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d623a36ca4900dd839c7f5e63fcae3a2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_5 | Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$ | We can notice that the number of purple candy times $20$ has to be divisible by $7$ , because of the $14$ green candies, and $3$ , because of the $12$ red candies. $7\cdot3=21$ , so the answer has to be $\boxed{21}$ | B | 21 |
e0b3c31612a40c13e296f56957dd64ab | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_6 | In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}$ | Notice that whatever point we pick for $C$ $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$ , since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point $C$ , we have to make sure that its $y$ -coordinate is $\pm20$ , because that's the only way the area of the triangle can be $100$
Now when the perimeter is minimized, by symmetry, we put $C$ in the middle, at $(5, 20)$ . We can easily see that $AC$ and $BC$ will both be $\sqrt{20^2+5^2} = \sqrt{425}$ . The perimeter of this minimal triangle is $2\sqrt{425} + 10$ , which is larger than $50$ . Since the minimum perimeter is greater than $50$ , there is no triangle that satisfies the condition, giving us $\boxed{0}$ | A | 0 |
e0b3c31612a40c13e296f56957dd64ab | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_6 | In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}$ | Without loss of generality, let $AB$ be a horizontal segment of length $10$ . Now realize that $C$ has to lie on one of the lines parallel to $AB$ and vertically $20$ units away from it. But $10+20+20$ is already 50, and this doesn't form a triangle. Otherwise, without loss of generality, $AC<20$ . Dropping altitude $CD$ , we have a right triangle $ACD$ with hypotenuse $AC<20$ and leg $CD=20$ , which is clearly impossible, again giving the answer as $\boxed{0}$ | A | 0 |
fd474a661a8c8474aa5000a0cf177da2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_7 | What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?
$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}$ | The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$
There are three possibilities for the median: it is either $6$ $8$ , or $x$
Let's start with $6$
$\frac{35+x}{5}=6$ has solution $x=-5$ , and the sequence is $-5, 4, 6, 8, 17$ , which does have median $6$ , so this is a valid solution.
Now let the median be $8$
$\frac{35+x}{5}=8$ gives $x=5$ , so the sequence is $4, 5, 6, 8, 17$ , which has median $6$ , so this is not valid.
Finally we let the median be $x$
$\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75$ , and the sequence is $4, 6, 8, 8.75, 17$ , which has median $8$ . This case is therefore again not valid.
Hence the only possible value of $x$ is $\boxed{5}.$ | A | 5 |
f4e8b32a0cd1cc8c6ec0a3e98a607266 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_8 | Let $f(x) = x^{2}(1-x)^{2}$ . What is the value of the sum \[f \left(\frac{1}{2019} \right)-f \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?\]
$\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}\qquad\textbf{(C) }\frac{2018^{2}}{2019^{4}}\qquad\textbf{(D) }\frac{2020^{2}}{2019^{4}}\qquad\textbf{(E) }1$ | First, note that $f(x) = f(1-x)$ . We can see this since \[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\left(1-\left(1-x\right)\right)^{2} = f(1-x)\] Using this result, we regroup the terms accordingly: \[\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) + \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)\] \[= \left( f \left(\frac{1}{2019} \right) - f \left(\frac{1}{2019} \right) \right) + \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)\] Now it is clear that all the terms will cancel out (the series telescopes), so the answer is $\boxed{0}$ | A | 0 |
0a29a31da282e93226777482a0e0916a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_9 | For how many integral values of $x$ can a triangle of positive area be formed having side lengths $\log_{2} x, \log_{4} x, 3$
$\textbf{(A) } 57 \qquad\textbf{(B) } 59 \qquad\textbf{(C) } 61 \qquad\textbf{(D) } 62 \qquad\textbf{(E) } 63$ | For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need \[\log_2{x} + \log_4{x} > 3\] \[\log_2{x} + 3 > \log_4{x}\] \[\log_4{x} + 3 > \log_2{x}.\] The second inequality is redundant, as it's always less restrictive than the last inequality.
Let's raise the first inequality to the power of $4$ . This gives $4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow \left(2^2\right)^{\log_2{x}} \cdot x > 64 \Rightarrow x^2 \cdot x > 64$ . Thus, $x > 4$
Doing the same for the second inequality gives $4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64$ (where we are allowed to divide both sides by $x$ since $x$ must be positive in order for the logarithms given in the problem statement to even have real values).
Combining our results, $x$ is an integer strictly between $4$ and $64$ , so the number of possible values of $x$ is $64 - 4 - 1 = \boxed{59}$ | B | 59 |
0a29a31da282e93226777482a0e0916a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_9 | For how many integral values of $x$ can a triangle of positive area be formed having side lengths $\log_{2} x, \log_{4} x, 3$
$\textbf{(A) } 57 \qquad\textbf{(B) } 59 \qquad\textbf{(C) } 61 \qquad\textbf{(D) } 62 \qquad\textbf{(E) } 63$ | Using the triangle inequality, you get $\log_2{x}+\log_4{x} > 3$ . Solving for $x$ , you get $x > 4$ . Now we need an upper-bound for $x$ and since we're dealing with bases of $2$ and $4$ , we're looking for answer choices close to a power of $2$ and $4$ . All the answer choices seem to be around $64$ , and plugging that into the inequality $3+\log_4{x} > \log_2{x}$ we see $64$ is the correct number. Now we have $64 > x > 4$ and the number of integers in between is $64-4-1 = \boxed{59}$ --OGBooger | B | 59 |
612fcc7b9efaacd652de8f70c6dc8f0a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_10 | The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); [/asy]
How many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$ | Note that of the $12$ cities, $6$ of them ( $2$ on the top, $2$ on the bottom, and $1$ on each side) have $3$ edges coming into/out of them (i.e., in graph theory terms, they have degree $3$ ). Therefore, at least $1$ edge connecting to each of these cities cannot be used. Additionally, the same applies to the start and end points, since we don't want to return to them. Thus there are $6+2=8$ vertices that we know have $1$ unused edge, and we have $17-13=4$ unused edges to work with (since there are $17$ edges in total, and we must use exactly $13$ of them). It is not hard to find that there is only one configuration satisfying these conditions:
Note: $\text{X}$ s represent unused edges.
[asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); label("X", (0, 1.5)); label("X", (1.5, 2)); label("X", (1.5, 0)); label("X", (3, 0.5)); label("O", (1,1), NE); label("O", (2,1), NE); [/asy]
Observe that at each of the $2$ cities marked with an $\text{O}$ on a path, there are $2$ possibilities: we can either continue straight and cross back over the path later, or we can make a left turn, then turn right when we approach the junction again. This gives us a total of $2\cdot 2 = \boxed{4}$ valid paths. | E | 4 |
612fcc7b9efaacd652de8f70c6dc8f0a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_10 | The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); [/asy]
How many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$ | Let the bottom-left vertex be $(0,0)$ , and let each of the edges have length $1$ , so that all of the vertices are at lattice points.
Firstly, notice that for any vertex $V$ on the graph (other than $A$ or $L$ ), we can visit it at most $M(V) = \left\lfloor \frac{\text{deg}(V)}{2} \right\rfloor$ times, where $\text{deg}(V)$ is, as usual, the degree of $V$ (i.e. the number of edges coming into/out of $V$ ). This is because to visit any of these vertices, we would have to enter and exit it by two different edges, in order to avoid using any portion of a road more than once. (Those who know some graph theory will recognise this a well-known principle.) We will label each vertex with this number. [asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); label("1", (1,2), N); label("1", (2,2), N); label("1", (3,2), N); label("1", (0,1), W); label("2", (1,1), NW); label("2", (2,1), NW); label("1", (3,1), E); label("1", (0,0),S); label("1", (1,0),S); label("1", (2,0),S); [/asy] Additionally, notice that if we visit $n$ vertices (not necessarily distinct, i.e. counting a vertex which is visited twice as two vertices) along our path, we must traverse $n-1$ edges (this is quite easy to prove). Thus, if we want to traverse $13$ edges in total, we have to visit $14$ vertices. We must visit $A$ and $L$ , leaving $12$ vertices from the rest of the graph to visit.
If we sum the maximum numbers of visits to each vertex, we find that this is exactly equal to the $12$ found above. This means that we have to visit each vertex $M(V)$ times, and must traverse $2\cdot M(V)$ edges connected to each vertex. Specifically, we must traverse all $4$ of the edges connected to the two central vertices at $(1,1)$ and $(1,2)$ , as well as both edges connected to the $2$ corner vertices (excluding $A$ and $L$ ), and any $2$ edges connected to the other vertices along the outside edge of the rectangle.
With this information, we can now proceed by dividing into cases.
Case 1 : We first move down from $A$
In this case, we see that we must immediately move right to $(1,1)$ in order to traverse the edge from $(0,1)$ to $(1,1)$ , since we can never revisit $(0,1)$ . However, by the same logic, we must immediately move to $(0,0)$ . This is a contradiction, so there are no possible paths in this case.
Case 2 : We first move right from $A$
Similar to the last case, we see that we must immediately move to $(1,1)$ . By symmetry, we conclude that our last two moves must be $(2,1) \rightarrow (2,0) \rightarrow (3,0)$ . With this information, we have reduced the problem to traveling from $(1,1)$ to $(2,1)$ with the same constraints as before. We redraw the graph, removing the edges we have already used and updating our labels (note that $(1,1)$ and $(2,1)$ are still labeled with $2$ since we will pass through them twice, at the start and the end). Then, we remove the vertices with label $0$ and the edges we know we can never traverse, giving:
[asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if(j==1 && i==2) {continue;} if(j==2 && i==0) {continue;} pair A = (j,i); dot(A); } } draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((2,1)--(3,1)--(3,2)--(2,2)--cycle); draw((1,1)--(2,1)); label("$A$", (0,2), W); label("$L$", (3,0), E); label("1", (2,2), N); label("1", (3,2), N); label("1", (0,1), W); label("2", (1,1), NW); label("2", (2,1), NW); label("1", (3,1), E); label("1", (0,0),S); label("1", (1,0),S); [/asy] Now, it is clear that we must complete a cycle in the lower left square, return to $(1,1)$ , go to $(2,1)$ , and complete a cycle in the top right square, returning to $(2,1)$ .
There are two ways to traverse each cycle (clockwise and anti-clockwise), giving us a total of $2\cdot 2 = \boxed{4}$ paths of length 13 from $A$ to $L$ | E | 4 |
612fcc7b9efaacd652de8f70c6dc8f0a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_10 | The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); [/asy]
How many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$ | Observe that only the two central vertices can be visited twice. Since the path is of length 13, we need to repeat a vertex. Using casework on each vertex, we can find there are two paths that go through each central vertex twice, for an answer of $\boxed{4}.$ | null | 4 |
612fcc7b9efaacd652de8f70c6dc8f0a | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_10 | The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); [/asy]
How many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$ | Looking at the answer choices we see that it would probably be easy enough to count the different paths. After some experimenting, we find that we must cross one of the middle vertexes twice, making counting easier.
We finally find that there are $\boxed{4}$ paths that satisfy the conditions. Luckily, $4$ is the largest answer choice so we know it must be correct. | null | 4 |
aa4975611de8ccfe8699665dda420591 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_11 | How many unordered pairs of edges of a given cube determine a plane?
