https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Russellk&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T15:30:54ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_13&diff=1805172019 AMC 12B Problems/Problem 132022-11-09T03:42:14Z<p>Russellk: /* Solution 11 (This method might be a coincidence) */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #17]] and [[2019 AMC 12B Problems|2019 AMC 12B #13]]}}<br />
==Problem==<br />
<br />
A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin <math>k</math> is <math>2^{-k}</math> for <math>k = 1,2,3....</math> What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?<br><br />
<math>\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}</math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin. Clearly, the probability of both landing in the same bin is <math>\sum_{k=1}^{\infty}{2^{-k} \cdot 2^{-k}} = \sum_{k=1}^{\infty}2^{-2k} = \frac{1}{3}</math> (by the geometric series sum formula). Therefore, since the other two probabilities have to both the same, they have to be <math>\frac{1-\frac{1}{3}}{2} = \boxed{\textbf{(C) } \frac{1}{3}}</math>.<br />
<br />
<br />
Note: the formula is <math>\frac{a_{1}}{1-r}</math> where <math>a_{1}</math> is the first term and <math>r</math> is the common ratio.<br />
Derivation of the geometric series sum formula:<br />
Let <math>S = S=a_{1}+a_{1}r+a_{1}r^{2}+a_{1}r^{3}+...</math> and so on to infinity.<br />
Then <math>rS=a_{1}r+a_{1}r^{2}+a_{1}r^{3}+...</math> and so on to infinity. Notice that the terms in the second expression are the same as all the terms in the first EXCEPT for <math>a_{1}</math>.<br />
Subtract <math>S-rS=a_{1}</math>, factor <math>S\left(1-r\right)=a_{1}</math>, and finally <math>S=\frac{a_{1}}{1-r}</math>.<br />
<br />
Post scriptum. The formula only works if r is less than 1; otherwise, the series will diverge to infinity or negative infinity.<br />
~JH. L<br />
<br />
=== Solution 2 ===<br />
Suppose the green ball goes in bin <math>i</math>, for some <math>i \ge 1</math>. The probability of this occurring is <math>\frac{1}{2^i}</math>. Given that this occurs, the probability that the red ball goes in a higher-numbered bin is <math>\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}</math> (by the geometric series sum formula). Thus the probability that the green ball goes in bin <math>i</math>, and the red ball goes in a bin greater than <math>i</math>, is <math>\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}</math>. Summing from <math>i=1</math> to infinity, we get<br />
<br />
<cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath><br />
where we again used the geometric series sum formula. (Alternatively, if this sum equals <math>n</math>, then by writing out the terms and multiplying both sides by <math>4</math>, we see <math>4n = n+1</math>, which gives <math>n = \frac{1}{3}</math>.)<br />
<br />
=== Solution 3 ===<br />
For red ball in bin <math>k</math>, <math>\Pr(\text{Green Below Red})=\sum\limits_{i=1}^{k-1}2^{-i}</math> (GBR) and <math>\Pr(\text{Red in Bin k}=2^{-k}</math> (RB). <br />
<cmath>\Pr(\text{GBR}|\text{RB})=\sum\limits_{k=1}^{\infty}2^{-k}\sum\limits_{i=1}^{k-1}2^{-i}=\sum\limits_{k=1}^{\infty}2^{-k}\cdot\frac{1}{2}(\frac{1-(1/2)^{k-1}}{1-1/2})</cmath><br />
<cmath>\sum\limits_{k=1}^{\infty}\frac{1}{2^{-k}}-2\sum\limits_{k=1}^\infty\frac{1}{(2^2)^{-k}}\implies 1-2/3=\boxed{(\textbf{C}) \frac{1}{3}}</cmath><br />
<br />
=== Solution 4 ===<br />
The probability that the two balls will go into adjacent bins is <math>\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots = \frac{1}{6}</math> by the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of <math>2</math> from each other is <math>\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} + \cdots = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{12}</math> (again recognizing a geometric series). We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is <math>\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots</math>, which, by the geometric series sum formula, is <math>\boxed{\textbf{(C) } \frac{1}{3}}</math>.<br />
-fidgetboss_4000<br />
<br />
=== Solution 5 (quick, conceptual) ===<br />
Define a win as a ball appearing in higher numbered box.<br />
<br />
Start from the first box. <br />
<br />
There are <math>4</math> possible results in the box: Red, Green, Red and Green, or none, with an equal probability of <math>\frac{1}{4}</math> for each. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution, so if <math>p</math> is the probability that Red wins, we can write <math>p = \frac{1}{4} + \frac{1}{4}p</math>: there is a <math>\frac{1}{4}</math> probability that "Red" wins immediately, a <math>0</math> probability in the cases "Green" or "Red and Green", and in the "None" case (which occurs with <math>\frac{1}{4}</math> probability), we then start again, giving the same probability <math>p</math>. Hence, solving the equation, we get <math>p = \boxed{\textbf{(C) } \frac{1}{3}}</math>.<br />
<br />
=== Solution 6 ===<br />
Write out the infinite geometric series as <math>\frac{1}{2}</math>, <math>\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \cdots</math>. To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms (i.e term <math>1</math>, term <math>3</math>, etc.), and then sum the remaining terms - this is in fact precisely equivalent to the method of Solution 2. Writing this out as another infinite geometric sequence, we are left with <math>\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \cdots</math>. Summing, we get <cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath><br />
<br />
=== Solution 7 ===<br />
Fixing the green ball to fall into bin <math>1</math> gives a probability of <math>\frac{1}{2}\left(\frac{1}{2^2}+\frac{1}{2^3} +...\right)</math> for the red ball to fall into a higher bin. Fixing the green ball to fall into bin <math>2</math> gives a probability of <math>\frac{1}{2^2}\left(\frac{1}{2^3}+\frac{1}{2^4} +...\right)</math>. Factoring out the denominator of the first fraction in each probability gives <math>\frac{1}{2^3}\left(1+\frac{1}{2}+\frac{1}{2^2}+...\right)+\frac{1}{2^5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...\right)+...</math> so factoring out <math>\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\right)</math> results in the probability simplifying to <math>\left(\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+...\right)\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\right)</math> and using the formula <math>\frac{a}{1-r}</math> to find both series, we obtain <math>\left(\frac{\frac{1}{2^3}}{1-\frac{1}{4}}\right)\left(\frac{1}{1-\frac{1}{2}}\right)</math> which simplifies to <math>\boxed{\textbf{(C) } \frac{1}{3}}</math> -- OGBooger<br />
<br />
=== Solution 8 ===<br />
We can think of this problem as "what is the probability that the green ball's bin is less than the red ball's bin". We do not consider the case where the red ball goes into bin <math>1</math> because the green ball has no where to go then. The chance that the green one is below the red one if the red one goes to bin <math>2</math> is <math>\frac{1}{4}</math> chance that the red ball even goes in bin <math>2</math> and <math>\frac{1}{2}</math> chance that the green ball goes into any bin less than <math>2</math>. Similarly, if the red goes into bin <math>3</math>, there is a <math>\frac{1}{8} \cdot \left(\frac{1}{4} + \frac{1}{2}\right)</math> chance, or <math>\frac{3}{32}</math>, continuing like this, we get this sequence:<br />
<br />
<math>\frac{1}{8}, \frac{3}{32}, \frac{7}{128}, ...</math><br />
<br />
Let <math>S</math> equal the sum of our series:<br />
<br />
<math>S = \frac{1}{8} + \frac{3}{32} + \frac{7}{128} + ...</math>. That means we can write another equation:<br />
<math>\frac{S}{4} = \frac{1}{32} + \frac{3}{128} + ...</math><br />
<br />
Subtracting <math>\frac{S}{4}</math> from <math>S</math>, yields:<br />
<br />
<math>S - \frac{S}{4} = \frac{1}{8} + \frac{2}{32} + \frac{4}{128} + ...</math><br />
<br />
We see that the above series is a infinite geometric sequence with common ratio <math>\frac{1}{2}</math>. Therefore, the sum of that infinite series is <math>\frac{\frac{1}{8}}{\frac{1}{2}}</math>, which equals <math>\frac{1}{4}</math>. Our equation is now <math>S - \frac{S}{4} = \frac{1}{4}</math>. Solving for <math>S</math> shows that <math>S = \frac{1}{3}</math>.<br />
<br />
Our answer is <math>\boxed{\textbf{(C) }\frac{1}{3}}</math><br />
<br />
~ericshi1685<br />
<br />
=== Solution 9 (quick, symmetry) ===<br />
<br />
Denote <math>G,R</math> the bin numbers of the green and red balls, respectively. The common probability distribution of <math>G,B</math> can be constructed by keep splitting the remaining unassigned probability into two halves: one goes to the smallest number that has not been assigned, and the other goes to the rest. In other words, <math>\Pr(G=k) = \Pr (G>k), \forall k \in \mathbb{N}</math>. Then,<br />
<br />
<cmath><br />
\Pr(G>R)=\sum_{k=1}^\infty \Pr(G>k) \Pr(R=k) = \sum_{k=1}^\infty \Pr(G=k) \Pr(R=k) = \Pr (G=R)<br />
</cmath><br />
<br />
Similarly <math>\Pr(G<R)=\Pr(G=R)</math>. Therefore all three probabilities equal <math>\boxed{\textbf{(C) }\frac{1}{3}}</math>.<br />
<br />
~asops<br />
<br />
=== Solution 10 ===<br />
<br />
The probability of the red ball falling ahead of the green ball is the same as the probability of the green ball falling ahead of the red ball. Therefore, if we calculate the probability of the red ball and the green ball falling inside the same box, we get the answer by subtracting that probability from 1. <math>P(same) = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \cdots = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} = \frac{1}{3}</math>. Therefore, our answer is <math>\frac{1-\frac{1}{3}}{2} = \boxed{\frac{1}{3}}</math><br />
<br />
-NL008<br />
<br />
=== Solution 11 (This method might be a coincidence)===<br />
My method could be completely coincidental, but, there are 3 different cases we have to consider: Red in a higher numbered bin than Green, Red in the same numbered bin as Green, or Green in a higher bin than Red. Since only the first case satisfies these conditions, the answer is <math>\boxed{\textbf{(C) }\frac{1}{3}}.</math><br />
- ethancui0529<br />
<br />
This is a coincidence, as the infinite geometric sum of the series that starts with <math>\dfrac{1}{2}</math> and has a common ratio of <math>\dfrac{1}{2}</math> is <math>\dfrac{1}{3}</math>. With any other starting term or ratio, this answer would not be valid.<br />
<br />
~russellk<br />
<br />
== Video Solutions ==<br />
=== Video Solution 1 ===<br />
For those who want a video solution: https://youtu.be/VP7ltu-XEq8<br />
<br />
=== Video Solution 2 ===<br />
https://youtu.be/_0YaCyxiMBo?t=353<br />
<br />
~IceMatrix<br />
<br />
===Video Solution 3, by OmegaLearn ===<br />
https://youtu.be/IRyWOZQMTV8?t=2484<br />
<br />
~ pi_is_3.14<br />
<br />
