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ebcbdab40a27280404bf0c0d9b9ff113 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_5 | Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating 10 pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50$ | From the problem, we see that 10 less than one of the answer choices must be a multiple of 3 and positive. The only answer choice satisfying this is $\boxed{40}$ . We can check that 40 blueberry and 20 cherry jelly beans indeed does work. | D | 40 |
ebcbdab40a27280404bf0c0d9b9ff113 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_5 | Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating 10 pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50$ | Note that the number of jellybeans minus $2 \cdot 10 = 20$ is a multiple of $4$ and positive. Therefore, the number of jellybeans is a multiple of $4$ and $>20$ . The only answer choice satisfying this is $\boxed{40}$ | D | 40 |
210361ff7b11099635fd2cebff9c4f14 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_6 | What is the largest number of solid $2\text{-in} \times 2\text{-in} \times 1\text{-in}$ blocks that can fit in a $3\text{-in} \times 2\text{-in}\times3\text{-in}$ box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | We find that the volume of the larger block is $18$ , and the volume of the smaller block is $4$ . Dividing the two, we see that only a maximum of four $2$ by $2$ by $1$ blocks can fit inside the $3$ by $3$ by $2$ block. Drawing it out, we see that such a configuration is indeed possible. Therefore, the answer is $\boxed{4}$ | B | 4 |
31231826acbe22d0c29c5f5d1682f984 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_7 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$ | Let's call the distance that Samia had to travel in total as $2x$ , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both $\frac{2x}{2}$ , or $x$ \[\] She bikes at a rate of $17$ kph, so she travels the distance she bikes in $\frac{x}{17}$ hours. She walks at a rate of $5$ kph, so she travels the distance she walks in $\frac{x}{5}$ hours. \[\] The total time is $\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}$ . This is equal to $\frac{44}{60} = \frac{11}{15}$ of an hour. Solving for $x$ , we have: \[\] \[\frac{22x}{85} = \frac{11}{15}\] \[\frac{2x}{85} = \frac{1}{15}\] \[30x = 85\] \[6x = 17\] \[x = \frac{17}{6}\] \[\] Since $x$ is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about $\boxed{2.8}$ | C | 2.8 |
31231826acbe22d0c29c5f5d1682f984 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_7 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$ | Notice that Samia walks $\frac {17}{5}$ times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is \[\frac{\frac{17}{5}}{\frac{17}{5}+\frac{5}{5}} = \frac{17}{22}\] Then, multiply this by the time \[\frac{17}{22} \cdot 44 \text{minutes} = 34 \text{minutes}\] 34 minutes is a little greater than $\frac {1}{2}$ of an hour so Samia traveled \[\sim \frac {1}{2} \cdot 5 = 2.5 \text{kilometers}\] The answer choice a little greater than 2.5 is $\boxed{2.8}$ by $5$ and gotten the exact answer as well) | C | 2.8 |
31231826acbe22d0c29c5f5d1682f984 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_7 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$ | Since the distance between biking and walking is equal, we can use the given rates' harmonic mean to find the average speed.
\[\frac{2ab}{a+b}\implies\frac{2 \cdot 17 \cdot 5}{17+5}=\frac{85}{11}\text{kph}\]
Then, multiplying by $\frac{11}{15}$ hours gives the overall distance $\frac{17}{3}$ kilometers. Samia only walks half of that, so $\frac{17}{6}\approx \boxed{2.8}$ kilometers. | C | 2.8 |
31231826acbe22d0c29c5f5d1682f984 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_7 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$ | Since the distance traveled by bicycle and foot are equal, we can substitute the time traveled by bike, or $b$ , as $\frac{5}{17}$ of the time traveled by foot, or $f$
Accordingly, we have $\frac{22}{17}f=44$ This comes out to $f=34$ This means that Samia traveled $34$ minutes on foot, and hence, $44-34=10$ minutes on bicycle
Because $10$ minutes on bike yields $\frac{17}{6}$ kilometers, and distance on bike = distance on foot, we have the final answer of $\frac{17}{6} \approx \boxed{2.8}$ | null | 2.8 |
bac97b83bdb94e9bfde1b179cf098089 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_8 | Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$ | Since $AB = AC$ , then $\triangle ABC$ is isosceles, so $BD = CD$ . Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = \boxed{4,9}$ | C | 4,9 |
bac97b83bdb94e9bfde1b179cf098089 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_8 | Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$ | Calculating the equation of the line running between points $B$ and $D$ $y = -2x + 1$ . The only coordinate of $C$ that is also on this line is $\boxed{4,9}$ | C | 4,9 |
bac97b83bdb94e9bfde1b179cf098089 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_8 | Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$ | Similar to the first solution, because the triangle is isosceles, then the line drawn in the middle separates the triangle into two smaller congruent triangles. To get from $B$ to $D$ , we go to the right $3$ and up $6$ . Then to get to point $C$ from point $D$ , we go to the right $3$ and up $6$ , getting us the coordinates $\boxed{4,9}$ | C | 4,9 |
bac97b83bdb94e9bfde1b179cf098089 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_8 | Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$ | As stated in solution 1, the triangle is isosceles.
[asy] pair A,B,C,D; A=(11,9); B=(2,-3); C=(-4,9); D=(-1,3); draw(A--B--C--cycle); draw(A--D); draw(rightanglemark(A,D,B)); label("$A$",A,E); label("$B$",B,S); label("$D$",D,W); label("$C$",C,N); [/asy]
This means that $D(-1, 3)$ is the midpoint of $B(2, -3)$ and $C(x,y)$ . So $\frac{x+2}{2}$ $= -1$ and so $x = -4$ . Similarly for $y$ , we have $\frac{y-3}{2}$ $=3$ and so $y=9$ . So our final answer is $\boxed{4,9}$ | C | 4,9 |
1d63433f4582f2df9dc4456aa3580dac | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_10 | The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, -5)$ . What is $c$
$\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$ | Writing each equation in slope-intercept form, we get $y=\frac{a}{2}x-\frac{1}{2}c$ and $y=-\frac{2}{b}x-\frac{c}{b}$ . We observe the slope of each equation is $\frac{a}{2}$ and $-\frac{2}{b}$ , respectively. Because the slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$ , we see that $\frac{a}{2}=-\frac{1}{-\frac{2}{b}}$ because it is given that the two lines are perpendicular. This equation simplifies to $a=b$
Because $(1, -5)$ is a solution of both equations, we deduce $a \times 1-2 \times -5=c$ and $2 \times 1+b \times -5=-c$ . Because we know that $a=b$ , the equations reduce to $a+10=c$ and $2-5a=-c$ . Solving this system of equations, we get $c=\boxed{13}$ | E | 13 |
613f41241e4a492fb8fc2f1dba525fa9 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_11 | At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%$ | $60\% \cdot 20\% = 12\%$ of the people that claim that they like dancing actually dislike it, and $40\% \cdot 90\% = 36\%$ of the people that claim that they dislike dancing actually dislike it. Therefore, the answer is $\frac{12\%}{12\%+36\%} = \boxed{25}$ | D | 25 |
613f41241e4a492fb8fc2f1dba525fa9 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_11 | At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%$ | Assume WLOG that there are 100 people. Then 60 of them like dancing, 40 dislike dancing. Of the ones that like dancing, 48 say they like dancing and 12 say they dislike it. Of the ones who dislike dancing, 36 say they dislike dancing and 4 say they like it. We want the ratio of students like it but say they dislike it to the total amount of students that say they dislike it. This is $\frac{12}{12+36}=\frac{12}{48}=\frac{1}{4}$ . We choose $\boxed{25}$ | D | 25 |
613f41241e4a492fb8fc2f1dba525fa9 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_11 | At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%$ | WLOG, assume that there are a total of $100$ students at Typico High School. We make a chart:
\[\begin{tabular}[t]{|c|c|c|c|}\hline & \text{Likes dancing} & \text{Doesn't like dancing} & \text{Total} \\\hline \text{Says they like dancing} & & & \\\hline \text{Says they don't like dancing} & & & \\\hline \text{Total} & & & 100 \\\hline \end{tabular}\]
We know that $60$ of the students like dancing (since $60\%$ of $100$ is $60$ ), so we fill that in:
\[\begin{tabular}[t]{|c|c|c|c|}\hline & \text{Likes dancing} & \text{Doesn't like dancing} & \text{Total} \\\hline \text{Says they like dancing} & & & \\\hline \text{Says they don't like dancing} & & & \\\hline \text{Total} & 60 & & 100 \\\hline \end{tabular}\]
$80\%$ of those $60$ kids say that they like dancing, so that's $48$ kids who like dancing and say that they like dancing. The other $12$ kids like dancing and say that they do not.
\[\begin{tabular}[t]{|c|c|c|c|}\hline & \text{Likes dancing} & \text{Doesn't like dancing} & \text{Total} \\\hline \text{Says they like dancing} & 48 & & \\\hline \text{Says they don't like dancing} & 12 & & \\\hline \text{Total} & 60 & & 100 \\\hline \end{tabular}\]
$40$ students do not like dancing. $90\%$ of those $40$ students say that they do not like it, which is $36$ of them.
\[\begin{tabular}[t]{|c|c|c|c|}\hline & \text{Likes dancing} & \text{Doesn't like dancing} & \text{Total} \\\hline \text{Says they like dancing} & 48 & & \\\hline \text{Says they don't like dancing} & 12 & 36 & \\\hline \text{Total} & 60 & 40 & 100 \\\hline \end{tabular}\]
At this point, one can see that there are $12+36=48$ total students who say that they do not like dancing. $12$ of those actually like it, so that is $\dfrac{12}{48}=\dfrac14=\boxed{25}.$ | D | 25 |
41dbed30340ab4456900aa509fffef05 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_12 | Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?
