<math> \textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 1 | Solution]]<br />
<br />
==Problem 2==<br />
If <math>a @ b = \frac{a\times b}{a+b}</math> for <math>a,b</math> positive integers, then what is <math>5 @10</math>? <br />
<br />
<math>\textbf{(A)}\ \frac{3}{10} \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{10}{3} \qquad\textbf{(E)}\ 50</math><br />
<br />
[[2010 AMC 8 Problems/Problem 2 | Solution]]<br />
<br />
==Problem 3==<br />
The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price?<br />
<br />
<asy><br />
import graph; size(16.38cm); real lsf=2; pathpen=linewidth(0.7); pointpen=black; pen fp = fontsize(10); pointfontpen=fp; real xmin=-1.33,xmax=11.05,ymin=-9.01,ymax=-0.44;<br />
pen ycycyc=rgb(0.55,0.55,0.55);<br />
pair A=(1,-6), B=(1,-2), D=(1,-5.8), E=(1,-5.6), F=(1,-5.4), G=(1,-5.2), H=(1,-5), J=(1,-4.8), K=(1,-4.6), L=(1,-4.4), M=(1,-4.2), N=(1,-4), P=(1,-3.8), Q=(1,-3.6), R=(1,-3.4), S=(1,-3.2), T=(1,-3), U=(1,-2.8), V=(1,-2.6), W=(1,-2.4), Z=(1,-2.2), E_1=(1.4,-2.6), F_1=(1.8,-2.6), O_1=(14,-6), P_1=(14,-5), Q_1=(14,-4), R_1=(14,-3), S_1=(14,-2), C_1=(1.4,-6), D_1=(1.8,-6), G_1=(2.4,-6), H_1=(2.8,-6), I_1=(3.4,-6), J_1=(3.8,-6), K_1=(4.4,-6), L_1=(4.8,-6), M_1=(5.4,-6), N_1=(5.8,-6), T_1=(6.4,-6), U_1=(6.8,-6), V_1=(7.4,-6), W_1=(7.8,-6), Z_1=(8.4,-6), A_2=(8.8,-6), B_2=(9.4,-6), C_2=(9.8,-6), D_2=(10.4,-6), E_2=(10.8,-6), L_2=(2.4,-3.2), M_2=(2.8,-3.2), N_2=(3.4,-4), O_2=(3.8,-4), P_2=(4.4,-3.6), Q_2=(4.8,-3.6), R_2=(5.4,-3.6), S_2=(5.8,-3.6), T_2=(6.4,-3.4), U_2=(6.8,-3.4), V_2=(7.4,-3.8), W_2=(7.8,-3.8), Z_2=(8.4,-2.8), A_3=(8.8,-2.8), B_3=(9.4,-3.2), C_3=(9.8,-3.2), D_3=(10.4,-3.8), E_3=(10.8,-3.8);<br />
filldraw(C_1--E_1--F_1--D_1--cycle,ycycyc); filldraw(G_1--L_2--M_2--H_1--cycle,ycycyc); filldraw(I_1--N_2--O_2--J_1--cycle,ycycyc); filldraw(K_1--P_2--Q_2--L_1--cycle,ycycyc); filldraw(M_1--R_2--S_2--N_1--cycle,ycycyc); filldraw(T_1--T_2--U_2--U_1--cycle,ycycyc); filldraw(V_1--V_2--W_2--W_1--cycle,ycycyc); filldraw(Z_1--Z_2--A_3--A_2--cycle,ycycyc); filldraw(B_2--B_3--C_3--C_2--cycle,ycycyc); filldraw(D_2--D_3--E_3--E_2--cycle,ycycyc); D(B--A,linewidth(0.4)); D(H--(8,-5),linewidth(0.4)); D(N--(8,-4),linewidth(0.4)); D(T--(8,-3),linewidth(0.4)); D(B--(8,-2),linewidth(0.4)); D(B--S_1); D(T--R_1); D(N--Q_1); D(H--P_1); D(A--O_1); D(C_1--E_1); D(E_1--F_1); D(F_1--D_1); D(D_1--C_1); D(G_1--L_2); D(L_2--M_2); D(M_2--H_1); D(H_1--G_1); D(I_1--N_2); D(N_2--O_2); D(O_2--J_1); D(J_1--I_1); D(K_1--P_2); D(P_2--Q_2); D(Q_2--L_1); D(L_1--K_1); D(M_1--R_2); D(R_2--S_2); D(S_2--N_1); D(N_1--M_1); D(T_1--T_2); D(T_2--U_2); D(U_2--U_1); D(U_1--T_1); D(V_1--V_2); D(V_2--W_2); D(W_2--W_1); D(W_1--V_1); D(Z_1--Z_2); D(Z_2--A_3); D(A_3--A_2); D(A_2--Z_1); D(B_2--B_3); D(B_3--C_3); D(C_3--C_2); D(C_2--B_2); D(D_2--D_3); D(D_3--E_3); D(E_3--E_2); D(E_2--D_2); label("0",(0.88,-5.91),SE*lsf,fp); label("\$ 5",(0.3,-4.84),SE*lsf,fp); label("\$ 10",(0.2,-3.84),SE*lsf,fp); label("\$ 15",(0.2,-2.85),SE*lsf,fp); label("\$ 20",(0.2,-1.85),SE*lsf,fp); label("$\mathrm{Price}$",(0.16,-3.45),SE*lsf,fp); label("$1$",(1.54,-5.97),SE*lsf,fp); label("$2$",(2.53,-5.95),SE*lsf,fp); label("$3$",(3.53,-5.94),SE*lsf,fp); label("$4$",(4.55,-5.94),SE*lsf,fp); label("$5$",(5.49,-5.95),SE*lsf,fp); label("$6$",(6.53,-5.95),SE*lsf,fp); label("$7$",(7.55,-5.95),SE*lsf,fp); label("$8$",(8.52,-5.95),SE*lsf,fp); label("$9$",(9.57,-5.97),SE*lsf,fp); label("$10$",(10.56,-5.94),SE*lsf,fp); label("Month",(7.14,-6.43),SE*lsf,fp);<br />