===Related Video===<br />
https://www.youtube.com/watch?v=RBf1s4TassI<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2019|ab=B|num-b=16|num-a=18}}<br />
{{AMC12 box|year=2019|ab=B|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_25&diff=1748142014 AMC 8 Problems/Problem 252022-06-13T17:26:37Z<p>Russellk: /* Problem */</p>
<hr />
<div>==Problem==<br />
A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5 miles per hour, how many hours will it take to cover the one-mile stretch?<br />
<asy>size(10cm); pathpen=black; pointpen=black;<br />
D(arc((-2,0),1,300,360));<br />
D(arc((0,0),1,0,180));<br />
D(arc((2,0),1,180,360));<br />
D(arc((4,0),1,0,180));<br />
D(arc((6,0),1,180,240));<br />
D((-1.5,1)--(5.5,1));<br />
D((-1.5,0)--(5.5,0),dashed);<br />
D((-1.5,-1)--(5.5,-1));</asy><br />
Note: 1 mile = 5280 feet<br />
<br />
<math> \textbf{(A) }\frac{\pi}{11}\qquad\textbf{(B) }\frac{\pi}{10}\qquad\textbf{(C) }\frac{\pi}{5}\qquad\textbf{(D) }\frac{2\pi}{5}\qquad\textbf{(E) }\frac{2\pi}{3} </math><br />
<br />
==Video Solution for Problems 21-25==<br />
https://www.youtube.com/watch?v=6S0u_fDjSxc<br />
<br />
==Video Solution==<br />
https://youtu.be/zi01koyAVhM ~savannahsolver<br />
<br />
==Solution(s)==<br />
===Solution 1===<br />
There are two possible interpretations of the problem: that the road as a whole is <math>40</math> feet wide, or that each lane is <math>40</math> feet wide. Both interpretations will arrive at the same result. However, let us stick with the first interpretation for simplicity. Each lane must then be <math>20</math> feet wide, so Robert must be riding his bike in semicircles with radius <math>20</math> feet and diameter <math>40</math> feet. Since the road is <math>5280</math> feet long, over the whole mile, Robert rides <math>\frac{5280}{40} =132</math> semicircles in total. Were the semicircles full circles, their circumference would be <math>2\pi\cdot 20=40\pi</math> feet; as it is, the circumference of each is half that, or <math>20\pi</math> feet. Therefore, over the stretch of highway, Robert rides a total of <math>132\cdot 20\pi =2640\pi</math> feet, equivalent to <math>\frac{\pi}{2}</math> miles. Robert rides at 5 miles per hour, so divide the <math>\frac{\pi}{2}</math> miles by <math>5</math> mph (because <math>t = \frac{d}{r}</math> and time = distance/rate) to arrive at <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours.<br />
<br />
===Solution 2===<br />
If Robert rides in a straight line, it will take him <math>\frac{1}{5}</math> hours. When riding in semicircles, let the radius of the semicircle <math>r</math>, then the circumference of a semicircle is <math>\pi r</math>. The ratio of the circumference of the semicircle to its diameter is <math>\frac{\pi}{2}</math>, so the time Robert takes is <math>\frac{1}{5} \cdot \frac{\pi}{2}</math>, which equals to <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2014|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_14&diff=1733652011 AMC 12A Problems/Problem 142022-04-15T03:06:57Z<p>Russellk: /* Video Solution */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>a</math> and <math>b</math> are single-digit positive integers chosen independently and at random. What is the probability that the point <math>(a,b)</math> lies above the parabola <math>y=ax^2-bx</math>?<br />
<br />
<math><br />
\textbf{(A)}\ \frac{11}{81} \qquad<br />
\textbf{(B)}\ \frac{13}{81} \qquad<br />
\textbf{(C)}\ \frac{5}{27} \qquad<br />
\textbf{(D)}\ \frac{17}{81} \qquad<br />
\textbf{(E)}\ \frac{19}{81} </math><br />
<br />
== Solution ==<br />
If <math>(a,b)</math> lies above the parabola, then <math>b</math> must be greater than <math>y(a)</math>. We thus get the inequality <math>b>a^3-ba</math>. Solving this for <math>b</math> gives us <math>b>\frac{a^3}{a+1}</math>. Now note that <math>\frac{a^3}{a+1}</math> constantly increases when <math>a</math> is positive. Then since this expression is greater than <math>9</math> when <math>a=4</math>, we can deduce that <math>a</math> must be less than <math>4</math> in order for the inequality to hold, since otherwise <math>b</math> would be greater than <math>9</math> and not a single-digit integer. The only possibilities for <math>a</math> are thus <math>1</math>, <math>2</math>, and <math>3</math>.<br />
<br />
For <math>a=1</math>, we get <math>b>\frac{1}{2}</math> for our inequality, and thus <math>b</math> can be any integer from <math>1</math> to <math>9</math>.<br />
<br />
For <math>a=2</math>, we get <math>b>\frac{8}{3}</math> for our inequality, and thus <math>b</math> can be any integer from <math>3</math> to <math>9</math>.<br />
<br />
For <math>a=3</math>, we get <math>b>\frac{27}{4}</math> for our inequality, and thus <math>b</math> can be any integer from <math>7</math> to <math>9</math>.<br />
<br />
Finally, if we total up all the possibilities we see there are <math>19</math> points that satisfy the condition, out of <math>9 \times 9 = 81</math> total points. The probability of picking a point that lies above the parabola is thus <math>\frac{19}{81} \rightarrow \boxed{\textbf{E}}</math><br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=u23iWcqbJlE<br />
~Shreyas S<br />
<br />
this links to problem 11...<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=13|num-a=15|ab=A}}<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=1733642011 AMC 12A Problems/Problem 122022-04-15T03:01:40Z<p>Russellk: /* Video Solution */</p>
<hr />
<div>== Problem ==<br />
A power boat and a raft both left dock <math>A</math> on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock <math>B</math> downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock <math>A.</math> How many hours did it take the power boat to go from <math>A</math> to <math>B</math>?<br />
<br />
<math><br />
\textbf{(A)}\ 3 \qquad<br />
\textbf{(B)}\ 3.5 \qquad<br />
\textbf{(C)}\ 4 \qquad<br />
\textbf{(D)}\ 4.5 \qquad<br />
\textbf{(E)}\ 5 </math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
WLOG let the speed of the river be 0. This is allowed because the problem never states that the speed of the current has to have a magnitude greater than 0. In this case, when the powerboat travels from <math>A</math> to <math>B</math>, the raft remains at <math>A</math>. Thus the trip from <math>A</math> to <math>B</math> takes the same time as the trip from <math>B</math> to the raft. Since these times are equal and sum to <math>9</math> hours, the trip from <math>A</math> to <math>B</math> must take half this time, or <math>4.5</math> hours. The answer is thus <math>\boxed{\textbf{D}}</math>.<br />
<br />
Remark: This is equivalent to viewing the problem from the raft's perspective.<br />
<br />
=== Solution 2 ===<br />
<br />
What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Think of the blue arrow as the power boat and the red arrow as the raft in the following three diagrams, which represent different time intervals of the problem.<br />
<br />
<asy><br />
size(8cm,8cm);<br />
pair A, B;<br />
A=(-3,4);<br />
B=(3,4);<br />
draw(A--B);<br />
label("$A$",A,S);<br />
label("$B$",B,S);<br />
arrow((-2.5,6.4),dir(180),blue);<br />
arrow((-2.5,6.2),dir(180),red);<br />
pair A, B;<br />
A=(-3,5);<br />
B=(3,5);<br />
draw(A--B);<br />
label("$A$",A,S);<br />
label("$B$",B,S);<br />
arrow((2.6,5.4),dir(360),blue);<br />
arrow((-0.5,5.2),dir(180),red);<br />
pair A, B;<br />
A=(-3,6);<br />
B=(3,6);<br />
draw(A--B);<br />
label("$A$",A,S);<br />
label("$B$",B,S);<br />
arrow((.3,4.4),dir(360),blue);<br />
arrow((1.1,4.2),dir(180),red);</asy><br />
<br />
Thinking about the distance covered as their distances with respect to each other, they are <math>0</math> distance apart in the first diagram when they haven't started to move yet, some distance <math>d</math> apart in the second diagram when the power boat reaches <math>B</math>, and again <math>0</math> distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of <math>d</math> on the way there, and again cover a distance of <math>d</math> on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.<br />
<br />
Let <math>b</math> denote the speed of the power boat (only the power boat, not factoring in current) and <math>r</math> denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from <math>A</math> to <math>B</math>, the boat travels at a velocity of <math>b+r</math>, and on the way back, travels at a velocity of <math>-(b-r)=r-b</math>, since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from <math>A</math> to <math>B</math> becomes <math>(r+b)-r=b</math>, and on the way back it becomes <math>(r-b)-r=-b</math>. Since the boat's velocities with respect to the raft are exact opposites, <math>b</math> and <math>-b</math>, we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.<br />
<br />
From this, we have that the boat travels a distance <math>d</math> at rate <math>b</math> with respect to the raft both on the way to <math>B</math> and on the way back. Thus, using <math>\dfrac{distance}{speed}=time</math>, we have <math>\dfrac{2d}{b}=9\text{ hours}</math>, and to see how long it took to travel half the distance, we have <math>\dfrac{d}{b}=4.5\text{ hours}\implies\boxed{\textbf{D}}</math><br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>t</math> be the time it takes the power boat to go from <math>A</math> to <math>B</math> in hours, <math>r</math> be the speed of the river current (and thus also the raft), and <math>p</math> to be the speed of the power boat with respect to the river. <br />
<br />
Using <math>d = rt</math>, the raft covers a distance of <math>9r</math>, the distance from <math>A</math> to <math>B</math> is <math>(p + r)t</math>, and the distance from <math>B</math> to where the raft and power boat met up is <math>(9 - t)(p - r)</math>. <br />
<br />
Then, <math>9r + (9 - t)(p - r) = (p + r)t</math>. Solving for <math>t</math>, we get <math>t = 4.5</math>, which is <math>\boxed{\textbf{D}}</math>.<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=u23iWcqbJlE<br />
~Shreyas S<br />
<br />
fyi, this links to the wrong problem, it links to problem 11<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_25&diff=1408042002 AMC 10A Problems/Problem 252020-12-28T03:52:41Z<p>Russellk: /* Quick Time Trouble Solution 5 */</p>
<hr />
<div>== Problem ==<br />
In [[trapezoid]] <math>ABCD</math> with bases <math>AB</math> and <math>CD</math>, we have <math>AB = 52</math>, <math>BC = 12</math>, <math>CD = 39</math>, and <math>DA = 5</math>. The area of <math>ABCD</math> is<br />
<br />
<math>\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
It shouldn't be hard to use [[trigonometry]] to bash this and find the height, but there is a much easier way. Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at point <math>E</math>:<br />
<br />
<center><asy><br />
size(250);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13);<br />
draw(A--B--C--D--cycle);<br />
draw(D--F--C,dashed);<br />