$\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$ | Suppose that his old car runs at $x$ km per liter. Then his new car runs at $\frac{3}{2}x$ km per liter, or $x$ km per $\frac{2}{3}$ of a liter. Let the cost of the old car's fuel be $c$ , so the trip in the old car takes $xc$ dollars, while the trip in the new car takes $\frac{2}{3}\cdot\frac{6}{5}xc = \frac{4}{5}xc$ . He saves $\frac{\frac{1}{5}xc}{xc} = \boxed{20}$ | A | 20 |
41dbed30340ab4456900aa509fffef05 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_12 | Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?
$\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$ | Because they do not give you a given amount of distance, we'll just make that distance $3x$ miles. Then, we find that the new car will use $2*1.2=2.4x$ . The old car will use $3x$ . Thus the answer is $(3-2.4)/3=.6/3=20/100= \boxed{20}$ | A | 20 |
41dbed30340ab4456900aa509fffef05 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_12 | Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?
$\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$ | You can find that the ratio of fuel used by the old car and the new car for the same amount of distance is $3 : 2$ , and the ratio between the fuel price of these two cars is $5 : 6$ . Therefore, by multiplying these two ratios, we get that the costs of using these two cars is \[15 : 12 = 5 : 4\] So the percentage of money saved is $1 - \frac{4}{5} = \boxed{20}$ | A | 20 |
41dbed30340ab4456900aa509fffef05 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_12 | Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?
$\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$ | Assume WLOG that Elmer's old car's range is $100$ miles. So, Elmer's new car's range is $100 \times 1.5 = 150$ miles. Also, assume that the gas Elmer's old car uses is $$10$ , which means that diesel will cost $$12$ . Now we can deduce that Elmer's old car uses $10 \div 100 = $0.10$ per mile, and Elmer's new car uses $12 \div 150 = $0.08$ per mile. Therefore, Elmer's new car saves $\boxed{20}$ more money than his old car. | A | 20 |
6a986c970315c125a402938fe0418e77 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_13 | There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | By PIE (Property of Inclusion/Exclusion), we have
$|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|.$ Number of people in at least two sets is $\sum |A_i \cap A_j| - 2|A_1 \cap A_2 \cap A_3| = 9.$ So, $20 = (10 + 13 + 9) - (9 + 2x) + x,$ which gives $x = \boxed{3}.$ | C | 3 |
6a986c970315c125a402938fe0418e77 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_13 | There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | The total number of classes taken among the 20 students is $10 + 13 + 9 = 32$ . Each student is taking at least one class so let's subtract the $20$ classes ( $1$ per each of the $20$ students) from $32$ classes to get $12$ $12$ classes is the total number of extra classes taken by the students who take $2$ or $3$ classes. Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \boxed{3}$ students that are taking all $3$ classes. | C | 3 |
6a986c970315c125a402938fe0418e77 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_13 | There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | Total class count is 32. Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$ . Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$ . There are $9$ students taking two or three classes, so $b+c = 9$ . Solving this system of equations gives us $c=\boxed{3}$ | C | 3 |
6a986c970315c125a402938fe0418e77 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_13 | There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | Let us assign the following variables and put them in our Venn Diagram [1] $a$ which designates the number of people taking exactly Bridge and Yoga. $b$ which designates the number of people taking exactly Bridge and Painting. $c$ which designates the number of people that took all $3$ classes or what we want to find. $d$ which designates the number of people taking exactly Yoga and Painting.
Let's now recall what information we have given: There are exactly $9$ people that are taking at least $2$ classes meaning in other words, $9$ people total are taking strictly $2$ classes or strictly all the available classes meaning that $a+b+c+d=9$
Let's now start filling out the Venn Diagram:
Strictly taking Bridge, no other classes: We know in total, the number is $13$ , however this includes the people taking other classes too meaning we'd need to do some subtraction. From our Venn Diagram we see that we'd need to subtract the following variables to get our wanted outcome here, $a, b, c$ . Giving our answer as $13-(a+b+c)$
However, this equation seems complicated as it has $3$ different variables, so to make this look a lot less complicated we can use our earlier equation: $a+b+c+d=9$ to see that $a+b+c=9-d$ . This means that this can also be written as $13-(9-d)=4+d$
Strictly taking Yoga Only: The total number of people is $10$ , but this would also count people taking other classes too along with it, so we need to subtract this overcount which is visible in the Venn Diagram giving us: $10-(a+c+d)$
Again, we can use substitution to see that $a+c+d=9-b$ . This simplifies our equation to $10-(9-b)=1+b$
Strictly taking Painting Only: We know again, in total this number is $9$ , which also accounts for the people taking other classes too. From our Venn Diagram it is visible that we need to subtract: $b, c, d$ giving $9-(b+c+d)$
Again, through substitution of our first equation we see that $b+c+d=9-a$ meaning we can simplify this equation to $9-(9-a)=a$
If we add these newly made equations of strictly taking one class, we get the total number of people taking exactly one class as these equations each were a subcase for it. We can also find the exact number for this because we are given that there are exactly $20$ students in total, and $9$ students are taking exactly $2$ or $3$ classes, meaning that if we do $20-9$ we get our answer for the number of students taking exactly $1$ class because those taking exactly one class have no overlap with those taking exactly $2$ or exactly $3$ classes as shown in our Venn Diagram and because $1$ $2$ , and $3$ classes are subcases for finding the total number of students as we know that each student is in exactly $1, 2$ , or $3$ classes. This means that exactly $11$ students took strictly $1$ class.
We can add up our equations we found to equal $11$ because those equations were for subcases of having exactly $1$ class giving: $(4+d)+(1+b)+a=11$ $=5+a+b+d=11$
From our equation $a+b+c+d=9$ , we can substitute $9-c$ for $a+b+d$ giving us: $5+9-c=11$
This gives $c=3$ . We assigned $c$ for the number of people taking exactly $3$ classes meaning that when we find $c$ , we find the answer. This means our answer is $\boxed{3}$ | C | 3 |
6a986c970315c125a402938fe0418e77 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_13 | There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We are told that there are $20$ students in all, and $10$ take yoga, $13$ take bridge, and $9$ take painting.
Representing each student as a number from $1$ to $20$ , we can then make a list of which classes they are taking. Students 1-10 take yoga, and students $11$ to $20$ take bridge. However, this means that only $10$ students are taking bridge. To make up for it, we go back to the top of the list and start over from student $1$ . Students $1$ $2$ , and $3$ will also take bridge giving the desired count of $13$ total bridge students.
Now, all that is left are the students who take painting. There are $9$ students who take painting, so students $4$ through students $9$ take painting. Note that because $9$ people take at least two classes, students $10$ through $20$ are unable to take more than one class. This means that we must once more start over from the top. Already $6$ painting slots have been filled, so students $1$ $2$ , and $3$ will also take painting. This gives a total of $3$ students (students $1$ $2$ , and $3$ ) who take all three classes. Therefore, our answer is $\boxed{3}$ | C | 3 |
391d32ea6745115f1b5a399e3d927b75 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_14 | An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1$
$\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{4}{5}\qquad\textbf{(E)}\ 1$ | Notice that we can rewrite $N^{16}$ as $(N^{4})^4$ . By Fermat's Little Theorem , we know that $N^{(5-1)} \equiv 1 \pmod {5}$ if $N \not \equiv 0 \pmod {5}$ . Therefore for all $N \not \equiv 0 \pmod {5}$ we have $N^{16} \equiv (N^{4})^4 \equiv 1^4 \equiv 1 \pmod 5$ . Since $1\leq N \leq 2020$ , and $2020$ is divisible by $5$ $\frac{1}{5}$ of the possible $N$ are divisible by $5$ . Therefore, $N^{16} \equiv 1 \pmod {5}$ with probability $1-\frac{1}{5},$ or $\boxed{45}$ | D | 45 |
391d32ea6745115f1b5a399e3d927b75 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_14 | An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1$
$\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{4}{5}\qquad\textbf{(E)}\ 1$ | Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits $0-9$ .