D(A,linewidth(1pt)); D(B,linewidth(1pt)); D(D,linewidth(1pt)); D(E,linewidth(1pt)); D(F,linewidth(1pt)); D(G,linewidth(1pt)); D(H,linewidth(1pt)); D(J,linewidth(1pt)); D(K,linewidth(1pt)); D(L,linewidth(1pt)); D(M,linewidth(1pt)); D(N,linewidth(1pt)); D(P,linewidth(1pt)); D(Q,linewidth(1pt)); D(R,linewidth(1pt)); D(S,linewidth(1pt)); D(T,linewidth(1pt)); D(U,linewidth(1pt)); D(V,linewidth(1pt)); D(W,linewidth(1pt)); D(Z,linewidth(1pt)); D(E_1,linewidth(1pt)); D(F_1,linewidth(1pt)); D(O_1,linewidth(1pt)); D(P_1,linewidth(1pt)); D(Q_1,linewidth(1pt)); D(R_1,linewidth(1pt)); D(S_1,linewidth(1pt)); D(C_1,linewidth(1pt)); D(D_1,linewidth(1pt)); D(G_1,linewidth(1pt)); D(H_1,linewidth(1pt)); D(I_1,linewidth(1pt)); D(J_1,linewidth(1pt)); D(K_1,linewidth(1pt)); D(L_1,linewidth(1pt)); D(M_1,linewidth(1pt)); D(N_1,linewidth(1pt)); D(T_1,linewidth(1pt)); D(U_1,linewidth(1pt)); D(V_1,linewidth(1pt)); D(W_1,linewidth(1pt)); D(Z_1,linewidth(1pt)); D(A_2,linewidth(1pt)); D(B_2,linewidth(1pt)); D(C_2,linewidth(1pt)); D(D_2,linewidth(1pt)); D(E_2,linewidth(1pt)); D(L_2,linewidth(1pt)); D(M_2,linewidth(1pt)); D(N_2,linewidth(1pt)); D(O_2,linewidth(1pt)); D(P_2,linewidth(1pt)); D(Q_2,linewidth(1pt)); D(R_2,linewidth(1pt)); D(S_2,linewidth(1pt)); D(T_2,linewidth(1pt)); D(U_2,linewidth(1pt)); D(V_2,linewidth(1pt)); D(W_2,linewidth(1pt)); D(Z_2,linewidth(1pt)); D(A_3,linewidth(1pt)); D(B_3,linewidth(1pt)); D(C_3,linewidth(1pt)); D(D_3,linewidth(1pt)); D(E_3,linewidth(1pt));<br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
<math>\textbf{(A)}\ 50 \qquad<br />
\textbf{(B)}\ 62 \qquad<br />
\textbf{(C)}\ 70 \qquad<br />
\textbf{(D)}\ 89 \qquad<br />
\textbf{(E)}\ 100</math><br />
<br />
[[2010 AMC 8 Problems/Problem 3 | Solution]]<br />
<br />
==Problem 4==<br />
What is the sum of the mean, medium, and mode of the numbers <math>2,3,0,3,1,4,0,3</math>? <br />
<br />
<math> \textbf{(A)}\ 6.5 \qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 7.5\qquad\textbf{(D)}\ 8.5\qquad\textbf{(E)}\ 9 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 4 | Solution]]<br />
<br />
==Problem 5==<br />
Alice needs to replace a light bulb located <math>10</math> centimeters below the ceiling in her kitchen. The ceiling is <math>2.4</math> meters above the floor. Alice is <math>1.5</math> meters tall and can reach <math>46</math> centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?<br />
<br />
<math> \textbf{(A)}\ 32 \qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 38\qquad\textbf{(E)}\ 40 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 5 | Solution]]<br />
<br />
==Problem 6==<br />
Which of the following figures has the greatest number of lines of symmetry? <br />
<br />
<math> \textbf{(A)}\ \text{equilateral triangle}</math><br />
<math>\textbf{(B)}\ \text{non-square rhombus} </math><br />
<math>\textbf{(C)}\ \text{non-square rectangle}</math><br />
<math>\textbf{(D)}\ \text{isosceles trapezoid}</math><br />
<math>\textbf{(E)}\ \text{square} </math><br />
<br />
[[2010 AMC 8 Problems/Problem 6 | Solution]]<br />
<br />
==Problem 7==<br />
Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar? <br />
<br />
<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 7 | Solution]]<br />
<br />
==Problem 8==<br />
As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction <math>1/2</math> mile in front of her. After she passes him, she can see him in her rear mirror until he is <math>1/2</math> mile behind her. Emily rides at a constant rate of <math>12</math> miles per hour, and Emerson skates at a constant rate of <math>8</math> miles per hour. For how many minutes can Emily see Emerson? <br />
<br />
<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 8 | Solution]]<br />
<br />
==Problem 9==<br />
Ryan got <math>80\%</math> of the problems correct on a <math>25</math>-problem test, <math>90\%</math> on a <math>40</math>-problem test, and <math>70\%</math> on a <math>10</math>-problem test. What percent of all the problems did Ryan answer correctly? <br />
<br />
<math> \textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 9 | Solution]]<br />
<br />
==Problem 10==<br />
Six pepperoni circles will exactly fit across the diameter of a <math>12</math>-inch pizza when placed. If a total of <math>24</math> circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?<br />
<br />