label("\(A\)",A,S);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,W);<br />
label("\(E\)",F,N);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,E);<br />
label("12",(B+C)/2,WSW);<br />
</asy></center><br />
<br />
Since <math>\overline{AB} || \overline{CD}</math> we have <math>\triangle AEB \sim \triangle DEC</math>, with the ratio of [[proportion]]ality being <math>\frac {39}{52} = \frac {3}{4}</math>. Thus<br />
<cmath><br />
\begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\<br />
\frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*}<br />
</cmath><br />
So the sides of <math>\triangle CDE</math> are <math>15,36,39</math>, which we recognize to be a <math>5 - 12 - 13</math> [[right triangle]]. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),<br />
<cmath><br />
[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}</cmath><br />
<br />
=== Solution 2 ===<br />
<br />
Draw altitudes from points <math>C</math> and <math>D</math>:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E,dashed);<br />
draw(D--F,dashed);<br />
label("\(A\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,N);<br />
label("\(D'\)",F,SSE);<br />
label("\(C'\)",E,S);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
Translate the triangle <math>ADD'</math> so that <math>DD'</math> coincides with <math>CC'</math>. We get the following triangle:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0);<br />
draw(A--B--C--cycle);<br />
draw(C--F,dashed);<br />
label("\(A'\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,N);<br />
label("\(C'\)",F,SE);<br />
label("5",(A+C)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
The length of <math>A'B</math> in this triangle is equal to the length of the original <math>AB</math>, minus the length of <math>CD</math>.<br />
Thus <math>A'B = 52 - 39 = 13</math>.<br />
<br />
Therefore <math>A'BC</math> is a well-known <math>(5,12,13)</math> right triangle. Its area is <math>[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30</math>, and therefore its altitude <math>CC'</math> is <math>\frac{[A'BC]}{A'B} = \frac{60}{13}</math>.<br />
<br />
Now the area of the original trapezoid is <math>\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}</math><br />
<br />
=== Solution 3 ===<br />
<br />
Draw altitudes from points <math>C</math> and <math>D</math>:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E,dashed);<br />
draw(D--F,dashed);<br />
label("\(A\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,N);<br />
label("\(D'\)",F,SSE);<br />
label("\(C'\)",E,S);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
Call the length of <math>AD'</math> to be <math>y</math>, the length of <math>BC'</math> to be <math>z</math>, and the height of the trapezoid to be <math>x</math>.<br />
By the Pythagorean Theorem, we have:<br />
<cmath>z^2 + x^2 = 144</cmath><br />
<cmath>y^2 + x^2 = 25</cmath><br />
<br />
Subtracting these two equation yields:<br />
<cmath>z^2-y^2=119 \implies (z+y)(z-y)=119</cmath><br />
<br />
We also have: <math>z+y=52-39=13</math>.<br />
<br />
We can substitute the value of <math>z+y</math> into the equation we just obtained, so we now have:<br />
<br />
<cmath>(13) (z-y)=119 \implies z-y=\frac{119}{13}</cmath>.<br />
<br />
We can add the <math>z+y</math> and the <math>z-y</math> equation to find the value of <math>z</math>, which simplifies down to be <math>\frac{144}{13}</math>. Finally, we can plug in <math>z</math> and use the Pythagorean theorem to find the height of the trapezoid.<br />
<br />
<cmath>\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2} \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}</cmath><br />
<br />
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.<br />
<br />
The median of the trapezoid is <math>\frac{39+52}{2} = \frac{91}{2}</math>, and multiplying this and the height of the trapezoid gets us:<br />
<br />
<cmath>\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}</cmath><br />
<br />
=== Solution 4 ===<br />
<br />
We construct a line segment parallel to <math>\overline{AD}</math> from point <math>C</math> to line <math>\overline{AB},</math> and label the intersection of this segment with line <math>\overline{AB}</math> as point <math>E.</math> Then quadrilateral <math>AECD</math> is a parallelogram, so <math>CE=5, AE=39,</math> and <math>EB=13.</math> Triangle <math>EBC</math> is therefore a right triangle, with area <math>\frac12 \cdot 5 \cdot 12 = 30.</math><br />
<br />
By continuing to split <math>\overline{AB}</math> and <math>\overline{CD}</math> into segments of length <math>13,</math> we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides <math>5,12,</math> and <math>13,</math> and each with area <math>30.</math> The total area is therefore <math>7 \cdot 30 = \boxed{\textbf{(C)} 210}.</math><br />
<br />
=== Solution 2 but quicker ===<br />
From Solution <math>2</math> we know that the the altitude of the trapezoid is <math>\frac{60}{13}</math> and the triangle's area is <math>30</math>.<br />
Note that once we remove the triangle we get a rectangle with length <math>39</math> and height <math>\frac{60}{13}</math>.<br />
The numbers multiply nicely to get <math>180+30=\boxed{(C) 210}</math><br />
-harsha12345<br />
<br />
=== Quick Time Trouble Solution 5 ===<br />
<br />
First note how the answer choices are all integers.<br />
The area of the trapezoid is <math>\frac{39+52}{2} * h = \frac{91}{2} h</math>. So h divides 2. Let <math>x</math> be <math>2h</math>. The area is now <math>91x</math>. <br />
Trying <math>x=1</math> and <math>x=2</math> can easily be seen to not work. Those make the only integers possible so now you know x is a fraction. <br />
Since the area is an integer the denominator of x must divide either 13 or 7 since <math>91 = 13*7</math>.<br />
Seeing how <math>39 = 3*13</math> and <math>52 = 4*13</math> assume that the denominator divides 13. Letting <math>y = \frac{x}{13}</math> the area is now <math>7y</math>.<br />
Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean <math>h</math> is <math>4</math> which we ruled out. <br />
So the answer is <math>\boxed{\textbf{(C)} 210}</math>. - megateleportingrobots<br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|num-b=24|after=-(Last question)|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=1307322011 AMC 10A Problems2020-08-06T03:54:10Z<p>Russellk: /* Problem 8 */</p>
<hr />
<div>{{AMC10 Problems|year=2011|ab=A}}<br />
<br />
== Problem 1 ==<br />
<br />
A cell phone plan costs <math>\textdollar 20</math> each month, plus <math>5</math>¢ per text message sent, plus <math>10</math>¢ for each minute used over <math>30</math> hours. In January Juan sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did he have to pay?<br />
<br />
<math> \textbf{(A)}\ \textdollar 24.00 \qquad\textbf{(B)}\ \textdollar 24.50 \qquad\textbf{(C)}\ \textdollar 25.50\qquad\textbf{(D)}\ \textdollar 28.00\qquad\textbf{(E)}\ \textdollar 30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math> \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Suppose <math>[a\ b]</math> denotes the average of <math>a</math> and <math>b</math>, and <math>\{a\ b\ c\}</math> denotes the average of <math>a, b</math>, and <math>c</math>. What is <math>\{\{1\ 1\ 0\}\ [0\ 1]\ 0\}</math>?<br />
<br />
<math> \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let <math>X</math> and <math>Y</math> be the following sums of arithmetic sequences: <cmath> \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} </cmath> What is the value of <math>Y - X</math>?<br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
Set <math>A </math> has 20 elements, and set <math>B </math> has 15 elements. What is the smallest possible number of elements in <math>A \cup B </math>, the union of <math>A </math> and <math>B </math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following equations does NOT have a solution?<br />
<br />
<math>\textbf{(A)}\:(x+7)^2=0</math><br />
<br />
<math>\textbf{(B)}\:\left|-3x\right|+5=0</math><br />
<br />
<math>\textbf{(C)}\:\sqrt{-x}-2=0</math><br />
<br />
<math>\textbf{(D)}\:\sqrt{x}-8=0</math><br />
<br />
<math>\textbf{(E)}\:\left|-3x\right|-4=0 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<math> \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A rectangular region is bounded by the graphs of the equations <math>y=a, y=-b, x=-c,</math> and <math>x=d</math>, where <math>a,b,c,</math> and <math>d</math> are all positive numbers. Which of the following represents the area of this region?<br />
<br />
<math> \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad</math> <math>+bc-bd\qquad\textbf{(C)}\ ac+ad</math> <math>-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad</math> <math>+bc+bd\qquad\textbf{(E)}\ ac-ad</math> <math> -bc+bd </math><br />
<br />
[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math> \textdollar 17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\textbf{(A)}\,7 \qquad\textbf{(B)}\,11 \qquad\textbf{(C)}\,17 \qquad\textbf{(D)}\,23 \qquad\textbf{(E)}\,77</math><br />
<br />
[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\textbf{(A)}\,\frac{49}{64} \qquad\textbf{(B)}\,\frac{25}{32} \qquad\textbf{(C)}\,\frac78 \qquad\textbf{(D)}\,\frac{5\sqrt{2}}{8} \qquad\textbf{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\textbf{(A)}\,13 \qquad\textbf{(B)}\,14 \qquad\textbf{(C)}\,15 \qquad\textbf{(D)}\,16 \qquad\textbf{(E)}\,17</math><br />
<br />
[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1, 2, 5, 7, 8, 9}?<br />
<br />
<math>\textbf{(A)}\,12 \qquad\textbf{(B)}\,20 \qquad\textbf{(C)}\,72 \qquad\textbf{(D)}\,120 \qquad\textbf{(E)}\,200</math><br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\textbf{(A)}\,\frac{1}{36} \qquad\textbf{(B)}\,\frac{1}{12} \qquad\textbf{(C)}\,\frac{1}{6} \qquad\textbf{(D)}\,\frac{1}{4} \qquad\textbf{(E)}\,\frac{5}{18}</math><br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
<math>\textbf{(A)}\,140 \qquad\textbf{(B)}\,240 \qquad\textbf{(C)}\,440 \qquad\textbf{(D)}\,640 \qquad\textbf{(E)}\,840</math><br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\textbf{(A)}\,3\sqrt2 \qquad\textbf{(B)}\,2\sqrt6 \qquad\textbf{(C)}\,\frac{7\sqrt2}{2} \qquad\textbf{(D)}\,3\sqrt3 \qquad\textbf{(E)}\,6</math><br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\textbf{(A)}\,17 \qquad\textbf{(B)}\,18 \qquad\textbf{(C)}\,25 \qquad\textbf{(D)}\,26 \qquad\textbf{(E)}\,43</math><br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
<math><br />
\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad<br />
\textbf{(B)}\ \frac{\pi}{2} \qquad<br />
\textbf{(C)}\ 2 \qquad<br />
\textbf{(D)}\ \frac{3\pi}{4} \qquad<br />
\textbf{(E)}\ 1+\frac{\pi}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br />
<br />
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Two points on the circumference of a circle of radius <math>r</math> are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Seven students count from 1 to 1000 as follows:<br />