$\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66$ | Without loss of generality, choose one of the $12$ edges of the cube to be among the two selected. We now calculate the probability that a randomly-selected second edge makes the pair satisfy the condition in the problem statement.
For two lines in space to determine a common plane, they must either intersect or be parallel (in other words, they cannot be skew lines). If all $12$ line segments are extended to lines, the first (arbitrarily chosen) edge's line intersects $4$ lines and is parallel to another $3$ . Thus $4+3=7$ of the $12-1=11$ remaining line segments (which could be chosen for the second edge) give a pair of lines determining a common plane. To see this, observe that in the diagram below, the red edge is parallel to the $3$ green edges and intersects with the $4$ blue edges.
[asy] import three; import three; unitsize(1cm); size(200); currentprojection=perspective(-6/5,-8/5,7/8); draw((0,1,0)--(0,0,0)--(1,0,0), blue); draw((1,0,0)--(1,1,0)--(0,1,0)); draw((0,0,0)--(0,0,1), red); draw((0,1,0)--(0,1,1), green); draw((1,1,0)--(1,1,1), green); draw((1,0,0)--(1,0,1), green); draw((0,1,1)--(0,0,1)--(1,0,1), blue); draw((1,0,1)--(1,1,1)--(0,1,1)); [/asy]
This means that the probability that a randomly-selected pair of edges determine a plane is $\frac{7}{11}$ , and we calculate that there are ${12 \choose 2} = 66$ total pairs of edges that could be chosen (without the restriction). Thus the answer is $\frac{7}{11} \cdot 66 =\boxed{42}$ | D | 42 |
aa4975611de8ccfe8699665dda420591 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_11 | How many unordered pairs of edges of a given cube determine a plane?
$\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66$ | As in Solution 1, we observe that the two edges must either be parallel or intersect. Clearly the edges will intersect if and only if they are part of the same face. We can thus divide into two cases:
Case 1 : The two edges are part of the same face. There are $6$ faces, and ${4 \choose 2}=6$ ways to choose $2$ of the $4$ edges of the square, giving a total of $6 \cdot 6 = 36$ possibilities.
Case 2 : The two edges are parallel and not part of the same face. Observe that each of the $12$ edges is parallel to exactly $1$ edge that is not part of its face. The edges can thus be paired up, giving $\frac{12}{2} = 6$ possibilities for this case.
Adding the two cases, the answer is hence $36 + 6 = \boxed{42}$ | D | 42 |
aa4975611de8ccfe8699665dda420591 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_11 | How many unordered pairs of edges of a given cube determine a plane?
$\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66$ | As in solution 1, we see that for an arbitrary edge of the cube, only $7$ of the other edges can be chosen to form a plane.
For each of the $12$ edges, this gives $(7)(12) = 84$ total possible pairings. However each pair has been counted twice (e.g edge $A$ with $B$ , and edge $B$ with $A$ ) so we divide by 2 to get $\boxed{42}$ | D | 42 |
6878a36bf1fd2f542bf58438566a3694 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_14 | Let $S$ be the set of all positive integer divisors of $100,000.$ How many numbers are the product of two distinct elements of $S?$
$\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$ | The prime factorization of $100,000$ is $2^5 \cdot 5^5$ . Thus, we choose two numbers $2^a5^b$ and $2^c5^d$ where $0 \le a,b,c,d \le 5$ and $(a,b) \neq (c,d)$ , whose product is $2^{a+c}5^{b+d}$ , where $0 \le a+c \le 10$ and $0 \le b+d \le 10$
Notice that this is similar to choosing a divisor of $100,000^2 = 2^{10}5^{10}$ , which has $(10+1)(10+1) = 121$ divisors. However, some of the divisors of $2^{10}5^{10}$ cannot be written as a product of two distinct divisors of $2^5 \cdot 5^5$ , namely: $1 = 2^05^0$ $2^{10}5^{10}$ $2^{10}$ , and $5^{10}$ . The last two cannot be written because the maximum factor of $100,000$ containing only $2$ s or $5$ s (and not both) is only $2^5$ or $5^5$ . Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require $2^5 \cdot 2^5$ or $5^5 \cdot 5^5$ . The first two would require $1 \cdot 1$ and $2^{5}5^{5} \cdot 2^{5}5^{5}$ , respectively. This gives $121-4 = 117$ candidate numbers. It is not too hard to show that every number of the form $2^p5^q$ , where $0 \le p, q \le 10$ , and $p,q$ are not both $0$ or $10$ , can be written as a product of two distinct elements in $S$ . Hence the answer is $\boxed{117}$ | C | 117 |
9f73c3b623378fb5c5d99ffc4f61679c | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_15 | As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\overarc{AEB},\overarc{BFC},$ and $\overarc{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F.$ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form \[\frac{a}{b}\cdot\pi-\sqrt{c}+d,\] where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d$
[asy] size(6cm); filldraw(circle((0,0),2), grey); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label("$A$",(-3,-1),S); dot((-2,0)); label("$E$",(-2,0),NW); dot((-1,-1)); label("$B$",(-1,-1),S); dot((0,0)); label("$F$",(0,0),N); dot((1,-1)); label("$C$",(1,-1), S); dot((2,0)); label("$G$", (2,0),NE); dot((3,-1)); label("$D$", (3,-1), S); [/asy]
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$ | Divide the circle into four parts: the top semicircle by connecting E, F, and G( $A$ ); the bottom sector ( $B$ ), whose arc angle is $120^{\circ}$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$ , giving a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle; the triangle formed by the radii of $A$ and the chord ( $C$ ); and the four parts which are the corners of a circle inscribed in a square ( $D$ ). Then the area is $A + B - C + D$ (in $B-C$ , we find the area of the bottom shaded region, and in $D$ we find the area of the shaded region above the semicircles but below the diameter).
The area of $A$ is $\frac{1}{2} \pi \cdot 2^2 = 2\pi$
The area of $B$ is $\frac{120^{\circ}}{360^{\circ}} \pi \cdot 2^2 = \frac{4\pi}{3}$
For the area of $C$ , the radius of $2$ , and the distance of $1$ (the smaller semicircles' radius) to $BC$ , creates two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, so $C$ 's area is $2 \cdot \frac{1}{2} \cdot 1 \cdot \sqrt{3} = \sqrt{3}$
The area of $D$ is $4 \cdot 1-\frac{1}{4}\pi \cdot 2^2=4-\pi$
Hence, finding $A+B-C+D$ , the desired area is $\frac{7\pi}{3}-\sqrt{3}+4$ , so the answer is $7+3+3+4=\boxed{17}$ | E | 17 |
9f73c3b623378fb5c5d99ffc4f61679c | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_15 | As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\overarc{AEB},\overarc{BFC},$ and $\overarc{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F.$ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form \[\frac{a}{b}\cdot\pi-\sqrt{c}+d,\] where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d$
[asy] size(6cm); filldraw(circle((0,0),2), grey); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label("$A$",(-3,-1),S); dot((-2,0)); label("$E$",(-2,0),NW); dot((-1,-1)); label("$B$",(-1,-1),S); dot((0,0)); label("$F$",(0,0),N); dot((1,-1)); label("$C$",(1,-1), S); dot((2,0)); label("$G$", (2,0),NE); dot((3,-1)); label("$D$", (3,-1), S); [/asy]
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$ | First we have to solve the area of the non-shaded area(the semicircles) that are in Circle $F$ .The middle semicircle has area $\frac12\pi$ and the other two have about half of their are inside the circle = $\frac14\pi\ + \frac14\pi\ + \frac12\pi\ = \pi$ . Then we subtract the part of the quartercircle that isn't in Circle $F$ . This is an area equal to that of a triangle minus an minor segment. The height of the triangle is the radius of the semicircles, which is $1$ . The length is the radius of Circle $F$ minus the length from the center of the middle semicircle up to until it is on the edge of the circle. Using the Pythagorean Theorem we can figure out that the length is: \[\sqrt{2^2 - 1^2} = \sqrt{3}.\] This means that the length of the triangle is $2 - \sqrt{3}$ and so the area of the triangle is $\frac{2 - \sqrt{3}}{2}$ . For the area of the segment, it's the area of the sector minus the area of the triangle. The triangle's length is the radius of $F$ $2$ , while its height is the radius of the semicircles: $1$ , so the area is 1. The angle is $30^{\circ}$ as the hypotenuse is the radius of $F$ and the opposite side is the radius of the semicircles, which means the area is $\frac{1}{12}$ of the whole area, which is $4\pi$ so the area of the sector is $\frac{\pi}{3}$ and the area of the segment is $\frac{\pi}{3} - 1$ and so the area of the part of the quartercircles that stick out of Circle $F$ is: \[(\frac{2 - \sqrt{3}}{2})-(\frac{\pi}{3} - 1) = \frac{2 - \sqrt{3}}{2} + 1 - \frac{\pi}{3} = \frac{4 - \sqrt{3}}{2} - \frac{\pi}{3}.\]
Since there are two, one for each side, we have to multiply it by 2, so we have ${4 - \sqrt{3}} - \frac{2\pi}{3}$ , which we subtract from $\pi$ which gets us $\frac{5\pi}{3} - 4 + \sqrt{3}$ which we subtract from $4\pi$ $=$ $\frac{12\pi}{3}$ , which is $\frac{7\pi}{3} + 4 - \sqrt{3}$ so we get $7+3+4+3=\boxed{17}$ | E | 17 |
71908b00fb27608f596e79c8599dc08b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_17 | How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}$ | Convert $z$ and $z^3$ into modulus-argument (polar) form, giving $z=r\text{cis}(\theta)$ for some $r$ and $\theta$ . Thus, by De Moivre's Theorem, $z^3=r^3\text{cis}(3\theta)$ . Since the distance from $0$ to $z$ is $r$ , and the triangle is equilateral, the distance from $0$ to $z^3$ must also be $r$ , so $r^3=r$ , giving $r=1$ . (We know $r \neq 0$ since the problem statement specifies that $z$ must be nonzero.)