The pattern for $0$ is $0$ , no matter what power, so $0$ doesn't work. Likewise, the pattern for $5$ is always $5$ . Doing the same for the rest of the digits, we find that the units digits of $1^{16}$ $2^{16}$ $3^{16}$ $4^{16}$ $6^{16}$ $7^{16}$ $8^{16}$ and $9^{16}$ all have the remainder of $1$ when divided by $5$ , so $\boxed{45}$ | D | 45 |
73cbcd45829db2c5e8aecf3bce45df29 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_16 | How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0$
$\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481$ | We can use complementary counting. There are $2017$ positive integers in total to consider, and there are $9$ one-digit integers, $9 \cdot 9 = 81$ two digit integers without a zero, $9 \cdot 9 \cdot 9 = 729$ three digit integers without a zero, and $9 \cdot 9 \cdot 9 = 729$ four-digit integers starting with a 1 without a zero. Therefore, the answer is $2017 - 9 - 81 - 729 - 729 = \boxed{469}$ | A | 469 |
73cbcd45829db2c5e8aecf3bce45df29 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_16 | How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0$
$\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481$ | First, we notice there are no one-digit numbers that contain a zero. There are $9$ two-digit integers and $9 \cdot 9 + 9 \cdot 9 + 9 = 171$ three-digit integers containing at least one zero. Next, we consider the four-digit integers beginning with one. There are $3 \cdot 9 \cdot 9 = 243$ of these four-digit integers with one zero, ${3 \choose 2} \cdot 9 = 27$ with two zeros, and $1$ with three zeros $(1000)$ . Finally, we consider the numbers $2000$ to $2017$ which all contain at least one zero. Adding all of these together we get $9 + 171 + 243 + 27 + 1 + 18 = \boxed{469}$ | A | 469 |
db8a18947f72965d805cb827c43018fd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_17 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | Case 1: monotonous numbers with digits in ascending order
There are $\sum_{n=1}^{9} \binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, $\emptyset$ (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.$
Case 2: monotonous numbers with digits in descending order
There are $\sum_{n=1}^{10} \binom{10}{n}$ ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. However, $\emptyset$ (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.$ We discard the number 0 since it is not positive. Thus there are $1022$ here.
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are $511+1022-9=\boxed{1524}$ monotonous numbers. | B | 1524 |
db8a18947f72965d805cb827c43018fd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_17 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | Like Solution 1, divide the problem into an increasing and decreasing case:
$\bullet$ Case 1: Monotonous numbers with digits in ascending order.
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.
To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are $2^9 = 512$ ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get $512-1=511$ monotonous numbers for this case.
$\bullet$ Case 2: Monotonous numbers with digits in descending order.
This time, we arrange all 10 digits in decreasing order and repeat the process to find $2^{10} = 1024$ ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get $1024-2=1022$ monotonous numbers for this case.
At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.
Thus our final answer is $511+1022-9 = \boxed{1524}$ | B | 1524 |
db8a18947f72965d805cb827c43018fd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_17 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | Unlike the first two solutions, we can do our casework based off of whether or not the number contains a $0$
If it does, then we know the $0$ must be the last digit in the number (and hence, the number has to be decreasing). Because $0$ is not positive, $0$ is not monotonous. So, we need to pick at least $1$ number in the set $[1, 9].$ After choosing our numbers, there will be just $1$ way to arrange them so that the overall number is monotonous.
In total, each of the $9$ digits can either be in the monotonous number or not, yielding $2^9 = 512$ total solutions. However, we said earlier that $0$ cannot be by itself so we have to subtract out the case in which we picked none of the numbers $1-9$ . So, this case gives us $511$
Onto the second case, if there are no $0$ s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits $1- 9$ , there are $2$ solutions. This gives \[2 \cdot (2^9 - 1) = 2 \cdot 511 = 1022\] possibilities. Note that we subtracted out the $1$ because we cannot choose none of the numbers.
However, realize that if we pick just $1$ digit, then there is only $1$ arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out $9$ from there (because there are $9$ possible ways to select one number and for each case, we overcounted by $1$ ).
All in all, that gives $511 + 1022 - 9 = \boxed{1524}$ monotonous numbers. | B | 1524 |
db8a18947f72965d805cb827c43018fd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_17 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | Let $n$ be the number of digits of a monotonous number. Notice for an increasing monotonous number with $n \ge 2$ , we can obtain 2 more monotonous numbers that are decreasing by reversing its digits and adding a $0$ to the end of the reversed digits. Whenever $n$ digits are chosen, the order is fixed. There are $\binom{9}{n}$ ways to obtain an increasing monotonous number with $n$ digits. So, there are $3\cdot \sum_{n=2}^{9} \binom{9}{n}$ monotonous numbers when $n \ge 2$ . When $n=1$ , there is no reverse but we could add $0$ to the end, so there are $2 \cdot \binom{9}{1}$ monotonous numbers.
The answer is:
$3\cdot \sum_{n=2}^{9} \binom{9}{n} + 2 \cdot \binom{9}{1}$ $=3\cdot \sum_{n=1}^{9} \binom{9}{n} - \binom{9}{1}$ $=3\cdot \left( \sum_{n=0}^{9} \binom{9}{n} - \binom{9}{0} \right) - \binom{9}{1}$ $= 3 \cdot (2^9-1) - 9$ $=\boxed{1524}$ | B | 1524 |
db8a18947f72965d805cb827c43018fd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_17 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$ | We have 2 cases.
Case 1: Ascending order
We can set up a 1-1 Correspondence. For any subset of all the digits $1$ to $9$ $0$ cannot be a digit for ascending order), we can always rearrange them into an ascending monotonous number. Therefore, the number of subsets of the integers $1$ to $9$ is equivalent to the number of ascending integers. So, $2^9=512$ . However, the empty set ( $\emptyset$ ) is not an integer, so we must subtract 1. Thus, $512-1=511$
Case 2: Descending order
Similarly, any subset of the digits $0$ to $9$ can be rearranged into a descending monotonous number. So, $2^{10}=1024$ . However, $\emptyset$ and $0$ are not positive integers, so we must subtract 2. Thus, $1024-2=1022$
We have covered all the cases. We add $511$ to $1022$ , giving us $1533$ . So now we just innocently go ahead and choose $\textbf{(C) } 1533$ as our answer, right? No! We overcounted the $9$ single-digit integers . The answer is actually $1533-9=\boxed{1524}$ | B | 1524 |
89af124d6c36b7a8c3fcc57e408ad926 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_18 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$ | First we figure out the number of ways to put the $3$ blue disks. Denote the spots to put the disks as $1-6$ from left to right, top to bottom. The cases to put the blue disks are $(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)$ . For each of those cases we can easily figure out the number of ways for each case, so the total amount is $2+2+3+3+1+1 = \boxed{12}$ | D | 12 |
89af124d6c36b7a8c3fcc57e408ad926 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_18 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$ | Denote the $6$ discs as in the first solution. Ignoring reflections or rotations, there are $\binom{6}{3} \cdot \binom{3}{2} = 60$ colorings. Now we need to count the number of fixed points under possible transformations:
1. The identity transformation. Since this doesn't change anything, there are $60$ fixed points.
2. Reflect on a line of symmetry. There are $3$ lines of reflections. Take the line of reflection going through the centers of circles $1$ and $5$ . Then, the colors of circles $2$ and $3$ must be the same, and the colors of circles $4$ and $6$ must be the same. This gives us $4$ fixed points per line of reflection.
3. Rotate by $120^\circ$ counterclockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles $1$ $4$ , and $6$ will be the same. Similarly, the colors of circles $2$ $3$ , and $5$ will be the same. This is impossible, so this case gives us $0$ fixed points per rotation.
By Burnside's Lemma , the total number of colorings is $(1 \cdot 60+3 \cdot 4+2 \cdot 0)/(1+3+2) = \boxed{12}$ | D | 12 |
89af124d6c36b7a8c3fcc57e408ad926 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_18 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$ | Note that the green disk has two possibilities; in a corner or on the side. WLOG, we can arrange these as
[asy] filldraw(circle((0,0),1),green); draw(circle((2,0),1)); draw(circle((4,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((3,sqrt(3)),1)); draw(circle((2,2sqrt(3)),1)); draw(circle((8,0),1)); filldraw(circle((10,0),1),green); draw(circle((12,0),1)); draw(circle((9,sqrt(3)),1)); draw(circle((11,sqrt(3)),1)); draw(circle((10,2sqrt(3)),1)); [/asy]
Take the first case. Now, we must pick two of the five remaining circles to fill in the red. There are $\dbinom{5}{2}=10$ of these. However, due to reflection we must divide this by two. But, in two of these cases, the reflection is itself, so we must subtract these out before dividing by 2, and add them back afterwards, giving $\frac{10-2}{2}+2=6$ arrangements in this case.
Now, look at the second case. We again must pick two of the five remaining circles, and like in the first case, two of the reflections give the same arrangement. Thus, there are also $6$ arrangements in this case.
In total, we have $6+6=\boxed{12}$ | D | 12 |
89af124d6c36b7a8c3fcc57e408ad926 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_18 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$ | We note that the group $G$ acting on the possible colorings is $D_3 = \{e, r, r^2, s, sr, sr^2\}$ , where $r$ is a $120^\circ$ rotation and $s$ is a reflection. In particular, the possible actions are the identity, the $120^\circ$ and $240^\circ$ rotations, and the three reflections.
We will calculate the number of colorings that are fixed under each action. Every coloring is fixed under the identity, so we count $\dfrac{6!}{3!2!1!} = 60$ fixed colorings. Note that no colorings are fixed under the rotations, since then the outer three and inner three circle must be the same color, which is impossible in our situation.
Finally, consider the reflection with a line of symmetry going through the top circle. Every fixed coloring is determined by the color of the top circle (either green or blue), and the color of the middle circles (either blue or red). Hence, there are $2\cdot 2 = 4$ colorings fixed under this reflection action. The other two actions are symmetric, so they also have $4$ fixed colorings. Hence, by Burnside's lemma, the number of unique colorings up to reflections and rotations is \[\dfrac{1}{|D_3|} (1\cdot 60 + 2\cdot 0 + 3\cdot 4) = \dfrac{1}{6}\cdot 72 = \boxed{12}.\] | D | 12 |
8d75303bc75ed6ed8d2022fc9678e64c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_23 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | We only need to find the remainders of N when divided by 5 and 9 to determine the answer.