<br />
•Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br />
<br />
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br />
<br />
•Finally, George says the only number that no one else says.<br />
<br />
What number does George say?<br />
<br />
<math> \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br />
<br />
<math>\textbf{(A)}\,1500 \qquad\textbf{(B)}\,1560 \qquad\textbf{(C)}\,2320 \qquad\textbf{(D)}\,2480 \qquad\textbf{(E)}\,2500</math><br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2011|ab=A|before=[[2010 AMC 10B Problems]]|after=[[2011 AMC 10B Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems&diff=1282732013 AMC 10B Problems2020-07-14T03:23:10Z<p>Russellk: /* Problem 1 */</p>
<hr />
<div>{{AMC10 Problems|year=2013|ab=B}}<br />
==Problem 1==<br />
What is <math>\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}</math>?<br />
<br />
<math> \textbf{(A)}\ -1 \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{7}{12} \qquad\textbf{(D)}\ \frac{49}{20} \qquad\textbf{(E)}\ \frac{43}{3} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Mr. Green measures his rectangular garden by walking two of the sides and finding that it is <math>15</math> steps by <math>20</math> steps. Each of Mr. Green's steps is <math>2</math> feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?<br />
<br />
<math> \textbf{(A)}\ 600 \qquad\textbf{(B)}\ 800 \qquad\textbf{(C)}\ 1000 \qquad\textbf{(D)}\ 1200 \qquad\textbf{(E)}\ 1400 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
On a particular January day, the high temperature in Lincoln, Nebraska, was <math>16</math> degrees higher than the low temperature, and the average of the high and the low temperatures was <math>3\,^\circ</math>. In degrees, what was the low temperature in Lincoln that day?<br />
<br />
<math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ -5\qquad\textbf{(D)}\ -3\qquad\textbf{(E)}\ 11</math><br />
<br />
[[2013 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
When counting from <math>3</math> to <math>201</math>, <math>53</math> is the <math>51^\mathrm{st}</math> number counted. When counting backwards from <math>201</math> to <math>3</math>, <math>53</math> is the <math>n^\mathrm{th}</math> number counted. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 146\qquad\textbf{(B)}\ 147\qquad\textbf{(C)}\ 148\qquad\textbf{(D)}\ 149\qquad\textbf{(E)}\ 150</math><br />
<br />
[[2013 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Positive integers <math>a</math> and <math>b</math> are each less than <math>6</math>. What is the smallest possible value for <math>2 \cdot a - a \cdot b</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2</math><br />
<br />
[[2013 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?<br />
<br />
<math>\textbf{(A)}\ 22\qquad\textbf{(B)}\ 23.25\qquad\textbf{(C)}\ 24.75\qquad\textbf{(D)}\ 26.25\qquad\textbf{(E)}\ 28</math><br />
<br />
[[2013 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?<br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2} </math><br />
<br />
<br />
[[2013 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }16\qquad\textbf{(C) }25\qquad\textbf{(D) }30\qquad\textbf{(E) }40 </math><br />
<br />
<br />
[[2013 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Three positive integers are each greater than <math>1</math>, have a product of <math> 27000 </math>, and are pairwise relatively prime. What is their sum?<br />
<br />
<math> \textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
Real numbers <math>x</math> and <math>y</math> satisfy the equation <math>x^2 + y^2 = 10x - 6y - 34</math>. What is <math>x+y</math>?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let <math> S </math> be the set of sides and diagonals of a regular pentagon. A pair of elements of <math> S </math> are selected at random without replacement. What is the probability that the two chosen segments have the same length?<br />
<br />
<math> \textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qquad\textbf{(E) }\frac{4}5 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Jo and Blair take turns counting from <math>1</math> to one more than the last number said by the other person. Jo starts by saying "<math>1</math>", so Blair follows by saying "<math>1, 2</math>" . Jo then says "<math>1, 2, 3</math>" , and so on. What is the 53rd number said?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Define <math> a\clubsuit b=a^2b-ab^2 </math>. Which of the following describes the set of points <math> (x, y) </math> for which <math> x\clubsuit y=y\clubsuit x </math>?<br />
<br />
<math> \textbf{(A)}\ \text{a finite set of points}\\ \qquad\textbf{(B)}\ \text{one line}\\ \qquad\textbf{(C)}\ \text{two parallel lines}\\ \qquad\textbf{(D)}\ \text{two intersecting lines}\\ \qquad\textbf{(E)}\ \text{three lines} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
A wire is cut into two pieces, one of length <math>a</math> and the other of length <math>b</math>. The piece of length <math>a</math> is bent to form an equilateral triangle, and the piece of length <math>b</math> is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is <math>\frac{a}{b}</math>?<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
In triangle <math>ABC</math>, medians <math>AD</math> and <math>CE</math> intersect at <math>P</math>, <math>PE=1.5</math>, <math>PD=2</math>, and <math>DE=2.5</math>. What is the area of <math>AEDC</math>?<br />
<br />
<math>\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}15\qquad</math><br />
<br />
[[2013 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Alex has <math>75</math> red tokens and <math>75</math> blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?<br />
<br />
<math>\textbf{(A)} 62 \qquad \textbf{(B)} 82 \qquad \textbf{(C)} 83 \qquad \textbf{(D)} 102 \qquad \textbf{(E)} 103</math><br />
<br />
[[2013 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
The number <math>2013</math> has the property that its units digit is the sum of its other digits, that is <math>2+0+1=3</math>. How many integers less than <math>2013</math> but greater than <math>1000</math> share this property?<br />
<br />
<math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
The real numbers <math>c,b,a</math> form an arithmetic sequence with <math>a\ge b\ge c\ge 0</math>. The quadratic <math>ax^2+bx+c</math> has exactly one root. What is this root?<br />
<br />
<math> \textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
The number <math> 2013 </math> is expressed in the form <cmath> 2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!}, </cmath> where <math> a_1\ge a_2\ge\cdots\ge a_m </math> and <math> b_1\ge b_2\ge\cdots\ge b_n </math> are positive integers and <math> a_1+b_1 </math> is as small as possible. What is <math>|a_1-b_1|</math>?<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 />
[[2013 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is <math>N</math>. What is the smallest possible value of <math>N</math>?<br />
<br />
<math> \textbf{(A)}\ 55 \qquad\textbf{(B)}\ 89 \qquad\textbf{(C)}\ 104 \qquad\textbf{(D)}\ 144 \qquad\textbf{(E)}\ 273 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
The regular octagon <math>ABCDEFGH</math> has its center at <math>J</math>. Each of the vertices and the center are to be associated with one of the digits <math>1</math> through <math>9</math>, with each digit used once, in such a way that the sums of the numbers on the lines <math>AJE</math>, <math>BJF</math>, <math>CJG</math>, and <math>DJH</math> are all equal. In how many ways can this be done?<br />
<br />
<math> \textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 </math><br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,J;<br />
A=(20,20(2+sqrt(2)));<br />
B=(20(1+sqrt(2)),20(2+sqrt(2)));<br />
C=(20(2+sqrt(2)),20(1+sqrt(2)));<br />
D=(20(2+sqrt(2)),20);<br />
E=(20(1+sqrt(2)),0);<br />
F=(20,0);<br />
G=(0,20);<br />
H=(0,20(1+sqrt(2)));<br />
J=(10(2+sqrt(2)),10(2+sqrt(2)));<br />
draw(A--B);<br />
draw(B--C);<br />
draw(C--D);<br />
draw(D--E);<br />
draw(E--F);<br />
draw(F--G);<br />
draw(G--H);<br />
draw(H--A);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(E);<br />
dot(F);<br />
dot(G);<br />
dot(H);<br />
dot(J);<br />
label("A",A,NNW);<br />
label("B",B,NNE);<br />
label("C",C,ENE);<br />
label("D",D,ESE);<br />
label("E",E,SSE);<br />
label("F",F,SSW);<br />
label("G",G,WSW);<br />
label("H",H,WNW);<br />
label("J",J,SE);<br />
</asy><br />
[[2013 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
In triangle <math>ABC</math>, <math>AB = 13</math>, <math>BC = 14</math>, and <math>CA = 15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD} \perp \overline{BC}</math>, <math>\overline{DE} \perp \overline{AC}</math>, and <math>\overline{AF} \perp \overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?<br />
<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 />
[[2013 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<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 />
[[2013 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Bernardo chooses a three-digit positive integer <math>N</math> and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer <math>S</math>. For example, if <math>N = 749</math>, Bernardo writes the numbers <math>10,444</math> and <math>3,245</math>, and LeRoy obtains the sum <math>S = 13,689</math>. For how many choices of <math>N</math> are the two rightmost digits of <math>S</math>, in order, the same as those of <math>2N</math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25</math><br />
<br />
[[2013 AMC 10B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|before=[[2013 AMC 10A Problems ]]|after=[[2014 AMC 10A Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2013 AMC 10B]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems&diff=1282722013 AMC 10B Problems2020-07-14T03:22:51Z<p>Russellk: /* Problem 1 */</p>
<hr />
<div>{{AMC10 Problems|year=2013|ab=B}}<br />
==Problem 1==<br />
What is <math>\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}</math>?<br />
<br />
<math> \textbf{(A)}\ -1 \qquad\textbf{(C)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{7}{12} \qquad\textbf{(D)}\ \frac{49}{20} \qquad\textbf{(E)}\ \frac{43}{3} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Mr. Green measures his rectangular garden by walking two of the sides and finding that it is <math>15</math> steps by <math>20</math> steps. Each of Mr. Green's steps is <math>2</math> feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?<br />
<br />
<math> \textbf{(A)}\ 600 \qquad\textbf{(B)}\ 800 \qquad\textbf{(C)}\ 1000 \qquad\textbf{(D)}\ 1200 \qquad\textbf{(E)}\ 1400 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
On a particular January day, the high temperature in Lincoln, Nebraska, was <math>16</math> degrees higher than the low temperature, and the average of the high and the low temperatures was <math>3\,^\circ</math>. In degrees, what was the low temperature in Lincoln that day?<br />