Now, to get from $z$ to $z^3$ , which should be a rotation of $60^{\circ}$ if the triangle is equilateral, we multiply by $z^2 = r^2\text{cis}(2\theta)$ , again using De Moivre's Theorem. Thus we require $2\theta=\pm\frac{\pi}{3} + 2\pi k$ (where $k$ can be any integer). If $0 < \theta < \frac{\pi}{2}$ , we must have $\theta=\frac{\pi}{6}$ , while if $\frac{\pi}{2} \leq \theta < \pi$ , we must have $\theta = \frac{5\pi}{6}$ . Hence there are $2$ values that work for $0 < \theta < \pi$ . By symmetry, the interval $\pi \leq \theta < 2\pi$ will also give $2$ solutions. The answer is thus $2 + 2 = \boxed{4}$ | D | 4 |
71908b00fb27608f596e79c8599dc08b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_17 | How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}$ | As before, $r=1$ . Represent $z$ in polar form. By De Moivre's Theorem, $z^3=\text{cis}(3\theta)$ . To form an equilateral triangle, their difference in angle must be $\frac{\pi}{3}$ , so \[\frac{\text{cis}(3\theta)}{\text{cis}(\theta)}=\text{cis}(2\theta)=\text{cis}(\pm\frac{\pi}{3})\] From the polar form of $z$ , we know that $0\geq\theta\leq2\pi$ , so $\text{cis}(2\theta)$ cycles in a circle twice. By contrast, $\pm\frac{\pi}{3}$ represent $2$ fixed, distinct points. Thus, $\text{cis}(2\theta)$ intersects these points twice each $\implies\boxed{4}$ | D | 4 |
71908b00fb27608f596e79c8599dc08b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_17 | How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}$ | For the triangle to be equilateral, the vector from $z$ to $z^3$ , i.e $z^3 - z$ , must be a $60^{\circ}$ rotation of the vector from $0$ to $z$ , i.e. just $z$ . Thus we must have
\[\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)\]
Simplifying gives \[z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)\] so \[z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)\]
Since any nonzero complex number will have two square roots, each equation gives two solutions. Thus, as before, the total number of possible values of $z$ is $\boxed{4}$ | D | 4 |
71908b00fb27608f596e79c8599dc08b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_17 | How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}$ | Since the complex numbers $0,z,$ and $z^3$ form an equilateral triangle in the complex plane, we note that either $z^3$ is a 60 degrees counterclockwise rotation about the origin from $z$ or $z$ is a 60 degrees counterclockwise rotation about the origin from $z^3$
Therefore, we note that either $z^3 = z \text{cis} 60^\circ{}$ or $z \text{cis}(-60^\circ{}) = z^3$
The first equation in $z$ (meaning $z^3 = z \text{cis} 60^\circ{}$ ) gives us: $z^2 = cis 60^\circ{}$ , which gives 2 solutions in $z$
The second equation in $z$ (which is $z \text{cis} (-60^\circ{}) = z^3$ ) gives us $z^2 = \text{cis} (-60^\circ{})$ , which must give another 2 solutions in $z$
Therefore, there are $\boxed{4}$ solutions for $z$ . (Professor-Mom) | D | 4 |
71908b00fb27608f596e79c8599dc08b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_17 | How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}$ | Let the $z = re^{i\theta}$ , so $z^3 = r^3e^{3i\theta}$ . To have an equilateral triangle, we must have $|z^3| = |z|$ , so $r^3 = r$ , so $r= 1$
Note that the angle between $z^3$ and $z$ is $3\theta - \theta = 2\theta$ . Then, by the Law of Cosines,
\[|z^3 - z|^2 = |z|^2 + |z^3|^2 - 2|z||z^3|\cos 2\theta.\]
Since we have an equilateral triangle, it must be that $|z^3 - z| =|z| = |z^3| = r = 1$ . Hence, \[1 = 2 - 2\cos 2\theta = 2 - 2(1-2\sin^2\theta) = 4\sin^2\theta,\] \[\sin^2\theta = \frac{1}{4}\] \[\sin\theta = \pm\frac{1}{2}\] \[\theta = \pm\frac{\pi}{6}, \pm\frac{5\pi}{6}.\]
These $4$ values of $\theta$ correspond to $\boxed{4}$ distinct values of $z$ .
~ brainfertilzer | D | 4 |
be8d1d0bf9a8fa3cf4be1d034cf2005f | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_21 | How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is $ax^2+bx+c,a\neq 0,$ and the roots are $r$ and $s,$ then the requirement is that $\{a,b,c\}=\{r,s\}$ .)
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}$ | Firstly, if $r=s$ , then $a=b=c$ , so the equation becomes $ax^2 + ax + a = 0 \Rightarrow x^2 + x + 1=0$ , which has no real roots.
Hence there are three cases we need to consider:
Case 1 $a=b=r$ and $c=s \neq r$ :
The equation becomes $ax^2+ax+c=0$ , and by Vieta's Formulas, we have $a+c=-1$ and $ac = \frac{c}{a}$ . This second equation becomes $(a^2-1)c=0$ . Hence one possibility is $c=0$ , in which case $a=-1$ , giving the equation $-x^2 - x = 0$ , which has roots $0$ and ${-}1$ . This gives one valid solution. On the other hand, if $c\neq 0$ , then $a^2-1=0$ , so $a = \pm 1$ . If $a=1$ , we have $c=-2$ , and the equation is $x^2 + x - 2 = 0$ , which clearly works, giving a second valid solution. If $a=-1$ , then we have $c=0$ , which has already been considered, so this possibility gives no further valid solutions.
Case 2 $a=c=r$ $b=s \neq r$ :
The equation becomes $ax^2 + bx + a = 0$ , so by Vieta's Formulas, we have $a+b = -\frac{b}{a}$ and $ab = 1$ . These equations reduce to $a^3 + a + 1 = 0$ . By sketching a graph of this function, we see that there is exactly one real root. (Alternatively, note that as it is of odd degree, there is at least one real root, and by differentiation, it has no stationary points, so there is at most one real root. Combining these gives exactly one real root.) This gives a third valid solution.
Case 3 $a=r$ $b=c=s \neq r$ :
The equation becomes $ax^2 + bx+b=0$ , so by Vieta's Formulas, we have $a+b = -\frac{b}{a}$ and $ab = \frac{b}{a}$ .
Observe that $b \neq 0$ , as if it were $0$ , the equation would just have one real root, $0$ , so this would not give a valid solution. Thus, taking the second equation and dividing both sides by $b$ , we deduce have $a=\frac{1}{a}$ , so $a=\pm 1$ . If $a=1$ , we have $1+b=-b$ , giving $b=-\frac{1}{2}$ , so the equation is $x^2 - \frac{1}{2}x - \frac{1}{2} = 0$ , which is a fourth valid solution. If $a=-1$ , we have $1+b=b$ , which is a contradiction, so this case gives no further valid solutions.
Hence the total number of valid solutions is $\boxed{4}$ | B | 4 |
7638b8d19159e6b7d6d136dd85888b20 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_22 | Define a sequence recursively by $x_0=5$ and \[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\] for all nonnegative integers $n.$ Let $m$ be the least positive integer such that
\[x_m\leq 4+\frac{1}{2^{20}}.\]
In which of the following intervals does $m$ lie?
$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$ | We first prove that $x_n > 4$ for all $n \ge 0$ , by induction. Observe that \[x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}.\] so (since $x_n$ is clearly positive for all $n$ , from the initial definition), $x_{n+1} > 4$ if and only if $x_{n} > 4$
We similarly prove that $x_n$ is decreasing: \[x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0.\]
Now we need to estimate the value of $x_{n+1}-4$ , which we can do using the rearranged equation: \[x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}.\] Since $x_n$ is decreasing, $\frac{x_n + 5}{x_n+6}$ is also decreasing, so we have \[\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}\] and \[\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4).\]
This becomes \[\left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^n \left(x_0-4\right) < x_{n} - 4 \le \left(\frac{10}{11}\right)^n \left(x_0-4\right) = \left(\frac{10}{11}\right)^n.\] The problem thus reduces to finding the least value of $n$ such that \[\left(\frac{9}{10}\right)^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{ and } \left(\frac{10}{11}\right)^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}.\]
Taking logarithms, we get $n \ln \frac{9}{10} < -20 \ln 2$ and $(n-1)\ln \frac{10}{11} > -20 \ln 2$ , i.e.
\[n > \frac{20\ln 2}{\ln\frac{10}{9}} \text{ and } n-1 < \frac{20\ln 2}{\ln\frac{11}{10}} .\]
As approximations, we can use $\ln\frac{10}{9} \approx \frac{1}{9}$ $\ln\frac{11}{10} \approx \frac{1}{10}$ , and $\ln 2\approx 0.7$ . These approximations allow us to estimate \[126 < n < 141,\] which gives $\boxed{81,242}$ | C | 81,242 |
7638b8d19159e6b7d6d136dd85888b20 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_22 | Define a sequence recursively by $x_0=5$ and \[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\] for all nonnegative integers $n.$ Let $m$ be the least positive integer such that
\[x_m\leq 4+\frac{1}{2^{20}}.\]
In which of the following intervals does $m$ lie?
$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$ | The condition where $x_m\leq 4+\frac{1}{2^{20}}$ gives the motivation to make a substitution to change the equilibrium from $4$ to $0$ . We can substitute $x_n = y_n + 4$ to achieve that. Now, we need to find the smallest value of $m$ such that $y_m\leq \frac{1}{2^{20}}$ given that $y_0 = 1$
Factoring the recursion $x_{n+1} = \frac{x_n^2 + 5x_n+4}{x_n + 6}$ , we get:
$x_{n+1}=\dfrac{(x_n + 4)(x_n + 1)}{x_n + 6} \Rightarrow y_{n+1}+4=\dfrac{(y_n+8)(y_n+5)}{y_n+10}$
$y_{n+1}+4=\dfrac{y_n^2+13y_n+40}{y_n+10} = \dfrac{y_n^2+9y_n +(4y_n+40)}{y_n+10}$
$y_{n+1}+4=\dfrac{y_n^2+9y_n}{y_n+10} + 4$
$y_{n+1}=\dfrac{y_n^2+9y_n}{y_n+10}$
Using wishful thinking, we can simplify the recursion as follows:
$y_{n+1} = \frac{y_n^2 + 9y_n + y_n - y_n}{y_n + 10}$
$y_{n+1} = \frac{y_n(y_n + 10) - y_n}{y_n + 10}$
$y_{n+1} = y_n - \frac{y_n}{y_n + 10}$
$y_{n+1} = y_n\left(1 - \frac{1}{y_n + 10}\right)$
The recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the $y_n$ sequence is strictly decreasing, so all the terms after $y_0$ will be less than 1. Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction.
With both of those observations in mind, $\frac{9}{10} < 1 - \frac{1}{y_n + 10} \leq \frac{10}{11}$ . Combining this with the fact that the recursion resembles a geometric sequence, we conclude that $\left(\frac{9}{10}\right)^n < y_n \leq \left(\frac{10}{11}\right)^n.$
$\frac{9}{10}$ is approximately equal to $\frac{10}{11}$ and the ranges that the answer choices give us are generous, so we should use either $\frac{9}{10}$ or $\frac{10}{11}$ to find a rough estimate for $m$
Since $\dfrac{1}{2}=0.5$ , that means $\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}} \approx 0.7$ . Additionally, $\left(\frac{9}{10}\right)^3=0.729$
Therefore, we can estimate that $2^{-\frac{1}{2}} < y_3$
Raising both sides to the 40th power, we get $2^{-20} < (y_3)^{40}$
But $y_3 = (y_0)^3$ , so $2^{-20} < (y_0)^{120}$ and therefore, $2^{-20} < y_{120}$
This tells us that $m$ is somewhere around 120, so our answer is $\boxed{81,242}$ | C | 81,242 |
7638b8d19159e6b7d6d136dd85888b20 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_22 | Define a sequence recursively by $x_0=5$ and \[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\] for all nonnegative integers $n.$ Let $m$ be the least positive integer such that
\[x_m\leq 4+\frac{1}{2^{20}}.\]
In which of the following intervals does $m$ lie?