By inspection, $N \equiv 4 \text{ (mod 5)}$ .
The remainder when $N$ is divided by $9$ is $1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4$ , but since $10 \equiv 1 \text{ (mod 9)}$ , we can also write this as $1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45$ , which has a remainder of 0 mod 9. Solving these modular congruence using the Chinese Remainder Theorem we get the remainder to be $9 \pmod{45}$ . Therefore, the answer is $\boxed{9}$ | C | 9 |
8d75303bc75ed6ed8d2022fc9678e64c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_23 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | Once we find our 2 modular congruences, we can narrow our options down to ${C}$ and ${D}$ because the remainder when $N$ is divided by $45$ should be a multiple of 9 by our modular congruence that states $N$ has a remainder of $0$ when divided by $9$ . Also, our other modular congruence states that the remainder when divided by $45$ should have a remainder of $4$ when divided by $5$ . Out of options $C$ and $D$ , only $\boxed{9}$ | C | 9 |
8d75303bc75ed6ed8d2022fc9678e64c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_23 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | Realize that $10 \equiv 10 \cdot 10 \equiv 10^{k} \pmod{45}$ for all positive integers $k$
Apply this on the expanded form of $N$ \[N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv\] \[10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv\] \[10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{9}\] | C | 9 |
8d75303bc75ed6ed8d2022fc9678e64c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_23 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | We know that $45 = 5 \cdot 9$ , so we can apply our restrictions to that. We know that the units digit must be $5$ or $0$ , and the digits must add up to a multiple of $9$ $1+2+3+4+\cdots + 44 = \frac{44 \cdot 45}{2}$ . We can quickly see this is a multiple of $9$ because $\frac{44}{2} \cdot 45 = 22 \cdot 45$ . We know $123 \ldots 4344$ is not a multiple of $5$ because the units digit isn't $5$ or $0$ . We can just subtract by 9 until we get a number whose units digit is 5 or 0.
We have $123 \ldots 4344$ is divisible by $9$ , so we can subtract by $9$ to get $123 \ldots 4335$ and we know that this is divisible by 5. So our answer is $\boxed{9}$ | C | 9 |
fdb98dda499770f48d26547d917bbc88 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_24 | The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$ | WLOG, let the centroid of $\triangle ABC$ be $I = (-1,-1)$ . The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, $A = (1,1)$ , so $AI = BI = CI = 2\sqrt{2}$ , so since $\triangle AIB$ is isosceles and $\angle AIB = 120^{\circ}$ , then by the Law of Cosines $AB = 2\sqrt{6}$ . Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to $\frac {s}{\sqrt{3}}$ . Therefore, the area of the triangle is $\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}$ , so the square of the area of the triangle is $\boxed{108}$ | C | 108 |
fdb98dda499770f48d26547d917bbc88 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_24 | The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$ | Without loss of generality, let the centroid of $\triangle ABC$ be $G = (-1,-1)$ . Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let $A = (1,1)$ . Then, point $B$ must be the reflection of $C$ across the line $y=x$ , so let $B = \left(a,\frac{1}{a}\right)$ and $C=\left(\frac{1}{a},a\right)$ , where $a <-1$ . Because $G$ is the centroid, the average of the $x$ -coordinates of the vertices of the triangle is $-1$ . So we know that $a + 1/a+ 1 = -3$ . Multiplying by $a$ and solving gives us $a=-2-\sqrt{3}$ . So $B=(-2-\sqrt{3},-2+\sqrt{3})$ and $C=(-2+\sqrt{3},-2-\sqrt{3})$ . So $BC=2\sqrt{6}$ , and finding the square of the area gives us $\boxed{108}$ | C | 108 |
fdb98dda499770f48d26547d917bbc88 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_24 | The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$ | Without loss of generality, let the centroid of $\triangle ABC$ be $G = (1, 1)$ and let point $A$ be $(-1, -1)$ . It is known that the centroid is equidistant from the three vertices of $\triangle ABC$ . Because we have the coordinates of both $A$ and $G$ , we know that the distance from $G$ to any vertice of $\triangle ABC$ is $\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}$ . Therefore, $AG=BG=CG=2\sqrt{2}$ . It follows that from $\triangle ABG$ , where $AG=BG=2\sqrt{2}$ and $\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}$ $[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$ using the formula for the area of a triangle with sine $\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)$ . Because $\triangle ACG$ and $\triangle BCG$ are congruent to $\triangle ABG$ , they also have an area of $2\sqrt{3}$ . Therefore, $[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}$ . Squaring that gives us the answer of $\boxed{108}$ | C | 108 |
fdb98dda499770f48d26547d917bbc88 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_24 | The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$ | Without loss of generality, let the centroid of $\triangle ABC$ be $G = (1, 1)$ . Assuming we don't know one vertex is $(-1, -1)$ we let the vertices be $A\left(x_1, \frac{1}{x_1}\right), B\left(x_2, \frac{1}{x_2}\right), C\left(x_3, \frac{1}{x_3}\right).$
Since the centroid coordinates are the average of the vertex coordinates, we have that $\frac{x_1+x_2+x_3}{3}=1$ and $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1.$
We also know that the centroid is the orthocenter in an equilateral triangle, so $CG \perp AB.$ Examining slopes, we simplify the equation to $x_1x_2x_3 = -1$ . From the equation $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1,$ we get that $x_1x_2+x_1x_3+x_2x_3 = -3$ . These equations are starting to resemble Vieta's:
$x_1+x_2+x_3=3$
$x_1x_2+x_1x_3+x_2x_3 = -3$
$x_1x_2x_3=-1$
$x_1,x_2,x_3$ are the roots of the equation $x^3 - 3x^2 - 3x + 1 = 0$ . This factors as $(x+1)(x^2-4x+1)=0 \implies x = -1, 2 \pm \sqrt3,$ for the points $(-1, -1), (2+\sqrt3, 2-\sqrt3), (2-\sqrt3, 2+\sqrt3)$ . The side length is clearly $\sqrt{24}$ , so the square of the area is $\boxed{108}.$ | null | 108 |
8bbdf1a795124500210fc8a16aedfadd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_25 | Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | Let the sum of the scores of Isabella's first $6$ tests be $S$ . Since the mean of her first $7$ scores is an integer, then $S + 95 \equiv 0 \text{ (mod 7)}$ , or $S \equiv3 \text{ (mod 7)}$ . Also, $S \equiv 0 \text{ (mod 6)}$ , so by the CRT $S \equiv 24 \text{ (mod 42)}$ . We also know that $91 \cdot 6 \leq S \leq 100 \cdot 6$ , so by inspection, $S = 570$ . However, we also have that the mean of the first $5$ test scores must be an integer, so the sum of the first $5$ test scores must be an multiple of $5$ , which implies that the $6$ th test score is $\boxed{100}$ | E | 100 |
8bbdf1a795124500210fc8a16aedfadd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_25 | Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | First, we find the largest sum of scores which is $100+99+98+97+96+95+94$ which equals $7(97)$ . Then we find the smallest sum of scores which is $91+92+93+94+95+96+97$ which is $7(94)$ . So the possible sums for the 7 test scores so that they provide an integer average are $7(97), 7(96), 7(95)$ and $7(94)$ which are $679, 672, 665,$ and $658$ respectively. Now in order to get the sum of the first 6 tests, we subtract $95$ from each sum producing $584, 577, 570,$ and $563$ . Notice only $570$ is divisible by $6$ so, therefore, the sum of the first $6$ tests is $570$ . We need to find her score on the $6th$ test so we have to find which number will give us a number divisible by $5$ when subtracted from $570.$ Since $95$ is the $7th$ test score and all test scores are distinct that only leaves $\boxed{100}$ | E | 100 |
8bbdf1a795124500210fc8a16aedfadd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_25 | Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | Since all of the scores are from $91 - 100$ , we can subtract 90 from all of the scores. Since the last score was a 95, the sum of the scores from the first six tests must be $3 \pmod 7$ and $0 \pmod 6$ . Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be $0 \pmod 5$ because $30\equiv 0\pmod5$ . The only possible test scores are $95$ and $100$ , and $95$ is already used, so the answer is $\boxed{100}$ | E | 100 |
8bbdf1a795124500210fc8a16aedfadd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_25 | Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | We work backwards to solve this problem. In the $n^{th}$ test, $\sum_{i=1}^{n} i$ (mod n) = 0. In the following table, the tests already taken are in bold, the latest test is underlined. We work from the row of (mod 7), (mod 6), and (mod 5) to determine the test order by trial and error.
\[\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline & 91 & 92 & 93 & 94 & 95 & 96 & 97 & 98 & 99 & 100 \\ \hline mod 7 & \textbf{0} & \textbf{1} & 2 & \textbf{3} & \underline{\textbf{4}} & \textbf{5} & \textbf{6} & 0 & 1 & \textbf{2} \\ \hline mod 6 & \textbf{1} & \textbf{2} & 3 & \textbf{4} & 5 & \textbf{0} & \textbf{1} & 2 & 3 & \underline{\textbf{4}} \\ \hline mod 5 & \textbf{1} & \textbf{2} & 3 & \underline{\textbf{4}} & 0 & \textbf{1} & \textbf{2} & 3 & 4 & 0 \\ \hline mod 4 & \underline{\textbf{3}} & \textbf{0} & 1 & 2 & 3 & \textbf{0} & \textbf{1} & 2 & 3 & 0 \\ \hline mod 3 & 1 & \textbf{2} & 0 & 1 & 2 & \textbf{0} & \underline{\textbf{1}} & 2 & 0 & 1 \\ \hline mod 2 & 1 & \underline{\textbf{0}} & 1 & 0 & 1 & \textbf{0} & 1 & 0 & 1 & 0 \\ \hline \end{tabular}\]
In the $4^{th}$ test, the test score could be 91 or 97.