<br />
<math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ -5\qquad\textbf{(D)}\ -3\qquad\textbf{(E)}\ 11</math><br />
<br />
[[2013 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
When counting from <math>3</math> to <math>201</math>, <math>53</math> is the <math>51^\mathrm{st}</math> number counted. When counting backwards from <math>201</math> to <math>3</math>, <math>53</math> is the <math>n^\mathrm{th}</math> number counted. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 146\qquad\textbf{(B)}\ 147\qquad\textbf{(C)}\ 148\qquad\textbf{(D)}\ 149\qquad\textbf{(E)}\ 150</math><br />
<br />
[[2013 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Positive integers <math>a</math> and <math>b</math> are each less than <math>6</math>. What is the smallest possible value for <math>2 \cdot a - a \cdot b</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2</math><br />
<br />
[[2013 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?<br />
<br />
<math>\textbf{(A)}\ 22\qquad\textbf{(B)}\ 23.25\qquad\textbf{(C)}\ 24.75\qquad\textbf{(D)}\ 26.25\qquad\textbf{(E)}\ 28</math><br />
<br />
[[2013 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?<br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2} </math><br />
<br />
<br />
[[2013 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }16\qquad\textbf{(C) }25\qquad\textbf{(D) }30\qquad\textbf{(E) }40 </math><br />
<br />
<br />
[[2013 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Three positive integers are each greater than <math>1</math>, have a product of <math> 27000 </math>, and are pairwise relatively prime. What is their sum?<br />
<br />
<math> \textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
Real numbers <math>x</math> and <math>y</math> satisfy the equation <math>x^2 + y^2 = 10x - 6y - 34</math>. What is <math>x+y</math>?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let <math> S </math> be the set of sides and diagonals of a regular pentagon. A pair of elements of <math> S </math> are selected at random without replacement. What is the probability that the two chosen segments have the same length?<br />
<br />
<math> \textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qquad\textbf{(E) }\frac{4}5 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Jo and Blair take turns counting from <math>1</math> to one more than the last number said by the other person. Jo starts by saying "<math>1</math>", so Blair follows by saying "<math>1, 2</math>" . Jo then says "<math>1, 2, 3</math>" , and so on. What is the 53rd number said?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Define <math> a\clubsuit b=a^2b-ab^2 </math>. Which of the following describes the set of points <math> (x, y) </math> for which <math> x\clubsuit y=y\clubsuit x </math>?<br />
<br />
<math> \textbf{(A)}\ \text{a finite set of points}\\ \qquad\textbf{(B)}\ \text{one line}\\ \qquad\textbf{(C)}\ \text{two parallel lines}\\ \qquad\textbf{(D)}\ \text{two intersecting lines}\\ \qquad\textbf{(E)}\ \text{three lines} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
A wire is cut into two pieces, one of length <math>a</math> and the other of length <math>b</math>. The piece of length <math>a</math> is bent to form an equilateral triangle, and the piece of length <math>b</math> is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is <math>\frac{a}{b}</math>?<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
In triangle <math>ABC</math>, medians <math>AD</math> and <math>CE</math> intersect at <math>P</math>, <math>PE=1.5</math>, <math>PD=2</math>, and <math>DE=2.5</math>. What is the area of <math>AEDC</math>?<br />
<br />
<math>\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}15\qquad</math><br />
<br />
[[2013 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Alex has <math>75</math> red tokens and <math>75</math> blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?<br />
<br />
<math>\textbf{(A)} 62 \qquad \textbf{(B)} 82 \qquad \textbf{(C)} 83 \qquad \textbf{(D)} 102 \qquad \textbf{(E)} 103</math><br />
<br />
[[2013 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
The number <math>2013</math> has the property that its units digit is the sum of its other digits, that is <math>2+0+1=3</math>. How many integers less than <math>2013</math> but greater than <math>1000</math> share this property?<br />
<br />
<math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
The real numbers <math>c,b,a</math> form an arithmetic sequence with <math>a\ge b\ge c\ge 0</math>. The quadratic <math>ax^2+bx+c</math> has exactly one root. What is this root?<br />
<br />
<math> \textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
The number <math> 2013 </math> is expressed in the form <cmath> 2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!}, </cmath> where <math> a_1\ge a_2\ge\cdots\ge a_m </math> and <math> b_1\ge b_2\ge\cdots\ge b_n </math> are positive integers and <math> a_1+b_1 </math> is as small as possible. What is <math>|a_1-b_1|</math>?<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 />
[[2013 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is <math>N</math>. What is the smallest possible value of <math>N</math>?<br />
<br />
<math> \textbf{(A)}\ 55 \qquad\textbf{(B)}\ 89 \qquad\textbf{(C)}\ 104 \qquad\textbf{(D)}\ 144 \qquad\textbf{(E)}\ 273 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
The regular octagon <math>ABCDEFGH</math> has its center at <math>J</math>. Each of the vertices and the center are to be associated with one of the digits <math>1</math> through <math>9</math>, with each digit used once, in such a way that the sums of the numbers on the lines <math>AJE</math>, <math>BJF</math>, <math>CJG</math>, and <math>DJH</math> are all equal. In how many ways can this be done?<br />
<br />
<math> \textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 </math><br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,J;<br />
A=(20,20(2+sqrt(2)));<br />
B=(20(1+sqrt(2)),20(2+sqrt(2)));<br />
C=(20(2+sqrt(2)),20(1+sqrt(2)));<br />
D=(20(2+sqrt(2)),20);<br />
E=(20(1+sqrt(2)),0);<br />
F=(20,0);<br />
G=(0,20);<br />
H=(0,20(1+sqrt(2)));<br />
J=(10(2+sqrt(2)),10(2+sqrt(2)));<br />
draw(A--B);<br />
draw(B--C);<br />
draw(C--D);<br />
draw(D--E);<br />
draw(E--F);<br />
draw(F--G);<br />
draw(G--H);<br />
draw(H--A);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(E);<br />
dot(F);<br />
dot(G);<br />
dot(H);<br />
dot(J);<br />
label("A",A,NNW);<br />
label("B",B,NNE);<br />
label("C",C,ENE);<br />
label("D",D,ESE);<br />
label("E",E,SSE);<br />
label("F",F,SSW);<br />
label("G",G,WSW);<br />
label("H",H,WNW);<br />
label("J",J,SE);<br />
</asy><br />
[[2013 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
In triangle <math>ABC</math>, <math>AB = 13</math>, <math>BC = 14</math>, and <math>CA = 15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD} \perp \overline{BC}</math>, <math>\overline{DE} \perp \overline{AC}</math>, and <math>\overline{AF} \perp \overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?<br />
<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 />
[[2013 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<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 />
[[2013 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Bernardo chooses a three-digit positive integer <math>N</math> and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer <math>S</math>. For example, if <math>N = 749</math>, Bernardo writes the numbers <math>10,444</math> and <math>3,245</math>, and LeRoy obtains the sum <math>S = 13,689</math>. For how many choices of <math>N</math> are the two rightmost digits of <math>S</math>, in order, the same as those of <math>2N</math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25</math><br />
<br />
[[2013 AMC 10B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|before=[[2013 AMC 10A Problems ]]|after=[[2014 AMC 10A Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2013 AMC 10B]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems&diff=1282712013 AMC 10B Problems2020-07-14T03:22:02Z<p>Russellk: /* See also */</p>
<hr />
<div>{{AMC10 Problems|year=2013|ab=B}}<br />
==Problem 1==<br />
What is <math>\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}</math>?<br />
<br />
<math> \textbf{(A)}\ -1 \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{7}{12} \qquad\textbf{(D)}\ \frac{49}{20} \qquad\textbf{(E)}\ \frac{43}{3} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Mr. Green measures his rectangular garden by walking two of the sides and finding that it is <math>15</math> steps by <math>20</math> steps. Each of Mr. Green's steps is <math>2</math> feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?<br />
<br />
<math> \textbf{(A)}\ 600 \qquad\textbf{(B)}\ 800 \qquad\textbf{(C)}\ 1000 \qquad\textbf{(D)}\ 1200 \qquad\textbf{(E)}\ 1400 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
On a particular January day, the high temperature in Lincoln, Nebraska, was <math>16</math> degrees higher than the low temperature, and the average of the high and the low temperatures was <math>3\,^\circ</math>. In degrees, what was the low temperature in Lincoln that day?<br />
<br />
<math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ -5\qquad\textbf{(D)}\ -3\qquad\textbf{(E)}\ 11</math><br />
<br />
[[2013 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
When counting from <math>3</math> to <math>201</math>, <math>53</math> is the <math>51^\mathrm{st}</math> number counted. When counting backwards from <math>201</math> to <math>3</math>, <math>53</math> is the <math>n^\mathrm{th}</math> number counted. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 146\qquad\textbf{(B)}\ 147\qquad\textbf{(C)}\ 148\qquad\textbf{(D)}\ 149\qquad\textbf{(E)}\ 150</math><br />
<br />
[[2013 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Positive integers <math>a</math> and <math>b</math> are each less than <math>6</math>. What is the smallest possible value for <math>2 \cdot a - a \cdot b</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2</math><br />
<br />
[[2013 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?<br />
<br />
<math>\textbf{(A)}\ 22\qquad\textbf{(B)}\ 23.25\qquad\textbf{(C)}\ 24.75\qquad\textbf{(D)}\ 26.25\qquad\textbf{(E)}\ 28</math><br />
<br />
[[2013 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?<br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2} </math><br />
<br />
<br />
[[2013 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }16\qquad\textbf{(C) }25\qquad\textbf{(D) }30\qquad\textbf{(E) }40 </math><br />
<br />
<br />
[[2013 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Three positive integers are each greater than <math>1</math>, have a product of <math> 27000 </math>, and are pairwise relatively prime. What is their sum?<br />
<br />
<math> \textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