$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$ | Since the choices are rather wide ranges, we can use approximation to make it easier. Notice that \[x_{n+1} - x_n = \frac{4-x_n}{x_n+6}\] And $x_0 =5$ , we know that $x_n$ is a declining sequence, and as it get close to 4 its decline will slow, never falling below 4. So we'll use 4 to approximate $x_n$ in the denominator so that we have a solvable difference equation: \[x_{n+1} - x_n = \frac{4-x_n}{10}\] \[x_{n+1} = \frac{9}{10}x_n + \frac{2}{5}\] Solve it with $x_0 = 5$ , we have \[x_n = 4 + (\frac{9}{10})^n\] Now we wish to find $n$ so that \[(\frac{9}{10})^n \approx \frac{1}{2^{20}}\] \[n \approx \frac{\log{2^{20}}}{\log{10}-\log9} \approx \frac{20*0.3}{0.05} = 120\] Since 120 is safely within the range of [81,242], we have the answer. $\boxed{81,242}$ | C | 81,242 |
7638b8d19159e6b7d6d136dd85888b20 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_22 | Define a sequence recursively by $x_0=5$ and \[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\] for all nonnegative integers $n.$ Let $m$ be the least positive integer such that
\[x_m\leq 4+\frac{1}{2^{20}}.\]
In which of the following intervals does $m$ lie?
$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$ | We start by simplifying the recursion: $x_{n+1} = x_n + \frac{4-x_n}{x_n+6}$ .
While $x_n > 4$ $x_n$ is a decreasing sequence. Let $x_n = 4 + f_n$ . Then \[x_{n} - x_{n+1} = \frac{x_n-4}{x_n+6} = \frac{f_n}{10 + f_n} \approx \frac{f_n}{10}.\] Now notice that we want to find $m$ , such that $x_m \leq 4 + \frac{1}{2^{20}}$ , and $x_0 = 4 + \frac{1}{2^0}$ .
We are going to find an approximate number of steps to go from $4 + \frac{1}{2^i}$ to $4 + \frac{1}{2^{i+1}}$ .
If $x_j = 4 + \frac{1}{2^i}$ , then $f_j = \frac{1}{2^i}$ and if $x_k = 4 + \frac{1}{2^{i+1}}$ , then $f_k = \frac{1}{2^{i+1}}$ . Then \[x_j - x_k = \frac{1}{2^{i+1}} \approx \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10}.\] Furthermore, \[\frac{1}{2^i} = f_j > f_{j+1} > f_{j+2} > ... > f_{k-1} > f_k = \frac{1}{2^{i+1}}.\] Therefore, \[(k-j) \cdot \frac{\frac{1}{2^i}}{10} > \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10} > (k-j) \cdot \frac{\frac{1}{2^{i+1}}}{10},\] so $5 < k-j < 10$ . Therefore, to go from $f_0 = \frac{1}{2^0}$ to $f_m = \frac{1}{2^{20}}$ , we will need to do this 20 times, which means $100 < m - 0 < 200$ , which is within the range of [81,242], so we have the answer. $\boxed{81,242}$ | C | 81,242 |
64e4a7bb65c8e7dc01d927411dd55b65 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_23 | How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | Let $f(n)$ be the number of valid sequences of length $n$ (satisfying the conditions given in the problem).
We know our valid sequence must end in a $0$ . Then, since we cannot have two consecutive $0$ s, it must end in a $10$ . Now, we only have two cases: it ends with $010$ , or it ends with $110$ which is equivalent to $0110$ . Thus, our sequence must be of the forms $0\ldots010$ or $0\ldots0110$ . In the first case, the first $n-2$ digits are equivalent to a valid sequence of length $n-2$ . In the second, the first $n-3$ digits are equivalent to a valid sequence of length $n-3$ . Therefore, it must be the case that $f(n) = f(n-3) + f(n-2)$ , with $n \ge 3$ (because otherwise, the sequence would contain only 0s and this is not allowed due to the given conditions).
It is easy to find $f(3) = 1$ since the only possible valid sequence is $010$ $f(4)=1$ since the only possible valid sequence is $0110$ $f(5)=1$ since the only possible valid sequence is $01010$
The recursive sequence is then as follows:
\[f(3)=1\] \[f(4)=1\] \[f(5) = 1\] \[f(6) = 1 + 1 = 2\] \[f(7) = 1 + 1 = 2\] \[f(8) = 1 + 2 = 3\] \[f(9) = 2 + 2 = 4\] \[f(10) = 2 + 3 = 5\] \[f(11) = 3 + 4 = 7\] \[f(12) = 4 + 5 = 9\] \[f(13) = 5 + 7 = 12\] \[f(14) = 7 + 9 = 16\] \[f(15) = 9 + 12 = 21\] \[f(16) = 12 + 16 = 28\] \[f(17) = 16 + 21 = 37\] \[f(18) = 21 + 28 = 49\] \[f(19) = 28 + 37 = 65\]
So, our answer is $\boxed{65}$ | C | 65 |
64e4a7bb65c8e7dc01d927411dd55b65 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_23 | How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | After any particular $0$ , the next $0$ in the sequence must appear exactly $2$ or $3$ positions down the line. In this case, we start at position $1$ and end at position $19$ , i.e. we move a total of $18$ positions down the line. Therefore, we must add a series of $2$ s and $3$ s to get $18$ . There are a number of ways to do this:
Case 1 : nine $2$ s - there is only $1$ way to arrange them.
Case 2 : two $3$ s and six $2$ s - there are ${8\choose2} = 28$ ways to arrange them.
Case 3 : four $3$ s and three $2$ s - there are ${7\choose4} = 35$ ways to arrange them.
Case 4 : six $3$ s - there is only $1$ way to arrange them.
Summing the four cases gives $1+28+35+1=\boxed{65}$ | C | 65 |
64e4a7bb65c8e7dc01d927411dd55b65 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_23 | How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | We can simplify the original problem into a problem where there are $2^{17}$ binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of $0$ s, $1$ s, and $11$ s. Now, we use casework:
Case 1 : Alternating 1s and 0s. There is simply 1 way to do this: $0101010101010101010$ .
Now, we note that there cannot be only one block of $11$ in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of $11$ s this cannot be satisfied. This is true for all odd numbers of $11$ blocks.
Case 2 : There are 2 $11$ blocks. Using the zeroes in the sequence as dividers, we have a sample as $0110110101010101010$ . We know there are 8 places for $11$ s, which will be filled by $1$ s if the $11$ s don't fill them. This is ${8\choose2} = 28$ ways.
Case 3 : Four $11$ blocks arranged. Using the same logic as Case 2, we have ${7\choose4} = 35$ ways to arrange four $11$ blocks.
Case 4 : No single $1$ blocks, only $11$ blocks. There is simply one case for this, which is $0110110110110110110$
Adding these four cases, we have $1+28+35+1=\boxed{65}$ as our final answer. | C | 65 |
64e4a7bb65c8e7dc01d927411dd55b65 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_23 | How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | Any valid sequence must start with a $0$ . We can then think of constructing a sequence as adding groups of terms to this $0$ , each ending in $0$ . (This is always possible because every valid string ends in $0$ .) For example, we can represent the string $01011010110110$ as: $0-10-110-10-110-110$ .
To not have any consecutive 0s, we must have at least one $1$ before the next $0$ . However, we cannot have three or more $1$ s before the next $0$ because we cannot have three consecutive $1$ s. Consequently, we can only have one or two $1$ s.
So we can have the groups: $10$ and $110$
After the initial $0$ , we have $18$ digits left to fill in the string. Let the number of $10$ blocks be $x$ , and $110$ be $y$ . Then $x$ and $y$ must satisfy $2x+3y=18$ . We recognize this as a Diophantine equation. Taking $\pmod{2}$ yields $y=0 \pmod{2}$ . Since $x$ and $y$ must both be nonnegative, we get the solutions $(9, 0)$ $(6, 2)$ $(3, 4)$ , and $(0, 6)$ . We now handle each of these cases separately.
$(9, 0)$ : Only one arrangement, namely all $10$ s.
$(6, 2)$ : We have 6 groups of $11$ , and $2$ groups of $110$ . This has $\binom{6+2}{2}=28$ cases.
$(3, 4)$ : This means we have 3 groups of $10$ , and 4 groups of $110$ . This has $\binom{3+4}{3}=35$ cases.
$(0, 6)$ : Only one arrangement, namely all $110$
Adding these, we have $1+28+35+1=65 \longrightarrow \boxed{65}$ .
~Math4Life2020 | C | 65 |
64e4a7bb65c8e7dc01d927411dd55b65 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_23 | How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | Suppose the number of $0$ s is $n$ . We can construct the sequence in two steps:
Step 1: put $n-1$ of $1$ s between the $0$ s;
Step 2: put the rest $19-n-(n-1)=20-2n$ of $1$ s in the $n-1$ spots where there is a $1$ . There are $\binom{n-1}{20-2n}$ ways of doing this.
Now we find the possible values of $n$
First of all $n+(n-1) \leq 19 \Rightarrow n\leq 10$ (otherwise there will be two consecutive $0$ s);
And secondly $20-2n \leq n-1\Rightarrow n\geq 7$ (otherwise there will be three consecutive $1$ s).
Therefore the answer is \[\sum_{n=7}^{10} \binom{n-1}{20-2n} = \binom{6}{6} + \binom{7}{4} + \binom{8}{2} + \binom{9}{0} = \boxed{65}.\] | C | 65 |
64e4a7bb65c8e7dc01d927411dd55b65 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_23 | How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | For a valid sequence of length $n$ , the sequence must be in the form of $01xx...xx10$ . By removing the $01$ at the start of the sequence and the $10$ at the end of the sequence, there are $n-4$ bits left. The $n-4$ bits left can be in the form of:
So, $f(n) = f(n-4) + 2f(n-5) + f(n-6)$
We will calculate $f(19)$ by Dynamic Programming
$f(3) = 1$
$f(4) = 1$
$f(5) = 1$
$f(6) = 2$
$f(7) = 2$
$f(8) = 3$
$f(9) = f(5) + 2 \cdot f(4) + f(3) = 1 + 2 \cdot 1 + 1 = 4$
$f(10) = f(6) + 2 \cdot f(5) + f(4) = 2 + 2 \cdot 1 + 1 = 5$
$f(11) = f(7) + 2 \cdot f(6) + f(5) = 2 + 2 \cdot 1 + 1 = 7$
$f(12) = f(8) + 2 \cdot f(7) + f(6) = 3 + 2 \cdot 2 + 2 = 9$
$f(13) = f(9) + 2 \cdot f(8) + f(7) = 4 + 2 \cdot 3 + 2 = 12$
$f(14) = f(10) + 2 \cdot f(9) + f(8) = 5 + 2 \cdot 4 + 3 = 16$
$f(15) = f(11) + 2 \cdot f(10) + f(9) = 7 + 2 \cdot 5 + 4 = 21$
$f(16) = f(12) + 2 \cdot f(11) + f(10) = 9 + 2 \cdot 7 + 5 = 28$
$f(17) = f(13) + 2 \cdot f(12) + f(11) = 12 + 2 \cdot 9 + 7 = 37$
$f(18) = f(14) + 2 \cdot f(13) + f(12) = 16 + 2 \cdot 12 + 9 = 49$
$f(19) = f(15) + 2 \cdot f(14) + f(13) = 21 + 2 \cdot 16 + 12 = \boxed{65}$ | C | 65 |
2db6da6821b72835402e820a42ec8945 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_1 | A large urn contains $100$ balls, of which $36 \%$ are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be $72 \%$ ? (No red balls are to be removed.)
