AMC12 2021 B

We see that $1, 2, 3,$ and $4$ each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$, we find that the sum is equal to \[10\cdot(1+10+100+1000)=\boxed{\textbf{(E)} \: 11{,}110}.\] Note that it is equally valid to manually add all four numbers together to get the answer.

The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$ Therefore, the area of the shaded figure is \[2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{\textbf{(B)} \: 6}.\]

At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$ At $4{:}00,$ we get \begin{align*} |(M-5)-(L+3)| &= 2 \\ |M-L-8| &= 2 \\ |N-8| &= 2. \end{align*} We have two cases: 1. If $N-8=2,$ then $N=10.$ 2. If $N-8=-2,$ then $N=6.$ Together, the product of all possible values of $N$ is $10\cdot6=\boxed{\textbf{(C)} \: 60}.$

We have \[n=8^{2022}= \left(8^\frac{2}{3}\right)^{3033}=4^{3033}.\] Therefore, \[\frac{n}4=\boxed{\textbf{(E)} \: 4^{3032}}.\]

The special fractions are \[\frac{1}{14},\frac{2}{13},\frac{3}{12},\frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3},\frac{13}{2},\frac{14}{1}.\] We rewrite them in the simplest form: \[\frac{1}{14},\frac{2}{13},\frac{1}{4},\frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},1\frac{1}{7},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once: \[\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] For the set $\{2,4,14\},$ two elements (not necessarily different) can sum to $4,6,8,16,18,28.$ For the set $\left\{\frac{1}{2},1\frac{1}{2},6\frac{1}{2}\right\},$ two elements (not necessarily different) can sum to $1,2,3,7,8,13.$ For the set $\left\{\frac{1}{4},2\frac{3}{4}\right\},$ two elements (not necessarily different) can sum to $3.$ Together, there are $\boxed{\textbf{(C)}\ 11}$ distinct integers that can be written as the sum of two, not necessarily different, special fractions: \[1,2,3,4,6,7,8,13,16,18,28.\]

We have \begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*} Since $129$ is composite, $127$ is the largest prime which can divide $16383$. The sum of $127$'s digits is $1+2+7=\boxed{\textbf{(C) }10}$. Note that you can quickly tell that $2^7 -1$ is prime because it is a Mersenne prime.

It is obvious $x$, $y$, and $z$ are symmetrical. We are going to solve the problem by Completing the Square. $x ^ 2 + y ^ 2 + z ^ 2 - xy - yz - zx = 1$ $2x ^ 2 + 2y ^ 2 + 2z ^ 2 - 2xy - 2yz - 2zx = 2$ $(x-y)^2 + (y-z)^2 + (z-x)^2 = 2$ Because $x, y, z$ are integers, $(x-y)^2$, $(y-z)^2$, and $(z-x)^2$ can only equal $0, 1, 1$. So one variable must equal another, and the third variable is $1$ different from those $2$ equal variables. So the answer is $\boxed{D}$.

Let the lengths of the two congruent sides of the triangle be $x$, then the product desired is $x^2$. Notice that the product of the base and twice the height is $4$ times the area of the triangle. Set the vertex angle to be $a$, we derive the equation: $x^2=4\left(\frac{1}{2}x^2\sin(a)\right)$ $\sin(a)=\frac{1}{2}$ As the triangle is obtuse, $a=150^\circ$ only. We get $\boxed{\textbf{(D)} \ 150}.$

Construct the circle that passes through $A$, $O$, and $C$, centered at $X$. Also notice that $\overline{OA}$ and $\overline{OC}$ are the angle bisectors of angle $\angle BAC$ and $\angle BCA$ respectively. We then deduce $\angle AOC=120^\circ$. Consider another point $M$ on Circle $X$ opposite to point $O$. As $AOCM$ is an inscribed quadrilateral of Circle $X$, $\angle AMC=180^\circ-120^\circ=60^\circ$. Afterward, deduce that $\angle AXC=2·\angle AMC=120^\circ$. By the Cosine Rule, we have the equation: (where $r$ is the radius of circle $X$) $2r^2(1-\cos(120^\circ))=6^2$ $r^2=12$ The area is therefore $\pi r^2 = \boxed{\textbf{(B)}\ 12\pi}$.