As you can see, the test score of the $6^{th}$ test is $\boxed{100}$ | E | 100 |
8bbdf1a795124500210fc8a16aedfadd | https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_25 | Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$ | Let's denote $x$ as $a_1+a_2+a_3+a_4+a_5+a_6$ , where $a_n$ is her $n$ th test score. We start by saying $\frac{x+95}{7} = [91,92,...100]$ . This results from the fact that her test averages are integers and her test scores are in the range of $91$ to $100$ . So $x+95=[637,644,...700] \implies x=[542,549,...605]$ . Next, we also see that $\frac{x}{6}=[91,92,...100]$ which becomes $x=[546,552,...600]$ . The only point of intersection of $[542,549,...605]$ and $[546,552,...600]$ is $570$ so $x=570$ . Now we can also see that $\frac{x-a_6}{5}=[91,92,...100]$ , so $x-a_6=[455,460,...500]$ , and since $x=570$ $570-a_6=[455,460,...500] \implies a_6=570-[455,460,...500]$ $a_6$ has to be in the range of $91$ and $100$ , so the only values in the set $[455,460,...500]$ that work are $470$ and $475$ which result in $a_6 = 100 \text{ or } 95$ respectively. But since her test scores are all distinct, and $95$ is already used for $a_7$ , our answer is $\boxed{100}$ | E | 100 |
b070cb422d81f83351252683890c0fbf | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_1 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | We can use subtraction of fractions to get \[\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{100}.\] | B | 100 |
b070cb422d81f83351252683890c0fbf | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_1 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | Factoring out $9!$ gives $\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{100}$ | B | 100 |
b070cb422d81f83351252683890c0fbf | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_1 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | We are given the equation $\frac{11!-10!}{9!}$
This is equivalent to $\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\frac{(11-1)(10!)}{9!}$ , which equals $10 \cdot 10$
Therefore, the answer is $10^2$ $\boxed{100}$ | B | 100 |
87b69252c2ca9f6dfbfbff8d7e711c33 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_2 | For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$ \[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\] Since the bases are equal, we can set the exponents equal, giving us $5x=15$ . Solving the equation gives us $x = \boxed{3}.$ | C | 3 |
87b69252c2ca9f6dfbfbff8d7e711c33 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_2 | For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We can rewrite this expression as $\log(10^x \cdot 100^{2x})=\log(1000^5)$ , which can be simplified to $\log(10^{x}\cdot10^{4x})=5\log(1000)$ , and that can be further simplified to $\log(10^{5x})=5\log(10^3)$ . This leads to $5x=15$ . Solving this linear equation yields $x = \boxed{3}.$ | C | 3 |
ad43651fa9188d415d220bce7958b58e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_3 | For every dollar Ben spent on bagels, David spent $25$ cents less. Ben paid $$12.50$ more than David. How much did they spend in the bagel store together?
$\textbf{(A)}\ $37.50 \qquad\textbf{(B)}\ $50.00\qquad\textbf{(C)}\ $87.50\qquad\textbf{(D)}\ $90.00\qquad\textbf{(E)}\ $92.50$ | If Ben paid $$ 12.50$ more than David, then he paid $\frac{12.5}{.25}= $ 50.00$ . Thus, David paid $$ 37.50$ , and they spent $50.00+37.50 =$ 87.50 \implies \boxed{87.50}$ | C | 87.50 |
9716e12f7b2a2ae1a1cdf0cc7461203c | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_5 | A rectangular box has integer side lengths in the ratio $1: 3: 4$ . Which of the following could be the volume of the box?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144$ | Let the smallest side length be $x$ . Then the volume is $x \cdot 3x \cdot 4x =12x^3$ . If $x=2$ , then $12x^3 = 96 \implies \boxed{96.}$ | D | 96. |
def29ff58c6fe4554c3ae1ba728068cd | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_6 | Star lists the whole numbers $1$ through $30$ once. Emilio copies Star's numbers, replacing each occurrence of the digit $2$ by the digit $1$ . Star adds her numbers and Emilio adds his numbers. How much larger is Star's sum than Emilio's?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110$ | For every tens digit 2, we subtract 10, and for every units digit 2, we subtract 1. Because 2 appears 10 times as a tens digit and 2 appears 3 times as a units digit, the answer is $10\cdot 10+1\cdot 3=\boxed{103.}$ | D | 103. |
f7d19823e67f471fec13d00937d24abc | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_7 | The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$ | Since $x$ is the mean, \begin{align*} x&=\frac{60+100+x+40+50+200+90}{7}\\ &=\frac{540+x}{7}. \end{align*}
Therefore, $7x=540+x$ , so $x=\boxed{90}.$ | D | 90 |
f7d19823e67f471fec13d00937d24abc | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_7 | The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$ | Note that $x$ must be the median so it must equal either $60$ or $90$ . You can see that the mean is also $x$ , and by intuition $x$ should be the greater one. $x=\boxed{90}.$ ~bjc | D | 90 |
0c8f5b186f1b6984b7728b58ebce58a9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_8 | Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays $40$ coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45$ | If you started backwards you would get: \[0\Rightarrow (+40)=40 , \Rightarrow \left(\frac{1}{2}\right)=20 , \Rightarrow (+40)=60 , \Rightarrow \left(\frac{1}{2}\right)=30 , \Rightarrow (+40)=70 , \Rightarrow \left(\frac{1}{2}\right)=\boxed{35}\] | C | 35 |
0c8f5b186f1b6984b7728b58ebce58a9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_8 | Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays $40$ coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45$ | If you have $x$ as the amount of money Foolish Fox started with we have $2(2(2x-40)-40)-40=0.$ Solving this we get $\boxed{35}$ | C | 35 |
c7946e6d34cb71f5351c00c9c014bbd0 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_9 | A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | We are trying to find the value of $N$ such that \[1+2+3\cdots+(N-1)+N=\frac{N(N+1)}{2}=2016.\] Noticing that $\frac{63\cdot 64}{2}=2016,$ we have $N=63,$ so our answer is $\boxed{9}.$ | D | 9 |
c7946e6d34cb71f5351c00c9c014bbd0 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_9 | A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get $2016$ .
Notice that $1 + 2 + 3 \cdots + 10 = 55.$ Knowing this, we can say that $11 + 12 \cdots + 20 = 155$ and $21 + \cdots +30 =255$ and so on. This is a quick way to get to the point that N is between 60 and 70. By subtracting from the sum of the number from 1 through 70, we learn that when we subtract $70, 69, 68, 67, 66, 65,$ and $64, N = 63.$ Adding those two digits, we get the answer $\boxed{9}.$ - CorgiARMY | D | 9 |
29718fd7f1aaa2fbbdd63bee83db1d15 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_10 | A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle?
[asy] size(6cm); defaultpen(fontsize(9pt)); path rectangle(pair X, pair Y){ return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle; } filldraw(rectangle((0,0),(7,5)),gray(0.5)); filldraw(rectangle((1,1),(6,4)),gray(0.75)); filldraw(rectangle((2,2),(5,3)),white); label("$1$",(0.5,2.5)); draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead)); draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead)); label("$1$",(1.5,2.5)); draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead)); draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead)); label("$1$",(4.5,2.5)); draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead)); draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead)); label("$1$",(4.1,1.5)); draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead)); draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead)); label("$1$",(3.7,0.5)); draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead)); draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead)); [/asy]
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8$ | Let the length of the inner rectangle be $x$
Then the area of that rectangle is $x\cdot1 = x$
The second largest rectangle has dimensions of $x+2$ and $3$ , making its area $3x+6$ . The area of the second shaded area, therefore, is $3x+6-x = 2x+6$
The largest rectangle has dimensions of $x+4$ and $5$ , making its area $5x + 20$ . The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is $(5x+20) - (3x+6) = 2x + 14$
The problem states that $x, 2x+6, 2x+14$ is an arithmetic progression, meaning that the terms in the sequence increase by the same amount each term.
Therefore, $(2x+6) - (x) = (2x+14) - (2x+6)\implies x+6 = 8\implies x =2\implies \boxed{2}$ | B | 2 |
d71137fcf98b1d5b4935e432e30ef94f | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_13 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for seats 2 and 3. This means that either Edie or Dee was originally in an edge seat. If Edie and Dee were in seats 1 and 2, then Bea must have been in seat 3, which would mean that seat 5 would now be occupied and the positioning would not work. So, Edie and Dee are in seats 4 and 5. This means that Bea was originally in seat 1. Ceci must have been in seat 3 to keep seat 1 open, which leaves seat 2.