$\textbf{(A)}\ 28 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 64$ | There are $36$ red balls; for these red balls to comprise $72 \%$ of the urn, there must be only $14$ blue balls. Since there are currently $64$ blue balls, this means we must remove $\boxed{50}$ | D | 50 |
2db6da6821b72835402e820a42ec8945 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_1 | A large urn contains $100$ balls, of which $36 \%$ are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be $72 \%$ ? (No red balls are to be removed.)
$\textbf{(A)}\ 28 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 64$ | There are $36$ red balls and $64$ blue balls. For the percentage of the red balls to double from $36 \%$ to $72 \%$ of the urn, half of the total number of balls must be removed. Therefore, the number of blue balls that need to be removed is $\boxed{50}$ | D | 50 |
2db6da6821b72835402e820a42ec8945 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_1 | A large urn contains $100$ balls, of which $36 \%$ are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be $72 \%$ ? (No red balls are to be removed.)
$\textbf{(A)}\ 28 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 64$ | There are $36$ red balls out of the total $100$ balls.
We want to continuously remove blue balls until the percentage of red balls in the urn is 72%.
Therefore, we want \[\frac{36}{100-x}=\frac{72}{100}.\] Solving for $x$ gives that we must remove $\boxed{50}$ blue balls. | D | 50 |
5ff08cca848e61a115962d4894d268f7 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_2 | While exploring a cave, Carl comes across a collection of $5$ -pound rocks worth $$14$ each, $4$ -pound rocks worth $$11$ each, and $1$ -pound rocks worth $$2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?
$\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52$ | The value of $5$ -pound rocks is $$14\div5=$2.80$ per pound, and the value of $4$ -pound rocks is $$11\div4=$2.75$ per pound. Clearly, Carl should not carry more than three $1$ -pound rocks. Otherwise, he can replace some $1$ -pound rocks with some heavier rocks, preserving the weight but increasing the total value.
We perform casework on the number of $1$ -pound rocks Carl can carry: \[\begin{array}{c|c|c||c} & & & \\ [-2.5ex] \boldsymbol{1}\textbf{-Pound Rocks} & \boldsymbol{4}\textbf{-Pound Rocks} & \boldsymbol{5}\textbf{-Pound Rocks} & \textbf{Total Value} \\ \textbf{(}\boldsymbol{$2}\textbf{ Each)} & \textbf{(}\boldsymbol{$11}\textbf{ Each)} & \textbf{(}\boldsymbol{$14}\textbf{ Each)} & \\ [0.5ex] \hline & & & \\ [-2ex] 0 & 2 & 2 & $50 \\ & & & \\ [-2.25ex] 1 & 3 & 1 & $49 \\ & & & \\ [-2.25ex] 2 & 4 & 0 & $48 \\ & & & \\ [-2.25ex] 3 & 0 & 3 & $48 \end{array}\] Clearly, the maximum value of the rocks Carl can carry is $\boxed{50}$ dollars. | C | 50 |
5ff08cca848e61a115962d4894d268f7 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_2 | While exploring a cave, Carl comes across a collection of $5$ -pound rocks worth $$14$ each, $4$ -pound rocks worth $$11$ each, and $1$ -pound rocks worth $$2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?
$\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52$ | Since each rock is worth $1$ dollar less than $3$ times its weight (in pounds), the answer is just $3\cdot 18=54$ minus the minimum number of rocks we need to make $18$ pounds. Note that we need at least $4$ rocks (two $5$ -pound rocks and two $4$ -pound rocks) to make $18$ pounds, so the answer is $54-4=\boxed{50}.$ | C | 50 |
52cc7f92e9be49162871792ee84324a0 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_3 | How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$ | We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.
Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:
Periods $1, 3, 5$
Periods $1, 3, 6$
Periods $1, 4, 6$
Periods $2, 4, 6$
There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.
Therefore, there are $4 \cdot 6 = \boxed{24}$ ways to choose the classes. | E | 24 |
52cc7f92e9be49162871792ee84324a0 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_3 | How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$ | Counting what we don't want is another slick way to solve this problem. Use PIE (Principle of Inclusion and Exclusion) to count two cases: 1. Two classes consecutive, 2. Three classes consecutive.
Case 1: Consider two consecutive periods as a "block" of which there are 5 places to put in(1,2; 2,3; 3,4; 4,5; 5,6). Then we simply need to place two classes within the block, $3 \cdot 2$ . Finally we have 4 periods remaining to place the final math class. Thus there are $5 \cdot 3 \cdot 2 \cdot 4$ ways to place two consecutive math classes with disregard to the third.
Case 2: Now consider three consecutive periods as a "block" of which there are now 4 places to put in(1,2,3; 2,3,4; 3,4,5; 4,5,6). We now need to arrange the math classes in the block, $3 \cdot 2 \cdot 1$ . Thus there are $4 \cdot 3 \cdot 2 \cdot 1$ ways to place all three consecutive math classes.
By PIE we subtract Case 1 by Case 2 in order to not overcount: $120-24$ . Then we subtract that answer from the total ways to place the classes with no restrictions: $(6 \cdot 5 \cdot 4) - 96= \boxed{24}$ | E | 24 |
52cc7f92e9be49162871792ee84324a0 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_3 | How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$ | We can tackle this problem with a stars-and-bars-ish approach. First, letting math class be 1 and non-math-class be 0, place 0s in between 3 1s: \[10101\] Now we need to place 1 additional 0. There are 4 places to put it: \[\underline{\hspace{0.3cm}}1\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}\] It can be placed in any 1 of the underscores.
Since there are $3!=6$ ways to order the math classes, the answer is $6\cdot 4=\boxed{24}$ | E | 24 |
52cc7f92e9be49162871792ee84324a0 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_3 | How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$ | Let $M$ represent a math class and $N$ represent a non-math class. We have _N_N_N_, where the spaces represent the possible spots the $M’s$ could go in. There are $4\choose 3$ ways to choose the spots for the math classes and $3!=6$ ways to order the classes. Hence, the answer is $4 \cdot 6 = \boxed{24}$ .
~azc1027 | E | 24 |
8968f9f7b1bbac7da38438618afa5eba | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_4 | Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least $6$ miles away," Bob replied, "We are at most $5$ miles away." Charlie then remarked, "Actually the nearest town is at most $4$ miles away." It turned out that none of the three statements were true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d$
$\textbf{(A) } (0,4) \qquad \textbf{(B) } (4,5) \qquad \textbf{(C) } (4,6) \qquad \textbf{(D) } (5,6) \qquad \textbf{(E) } (5,\infty)$ | For each of the false statements, we identify its corresponding true statement. Note that:
We construct the following table: \[\begin{array}{c||c|c} & & \\ [-2.5ex] \textbf{Hiker} & \textbf{False Statement} & \textbf{True Statement} \\ [0.5ex] \hline & & \\ [-2ex] \textbf{Alice} & [6,\infty) & [0,6) \\ & & \\ [-2.25ex] \textbf{Bob} & [0,5] & (5,\infty) \\ & & \\ [-2.25ex] \textbf{Charlie} & [0,4] & (4,\infty) \end{array}\] Taking the intersection of the true statements, we have \[[0,6)\cap(5,\infty)\cap(4,\infty)=(5,6)\cap(4,\infty)=\boxed{5,6}.\] ~MRENTHUSIASM | D | 5,6 |
8968f9f7b1bbac7da38438618afa5eba | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_4 | Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least $6$ miles away," Bob replied, "We are at most $5$ miles away." Charlie then remarked, "Actually the nearest town is at most $4$ miles away." It turned out that none of the three statements were true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d$
$\textbf{(A) } (0,4) \qquad \textbf{(B) } (4,5) \qquad \textbf{(C) } (4,6) \qquad \textbf{(D) } (5,6) \qquad \textbf{(E) } (5,\infty)$ | Think of the distances as if they are on a number line. Alice claims that $d > 6$ , Bob says $d < 5$ , while Charlie thinks $d < 4$ . This means that all possible numbers less than $5$ and greater than $6$ are included. However, since the three statements are actually false, the distance to the nearest town is one of the numbers not covered. Therefore, the answer is $\boxed{5,6}$ | D | 5,6 |
e09ae8e2c6557bb28885849afc16664c | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_5 | What is the sum of all possible values of $k$ for which the polynomials $x^2 - 3x + 2$ and $x^2 - 5x + k$ have a root in common?
$\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10$ | We factor $x^2-3x+2$ into $(x-1)(x-2)$ . Thus, either $1$ or $2$ is a root of $x^2-5x+k$ . If $1$ is a root, then $1^2-5\cdot1+k=0$ , so $k=4$ . If $2$ is a root, then $2^2-5\cdot2+k=0$ , so $k=6$ . The sum of all possible values of $k$ is $\boxed{10}$ | E | 10 |
424515a316bed2c1ba66042debb47e2e | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_6 | For positive integers $m$ and $n$ such that $m+10<n+1$ , both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$ . What is $m+n$
$\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24$ | The mean and median are \[\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,\] so $3m+17=2n$ and $m+11=n$ . Solving this gives $\left(m,n\right)=\left(5,16\right)$ for $m+n=\boxed{21}$ . (trumpeter) | null | 21 |
424515a316bed2c1ba66042debb47e2e | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_6 | For positive integers $m$ and $n$ such that $m+10<n+1$ , both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$ . What is $m+n$
$\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24$ | This is an alternate solution if you don't want to solve using algebra. First, notice that the median $n$ is the average of $m+10$ and $n+1$ . Therefore, $n=m+11$ , so the answer is $m+n=2m+11$ , which must be odd. This leaves two remaining options: ${(B) 21}$ and ${(D) 23}$ . Notice that if the answer is $(B)$ , then $m$ is odd, while $m$ is even if the answer is $(D)$ . Since the average of the set is an integer $n$ , the sum of the terms must be even. $4+10+1+2+2n$ is odd by definition, so we know that $3m+2n$ must also be odd, thus with a few simple calculations $m$ is odd. Because all other answers have been eliminated, $(B)$ is the only possibility left. Therefore, $m+n=\boxed{21}$ . ∎ --anna0kear | null | 21 |
424515a316bed2c1ba66042debb47e2e | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_6 | For positive integers $m$ and $n$ such that $m+10<n+1$ , both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$ . What is $m+n$
$\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24$ | Since the median is $n$ , then $\frac{m+10+n+1}{2} = n \Rightarrow m+11 = n$ , or $m= n-11$ . Plug this in for $m$ values to get $\frac{7n-16}{6} = n \Rightarrow 7n-16 = 6n \Rightarrow n= 16$ . Plug it back in to get $m = 5$ , thus $16 + 5 = \boxed{21}$ | null | 21 |
11ad48262799ab79b3355f99f362f899 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_7 | For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$ | Note that \[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\] Since this expression is an integer, we need:
Taking the intersection gives $-5\leq n\leq3.$ So, there are $3-(-5)+1=\boxed{9}$ integer values of $n.$ | E | 9 |
11ad48262799ab79b3355f99f362f899 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_7 | For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$ | Note that $4000\cdot \left(\frac{2}{5}\right)^n$ will be an integer if the denominator is a factor of $4000$ . We also know that the denominator will always be a power of $5$ for positive values and a power of $2$ for all negative values. So we can proceed to divide $4000$ by $5^n$ for each increasing positive value of $n$ until we get a non-factor of $4000$ and also divide $4000$ by $2^{-n}$ for each decreasing negative value of $n$ . For positive values we get $n= 1, 2, 3$ and for negative values we get $n= -1, -2, -3, -4, -5$ . Also keep in mind that the expression will be an integer for $n=0$ , which gives us a total of $\boxed{9}$ for $n.$ | E | 9 |
11ad48262799ab79b3355f99f362f899 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_7 | For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$ | The values for $n$ are $-5, -4, -3, -2, -1, 0, 1, 2,$ and $3.$
The corresponding values for $4000\cdot \left(\frac{2}{5}\right)^n$ are $390625, 156250, 62500, 25000, 10000, 4000, 1600, 640,$ and $256,$ respectively.