Real numbers <math>x</math> and <math>y</math> satisfy the equation <math>x^2 + y^2 = 10x - 6y - 34</math>. What is <math>x+y</math>?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let <math> S </math> be the set of sides and diagonals of a regular pentagon. A pair of elements of <math> S </math> are selected at random without replacement. What is the probability that the two chosen segments have the same length?<br />
<br />
<math> \textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qquad\textbf{(E) }\frac{4}5 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Jo and Blair take turns counting from <math>1</math> to one more than the last number said by the other person. Jo starts by saying "<math>1</math>", so Blair follows by saying "<math>1, 2</math>" . Jo then says "<math>1, 2, 3</math>" , and so on. What is the 53rd number said?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Define <math> a\clubsuit b=a^2b-ab^2 </math>. Which of the following describes the set of points <math> (x, y) </math> for which <math> x\clubsuit y=y\clubsuit x </math>?<br />
<br />
<math> \textbf{(A)}\ \text{a finite set of points}\\ \qquad\textbf{(B)}\ \text{one line}\\ \qquad\textbf{(C)}\ \text{two parallel lines}\\ \qquad\textbf{(D)}\ \text{two intersecting lines}\\ \qquad\textbf{(E)}\ \text{three lines} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
A wire is cut into two pieces, one of length <math>a</math> and the other of length <math>b</math>. The piece of length <math>a</math> is bent to form an equilateral triangle, and the piece of length <math>b</math> is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is <math>\frac{a}{b}</math>?<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
In triangle <math>ABC</math>, medians <math>AD</math> and <math>CE</math> intersect at <math>P</math>, <math>PE=1.5</math>, <math>PD=2</math>, and <math>DE=2.5</math>. What is the area of <math>AEDC</math>?<br />
<br />
<math>\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}15\qquad</math><br />
<br />
[[2013 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Alex has <math>75</math> red tokens and <math>75</math> blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?<br />
<br />
<math>\textbf{(A)} 62 \qquad \textbf{(B)} 82 \qquad \textbf{(C)} 83 \qquad \textbf{(D)} 102 \qquad \textbf{(E)} 103</math><br />
<br />
[[2013 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
The number <math>2013</math> has the property that its units digit is the sum of its other digits, that is <math>2+0+1=3</math>. How many integers less than <math>2013</math> but greater than <math>1000</math> share this property?<br />
<br />
<math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
The real numbers <math>c,b,a</math> form an arithmetic sequence with <math>a\ge b\ge c\ge 0</math>. The quadratic <math>ax^2+bx+c</math> has exactly one root. What is this root?<br />
<br />
<math> \textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
The number <math> 2013 </math> is expressed in the form <cmath> 2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!}, </cmath> where <math> a_1\ge a_2\ge\cdots\ge a_m </math> and <math> b_1\ge b_2\ge\cdots\ge b_n </math> are positive integers and <math> a_1+b_1 </math> is as small as possible. What is <math>|a_1-b_1|</math>?<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 />
[[2013 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is <math>N</math>. What is the smallest possible value of <math>N</math>?<br />
<br />
<math> \textbf{(A)}\ 55 \qquad\textbf{(B)}\ 89 \qquad\textbf{(C)}\ 104 \qquad\textbf{(D)}\ 144 \qquad\textbf{(E)}\ 273 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
The regular octagon <math>ABCDEFGH</math> has its center at <math>J</math>. Each of the vertices and the center are to be associated with one of the digits <math>1</math> through <math>9</math>, with each digit used once, in such a way that the sums of the numbers on the lines <math>AJE</math>, <math>BJF</math>, <math>CJG</math>, and <math>DJH</math> are all equal. In how many ways can this be done?<br />
<br />
<math> \textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 </math><br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,J;<br />
A=(20,20(2+sqrt(2)));<br />
B=(20(1+sqrt(2)),20(2+sqrt(2)));<br />
C=(20(2+sqrt(2)),20(1+sqrt(2)));<br />
D=(20(2+sqrt(2)),20);<br />
E=(20(1+sqrt(2)),0);<br />
F=(20,0);<br />
G=(0,20);<br />
H=(0,20(1+sqrt(2)));<br />
J=(10(2+sqrt(2)),10(2+sqrt(2)));<br />
draw(A--B);<br />
draw(B--C);<br />
draw(C--D);<br />
draw(D--E);<br />
draw(E--F);<br />
draw(F--G);<br />
draw(G--H);<br />
draw(H--A);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(E);<br />
dot(F);<br />
dot(G);<br />
dot(H);<br />
dot(J);<br />
label("A",A,NNW);<br />
label("B",B,NNE);<br />
label("C",C,ENE);<br />
label("D",D,ESE);<br />
label("E",E,SSE);<br />
label("F",F,SSW);<br />
label("G",G,WSW);<br />
label("H",H,WNW);<br />
label("J",J,SE);<br />
</asy><br />
[[2013 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
In triangle <math>ABC</math>, <math>AB = 13</math>, <math>BC = 14</math>, and <math>CA = 15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD} \perp \overline{BC}</math>, <math>\overline{DE} \perp \overline{AC}</math>, and <math>\overline{AF} \perp \overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?<br />
<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 />
[[2013 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<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 />
[[2013 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Bernardo chooses a three-digit positive integer <math>N</math> and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer <math>S</math>. For example, if <math>N = 749</math>, Bernardo writes the numbers <math>10,444</math> and <math>3,245</math>, and LeRoy obtains the sum <math>S = 13,689</math>. For how many choices of <math>N</math> are the two rightmost digits of <math>S</math>, in order, the same as those of <math>2N</math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25</math><br />
<br />
[[2013 AMC 10B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|before=[[2013 AMC 10A Problems ]]|after=[[2014 AMC 10A Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2013 AMC 10B]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems&diff=1282702013 AMC 10B Problems2020-07-14T03:21:44Z<p>Russellk: /* See also */</p>
<hr />
<div>{{AMC10 Problems|year=2013|ab=B}}<br />
==Problem 1==<br />
What is <math>\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}</math>?<br />
<br />
<math> \textbf{(A)}\ -1 \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{7}{12} \qquad\textbf{(D)}\ \frac{49}{20} \qquad\textbf{(E)}\ \frac{43}{3} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Mr. Green measures his rectangular garden by walking two of the sides and finding that it is <math>15</math> steps by <math>20</math> steps. Each of Mr. Green's steps is <math>2</math> feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?<br />
<br />
<math> \textbf{(A)}\ 600 \qquad\textbf{(B)}\ 800 \qquad\textbf{(C)}\ 1000 \qquad\textbf{(D)}\ 1200 \qquad\textbf{(E)}\ 1400 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
On a particular January day, the high temperature in Lincoln, Nebraska, was <math>16</math> degrees higher than the low temperature, and the average of the high and the low temperatures was <math>3\,^\circ</math>. In degrees, what was the low temperature in Lincoln that day?<br />
<br />
<math>\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ -5\qquad\textbf{(D)}\ -3\qquad\textbf{(E)}\ 11</math><br />
<br />
[[2013 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
When counting from <math>3</math> to <math>201</math>, <math>53</math> is the <math>51^\mathrm{st}</math> number counted. When counting backwards from <math>201</math> to <math>3</math>, <math>53</math> is the <math>n^\mathrm{th}</math> number counted. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 146\qquad\textbf{(B)}\ 147\qquad\textbf{(C)}\ 148\qquad\textbf{(D)}\ 149\qquad\textbf{(E)}\ 150</math><br />
<br />
[[2013 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Positive integers <math>a</math> and <math>b</math> are each less than <math>6</math>. What is the smallest possible value for <math>2 \cdot a - a \cdot b</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2</math><br />
<br />
[[2013 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?<br />
<br />
<math>\textbf{(A)}\ 22\qquad\textbf{(B)}\ 23.25\qquad\textbf{(C)}\ 24.75\qquad\textbf{(D)}\ 26.25\qquad\textbf{(E)}\ 28</math><br />
<br />
[[2013 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?<br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2} </math><br />
<br />
<br />
[[2013 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }16\qquad\textbf{(C) }25\qquad\textbf{(D) }30\qquad\textbf{(E) }40 </math><br />
<br />
<br />
[[2013 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Three positive integers are each greater than <math>1</math>, have a product of <math> 27000 </math>, and are pairwise relatively prime. What is their sum?<br />
<br />
<math> \textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
Real numbers <math>x</math> and <math>y</math> satisfy the equation <math>x^2 + y^2 = 10x - 6y - 34</math>. What is <math>x+y</math>?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let <math> S </math> be the set of sides and diagonals of a regular pentagon. A pair of elements of <math> S </math> are selected at random without replacement. What is the probability that the two chosen segments have the same length?<br />
<br />
<math> \textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qquad\textbf{(E) }\frac{4}5 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Jo and Blair take turns counting from <math>1</math> to one more than the last number said by the other person. Jo starts by saying "<math>1</math>", so Blair follows by saying "<math>1, 2</math>" . Jo then says "<math>1, 2, 3</math>" , and so on. What is the 53rd number said?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Define <math> a\clubsuit b=a^2b-ab^2 </math>. Which of the following describes the set of points <math> (x, y) </math> for which <math> x\clubsuit y=y\clubsuit x </math>?<br />
<br />
<math> \textbf{(A)}\ \text{a finite set of points}\\ \qquad\textbf{(B)}\ \text{one line}\\ \qquad\textbf{(C)}\ \text{two parallel lines}\\ \qquad\textbf{(D)}\ \text{two intersecting lines}\\ \qquad\textbf{(E)}\ \text{three lines} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
A wire is cut into two pieces, one of length <math>a</math> and the other of length <math>b</math>. The piece of length <math>a</math> is bent to form an equilateral triangle, and the piece of length <math>b</math> is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is <math>\frac{a}{b}</math>?<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