Thus, Ada was in seat $\boxed{2}$ | B | 2 |
d71137fcf98b1d5b4935e432e30ef94f | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_13 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Note that the person (out of A,B,C) that moves the most, moves the amount equal to the sum of what the other 2 move. They essentially make a cycle. D & E are seat fillers and can be ignored. A,B,C take up either seats 1,2,3 or 2,4,5. In each case you find A was originally in seat $\boxed{2}$ | null | 2 |
d71137fcf98b1d5b4935e432e30ef94f | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_13 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Note that the net displacements in the right direction sum up to $0$ . The sum of the net displacements of Bea, Ceci, Dee, Edie is $2-1 = 1$ , so Ada moved exactly $1$ place to the left. Since Ada ended on an end seat, she must have started on seat $\boxed{2}$ | null | 2 |
659ac53dc6de527844e9e9812ac8a09b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_14 | How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$ | The amount of twos in our sum ranges from $0$ to $1008$ , with differences of $3$ because $2 \cdot 3 = \operatorname{lcm}(2, 3)$
The possible amount of twos is $\frac{1008 - 0}{3} + 1 \Rightarrow \boxed{337}$ | C | 337 |
659ac53dc6de527844e9e9812ac8a09b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_14 | How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$ | You can also see that you can rewrite the word problem into an equation $2x + 3y$ $2016$ . Therefore the question is just how many multiples of $3$ subtracted from 2016 will be an even number. We can see that $x = 1008$ $y = 0$ all the way to $x = 0$ , and $y = 672$ works, with $y$ being incremented by $2$ 's.Therefore, between $0$ and $672$ , the number of multiples of $2$ is $\boxed{337}$ | C | 337 |
659ac53dc6de527844e9e9812ac8a09b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_14 | How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$ | We can utilize the stars-and-bars distribution technique to solve this problem.
We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have $\binom{2016-1}{2-1}$ "total" answers. We want every third x and second y, so we divide our previous total by 6, which will result in $2015/6$ . We have to round down to the nearest integer, and we have to add 2 because we did not consider the 2 solutions involving x or y being 0.
So, $2015/6\implies335\implies335+2=\boxed{337}$ | C | 337 |
659ac53dc6de527844e9e9812ac8a09b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_14 | How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$ | Note that $2016 = 6 \cdot 336$ . In other words, we can write $2016$ as the sum of $336$ sixes.
In turn, we can express each $6$ as either $2 + 2 + 2$ or $3 + 3$
Therefore, we can write $2016$ as $n (2 + 2 + 2) + (336 - n) (3 + 3)$ , where $n$ is an integer between $0$ and $336$ , inclusive. Since each value of $n$ corresponds to a unique way to write the sum, we get $336 + 1 = \boxed{337}$ | C | 337 |
9185495aa820991bb2ef08c580f53ab2 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_15 | Seven cookies of radius $1$ inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?
[asy] draw(circle((0,0),3)); draw(circle((0,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((-1,sqrt(3)),1)); draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); draw(circle((2,0),1)); draw(circle((-2,0),1)); [/asy]
$\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi$ | The big cookie has radius $3$ , since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is $3^2\pi-7\pi=9\pi-7\pi=2 \pi$ . The scrap cookie has this area, so its radius must be $\boxed{2}$ | A | 2 |
10595b3197aedf1fde50cfcac9dd084c | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_17 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$ | Let $n = \frac{N}{5}$ . Then, consider $5$ blocks of $n$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $N + 1$ positions between the green balls to insert the red ball. Less than $\frac{3}{5}$ of the green balls will be on the same side of the red ball if the red ball is inserted inside the middle block of $n$ balls, and there are $n - 1$ positions where this happens. Thus, $P(N) = 1 - \frac{n - 1}{N + 1} = \frac{4n + 2}{5n + 1}$ , so
\[P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.\]
Multiplying both sides of the inequality by $400(5n+1)$ , we have
\[400(4n+2)<321(5n+1),\]
and by the distributive property,
\[1600n+800<1605n+321.\]
Subtracting $1600n+321$ on both sides of the inequality gives us
\[479<5n.\]
Therefore, $N=5n>479$ , so the least possible value of $N$ is $480$ . The sum of the digits of $480$ is $\boxed{12}$ | A | 12 |
10595b3197aedf1fde50cfcac9dd084c | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_17 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$ | Let $N=5$ $P(N)=1$ (Given)
Let $N=10$ $P(N)=\frac{10}{11}$
Let $N=15$ $P(N)=\frac{14}{16}$
Notice that the fraction can be written as $1-\frac{\frac{N}{5}-1}{N+1}$
Now it's quite simple to write the inequality as $1-\frac{\frac{N}{5}-1}{N+1}<\frac{321}{400}$
We can subtract $1$ on both sides to obtain $-\frac{\frac{N}{5}-1}{N+1}<-\frac{79}{400}$
Dividing both sides by $-1$ , we derive $\frac{\frac{N}{5}-1}{N+1}>\frac{79}{400}$ . (Switch the inequality sign when dividing by $-1$
We then cross multiply to get $80N - 400 > 79N + 79$
Finally we get $N > 479$
To achieve $N = 480$
So the sum of the digits of $N$ $\boxed{12}$ | A | 12 |
10595b3197aedf1fde50cfcac9dd084c | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_17 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$ | We are trying to find the number of places to put the red ball, such that $\frac{3}{5}$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $N$ : Trying a few values, we see that the ball "works" in places $1$ to $\frac{2}{5}N + 1$ and spaces $\frac{3}{5}N+1$ to $N+1$ . This is a total of $\frac{4}{5}N + 2$ spaces, over a total possible $N + 1$ places to put the ball. So:
$\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.$ And we know that the next value is what we are looking for, so $N+1 = 480$ , and the sum of its digits is $\boxed{12}$ | A | 12 |
3c288b5d6747970212327a628ab58158 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_18 | Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$ | Note that the sum of the numbers on each face must be 18, because $\frac{1+2+\cdots+8}{2}=18$
So now consider the opposite edges (two edges which are parallel but not on same face of the cube);
they must have the same sum value too.
Now think about the points $1$ and $8$ . If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$ . However, this scenario would yield no solution for $[7,2]$ , which is a contradiction. (Try drawing out the cube if it doesn't make sense to you.)
The points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9$
Now we have $4$ parallel sides $1-8, 2-7, 3-6, 4-5$ .
Thinking about $4$ endpoints, we realize they need to sum to $18$ .
It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.
So if we fix one direction $1-8 ($ or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$
Now, the problem is the same as arranging $4$ points in a two-dimensional square, which is $\frac{4!}{4}=\boxed{6.}$ | C | 6. |
3c288b5d6747970212327a628ab58158 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_18 | Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$ | Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to $1$ , then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x<y<z.$ \[3,7,8\text{: Does not work.}\] \[4,6,8\text{: Works.}\] \[4,7,8\text{: Works.}\] \[5,6,7\text{: Does not work.}\] \[5,6,8\text{: Does not work.}\] \[5,7,8\text{: Does not work.}\] \[6,7,8\text{: Works.}\]
Keeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with $[x, y, z]$ and one with $[z, y, x]$ ) for each combination of $x$ $y$ , and $z$ from above. So, our answer is $3\cdot 2=\boxed{6.}$ | C | 6. |
3c288b5d6747970212327a628ab58158 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_18 | Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$ | We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$ , it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same edge, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly, if $8$ and $7$ were on the same edge, the only way to get the sum is with $1$ and $2$ . This means that $6$ and $7$ are not on the same edge as $8$ , or in other words they are diagonally across from it on the same face, or on the other end of the cube.
Now we look at three cases, each yielding two solutions which are reflections of each other:
1) $6$ and $7$ are diagonally opposite $8$ on the same face.
2) $6$ is diagonally across the cube from $8$ , while $7$ is diagonally across from $8$ on the same face.
3) $7$ is diagonally across the cube from $8$ , while $6$ is diagonally across from $8$ on the same face.
This means the answer is $3\cdot 2=\boxed{6.}$ | C | 6. |
00d57b859a5b81a728a5de00a263ab06 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$
$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$ | [asy] size(6cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); label("$Q$", extension(A,(6,1),B,D),dir(-90)); label("$P$", extension(A,(6,2),B,D), dir(90)); label("$F$", (6,1), dir(0)); label("$E$", (6,2), dir(0)); [/asy]
Use similar triangles. Our goal is to put the ratio in terms of ${BD}$ . Since $\triangle APD \sim \triangle EPB,$ $\frac{DP}{PB}=\frac{AD}{BE}=3.$ Therefore, $PB=\frac{BD}{4}$ . Similarly, $\frac{DQ}{QB}=\frac{3}{2}$ . This means that ${DQ}=\frac{3\cdot BD}{5}$ . Therefore, $r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,$ so $r+s+t=\boxed{20.}$ | E | 20. |
00d57b859a5b81a728a5de00a263ab06 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$
$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$ | [asy] size(9cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3), G=(2, 1), H=(9, 0); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); draw(A--H); draw((6,1)--G); draw(D--H); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); label("$Q$", extension(A,(6,1),B,D),dir(-90)); label("$P$", extension(A,(6,2),B,D), dir(90)); label("$F$", (6,1), dir(45)); label("$E$", (6,2), dir(45)); label("$H$", H, dir(0)); label("$G$", G, dir(135)); [/asy]
This problem breaks down into finding $QP:PB$ and $DQ:QB$ . We can find the first using mass points, and the second using similar triangles.