In total, there are $\boxed{9}$ values for $n.$ | E | 9 |
6b7eea1057b60e9c2e79ed9d0adf9827 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_8 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | Let $x$ be the area of $ADE$ . Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$ ). Thus, we have \[x=7+\dfrac{9}{16}x.\] This gives $x=16$ , so the area of $DBCE=40-x=\boxed{24}$ | E | 24 |
6b7eea1057b60e9c2e79ed9d0adf9827 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_8 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | Let the base length of the small triangle be $x$ . Then, there is a triangle $ADE$ encompassing the 7 small triangles and sharing the top angle with a base length of $4x$ . Because the area is proportional to the square of the side, let the base $BC$ be $\sqrt{40}x$ . The ratio of the area of triangle $ADE$ to triangle $ABC$ is $\left(\frac{4x}{\sqrt{40}x}\right)^2 = \frac{16}{40}$ . The problem says the area of triangle $ABC$ is $40$ , so the area of triangle $ADE$ is $16$ . So the area of trapezoid $DBCE$ is $40 - 16 = \boxed{24}$ | null | 24 |
6b7eea1057b60e9c2e79ed9d0adf9827 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_8 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | Notice $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$ .
Let the base of the small triangles of area 1 be $x$ , then the base length of $\Delta ADE=4x$ . Notice, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$ , then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]$ Thus, $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.$ | null | 24 |
6b7eea1057b60e9c2e79ed9d0adf9827 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_8 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$ . Therefore the area of the trapezoid is $40-16=\boxed{24}$ | null | 24 |
6b7eea1057b60e9c2e79ed9d0adf9827 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_8 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$ , so to find the area of such trapezoid $BCED$ , we just take $40-16=\boxed{24}$ , like so. | null | 24 |
6b7eea1057b60e9c2e79ed9d0adf9827 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_8 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | The combined area of the small triangles is $7$ , and from the fact that each small triangle has an area of $1$ , we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\frac{1}{3}$ , so will their areas have a proportion that is the square of the proportion of their sides, or $\frac {1}{9}$ ). Thus, the combined area of the top triangle and the trapezoid immediately below is $7 + 9 = 16$ . The area of trapezoid $BCED$ is thus the area of triangle $ABC-16 =\boxed{24}$ | null | 24 |
6b7eea1057b60e9c2e79ed9d0adf9827 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_8 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | You can assume for the base of one of the smaller triangles to be $\frac{1}{a}$ and the height to be $2a$ , giving an area of 1. The larger triangle above the 7 smaller ones then has base $\frac{3}{a}$ and height $6a$ , giving it an area of $9$ . Then the area of triangle $ADE$ is $16$ and $40-16=\boxed{24}$ | null | 24 |
6b7eea1057b60e9c2e79ed9d0adf9827 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_8 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence $7\cdot\left(1+\left(\frac{9}{16}\right)+\left(\frac{9}{16}\right)^2+\left(\frac{9}{16}\right)^3\dots\right)$ . Using the infinite geometric series sum formula gives us $7\cdot\left(\frac{1}{1-\frac{9}{16}}\right)=7\cdot\frac{16}{7}=16$ . The triangle's area would thus be $40-16=\boxed{24}$ | null | 24 |
6b7eea1057b60e9c2e79ed9d0adf9827 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_8 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is $1$ , and thus its base is $2.$
Let $h$ be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and $\triangle{ADE}$ . The top triangle has a base of $3\cdot{2}=6$ , and $DE=4\cdot{2}=8.$ The height of $\triangle{ADE}$ is $h+1$ , therefore our ratio is $\frac{h}{6}=\frac{h+1}{8}$ , which yields $h=3$ as our answer.
To find the area of the trapezoid, we can take the area of $\triangle{ABC}$ and subtract the area of $\triangle{ADE},$ whose base is $8$ and height $3+1=4$ . It follows that the area of $\triangle{ADE}=16$ , and subtracting this from $40$ gives us $40-16=\boxed{24}$ | D | 24 |
0513561a85f000e7896b404d3aa9d1d0 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_9 | Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which \[\sin(x+y)\leq \sin(x)+\sin(y)\] for every $x$ between $0$ and $\pi$ , inclusive?
$\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi$ | On the interval $[0, \pi]$ sine is nonnegative; thus $\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y$ for all $x, y \in [0, \pi]$ and equality only occurs when $\cos x = \cos y = 1$ , which is cosine's maximum value. The answer is $\boxed{0}$ . (CantonMathGuy) | E | 0 |
0513561a85f000e7896b404d3aa9d1d0 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_9 | Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which \[\sin(x+y)\leq \sin(x)+\sin(y)\] for every $x$ between $0$ and $\pi$ , inclusive?
$\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi$ | Expanding, \[\cos y \sin x + \cos x \sin y \le \sin x + \sin y\] Let $\sin x =a \ge 0$ $\sin y = b \ge 0$ . We have that \[(\cos y)a+(\cos x)b \le a+b\] Comparing coefficients of $a$ and $b$ gives a clear solution: both $\cos y$ and $\cos x$ are less than or equal to one, so the coefficients of $a$ and $b$ on the left are less than on the right. Since $a, b \ge 0$ , that means that this equality is always satisfied over this interval, or $\boxed{0}$ | E | 0 |
0513561a85f000e7896b404d3aa9d1d0 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_9 | Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which \[\sin(x+y)\leq \sin(x)+\sin(y)\] for every $x$ between $0$ and $\pi$ , inclusive?
$\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi$ | If we plug in $\pi$ , we can see that $\sin(x+\pi) \le \sin(x)$ . Note that since $\sin(x)$ is always nonnegative, $\sin(x+\pi)$ is always nonpositive. So, the inequality holds true when $y=\pi$ . The only interval that contains $\pi$ in the answer choices is $\boxed{0}$ | E | 0 |
0513561a85f000e7896b404d3aa9d1d0 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_9 | Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which \[\sin(x+y)\leq \sin(x)+\sin(y)\] for every $x$ between $0$ and $\pi$ , inclusive?
$\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi$ | Rewrite it as \[\cos y \sin x + \cos x \sin y \le \sin x + \sin y\] then move the terms such that it appears as this \[\sin x (\cos y -1) \le \sin y (1- \cos x)\] We know that on the interval of $[0, \pi]$ $0\le (1-\cos x)\le 2$ , and $0\le \sin y$ , such that the whole thing on the right hand side is always greater than or equal to $0$ . On the left hand side, $0\le \sin x$ and we want $(\cos y -1)\le 0$ , which $\boxed{0}$ satisfies.
~azure123456 | null | 0 |
52ebe81e1e0f41e35ed45a06345275fd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_10 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this: [asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); [/asy] Now, it becomes clear that there are $\boxed{3}$ intersection points. | C | 3 |
52ebe81e1e0f41e35ed45a06345275fd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_10 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | $x+3y=3$ can be rewritten to $x=3-3y$ . Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$ . Splitting this question into casework for the ranges of $y$ will give us the total number of solutions.
$\textbf{Case 1:}$ $y>1$ $3-3y$ will be negative so $|3-3y| = 3y-3.$ $|3y-3-y| = |2y-3| = 1$
$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$
$2y-3$ is negative so $|2y-3| = 3-2y = 1$ $2y = 2$ and so there are no solutions ( $y$ can't equal to $1$
$\textbf{Case 2:}$ $y = 1$ :
It is fairly clear that $x = 0.$
$\textbf{Case 3:}$ $y<1$ $3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$
$3-4y$ will be negative so $4y-3 = 1$ $\rightarrow$ $4y = 4$ . We already have this solution from Case 2 as $y = 1$
$3-4y$ will be positive so $3-4y = 1$ $\rightarrow$ $4y = 2$ $y = \frac{1}{2}$ and $x = \frac{3}{2}$ .
Thus, the solutions are: $(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)$ , and the answer is $\boxed{3}$ | C | 3 |
52ebe81e1e0f41e35ed45a06345275fd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_10 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | Note that $||x| - |y||$ can take on one of four values: $x + y$ $x - y$ $-x + y$ $-x -y$ . So we have 4 cases:
Case 1: ||x| - |y|| = x+y \[x+3y=3\] \[x+y=1\]
Subtracting:
$2y=2 \Rightarrow y=1$ and $x=0$
$\text{Result: } (0,1)$
Case 2: ||x| - |y|| = x-y \[x+3y=3\] \[x-y=1\]
Subtacting:
$4y=2 \Rightarrow y=\dfrac{1}{2}$ an $x=\dfrac{3}{2}$
$\text{Result: } \left(\dfrac{3}{2},\dfrac{1}{2}\right)$
Case 3: ||x| - |y|| = -x+y \[x+3y=3\] \[-x+y=1\]
Adding:
$4y=4 \Rightarrow y=1$ and $x=0$
$\text{Result: } (0,1)$ . Since this is the same solution as we got in Case 1, we can not count this (otherwise, we would have overcounted).
Case 4: ||x| - |y|| = -x-y \[x+3y=3\] \[-x-y=1\]
Adding:
$2y=4 \Rightarrow y=2$ and $x=-3$
$\text{Result: } (-3,2)$
Answer
We solved each case by elimination (either adding the two equations or subtracting), to obtain three solutions: $(0, 1)$ $(-3,2)$ $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$
Our answer is $\boxed{3}$ | C | 3 |
52ebe81e1e0f41e35ed45a06345275fd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_10 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | Just as in solution $2$ , we derive the equation $||3-3y|-|y||=1$ . If we remove the absolute values, the equation collapses into four different possible values. $3-2y$ $3-4y$ $2y-3$ , and $4y-3$ , each equal to either $1$ or $-1$ . Remember that if $P-Q=a$ , then $Q-P=-a$ . Because we have already taken $1$ and $-1$ into account, we can eliminate one of the conjugates of each pair, namely $3-2y$ and $2y-3$ , and $3-4y$ and $4y-3$ . Find the values of $y$ when $3-2y=1$ $3-2y=-1$ $3-4y=1$ and $3-4y=-1$ . We see that $3-2y=1$ and $3-4y=-1$ give us the same value for $y$ , so the answer is $\boxed{3}$ | C | 3 |
52ebe81e1e0f41e35ed45a06345275fd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_10 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | Just as in solution $2$ , we derive the equation $x=3-3y$ . Squaring both sides in the second equation gives $x^2+y^2-2|xy|=1$ . Putting $x=3-3y$ and doing a little calculation gives $10y^2-18y+9-2|3y-3y^2|=1$ . From here we know that $3y-3y^2$ is either positive or negative.