In triangle <math>ABC</math>, medians <math>AD</math> and <math>CE</math> intersect at <math>P</math>, <math>PE=1.5</math>, <math>PD=2</math>, and <math>DE=2.5</math>. What is the area of <math>AEDC</math>?<br />
<br />
<math>\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}15\qquad</math><br />
<br />
[[2013 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Alex has <math>75</math> red tokens and <math>75</math> blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?<br />
<br />
<math>\textbf{(A)} 62 \qquad \textbf{(B)} 82 \qquad \textbf{(C)} 83 \qquad \textbf{(D)} 102 \qquad \textbf{(E)} 103</math><br />
<br />
[[2013 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
The number <math>2013</math> has the property that its units digit is the sum of its other digits, that is <math>2+0+1=3</math>. How many integers less than <math>2013</math> but greater than <math>1000</math> share this property?<br />
<br />
<math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
The real numbers <math>c,b,a</math> form an arithmetic sequence with <math>a\ge b\ge c\ge 0</math>. The quadratic <math>ax^2+bx+c</math> has exactly one root. What is this root?<br />
<br />
<math> \textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} </math><br />
<br />
[[2013 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
The number <math> 2013 </math> is expressed in the form <cmath> 2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!}, </cmath> where <math> a_1\ge a_2\ge\cdots\ge a_m </math> and <math> b_1\ge b_2\ge\cdots\ge b_n </math> are positive integers and <math> a_1+b_1 </math> is as small as possible. What is <math>|a_1-b_1|</math>?<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 />
[[2013 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is <math>N</math>. What is the smallest possible value of <math>N</math>?<br />
<br />
<math> \textbf{(A)}\ 55 \qquad\textbf{(B)}\ 89 \qquad\textbf{(C)}\ 104 \qquad\textbf{(D)}\ 144 \qquad\textbf{(E)}\ 273 </math><br />
<br />
[[2013 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
The regular octagon <math>ABCDEFGH</math> has its center at <math>J</math>. Each of the vertices and the center are to be associated with one of the digits <math>1</math> through <math>9</math>, with each digit used once, in such a way that the sums of the numbers on the lines <math>AJE</math>, <math>BJF</math>, <math>CJG</math>, and <math>DJH</math> are all equal. In how many ways can this be done?<br />
<br />
<math> \textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 </math><br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,J;<br />
A=(20,20(2+sqrt(2)));<br />
B=(20(1+sqrt(2)),20(2+sqrt(2)));<br />
C=(20(2+sqrt(2)),20(1+sqrt(2)));<br />
D=(20(2+sqrt(2)),20);<br />
E=(20(1+sqrt(2)),0);<br />
F=(20,0);<br />
G=(0,20);<br />
H=(0,20(1+sqrt(2)));<br />
J=(10(2+sqrt(2)),10(2+sqrt(2)));<br />
draw(A--B);<br />
draw(B--C);<br />
draw(C--D);<br />
draw(D--E);<br />
draw(E--F);<br />
draw(F--G);<br />
draw(G--H);<br />
draw(H--A);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(E);<br />
dot(F);<br />
dot(G);<br />
dot(H);<br />
dot(J);<br />
label("A",A,NNW);<br />
label("B",B,NNE);<br />
label("C",C,ENE);<br />
label("D",D,ESE);<br />
label("E",E,SSE);<br />
label("F",F,SSW);<br />
label("G",G,WSW);<br />
label("H",H,WNW);<br />
label("J",J,SE);<br />
</asy><br />
[[2013 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
In triangle <math>ABC</math>, <math>AB = 13</math>, <math>BC = 14</math>, and <math>CA = 15</math>. Distinct points <math>D</math>, <math>E</math>, and <math>F</math> lie on segments <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{DE}</math>, respectively, such that <math>\overline{AD} \perp \overline{BC}</math>, <math>\overline{DE} \perp \overline{AC}</math>, and <math>\overline{AF} \perp \overline{BF}</math>. The length of segment <math>\overline{DF}</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?<br />
<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 />
[[2013 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?<br />
<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 />
[[2013 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Bernardo chooses a three-digit positive integer <math>N</math> and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer <math>S</math>. For example, if <math>N = 749</math>, Bernardo writes the numbers <math>10,444</math> and <math>3,245</math>, and LeRoy obtains the sum <math>S = 13,689</math>. For how many choices of <math>N</math> are the two rightmost digits of <math>S</math>, in order, the same as those of <math>2N</math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25</math><br />
<br />
[[2013 AMC 10B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|before=[[2013 AMC 10A Problems :)]]|after=[[2014 AMC 10A Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2013 AMC 10B]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi/Chess&diff=125524User:Piphi/Chess2020-06-15T21:18:07Z<p>Russellk: /* Signups */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
{{User:Piphi/Chess/Template:Header}}<br />
<br />
== Signups ==<br />
Put your username below if you want to play chess through the wiki. I will be going through this list periodically and add people to games. If you have any preference on who you want to play against then put their username in parenthesis next to your username.<br />
<br />
'''Note:''' You are allowed to be in a maximum of 3 games at once.<br />
<br />
* piphi <span style="color:green;">assigned to game 1</span><br />
* BorealBear (But I am not on often :|) <span style="color:green;">assigned to game 1</span><br />
* awesomeguy856 <span style="color:green;">assigned to game 2</span><br />
* Heavytoothpaste, I play a lot :) <span style="color:green;">assigned to game 2</span><br />
* FaerieDragon88, (I’m on frequently) <span style="color:green;">assigned to game 3</span><br />
* williamxiao (don't know if i will be any good at this) <span style="color:green;">assigned to game 3</span><br />
* Mark888 (I'll fail horribly, let's do this) <span style="color:green;">assigned to game 4</span><br />
* ARay10 (But I'm worse) <span style="color:green;">assigned to game 4</span><br />
* wuwang2002 (I will just try) <span style="color:green;">assigned to game 5</span><br />
* Mark888 (am I allowed to run 2 games at once?) <span style="color:green;">assigned to game 5</span><br />
* Juno (EPIC FAILER HERE) <span style="color:green;">assigned to game 6</span><br />
* ARay10 <span style="color:green;">assigned to game 6</span><br />
* Paul10 <span style="color:green;">assigned to game 7</span><br />
* Kmath1234 <span style="color:green;">assigned to game 7</span><br />
* ARay10 (wants to verse piphi) <span style="color:green;">assigned to game 8</span><br />
* piphi (sure, this is your last available chess game BTW, you can still play [[User:Piphi/Checkers|checkers]] or wait until one of your games end) <span style="color:green;">assigned to game 8</span><br />
* Aspiring_Mathletes <span style="color:green;">assigned to game 9</span><br />
* LJCoder619 <span style="color:green;">assigned to game 9</span><br />
* SamuraiA (not going to be too active) <span style="color:green;">assigned to game 10</span><br />
* rf20008 <span style="color:green;">assigned to game 10</span><br />
* BrightMonkey <span style="color:green;">assigned to game 11</span><br />
* Alpha_2 <span style="color:green;">assigned to game 11</span><br />
* HonestCat (I'm excited to play chess!) <span style="color:green;">assigned to game 12</span><br />
* aw2017 (I can play anyone) <span style="color:green;">assigned to game 12</span><br />
* CreativeHedgehog (I'm a beginner) <span style="color:green;">assigned to game 13</span><br />
* HonestCat <span style="color:green;">assigned to game 13</span><br />
* ARay10<br />
<br />
== Rules ==<br />
[[User:Piphi/Chess/Rules]]<br />
<br />
== Help ==<br />
[[User:Piphi/Chess/Help]]<br />
<br />
== Game #1 piphi v BorealBear ==<br />
Game is [[User:Piphi/Chess/Game_1|here]]. Discussion is [[User talk:Piphi/Chess/Game_1|here]].<br />
<br />
== Game #2 awesomeguy856 v Heavytoothpaste ==<br />
Game is [[User:Piphi/Chess/Game_2|here]]. Discussion is [[User talk:Piphi/Chess/Game_2|here]].<br />
<br />
== Game #3 FaerieDragon88 v williamxiao ==<br />
Game is [[User:Piphi/Chess/Game_3|here]]. Discussion is [[User talk:Piphi/Chess/Game_3|here]].<br />
<br />
== Game #4 Mark888 v ARay10 ==<br />
Game is [[User:Piphi/Chess/Game_4|here]]. Discussion is [[User talk:Piphi/Chess/Game_4|here]].<br />
<br />
== Game #5 wuwang2002 v Mark888 ==<br />
Game is [[User:Piphi/Chess/Game_5|here]]. Discussion is [[User talk:Piphi/Chess/Game_5|here]].<br />
<br />
== Game #6 Juno v ARay10 ==<br />
Game is [[User:Piphi/Chess/Game_6|here]]. Discussion is [[User talk:Piphi/Chess/Game_6|here]].<br />
<br />
== Game #7 Paul10 v Kmath1234 ==<br />
Game is [[User:Piphi/Chess/Game_7|here]]. Discussion is [[User talk:Piphi/Chess/Game_7|here]].<br />
<br />
== Game #8 ARay10 v piphi ==<br />
Game is [[User:Piphi/Chess/Game_8|here]]. Discussion is [[User talk:Piphi/Chess/Game_8|here]].<br />
<br />
== Game #9 Aspiring_Mathletes v LJCoder619 ==<br />
Game is [[User:Piphi/Chess/Game_9|here]]. Discussion is [[User talk:Piphi/Chess/Game_9|here]].<br />
<br />
== Game #10 SamuraiA v rf20008 ==<br />
Game is [[User:Piphi/Chess/Game_10|here]]. Discussion is [[User talk:Piphi/Chess/Game_10|here]].<br />
<br />
== Game #11 BrightMonkey v Alpha_2 ==<br />
Game is [[User:Piphi/Chess/Game_11|here]]. Discussion is [[User talk:Piphi/Chess/Game_11|here]].<br />
<br />
== Game #12 HonestCat v aw2017 ==<br />
Game is [[User:Piphi/Chess/Game_12|here]]. Discussion is [[User talk:Piphi/Chess/Game_12|here]].<br />
<br />
== Game #13 HonestCat v CreativeHedgehog ==<br />
Game is [[User:Piphi/Chess/Game_13|here]]. Discussion is [[User talk:Piphi/Chess/Game_13|here]].<br />
<br />
[[Category:Chess_Piphi]]</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi/Chess&diff=125523User:Piphi/Chess2020-06-15T21:17:11Z<p>Russellk: /* Signups */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
{{User:Piphi/Chess/Template:Header}}<br />
<br />
== Signups ==<br />
Put your username below if you want to play chess through the wiki. I will be going through this list periodically and add people to games. If you have any preference on who you want to play against then put their username in parenthesis next to your username.<br />
<br />
'''Note:''' You are allowed to be in a maximum of 3 games at once.<br />
<br />
* piphi <span style="color:green;">assigned to game 1</span><br />
* BorealBear (But I am not on often :|) <span style="color:green;">assigned to game 1</span><br />
* awesomeguy856 <span style="color:green;">assigned to game 2</span><br />
* Heavytoothpaste, I play a lot :) <span style="color:green;">assigned to game 2</span><br />
* FaerieDragon88, (I’m on frequently) <span style="color:green;">assigned to game 3</span><br />