Draw point $G$ on $DB$ such that $FG\parallel CD$ . Then, by similar triangles $FG=BF\cdot 2=4$ . Again, by similar triangles $AQB$ and $FQG$ $AQ:FQ=AB:FG=6:4=3:2$ . Now we begin Mass Points.
We will consider the triangle $ABF$ with center $P$ , so that $E$ balances $B$ and $F$ , and $Q$ balances $A$ and $F$ . Assign a mass of $1$ to $B$ . Then, $BE:FE=1:1$ so $F=B=1$ . By mass points addition, $E=B+F=2$ since $E$ balances $B$ and $F$ .
Also, $AQ:QF=3:2$ so $A=\frac{2}{3}F=\frac{2}{3}$ so $Q=\frac{5}{3}$ . Then, $QP:PB=1:\frac{5}{3}=3:5$
To calculate $DQ:BQ$ , extend $AF$ past $F$ to point $H$ such that $H$ lies on $BC$ . Then $AFB$ is similar to $HAD$ so $DH=3\cdot AD=9$ . Also, $AQB$ is similar to $HQD$ so $DQ:BQ=9:6=3:2$
Now, we wish to get $DQ:QP:PB$ . Observe that $BQ=QP+PB$ . So, $DQ:BQ=DQ:QP+PB=3:2$ so (since $QP:PB=3:5$ has sum $3+5=8$ ), $DQ:QP+PB=3:2=12:8$ . now, we may combine the two and get $DQ:QP:PB=12:3:5$ so $r+s+t=12+3+5=\boxed{20}$ | E | 20 |
00d57b859a5b81a728a5de00a263ab06 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$
$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$ | We can set coordinates for the points. $D=(0,0), C=(6,0), B=(6,3),$ and $A=(0,3)$ . The line $BD$ 's equation is $y = \frac{1}{2}x$ , line $AE$ 's equation is $y = -\frac{1}{6}x + 3$ , and line $AF$ 's equation is $y = -\frac{1}{3}x + 3$ . Adding the equations of lines $BD$ and $AE$ , we find that the coordinates of $P$ are $\left(\frac{9}{2},\frac{9}{4}\right)$ . Furthermore we find that the coordinates of $Q$ are $\left(\frac{18}{5}, \frac{9}{5}\right)$ . Using the Pythagorean Theorem , we get that the length of $QD$ is $\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}$ , and the length of $DP$ is $\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.$ $PQ = DP - DQ = \frac{9\sqrt{5}}{4} - \frac{9\sqrt{5}}{5} = \frac{9\sqrt{5}}{20}.$ The length of $DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}$ . Then $BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.$ The ratio $BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.$ Then $r, s,$ and $t$ is $5, 3,$ and $12$ , respectively. The problem tells us to find $r + s + t$ , so $5 + 3 + 12 = \boxed{20}$ ~ minor LaTeX edits by dolphin7 | E | 20 |
00d57b859a5b81a728a5de00a263ab06 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$
$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$ | Extend $AF$ to meet $CD$ at point $T$ . Since $FC=1$ and $BF=2$ $TC=3$ by similar triangles $\triangle TFC$ and $\triangle AFB$ . It follows that $\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}$ . Now, using similar triangles $\triangle BEP$ and $\triangle DAP$ $\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}$ . WLOG let $BP=1$ . Solving for $PQ, QD$ gives $PQ=\frac{3}{5}$ and $QD=\frac{12}{5}$ . So our desired ratio is $5:3:12$ and $5+3+12=\boxed{20}$ | E | 20 |
00d57b859a5b81a728a5de00a263ab06 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$
$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$ | Draw line segment $AC$ , and call the intersection between $AC$ and $BD$ point $K$ . In $\delta ABC$ , observe that $BE:EC=1:2$ and $AK:KC=1:1$ . Using mass points, find that $BP:PK=1:1$ . Again utilizing $\delta ABC$ , observe that $BF:FC=2:1$ and $AK:KC=1:1$ . Use mass points to find that $BQ:QK=4:1$ . Now, draw a line segment with points $B$ $P$ $Q$ , and $K$ ordered from left to right. Set the values $BP=x$ $PK=x$ $BQ=4y$ and $QK=y$ . Setting both sides segment $BK$ equal, we get $y= \frac{2}{5}x$ . Plugging in and solving gives $QK= \frac{2}{5}x$ $PQ=\frac{3}{5}x$ $BP=x$ . The question asks for $BP:PQ:QD$ , so we add $2x$ to $QK$ and multiply the ratio by $5$ to create integers. This creates $5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12$ . This sums up to $3+5+12=\boxed{20}$ | E | 20 |
00d57b859a5b81a728a5de00a263ab06 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$
$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$ | We set coordinates for the points. Let $A=(0,3), B=(6,3), C=(6,0)$ and $D=(0,0)$ . Then the equation of line $AE$ is $y = -\frac{1}{6}x + 3,$ the equation of line $AF$ is $y = -\frac{1}{3}x + 3,$ and the equation of line $BD$ is $y = \frac{1}{2}x$ . We find that the x-coordinate of point $P$ is $\frac 9 2$ by solving $-\frac{1}{6}x + 3=\frac{1}{2}x.$ Similarly we find that the x-coordinate of point $Q$ is $\frac {18} 5$ by solving $-\frac{1}{3}x + 3=\frac{1}{2}x.$ It follows that $BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.$ Hence $r,s,t=5,3,12$ and $r+s+t=5+3+12=\boxed{20}.$ ~ Solution by dolphin7 | E | 20 |
9a56485e4c7008f86bb962519846d24c | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_20 | For some particular value of $N$ , when $(a+b+c+d+1)^N$ is expanded and like terms are combined, the resulting expression contains exactly $1001$ terms that include all four variables $a, b,c,$ and $d$ , each to some positive power. What is $N$
$\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | All the desired terms are in the form $a^xb^yc^zd^w1^t$ , where $x + y + z + w + t = N$ (the $1^t$ part is necessary to make stars and bars work better.)
Since $x$ $y$ $z$ , and $w$ must be at least $1$ $t$ can be $0$ ), let $x' = x - 1$ $y' = y - 1$ $z' = z - 1$ , and $w' = w - 1$ , so $x' + y' + z' + w' + t = N - 4$ . Now, we use stars and bars (also known as ball and urn) to see that there are $\binom{(N-4)+4}{4}$ or $\binom{N}{4}$ solutions to this equation. We notice that $1001=7\cdot11\cdot13$ , which leads us to guess that $N$ is around these numbers. This suspicion proves to be correct, as we see that $\binom{14}{4} = 1001$ , giving us our answer of $\boxed{14.}$ | B | 14. |
9a56485e4c7008f86bb962519846d24c | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_20 | For some particular value of $N$ , when $(a+b+c+d+1)^N$ is expanded and like terms are combined, the resulting expression contains exactly $1001$ terms that include all four variables $a, b,c,$ and $d$ , each to some positive power. What is $N$
$\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | By the Hockey Stick Identity , the number of terms that have all $a,b,c,d$ raised to a positive power is $\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}$ . We now want to find some $N$ such that $\binom{N}{4} = 1001$ . As mentioned above, after noticing that $1001 = 7\cdot11\cdot13$ , and some trial and error, we find that $\binom{14}{4} = 1001$ , giving us our answer of $\boxed{14.}$ | B | 14. |
9a56485e4c7008f86bb962519846d24c | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_20 | For some particular value of $N$ , when $(a+b+c+d+1)^N$ is expanded and like terms are combined, the resulting expression contains exactly $1001$ terms that include all four variables $a, b,c,$ and $d$ , each to some positive power. What is $N$
$\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | The terms are in the form $a^xb^yc^zd^w1^t$ , where $x + y + z + w + t = N$ . The problem becomes distributing $N$ identical balls to $5$ different boxes $(x, y, z, w, t)$ such that each of the boxes $(x, y, z, w)$ has at least $1$ ball. The $N$ balls in a row have $N-1$ gaps among them. We are going to put $4$ or $3$ divisors into those $N-1$ gaps. There are $2$ cases of how to put the divisors.
Case $1$ :
Put 4 divisors into $N-1$ gaps. It corresponds to each of $(a, b, c, d, 1)$ has at least one term. There are $\binom{N-1}{4}$ terms.
Case $2$ :
Put 3 divisors into $N-1$ gaps. It corresponds to each of $(a, b, c, d)$ has at least one term. There are $\binom{N-1}{3}$ terms.
So, there are $\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}$ terms. $\binom{N}{4} = 1001$ , and since we have $\binom{14}{4} = 1001, N=\boxed{14.}$ | B | 14. |
79e97a8c7f5c5b5675364345daa7f16e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_22 | For some positive integer $n$ , the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$ . How many positive integer divisors does the number $81n^4$ have?