When positive, we get $2y^2-3y+1=0$ and then, $y=1/2$ or $y=1$ .
When negative, we get $y^2-3y+2=0$ and then, $y=2$ or $y=1$ . Clearly, there are $3$ different pairs of values and that gives us $\boxed{3}$ | C | 3 |
52ebe81e1e0f41e35ed45a06345275fd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_10 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | Since the absolute value is the square root of the square, we get that the first equation is quartic(degree $4$ ) and the other is linear. Subtract to get $\boxed{3}$ | C | 3 |
97669f36366d95230733fa7aa46906e8 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_12 | Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$ | We start with $2$ because $1$ is not an answer choice. We would have to include every odd number except $1$ to fill out the set, but then $3$ and $9$ would violate the rule, so that won't work.
Experimentation with $3$ shows it's likewise impossible. You can include $7,11,$ and either $5$ or $10$ (which are always safe). But after adding either $4$ or $8$ we have no more valid numbers.
Finally, starting with $4,$ we find that the sequence $4,5,6,7,9,11$ works, giving us $\boxed{4}.$ | C | 4 |
97669f36366d95230733fa7aa46906e8 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_12 | Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$ | We know that all odd numbers except $1,$ namely $3, 5, 7, 9, 11,$ can be used.
Now we have $7$ possibilities to choose from for the last number (out of $1, 2, 4, 6, 8, 10, 12$ ). We can eliminate $1, 2, 10,$ and $12,$ and we have $4, 6, 8$ to choose from. However, $9$ is a multiple of $3.$ Now we have to take out either $3$ or $9$ from the list. If we take out $9,$ none of the numbers would work, but if we take out $3,$ we get \[4, 5, 6, 7, 9, 11.\] The least number is $4,$ so the answer is $\boxed{4}.$ | C | 4 |
97669f36366d95230733fa7aa46906e8 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_12 | Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$ | We can get the multiples for the numbers in the original set with multiples in the same original set \begin{align*} 1&: \ \text{all elements of }\{1,2,\dots,12\} \\ 2&: \ 4,6,8,10,12 \\ 3&: \ 6,9,12 \\ 4&: \ 8,12 \\ 5&: \ 10 \\ 6&: \ 12 \end{align*} It will be safe to start with $5$ or $6$ since they have the smallest number of multiples as listed above, but since the question asks for the least, it will be better to try others.
Trying $4,$ we can get $4,5,6,7,9,11.$ So $4$ works.
Trying $3,$ it won't work, so the least is $4.$ This means the answer is $\boxed{4}.$ | C | 4 |
97669f36366d95230733fa7aa46906e8 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_12 | Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$ | We partition $\{1,2,\ldots,12\}$ into six nonempty subsets such that for every subset, each element is a multiple of all elements less than or equal to itself: \[\{1,2,4,8\}, \ \{3,6,12\}, \ \{5,10\}, \ \{7\}, \ \{9\}, \ \{11\}.\] Clearly, $S$ must contain exactly one element from each subset:
If $4\in S,$ then the possibilities of $S$ are $\{4,5,6,7,9,11\}$ or $\{4,6,7,9,10,11\}.$ So, the least possible value of an element in $S$ is $\boxed{4}.$ | C | 4 |
97669f36366d95230733fa7aa46906e8 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_12 | Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$ | We start with 2 as 1 is not an answer option. Our set would be $\{2,3,5,7,11\}$ . We realize we cannot add 12 to the set because 12 is a multiple of 3. Our set only has 5 elements, so starting with 2 won't work.
We try 3 next. Our set becomes $\{3,4,5,7,11\}$ . We run into the same issue as before so starting with 3 won't work.
We then try 4. Our set becomes $\{4,5,6,7,9,11\}$ . We see we have 6 elements with none being multiples of each other. Therefore our answer is $\boxed{4}.$ | C | 4 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of $|x|=3280.5$ , which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are $3280+1=\boxed{3281}$ | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all $a_i=0$ . The total number of ways to pick $a_i$ from $i=0, 1, 2, 3, ... 7$ is $3^8=6561$ $\frac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of non-negative integers, so subtracting from the total gives $6561-3280=\boxed{3281}$ . (RegularHexagon, KLBBC minor changes) | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | Note that the number of total possibilities (ignoring the conditions set by the problem) is $3^8=6561$ . So, E is clearly unrealistic.
Note that if $a_7$ is 1, then it's impossible for \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] to be negative. Therefore, if $a_7$ is 1, there are $3^7=2187$ possibilities. (We also must convince ourselves that these $2187$ different sets of coefficients must necessarily yield $2187$ different integer results.)
As A, B, and C are all less than 2187, the answer must be $\boxed{3281}$ | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | Note that we can do some simple casework:
If $a_7=1$ , then we can choose anything for the other 7 variables, so this give us $3^7$ .
If $a_7=0$ and $a_6=1$ , then we can choose anything for the other 6 variables, giving us $3^6$ .
If $a_7=0$ $a_6=0$ , and $a_5=1$ , then we have $3^5$ .
Continuing in this vein, we have $3^7+3^6+\cdots+3^1+3^0$ ways to choose the variables' values, except we have to add 1
because we haven't counted the case where all variables are 0. So our total sum is $\boxed{3281}$ .
Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative. | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | This solution has a similar idea to that of above.
We will start off with casework on $a_{7}$ :
If $a_7 = -1$ , we can show that there is no way to assign the values of $-1, 0, 1$ to the other $a_{i}$ 's for $0 \leq i \leq 6$ . This is because even if we make all of the other $a_{i}$ equal to $1$ , we will have that our number is $(-1)(3^{7}) + (3^{6}+3^{5}+...+3^{0}) = -3^{7} + \frac{3^{7}-1}{3-1}$ . However, note that since $\frac{3^{7}-1}{3-1} < 3^{7}$ $-3^{7} + \frac{3^{7}-1}{3-1} < 0$ .
If $a_{7}=0$ , we can ignore $a_{7}$ and pretend like it never even existed. This then simplifies the problem: the question remains the same, except that instead of having $a_{0}...a_{7}$ , we only have $a_{0}...a_{6}$ . Now, we introduce some notation: let the number of nonnegative integers that can be written in the form of $a_{n} \cdot 3^{n} + ... + a_{0} \cdot 3^{0}$ be denoted as $S_{n}$ . Then, the problem is asking us for $S_{7}$ , and the total number of nonnegative integers we get from this case is $S_{6}$ . Let's keep this in mind and revisit this idea later.
If $a_{7}=1$ , we can show that no matter what values we assign to the rest of the $a_{i}$ , we will always get a nonnegative integer (we can make this even stricter and say positive integer). This is because of something we figured out in the first case: $\frac{3^{7}-1}{3-1} < 3^{7}$ . So, even if we let $a_{i}=-1$ for $0 \leq i \leq 6$ , we will still have $3^{7} - (3^{6}+...+3^{0}) > 0$ . So, for this case, we have $3^7$ possible nonnegative integers.
Totalling up our values from the three cases, we see that $S_{7} = 0 + S_{6} + 3^{7}$ . In fact, using the reasoning from the three cases above, we can deduce that, for all positive integers $n$ $S_{n} = 3^{n} + S_{n-1}$ . We can now start building up our values for $S$ $S_{0} = 2$ $a_{0} = 0, 1$ ), $S_{1} = 3^{1}+S_{0} = 5, S_{2} = 14, ..., S_{6}=1094$ . So, we have $S_{7} = 2187 + 1094 = \boxed{3281}$ .
~advanture | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | The key is to realize that this question is basically taking place in $a\in\{0,1,2\}$ if each value of $a$ was increased by $1$ , essentially making it into base $3$ . Then the range would be from $0\cdot3^7+$ $0\cdot3^6+$ $0\cdot3^5+$ $0\cdot3^4+$ $0\cdot3^3+$ $0\cdot3^2+$ $0\cdot3^1+$ $0\cdot3^0=$ $0$ to $2\cdot3^7+$ $2\cdot3^6+$ $2\cdot3^5+$ $2\cdot3^4+$ $2\cdot3^3+$ $2\cdot3^2+$ $2\cdot3^1+$ $2\cdot3^0=$ $3^8-1=$ $6561-1=$ $6560$ , yielding $6561$ different values. Since the distribution for all $a_i\in \{-1,0,1\}$ the question originally gave is symmetrical, we retain the $3280$ positive integers and one $0$ but discard the $3280$ negative integers. Thus, we are left with the answer, $\boxed{3281}\qquad$ . --anna0kear | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | First, set $a_i=0$ for all $i\geq1$ . The range would be the integers for which $[-1,1]$ . If $a_i=0$ for all $i\geq2$ , our set expands to include all integers such that $-4\leq\mathbb{Z}\leq4$ . Similarly, when $i\geq3$ we get $-13\leq\mathbb{Z}\leq13$ , and when $i\geq4$ the range is $-40\leq\mathbb{Z}\leq40$ . The pattern continues until we reach $i=7$ , where $-3280\leq\mathbb{Z}\leq3280$ . Because we are only looking for positive integers, we filter out all $\mathbb{Z}<0$ , leaving us with all integers between $0\leq\mathbb{Z}\leq3280$ , inclusive. The answer becomes $\boxed{3281}$ . --anna0kear | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | To get the number of integers, we can get the highest positive integer that can be represented using \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
Note that the least nonnegative integer that can be represented is $0$ , when all $a_i=0$ . The highest number will be the number when all $a_i=1$ . That will be \[3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}\] \[=3280\]
Therefore, there are $3280$ positive integers and $(3280+1)$ nonnegative integers (while including $0$ ) that can be represented. Our answer is $\boxed{3281}$ | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | Notice that there are $3^8$ options for $a_7, a_6, \cdots a_0$ since each $a_i$ can take on the value $-1$ $0$ , or $1$ . Now we want to find how many of them are positive and then we can add one in the end to account for $0$ (they are asking for non-negative).
By symmetry (look out for these on the contest), we see that exactly half of them are positive. So $\lfloor{\tfrac{3^8}{2}}\rfloor = 3280.$ Now we will add $1$ because of the $0$ to account for the non-negative solutions.