Let $A = (\cos 40^{\circ}, \sin 40^{\circ}), B = (\cos 60^{\circ}, \sin 60^{\circ}),$ and $C = (\cos t^{\circ}, \sin t^{\circ}).$ We apply casework to the legs of isosceles $\triangle ABC:$ 1. $AB=AC$ Note that $A$ must be the midpoint of $\widehat{BC}.$ It follows that $C = (\cos 20^{\circ}, \sin 20^{\circ}),$ so $t=20.$ 2. $BA=BC$ Note that $B$ must be the midpoint of $\widehat{AC}.$ It follows that $C = (\cos 80^{\circ}, \sin 80^{\circ}),$ so $t=80.$ 3. $CA=CB$ Note that $C$ must be the midpoint of $\widehat{AB}.$ It follows that $C = (\cos 50^{\circ}, \sin 50^{\circ})$ or $C = (\cos 230^{\circ}, \sin 230^{\circ}),$ so $t=50$ or $t=230.$ Together, the sum of all such possible values of $t$ is $20+80+50+230=\boxed{\textbf{(E)} \: 380}.$ Remark The following diagram shows all possible locations of $C:$

We will use complementary counting to find the probability that the product is not divisible by $4$. Then, we can find the probability that we want by subtracting this from 1. We split this into $2$ cases. Case 1: The product is not divisible by $2$. We need every number to be odd, and since the chance we roll an odd number is $\frac12,$ our probability is $\left(\frac12\right)^6=\frac1{64}.$ Case 2: The product is divisible by $2$, but not by $4$. We need $5$ numbers to be odd, and one to be divisible by $2$, but not by $4$. There is a $\frac12$ chance that an odd number is rolled, a $\frac13$ chance that we roll a number satisfying the second condition (only $2$ and $6$ work), and $6$ ways to choose the order in which the even number appears. Our probability is $\left(\frac12\right)^5\left(\frac13\right)\cdot6=\frac1{16}.$ Therefore, the probability the product is not divisible by $4$ is $\frac1{64}+\frac1{16}=\frac{5}{64}$. Our answer is $1-\frac{5}{64}=\boxed{\textbf{(C)}\ \frac{59}{64}}$.

The prime factorizations of $768$ and $384$ are $2^8\cdot3$ and $2^7\cdot3,$ respectively. Note that $f(n)$ is the sum of all fractions of the form $\frac 1d,$ where $d$ is a positive divisor of $n.$ By geometric series, it follows that \begin{alignat*}{8} f(768)&=\left(\sum_{k=0}^{8}\frac{1}{2^k}\right)+\left(\sum_{k=0}^{8}\frac{1}{2^k\cdot3}\right)&&=\frac{511}{256}+\frac{511}{768}&&=\frac{2044}{768}, \\ f(384)&=\left(\sum_{k=0}^{7}\frac{1}{2^k}\right)+\left(\sum_{k=0}^{7}\frac{1}{2^k\cdot3}\right)&&=\frac{255}{128}+\frac{255}{384}&&=\frac{1020}{384}. \end{alignat*} Therefore, the answer is $f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.$

Plugging in $c$, we get \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.\] Since $\sin(x+2\pi)=\sin(x),$ $\sin(2\pi-x)=\sin(-x),$ and $\sin(-x)=-\sin(x),$ we get \[\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{\textbf{(E)}\ 1}.\]

The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z) \cdot Q(z)$ has degree $2 + 3 = 5,$ we conclude that $R(z) - P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant. It now suffices to illustrate an example for which $N = 1$: Take \begin{align*} P(z)&=z^2+1, \\ Q(z)&=z^3+2, \\ R(z)&=(z+1)^6 + P(z) \cdot Q(z). \end{align*} Note that $R(z)$ has degree $6$ and constant term $3,$ so it satisfies the conditions. We need to find the solutions to \begin{align*} P(z) \cdot Q(z) &= (z+1)^6 + P(z) \cdot Q(z) \\ 0 &= (z+1)^6. \end{align*} Clearly, the only distinct complex root is $-1,$ so our answer is $N=\boxed{\textbf{(B)} \: 1}.$

The $24$-sided polygon is made out of $24$ shapes like $\triangle ABC$. Then $\angle BAC=360^\circ/24=15^\circ$, and $\angle EAC = 45^\circ$, so $\angle{EAB} = 30^{\circ}$. Then $EB=AE\tan 30^\circ = \sqrt{3}$; therefore $BC=EC-EB=3-\sqrt{3}$. Thus \[[ABC] = \frac{BC}{EC}\cdot [ACE] = \frac{3-\sqrt{3}}{3}\cdot \frac 92\]and the required area is $24\cdot[ABC] =108-36\sqrt{3}$. Finally $108+36+3=\boxed{(\textbf{E})\ 147}$.