<math> \textbf{(A)}\ \frac 12 \qquad\textbf{(B)}\ \frac 23 \qquad\textbf{(C)}\ \frac 34 \qquad\textbf{(D)}\ \frac 56 \qquad\textbf{(E)}\ \frac 78 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 10 | Solution]]<br />
<br />
==Problem 11==<br />
The top of one tree is <math>16</math> feet higher than the top of another tree. The heights of the two trees are in the ratio <math>3:4</math>. In feet, how tall is the taller tree? <br />
<br />
<math> \textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 11 | Solution]]<br />
<br />
==Problem 12==<br />
<br />
Of the <math>500</math> balls in a large bag, <math>80\%</math> are red and the rest are blue. How many of the red balls must be removed from the bag so that <math>75\%</math> of the remaining balls are red?<br />
<br />
<math> \textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 12 | Solution]]<br />
<br />
==Problem 13==<br />
The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is <math>30\%</math> of the perimeter. What is the length of the longest side?<br />
<br />
<math> \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 13 | Solution]]<br />
<br />
==Problem 14==<br />
What is the sum of the prime factors of <math>2010</math>? <br />
<br />
<math> \textbf{(A)}\ 67 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 14 | Solution]]<br />
<br />
==Problem 15==<br />
A jar contains five different colors of gumdrops: <math>30\%</math> are blue, <math>20\%</math> are brown, <math>15\%</math> red, <math>10\%</math> yellow, and the other <math>30</math> gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?<br />
<br />
<math> \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 15 | Solution]]<br />
<br />
==Problem 16==<br />
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle? <br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{\pi}}{2} \qquad\textbf{(B)}\ \sqrt{\pi} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ 2\pi \qquad\textbf{(E)}\ \pi^{2}</math><br />
<br />
[[2010 AMC 8 Problems/Problem 16 | Solution]]<br />
<br />
==Problem 17==<br />
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\frac{XQ}{QY}</math>?<br />
<br />
<asy><br />
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1);<br />
draw((0,0)--(6,0),linewidth(1.2pt)); draw((0,0)--(0,1),linewidth(1.2pt)); draw((0,1)--(1,1),linewidth(1.2pt)); draw((1,1)--(1,2),linewidth(1.2pt)); draw((1,2)--(5,2),linewidth(1.2pt)); draw((5,2)--(5,1),linewidth(1.2pt)); draw((5,1)--(6,1),linewidth(1.2pt)); draw((6,1)--(6,0),linewidth(1.2pt)); draw((1,1)--(5,1),linewidth(1.2pt)+linetype("2pt 2pt")); draw((1,1)--(1,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((2,2)--(2,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((3,2)--(3,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((4,2)--(4,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((5,1)--(5,0),linewidth(1.2pt)+linetype("2pt 2pt")); draw((0,0)--(5,1.5),linewidth(1.2pt));<br />
dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); dot((0,1),ds); dot((1,1),ds); dot((1,2),ds); dot((5,2),ds); label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); dot((4,2),ds); dot((5,1.5),ds); label("$Q$", (5.14,1.51),NE*lsf); clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ \frac 25 \qquad<br />
\textbf{(B)}\ \frac 12 \qquad<br />
\textbf{(C)}\ \frac 35 \qquad<br />
\textbf{(D)}\ \frac 23 \qquad<br />
\textbf{(E)}\ \frac 34</math><br />
<br />
[[2010 AMC 8 Problems/Problem 17 | Solution]]<br />
<br />
==Problem 18==<br />
A decorative window is made up of a rectangle with semicircles on either end. The ratio of <math>AD</math> to <math>AB</math> is <math>3:2</math>, and <math>AB</math> is 30 inches. What is the ratio of the area of the rectangle to the combined areas of the semicircles? <br />
<br />
<asy><br />
import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0));<br />
dot((0,0),ds); label("$A$",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label("$B$",(2.61,-0.26),NE*lsf); dot((0,4),ds); label("$D$",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label("$C$",(2.64,3.98),NE*lsf);<br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<br />