So our final answer is $3280 + 1 = 3281$ which is $\boxed{3281}$ | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | Obviously, there are $3^8 = 6561$ possible, and one of them is 0, so other $6560$ are either positive or negative. By the symmetry, we can get the answer is $6560/2 + 1$ which is $\boxed{3281}$ | D | 3281 |
d68453cb73aa6603c00b9940b825f276 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_13 | How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$ | The solution has to obviously be smaller than $3^{7-0+1}=3^8$ but larger than $3^{(7-0+1)-1} = 3^7.$ The only remaining answer is $\boxed{3281}.$ ~mathboy282 | D | 3281 |
9f5c7ff62c3faf900549db081f968d06 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_14 | The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$ , where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$ , can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$
$\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35$ | We apply the Change of Base Formula, then rearrange: \begin{align*} \frac{\log_2{4}}{\log_2{(3x)}}&=\frac{\log_2{8}}{\log_2{(2x)}} \\ \frac{2}{\log_2{(3x)}}&=\frac{3}{\log_2{(2x)}} \\ 3\log_2{(3x)}&=2\log_2{(2x)}. \\ \end{align*} By the logarithmic identity $n\log_b{a}=\log_b{\left(a^n\right)},$ it follows that \begin{align*} \log_2{\left[(3x)^3\right]}&=\log_2{\left[(2x)^2\right]} \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{31}.$ | D | 31 |
9f5c7ff62c3faf900549db081f968d06 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_14 | The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$ , where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$ , can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$
$\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35$ | We will apply the following logarithmic identity: \[\log_{p^n}{\left(q^n\right)}=\log_{p}{q},\] which can be proven by the Change of Base Formula: \[\log_{p^n}{\left(q^n\right)}=\frac{\log_{p}{\left(q^n\right)}}{\log_{p}{\left(p^n\right)}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.\] We rewrite the original equation as $\log_{(3x)^3} 64 = \log_{(2x)^2} 64,$ from which \begin{align*} (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}. \end{align*} Therefore, the answer is $4+27=\boxed{31}.$ | D | 31 |
9f5c7ff62c3faf900549db081f968d06 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_14 | The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$ , where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$ , can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$
$\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35$ | By the logarithmic identity $n\log_b{a}=\log_b{\left(a^n\right)},$ the original equation becomes \[2\log_{3x} 2 = 3\log_{2x} 2.\] By the logarithmic identity $\log_b{a}\cdot\log_a{b}=1,$ we multiply both sides by $\log_2{(2x)},$ then apply the Change of Base Formula to the left side: \begin{align*} 2\left[\log_{3x}2\right]\left[\log_2{(2x)}\right] &= 3 \\ 2\left[\frac{\log_2 2}{\log_2{(3x)}}\right]\left[\frac{\log_2{(2x)}}{\log_2 2}\right] &= 3 \\ 2\left[\frac{\log_2{(2x)}}{\log_2{(3x)}}\right] &=3 \\ 2\left[\log_{3x}{(2x)}\right] &= 3 \\ \log_{3x}{\left[(2x)^2\right]} &= 3 \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}. \end{align*} Therefore, the answer is $4+27=\boxed{31}.$ | D | 31 |
9f5c7ff62c3faf900549db081f968d06 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_14 | The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$ , where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$ , can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$
$\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35$ | We can convert both $4$ and $8$ into $2^2$ and $2^3,$ respectively: \[2\log_{3x} (2) = 3\log_{2x} (2).\] Converting the bases of the right side, we get \begin{align*} \log_{2x} 2 &= \frac{\ln 2}{\ln (2x)} \\ \frac{2}{3}\cdot\log_{3x} (2) &= \frac{\ln 2}{\ln (2x)} \\ 2^\frac{2}{3} &= (3x)^\frac{\ln 2}{\ln (2x)} \\ \frac{2}{3} \cdot \ln 2 &= \frac{\ln 2}{\ln (2x)} \cdot \ln (3x). \end{align*} Dividing both sides by $\ln 2,$ we get $\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)},$ from which \[2\ln (2x) = 3\ln (3x).\] Expanding this equation gives \begin{align*} 2\ln 2 + 2\ln (x) &= 3\ln 3 + 3\ln (x) \\ \ln (x) &= 2\ln 2 - 3\ln 3. \end{align*} Thus, we have \[x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}} = \frac{2^2}{3^3} = \frac{4}{27},\] from which the answer is $4+27=\boxed{31}.$ | D | 31 |
9f5c7ff62c3faf900549db081f968d06 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_14 | The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$ , where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$ , can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$
$\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35$ | Let $y=\log_{3x} 4 = \log_{2x} 8.$ We convert the equations with $y$ to the exponential form: \begin{align*} (3x)^y&=4, \\ (2x)^y&=8. \end{align*} Cubing the first equation and squaring the second equation, we have \begin{align*} (3x)^{3y}&=64, \\ (2x)^{2y}&=64. \end{align*} Applying the Transitive Property, we get \begin{align*} (3x)^{3y}&=(2x)^{2y} \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{31}.$ | D | 31 |
df15273ad9ad14e8372d3af9441bc61a | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_15 | A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?
$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$ | [asy] size(100pt); draw((1,0)--(8,0),linewidth(0.5)); draw((1,2)--(6,2),linewidth(0.5)); draw((1,4)--(4,4),linewidth(0.5)); draw((1,6)--(2,6),linewidth(0.5)); draw((2,6)--(2,0),linewidth(0.5)); draw((4,4)--(4,0),linewidth(0.5)); draw((6,2)--(6,0),linewidth(0.5)); draw((1,0)--(1,7),dashed+linewidth(0.5)); draw((1,7)--(8,0),dashed+linewidth(0.5)); [/asy]
Imagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\boxed{1022.}$ | B | 1022. |
df15273ad9ad14e8372d3af9441bc61a | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_15 | A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?
$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$ | \[\begin{tabular}{|c|c|c|c|c|c|c|} \hline T & T & T & X & T & T & T \\ \hline T & T & T & Y& T & T & T \\ \hline T & T & T & Z & T & T & T \\ \hline X & Y & Z & W & Z & Y & X \\ \hline T & T & T & Z & T & T & T \\ \hline T & T & T & Y & T & T & T \\ \hline T & T & T & X & T & T & T \\ \hline \end{tabular}\]
There are $3 \times 3$ squares in the corners of this $7 \times 7$ square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)
\[\begin{tabular}{|c|c|c|} \hline A & B & C \\ \hline B & D & E \\ \hline C & E & F \\ \hline \end{tabular}\] (Note that coloring one $3 \times 3$ square will also determine the colorings of the other $3$ because the large $7 \times 7$ square must look the same when it is rotated by $90^\circ$
There are $6$ different letters and $2$ choices of color (black and white) for each letter, so there are $2^6=64$ colorings of a proper $3 \times 3$ square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are $4$ different letters, so there are $2^4=16$ different colorings. Multiplying them together gives $16 \times 64 = 1024$ . Going back to the question, we see that "there must be at least one square of each color in this grid of $49$ squares." We must then eliminate $2$ options: an all-black grid and an all-white grid. $1024-2= \boxed{1022}$ -hansenhe | B | 1022 |
0269dd79e6f75f0e2574c17f7aa77bf5 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_16 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get \[x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0\] Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root).
The other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$ . We must have $2a-1> 0;$ otherwise we are in the case where the parabola lies entirely above the circle (tangent at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see that $\boxed{12}$ | E | 12 |
0269dd79e6f75f0e2574c17f7aa77bf5 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_16 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | Substituting $y = x^2 - a$ gives $x^2 + (x^2 - a)^2 = a^2$ , which simplifies to $x^2 + x^4 - 2x^2a + a^2 = a^2$ . This further simplifies to $x^2(1 + x^2 - 2a) = 0$ . Thus, either $x^2 = 0$ , or $x^2 - 2a + 1 = 0$
Since we care about $a$ , we consider the second case. We solve in terms of $a$ , giving $a = \frac{x^2}{2} + \frac{1}{2}$ . We see that in order to find the range in which $a$ lies, we must find the vertex of this equation, which turns out to be $\left(0, \frac{1}{2}\right)$ . Hence, we know that the minimum is $\frac{1}{2}$ , which further implies that $\boxed{12}$ | E | 12 |
0269dd79e6f75f0e2574c17f7aa77bf5 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_16 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | [asy] Label f; f.p=fontsize(6); xaxis(-2,2,Ticks(f, 0.2)); yaxis(-2,2,Ticks(f, 0.2)); real g(real x) { return x^2-1; } draw(graph(g, 1.7, -1.7)); real h(real x) { return sqrt(1-x^2); } draw(graph(h, 1, -1)); real j(real x) { return -sqrt(1-x^2); } draw(graph(j, 1, -1)); [/asy]
Looking at a graph, it is obvious that the two curves intersect at $(0, -a)$ . We also see that if the parabola goes "in" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.
We have $x^2 - a = -\sqrt{a^2 - x^2}$ . Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$ . Since $x = 0$ is already accounted for, we only need to find one solution for $x^2 = 2a - 1$ , where the right hand side portion is obviously increasing. Since $a = \frac{1}{2}$ begets $x = 0$ (an overcount), we have $\boxed{12}$ as the right answer. | E | 12 |
0269dd79e6f75f0e2574c17f7aa77bf5 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_16 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is $2$ , the radius of the circle that matches it has a radius of $\frac{1}{2}$ . This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only $1$ point. When a larger circle is used, it is tangent to $3$ points because the points on either side are now separated from the vertex. Therefore, $\boxed{12}$ is correct. | E | 12 |
0269dd79e6f75f0e2574c17f7aa77bf5 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_16 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | Now, let's graph these two equations. We want the blue parabola to be inside this red circle. [asy] import graph; size(6cm); draw((0,0)--(0,10),EndArrow); draw((0,0)--(0,-10),EndArrow); draw((0,0)--(10,0),EndArrow); draw((0,0)--(-10,0),EndArrow); Label f; f.p=fontsize(6); xaxis(-10,10); yaxis(-10,10); real f(real x) { return x^2-5; } draw(graph(f,-4,4),blue+linewidth(1)); draw(circle((0,0),5),red); dot(scale(.7)*"$a$",(0,5),NE); dot(scale(.7)*"$-a$",(0,-5),N); dot(scale(.7)*"$a$",(5,0),NE); dot(scale(.7)*"$-a$",(-5,0),SE); [/asy] Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$ . Expanding, we get $x^4-2ax^2+x^2=0$ . Factoring out the $x^2$ , we get $x^2(x^2-2a+1)=0$ . Then we find that $x=0$ or $x=\pm\sqrt{2a-1}$ . Therefore, $2a-1>0$ , which means $\boxed{12}$ | E | 12 |
0269dd79e6f75f0e2574c17f7aa77bf5 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_16 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | In order to solve for the values of $a$ , we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that $\sqrt{a^2 - x^2} = x^2 - a$ . Now, we take square of both sides, and rearrange to obtain $x^4 - (2a - 1)x^2 = 0$ . Now, we may take the second derivative of the equation to obtain $6x^2 - (2a - 1) = 0$ . Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,
\[b^2 - 4ac > 0 \rightarrow 0 - 4(6)(-(2a - 1)) > 0 \rightarrow a > \frac{1}{2}\] The answer is $\boxed{12}$ and we are done. | E | 12 |
0269dd79e6f75f0e2574c17f7aa77bf5 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_16 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the $y$ -axis, thus, for them to intersect at exactly $3$ points, it suffices to find the $y$ solution.
First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than $0$ (It is the square root part of the quadratic formula). To find when it is $0$ , we find the roots: \[4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}\] Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$ , our range is $\boxed{12}$ | E | 12 |
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