Because $a + b + c$ is odd, $a$, $b$, $c$ are either one odd and two evens or three odds. $\textbf{Case 1}$: $a$, $b$, $c$ have one odd and two evens. Without loss of generality, we assume $a$ is odd and $b$ and $c$ are even. Hence, ${\rm gcd} \left( a , b \right)$ and ${\rm gcd} \left( a , c \right)$ are odd, and ${\rm gcd} \left( b , c \right)$ is even. Hence, ${\rm gcd} \left( a , b \right) + {\rm gcd} \left( b , c \right) + {\rm gcd} \left( c , a \right)$ is even. This violates the condition given in the problem. Therefore, there is no solution in this case. $\textbf{Case 2}$: $a$, $b$, $c$ are all odd. In this case, ${\rm gcd} \left( a , b \right)$, ${\rm gcd} \left( a , c \right)$, ${\rm gcd} \left( b , c \right)$ are all odd. Without loss of generality, we assume \[ {\rm gcd} \left( a , b \right) \leq {\rm gcd} \left( b , c \right) \leq {\rm gcd} \left( c , a \right) . \] $\textbf{Case 2.1}$: ${\rm gcd} \left( a , b \right) = 1$, ${\rm gcd} \left( b , c \right) = 1$, ${\rm gcd} \left( c , a \right) = 7$. The only solution is $(a, b, c) = (7, 9, 7)$. Hence, $a^2 + b^2 + c^2 = 179$. $\textbf{Case 2.2}$: ${\rm gcd} \left( a , b \right) = 1$, ${\rm gcd} \left( b , c \right) = 3$, ${\rm gcd} \left( c , a \right) = 5$. The only solution is $(a, b, c) = (5, 3, 15)$. Hence, $a^2 + b^2 + c^2 = 259$. $\textbf{Case 2.3}$: ${\rm gcd} \left( a , b \right) = 3$, ${\rm gcd} \left( b , c \right) = 3$, ${\rm gcd} \left( c , a \right) = 3$. There is no solution in this case. Therefore, putting all cases together, the answer is $179 + 259 = \boxed{\textbf{(B)} \: 438}$.

Let $S(n)$ be the number of paths of $n$ moves such that the bug never will have been more than $1$ unit away from the starting position. Clearly, by symmetry, there are two possible states here, the bug being on the center and the bug being on one of the vertices of the unit hexagon around the center. Let $C(n)$ be the number of paths with the aforementioned restriction that end on the center. Let $V(n)$ be the number of paths with the aforementioned restriction that end on a vertex of the surrounding unit hexagon. We have $S(n) = 6C(n-1) + 3V(n-1),$ since from the center, there are $6$ possible points to land to and from a vertex there are $3$ possible points to land to (the two adjacent vertices and the center). We also have $C(n) = V(n-1)$, since to get to the center the bug must have come from a vertex, and $V(n) = 2V(n-1) + 6C(n-1),$ since from a vertex there are two vertices to move to, and from the center there are $6$ vertices to move to. We can construct a recursion table using the base cases $V(1) = 6$ and $C(1) = 0$ and our recursive rules for $C(n)$ and $V(n)$ as follows: \[\begin{array}{c|c|c} n & V(n) & C(n) \\ \hline & & \\ [-2ex] 1 & 6 & 0 \\ 2 & 12 & 6 \\ 3 & 60 & 12 \\ 4 & 192 & 60 \\ \end{array}\] Then, $S(5) = 6C(4) + 3V(4) = 6 \cdot 60 + 3 \cdot 192 = 936,$ and the desired probability is thus $\frac{936}{6^5} = \boxed{\textbf{(A)}\ \frac{13}{108}}.$

Note that terms of the sequence $(u_k)$ lie in the interval $\left(0,\frac12\right)$ and are strictly increasing. Since the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$ The given equation becomes \[L=2L-2L^2,\] from which $L=\frac12.$ The given inequality becomes \[\frac12-\frac{1}{2^{1000}} \leq u_k \leq \frac12+\frac{1}{2^{1000}},\] and we only need to consider $\frac12-\frac{1}{2^{1000}} \leq u_k.$ We have \begin{alignat*}{8} u_0 &= \phantom{1}\frac14 &&= \frac{2^1-1}{2^2}, \\ u_1 &= \phantom{1}\frac38 &&= \frac{2^2-1}{2^3}, \\ u_2 &= \ \frac{15}{32} &&= \frac{2^4-1}{2^5}, \\ u_3 &= \frac{255}{512} &&= \frac{2^8-1}{2^9}, \\ & \phantom{1111} \vdots \end{alignat*} By induction, it can be proven that \[u_k=\frac{2^{2^k}-1}{2^{2^k+1}}=\frac12-\frac{1}{2^{2^k+1}}.\] We substitute this into the inequality, then solve for $k:$ \begin{align*} \frac12-\frac{1}{2^{1000}} &\leq \frac12-\frac{1}{2^{2^k+1}} \\ -\frac{1}{2^{1000}} &\leq -\frac{1}{2^{2^k+1}} \\ 2^{1000} &\leq 2^{2^k+1} \\ 1000 &\leq 2^k+1. \end{align*} Since $2^9+1 \leq 1000 \leq 2^{10}+1,$ the least such value of $k$ is $\boxed{\textbf{(A)}\: 10}.$