$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$ | Since the prime factorization of $110$ is $2 \cdot 5 \cdot 11$ , we have that the number is equal to $2 \cdot 5 \cdot 11 \cdot n^3$ . This has $2 \cdot 2 \cdot 2=8$ factors when $n=1$ . This needs a multiple of 11 factors, which we can achieve by setting $n=2^3$ , so we have $2^{10} \cdot 5 \cdot 11$ has $44$ factors. To achieve the desired $110$ factors, we need the number of factors to also be divisible by $5$ , so we can set $n=2^3 \cdot 5$ , so $2^{10} \cdot 5^4 \cdot 11$ has $110$ factors. Therefore, $n=2^3 \cdot 5$ . In order to find the number of factors of $81n^4$ , we raise this to the fourth power and multiply it by $81$ , and find the factors of that number. We have $3^4 \cdot 2^{12} \cdot 5^4$ , and this has $5 \cdot 13 \cdot 5=\boxed{325}$ factors. | D | 325 |
79e97a8c7f5c5b5675364345daa7f16e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_22 | For some positive integer $n$ , the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$ . How many positive integer divisors does the number $81n^4$ have?
$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$ | $110n^3$ clearly has at least three distinct prime factors, namely 2, 5, and 11.
The number of factors of $p_1^{n_1}\cdots p_k^{n_k}$ is $(n_1+1)\cdots(n_k+1)$ when the $p$ 's are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without $1$ s is $2\cdot 5\cdot 11$
We conclude that $110n^3$ has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of $n$ all of the form $n=p_1\cdot p_2^3$
$81n^4$ thus has prime factorization $81n^4=3^4\cdot p_1^4\cdot p_2^{12}$ and a factor count of $5\cdot5\cdot13=\boxed{325}$ | D | 325 |
1787618c91fb9d5209db93104c08f05e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | We see that $a \, \diamondsuit \, a = 1$ , and think of division. Testing, we see that the first condition $a \, \diamondsuit \, (b \, \diamondsuit \, c) = (a \, \diamondsuit \, b) \cdot c$ is satisfied, because $\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c$ . Therefore, division can be the operation $\diamondsuit$ . Solving the equation, \[\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100\implies x=\frac{100}{336} = \frac{25}{84},\] so the answer is $25 + 84 = \boxed{109}$ | A | 109 |
1787618c91fb9d5209db93104c08f05e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | If the given conditions hold for all nonzero numbers $a, b,$ and $c$
Let $a=b=c.$ From the first two givens, this implies that
\[a\diamondsuit\, (a\diamondsuit\, {a})=(a\diamondsuit\, a)\cdot{a}.\]
From $a\diamondsuit\,{a}=1,$ this equation simply becomes $a\diamondsuit\,{1}=a.$
Let $c=b.$ Substituting this into the first two conditions, we see that
\[a\diamondsuit\, (b\diamondsuit\, {c})=(a\diamondsuit\, {b})\cdot{c} \implies a\diamondsuit\, (b\diamondsuit\, {b})=(a\diamondsuit\, {b})\cdot{b}.\]
Substituting $b\diamondsuit\, {b} =1$ , the second equation becomes
\[a\diamondsuit\, {1}=(a\diamondsuit\, {b})\cdot{b} \implies a=(a\diamondsuit\,{b})\cdot{b}.\]
Since $a, b$ and $c$ are nonzero, we can divide by $b$ which yields,
\[\frac{a}{b}=(a\diamondsuit\, {b}).\]
Now we can find the value of $x$ straightforwardly:
\[\frac{2016}{(\frac{6}{x})}=100 \implies 2016=\frac{600}{x} \implies x=\frac{600}{2016} = \frac{25}{84}.\]
Therefore, $a+b=25+84=\boxed{109}$ | A | 109 |
1787618c91fb9d5209db93104c08f05e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | One way to eliminate the $\diamondsuit$ in this equation is to make $a = b$ so that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = c$ . In this case, we can make $b = 2016$
\[2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\implies (2016\, \diamondsuit\, 6) \cdot x = 100\]
By multiplying both sides by $\frac{6}{x}$ , we get:
\[(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}\implies 2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}\]
Because $6\, \diamondsuit\, 6 = 2016\, \diamondsuit\, 2016 = 1:$
\[2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}\implies (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}\implies 2016 = \frac{600}{x}\]
Therefore, $x = \frac{600}{2016} = \frac{25}{84}$ , so the answer is $25 + 84 = \boxed{109.}$ | A | 109. |
1787618c91fb9d5209db93104c08f05e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | We can manipulate the given identities to arrive at a conclusion about the binary operator $\diamondsuit$ . Substituting $b = c$ into the first identity yields \[( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\ b) = a\ \diamondsuit\ 1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a.\] Hence, $( a\ \diamondsuit\ b) \cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\ \diamondsuit\ b) = \frac{a}{b}.$
Hence, the given equation becomes $\frac{2016}{\frac{6}{x}} = 100$ . Solving yields $x=\frac{100}{336} = \frac{25}{84},$ so the answer is $25 + 84 = \boxed{109.}$ | A | 109. |
1787618c91fb9d5209db93104c08f05e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | $2016 \diamondsuit (6 \diamondsuit x) = (2016 \diamondsuit 6) \cdot x = 100$
$2016 \diamondsuit (2016 \diamondsuit 1) = (2016 \diamondsuit 2016) \cdot 1 = 1 \cdot 1 = 1$
$2016 \diamondsuit 2016 = 1$ $2016 \diamondsuit (2016 \diamondsuit 1) = 1$ , so $2016 \diamondsuit 1 = 2016$
$2016 \diamondsuit 1 = (2016 \diamondsuit 6) \cdot 6$ $2016 \diamondsuit 6 = \frac{2016 \diamondsuit 1}{6} = 336$
$x = \frac{100}{2016 \diamondsuit 6} = \frac{100}{336} = \frac{25}{84}$ $24 + 85 = \boxed{109}$ | A | 109 |
1787618c91fb9d5209db93104c08f05e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | Notice that $2016 \diamondsuit (6 \diamondsuit 6)=(2016 \diamondsuit 6) \cdot 6 = 2016$ . Hence, $2016 \diamondsuit 6 = 336$ . Thus, $2016 \diamondsuit (6 \diamondsuit x)=100 \implies (2016 \diamondsuit 6) \cdot x = 100 \implies 336x=100 \implies x=\frac{25}{84}$ . Therefore, the answer is $\boxed{109}$ | A | 109 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("$E$",E,NE); F=extension(C,O,A,D); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]
Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$ , thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$ . Since $\frac{OA}{AB} = \frac{OB}{AE}$ , we have that $AB = AE$ . Similarly, $CD = DF$
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{500}$ | E | 500 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(B,D,O,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$E$",E,WSW); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(B--D); draw(rightanglemark(C,E,D)); [/asy]
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Let the intersection of $BD$ and $OC$ be point $E$ . Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.
We set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem, \[\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}\]
We solve for $x$ \[1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2\] \[2\sqrt{(1-x^2)(2-x^2)}=2x^2-1\] \[4(1-x^2)(2-x^2)=(2x^2-1)^2\] \[8-12x^2+4x^4=4x^4-4x^2+1\] \[8x^2=7\] \[x=\frac{\sqrt{14}}{4}\]
By Ptolemy's Theorem \[AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2\]
Substituting values, \[1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2\] \[1+AD=\frac{7}{2}\] \[AD=\frac{5}{2}\]
Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{500}$ | E | 500 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); [/asy]
Let quadrilateral $ABCD$ be inscribed in circle $O$ , where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$
By the Pythagorean Theorem, the length of $OH$ is \begin{align*} \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} \\ &= \sqrt{80000 - 10000} \\ &= \sqrt{70000} \\ &= 100\sqrt{7}. \end{align*}
Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$ ; then we have that
$[AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$
Furthermore, \begin{align*} h &= \sqrt{OD^2 - GD^2} \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} \\ &= \sqrt{80000 - \frac{x^2}{4}} \end{align*}
Substituting this value of $h$ into the previous equation and evaluating for $x$ , we get: \[\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}\] \[\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}\] \[60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)\] \[40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}\] \[400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}\] \[(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}\] \[7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)\] \[7x^2 - 5600x + 1120000 = 320000 - x^2\] \[8x^2 - 5600x + 800000 = 0\] \[x^2 - 700x + 100000 = 0\]
The roots of this quadratic are found by using the quadratic formula: \begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}
If the length of $AD$ is $200$ , then $ABCD$ would be a square. Thus, the radius of the circle would be \[\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}\] Which is a contradiction. Therefore, our answer is $\boxed{500}.$ | null | 500 |
026f196b9b8ea005ebfc988e15ba8397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); [/asy]
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Apply the law of cosines on $\Delta BOC$ ; let $\theta = \angle BOC$ . We get the following equation: \[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta\] Substituting the values in, we get \[(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta\] Canceling out, we get \[\cos\theta=\frac{3}{4}\] Because $\angle AOB$ $\angle BOC$ , and $\angle COD$ are congruent, $\angle AOD = 3\theta$ . To find the remaining side ( $AD$ ), we simply have to apply the law of cosines to $\Delta AOD$ . Now, to find $\cos 3\theta$ , we can derive a formula that only uses $\cos\theta$ \[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta\] \[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)\] \[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\] It is useful to memorize the triple angle formulas ( $\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta$ ). Plugging in $\cos\theta=\frac{3}{4}$ , we get $\cos 3\theta= -\frac{9}{16}$ . Now, applying law of cosines on triangle $OAD$ , we get \[(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}\] \[\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}\] \[AD=200 \cdot \frac{5}{2}=\boxed{500}\] | null | 500 |
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