* williamxiao (don't know if i will be any good at this) <span style="color:green;">assigned to game 3</span><br />
* Mark888 (I'll fail horribly, let's do this) <span style="color:green;">assigned to game 4</span><br />
* ARay10 (But I'm worse) <span style="color:green;">assigned to game 4</span><br />
* wuwang2002 (I will just try) <span style="color:green;">assigned to game 5</span><br />
* Mark888 (am I allowed to run 2 games at once?) <span style="color:green;">assigned to game 5</span><br />
* Juno (EPIC FAILER HERE) <span style="color:green;">assigned to game 6</span><br />
* ARay10 <span style="color:green;">assigned to game 6</span><br />
* Paul10 <span style="color:green;">assigned to game 7</span><br />
* Kmath1234 <span style="color:green;">assigned to game 7</span><br />
* ARay10 (wants to verse piphi) <span style="color:green;">assigned to game 8</span><br />
* piphi (sure, this is your last available chess game BTW, you can still play [[User:Piphi/Checkers|checkers]] or wait until one of your games end) <span style="color:green;">assigned to game 8</span><br />
* Aspiring_Mathletes <span style="color:green;">assigned to game 9</span><br />
* LJCoder619 <span style="color:green;">assigned to game 9</span><br />
* SamuraiA (not going to be too active) <span style="color:green;">assigned to game 10</span><br />
* rf20008 <span style="color:green;">assigned to game 10</span><br />
* BrightMonkey <span style="color:green;">assigned to game 11</span><br />
* Alpha_2 <span style="color:green;">assigned to game 11</span><br />
* HonestCat (I'm excited to play chess!) <span style="color:green;">assigned to game 12</span><br />
* aw2017 (I can play anyone) <span style="color:green;">assigned to game 12</span><br />
* CreativeHedgehog (I'm a beginner) <span style="color:green;">assigned to game 13</span><br />
* HonestCat <span style="color:green;">assigned to game 13</span><br />
* ARay10<br />
* russellk (I just want to play someone)<br />
<br />
== Rules ==<br />
[[User:Piphi/Chess/Rules]]<br />
<br />
== Help ==<br />
[[User:Piphi/Chess/Help]]<br />
<br />
== Game #1 piphi v BorealBear ==<br />
Game is [[User:Piphi/Chess/Game_1|here]]. Discussion is [[User talk:Piphi/Chess/Game_1|here]].<br />
<br />
== Game #2 awesomeguy856 v Heavytoothpaste ==<br />
Game is [[User:Piphi/Chess/Game_2|here]]. Discussion is [[User talk:Piphi/Chess/Game_2|here]].<br />
<br />
== Game #3 FaerieDragon88 v williamxiao ==<br />
Game is [[User:Piphi/Chess/Game_3|here]]. Discussion is [[User talk:Piphi/Chess/Game_3|here]].<br />
<br />
== Game #4 Mark888 v ARay10 ==<br />
Game is [[User:Piphi/Chess/Game_4|here]]. Discussion is [[User talk:Piphi/Chess/Game_4|here]].<br />
<br />
== Game #5 wuwang2002 v Mark888 ==<br />
Game is [[User:Piphi/Chess/Game_5|here]]. Discussion is [[User talk:Piphi/Chess/Game_5|here]].<br />
<br />
== Game #6 Juno v ARay10 ==<br />
Game is [[User:Piphi/Chess/Game_6|here]]. Discussion is [[User talk:Piphi/Chess/Game_6|here]].<br />
<br />
== Game #7 Paul10 v Kmath1234 ==<br />
Game is [[User:Piphi/Chess/Game_7|here]]. Discussion is [[User talk:Piphi/Chess/Game_7|here]].<br />
<br />
== Game #8 ARay10 v piphi ==<br />
Game is [[User:Piphi/Chess/Game_8|here]]. Discussion is [[User talk:Piphi/Chess/Game_8|here]].<br />
<br />
== Game #9 Aspiring_Mathletes v LJCoder619 ==<br />
Game is [[User:Piphi/Chess/Game_9|here]]. Discussion is [[User talk:Piphi/Chess/Game_9|here]].<br />
<br />
== Game #10 SamuraiA v rf20008 ==<br />
Game is [[User:Piphi/Chess/Game_10|here]]. Discussion is [[User talk:Piphi/Chess/Game_10|here]].<br />
<br />
== Game #11 BrightMonkey v Alpha_2 ==<br />
Game is [[User:Piphi/Chess/Game_11|here]]. Discussion is [[User talk:Piphi/Chess/Game_11|here]].<br />
<br />
== Game #12 HonestCat v aw2017 ==<br />
Game is [[User:Piphi/Chess/Game_12|here]]. Discussion is [[User talk:Piphi/Chess/Game_12|here]].<br />
<br />
== Game #13 HonestCat v CreativeHedgehog ==<br />
Game is [[User:Piphi/Chess/Game_13|here]]. Discussion is [[User talk:Piphi/Chess/Game_13|here]].<br />
<br />
[[Category:Chess_Piphi]]</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi/Chess&diff=125522User:Piphi/Chess2020-06-15T21:16:51Z<p>Russellk: /* Signups */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
{{User:Piphi/Chess/Template:Header}}<br />
<br />
== Signups ==<br />
Put your username below if you want to play chess through the wiki. I will be going through this list periodically and add people to games. If you have any preference on who you want to play against then put their username in parenthesis next to your username.<br />
<br />
'''Note:''' You are allowed to be in a maximum of 3 games at once.<br />
<br />
* piphi <span style="color:green;">assigned to game 1</span><br />
* BorealBear (But I am not on often :|) <span style="color:green;">assigned to game 1</span><br />
* awesomeguy856 <span style="color:green;">assigned to game 2</span><br />
* Heavytoothpaste, I play a lot :) <span style="color:green;">assigned to game 2</span><br />
* FaerieDragon88, (I’m on frequently) <span style="color:green;">assigned to game 3</span><br />
* williamxiao (don't know if i will be any good at this) <span style="color:green;">assigned to game 3</span><br />
* Mark888 (I'll fail horribly, let's do this) <span style="color:green;">assigned to game 4</span><br />
* ARay10 (But I'm worse) <span style="color:green;">assigned to game 4</span><br />
* wuwang2002 (I will just try) <span style="color:green;">assigned to game 5</span><br />
* Mark888 (am I allowed to run 2 games at once?) <span style="color:green;">assigned to game 5</span><br />
* Juno (EPIC FAILER HERE) <span style="color:green;">assigned to game 6</span><br />
* ARay10 <span style="color:green;">assigned to game 6</span><br />
* Paul10 <span style="color:green;">assigned to game 7</span><br />
* Kmath1234 <span style="color:green;">assigned to game 7</span><br />
* ARay10 (wants to verse piphi) <span style="color:green;">assigned to game 8</span><br />
* piphi (sure, this is your last available chess game BTW, you can still play [[User:Piphi/Checkers|checkers]] or wait until one of your games end) <span style="color:green;">assigned to game 8</span><br />
* Aspiring_Mathletes <span style="color:green;">assigned to game 9</span><br />
* LJCoder619 <span style="color:green;">assigned to game 9</span><br />
* SamuraiA (not going to be too active) <span style="color:green;">assigned to game 10</span><br />
* rf20008 <span style="color:green;">assigned to game 10</span><br />
* BrightMonkey <span style="color:green;">assigned to game 11</span><br />
* Alpha_2 <span style="color:green;">assigned to game 11</span><br />
* HonestCat (I'm excited to play chess!) <span style="color:green;">assigned to game 12</span><br />
* aw2017 (I can play anyone) <span style="color:green;">assigned to game 12</span><br />
* CreativeHedgehog (I'm a beginner) <span style="color:green;">assigned to game 13</span><br />
* HonestCat <span style="color:green;">assigned to game 13</span><br />
* ARay10<br />
* russellk (I just want to play someone) <span style="color:green;"<br />
<br />
== Rules ==<br />
[[User:Piphi/Chess/Rules]]<br />
<br />
== Help ==<br />
[[User:Piphi/Chess/Help]]<br />
<br />
== Game #1 piphi v BorealBear ==<br />
Game is [[User:Piphi/Chess/Game_1|here]]. Discussion is [[User talk:Piphi/Chess/Game_1|here]].<br />
<br />
== Game #2 awesomeguy856 v Heavytoothpaste ==<br />
Game is [[User:Piphi/Chess/Game_2|here]]. Discussion is [[User talk:Piphi/Chess/Game_2|here]].<br />
<br />
== Game #3 FaerieDragon88 v williamxiao ==<br />
Game is [[User:Piphi/Chess/Game_3|here]]. Discussion is [[User talk:Piphi/Chess/Game_3|here]].<br />
<br />
== Game #4 Mark888 v ARay10 ==<br />
Game is [[User:Piphi/Chess/Game_4|here]]. Discussion is [[User talk:Piphi/Chess/Game_4|here]].<br />
<br />
== Game #5 wuwang2002 v Mark888 ==<br />
Game is [[User:Piphi/Chess/Game_5|here]]. Discussion is [[User talk:Piphi/Chess/Game_5|here]].<br />
<br />
== Game #6 Juno v ARay10 ==<br />
Game is [[User:Piphi/Chess/Game_6|here]]. Discussion is [[User talk:Piphi/Chess/Game_6|here]].<br />
<br />
== Game #7 Paul10 v Kmath1234 ==<br />
Game is [[User:Piphi/Chess/Game_7|here]]. Discussion is [[User talk:Piphi/Chess/Game_7|here]].<br />
<br />
== Game #8 ARay10 v piphi ==<br />
Game is [[User:Piphi/Chess/Game_8|here]]. Discussion is [[User talk:Piphi/Chess/Game_8|here]].<br />
<br />
== Game #9 Aspiring_Mathletes v LJCoder619 ==<br />
Game is [[User:Piphi/Chess/Game_9|here]]. Discussion is [[User talk:Piphi/Chess/Game_9|here]].<br />
<br />
== Game #10 SamuraiA v rf20008 ==<br />
Game is [[User:Piphi/Chess/Game_10|here]]. Discussion is [[User talk:Piphi/Chess/Game_10|here]].<br />
<br />
== Game #11 BrightMonkey v Alpha_2 ==<br />
Game is [[User:Piphi/Chess/Game_11|here]]. Discussion is [[User talk:Piphi/Chess/Game_11|here]].<br />
<br />
== Game #12 HonestCat v aw2017 ==<br />
Game is [[User:Piphi/Chess/Game_12|here]]. Discussion is [[User talk:Piphi/Chess/Game_12|here]].<br />
<br />
== Game #13 HonestCat v CreativeHedgehog ==<br />
Game is [[User:Piphi/Chess/Game_13|here]]. Discussion is [[User talk:Piphi/Chess/Game_13|here]].<br />
<br />
[[Category:Chess_Piphi]]</div>Russellkhttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=124924User:Piphi2020-06-11T02:58:43Z<p>Russellk: /* User Count */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
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<div style="border:2px solid black; -webkit-border-radius: 10px; background:#dddddd"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="101px">162</font></center><br />
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<div style="border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center"><br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">Piphi is legendary.<br><br />
<br />
Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.<br><br />
<br />
Piphi also started the signature trend at around May 2020.<br><br />
<br />
Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].<br><br />
<br />
Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].<br><br />
<br />
Piphi has been called OP by many AoPSers, including the legendary [[User:Radio2|Radio2]] himself [https://artofproblemsolving.com/community/c19451h1826745p15526800 here].<br><br />
<br />
Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.<br><br />
<br />
Piphi has also added a lot of the info that is in the [[Reaper Archives]].<br><br />
<br />
Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].<br><br />
<br />
Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].</font></div><br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center"><br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">[[User:Piphi/Asymptote|Asymptote]]</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px">For a complete list of my Asymptote drawings, go [[User:Piphi/Asymptote|here]].</div><br />
</div></div>Russellk