<math> \textbf{(A)}\ 2:3 \qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi \qquad\textbf{(D)}\ 9: \pi \qquad\textbf{(E)}\ 30 : \pi</math><br />
<br />
[[2010 AMC 8 Problems/Problem 18 | Solution]]<br />
<br />
==Problem 19==<br />
The two circles pictured have the same center <math>C</math>. Chord <math>\overline{AD}</math> is tangent to the inner circle at <math>B</math>, <math>AC</math> is <math>10</math>, and chord <math>\overline{AD}</math> has length <math>16</math>. What is the area between the two circles?<br />
<br />
<asy><br />
unitsize(45);<br />
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1);<br />
draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0));<br />
dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi </math><br />
<br />
[[2010 AMC 8 Problems/Problem 19 | Solution]]<br />
<br />
==Problem 20==<br />
In a room, <math>2/5</math> of the people are wearing gloves, and <math>3/4</math> of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove? <br />
<br />
<math> \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 20 | Solution]]<br />
<br />
==Problem 21==<br />
Hui is an avid reader. She bought a copy of the best seller ''Math is Beautiful''. On the first day, Hui read <math>1/5</math> of the pages plus <math>12</math> more, and on the second day she read <math>1/4</math> of the remaining pages plus <math>15</math> pages. On the third day she read <math>1/3</math> of the remaining pages plus <math>18</math> pages. She then realized that there were only <math>62</math> pages left to read, which she read the next day. How many pages are in this book? <br />
<br />
<math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 21 | Solution]]<br />
<br />
==Problem 22==<br />
The hundreds digit of a three-digit number is <math>2</math> more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?<br />
<br />
<math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 22 | Solution]]<br />
<br />
==Problem 23==<br />
<br />
Semicircles <math>POQ</math> and <math>ROS</math> pass through the center <math>O</math>. What is the ratio of the combined areas of the two semicircles to the area of circle <math>O</math>? <br />
<asy><br />
import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); <br />
dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<math> \textbf{(A)}\ \frac{\sqrt 2}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{\pi}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2} </math><br />
<br />
[[2010 AMC 8 Problems/Problem 23 | Solution]]<br />
<br />
==Problem 24==<br />
What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>?<br />
<br />
<math> \textbf{(A)}\ 2^2^4<10^8<5^1^2 </math><br />
<math> \textbf{(B)}\ 2^2^4<5^1^2<10^8 </math><br />
<math> \textbf{(C)}\ 5^1^2<2^2^4<10^8 </math><br />
<math> \textbf{(D)}\ 10^8<5^1^2<2^2^4</math><br />
<math> \textbf{(E)}\ 10^8<2^2^4<5^1^2 </math><br />
<br />
[[2010 AMC 8 Problems/Problem 24 | Solution]]<br />
<br />
==Problem 25==<br />
Everyday at school, Jo climbs a flight of <math>6</math> stairs. Joe can take the stairs <math>1</math>, <math>2</math>, or <math>3</math> at a time. For example, Jo could climb <math>3</math>, then <math>1</math>, then <math>2</math>. In how many ways can Jo climb the stairs?<br />
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<math> \textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24 </math><br />
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[[2010 AMC 8 Problems/Problem 25 | Solution]]<br />
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==See Also==<br />
{{AMC8 box|year=2010|before=[[2009 AMC 8 Problems|2009 AMC 8]]|after=[[2011 AMC 8 Problems|2011 AMC 8]]}}<br />
* [[AMC 8]]<br />
* [[AMC 8 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
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{{MAA Notice}}</div>Meljel