Imagine we have $2$ regular polygons with $m$ and $n$ sides and $m>n$ inscribed in a circle without sharing a vertex. We see that each side of the polygon with $n$ sides (the polygon with fewer sides) will be intersected twice. (We can see this because to have a vertex of the $m$-gon on an arc subtended by a side of the $n$-gon, there will be one intersection to “enter” the arc and one to “exit” the arc. ~KingRavi) This means that we will end up with $2$ times the number of sides in the polygon with fewer sides. If we have polygons with $5,$ $6,$ $7,$ and $8$ sides, we need to consider each possible pair of polygons and count their intersections. Throughout $6$ of these pairs, the $5$-sided polygon has the least number of sides $3$ times, the $6$-sided polygon has the least number of sides $2$ times, and the $7$-sided polygon has the least number of sides $1$ time. Therefore the number of intersections is $2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{(\textbf{E}) \:68}$.

This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory. We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube. Now the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence. For example, in Figure $(1)$, as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube. Here is how the $4$ blue unit cubes are arranged: In Figure $(1)$: $4$ blue unit cubes are on the same layer (horizontal or vertical). In Figure $(2)$: $4$ blue unit cubes are in $T$ shape. In Figure $(3)$ and $(4)$: $4$ blue unit cubes are in $S$ shape. In Figure $(5)$: $3$ blue unit cubes are in $L$ shape, and the other is isolated without a shared face. In Figure $(6)$: $2$ pairs of neighboring blue unit cubes are isolated from each other without a shared face. In Figure $(7)$: $4$ blue unit cubes are isolated from each other without a shared face. So the answer is $\boxed{\textbf{(A)}\ 7}$

Let $a=\cos(x)+i\sin(x)$. Now $P(a)=1+a-a^2+a^3$. $P(-1)=-2$ and $P(0)=1$ so there is a real root $a_1$ between $-1$ and $0$. The other $a$'s must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex $a$'s squared is $-\frac{1}{a_1}$, which is greater than $1$. If $x$ is real number then $a$ must have magnitude of $1$, but none of the solutions for $a$ have magnitude of $1$, so the answer is $\boxed{\textbf{(A)}\ 0 }$ ~lopkiloinm

Let $M$ be the midpoint of $AB$; so $BM=AM=5$. Let $D$ be the point such that $ABCD$ is a rectangle. Then $MO\perp AB$ and $MP\perp AB$. Let $\theta = \angle BAC$; so $\tan\theta = \tfrac 68 = \tfrac 34$. Then \[OP=MP-MO=AM\cot\theta - BM\tan\theta = 5(\tfrac 43 - \tfrac 34) = \boxed{\textbf{(C)}\ \tfrac{35}{12}}.\]

There are $29$ possible pairs of consecutive integers, namely $p_1=\{1,2\}, \cdots, p_{29}=\{29,30\}$. Define a random variable $X_i$, with $X_i=1$, if $p_i$ is part of the 5-element subset, and $0$ otherwise. Then the number of pairs of consecutive integers in a $5$-element selection is given by the sum $X_1+\cdots + X_{29}$. By linearity of expectation, the expected value is equal to the sum of the $\mathbb{E}[X_i]$: \[\mathbb{E}[X_1+\cdots +X_{29}]=\mathbb{E}[X_1]+\cdots + \mathbb{E}[X_{29}]\] To compute $\mathbb{E}[X_i]$, note that $X_i=1$ for a total of $_{28}C_3$ out of $_{30}C_5$ possible selections. Thus\[\mathbb{E}[X_i]=\frac{\binom{28}{3}}{\binom{30}{5}}= \frac 1{29}\cdot \frac 23, \quad \textrm{which implies} \quad \mathbb{E}[X_1+\cdots +X_{29}]= \boxed{\textbf{(A)}\ \frac{2}{3}}.\]

By the Law of Cosine $\cos A = \frac{AC^2 + AB^2 - BC^2}{ 2 \cdot AC \cdot AB} = \frac{20^2 + 11^2 - 24^2}{2\cdot20\cdot11} = -\frac18$ As $ABEC$ is a cyclic quadrilateral, $\angle CEA = \angle CBA$. As $BDEF$ is a cyclic quadrilateral, $\angle CBA = \angle FEA$. $\because \quad \angle CEA = \angle FEA \quad \text{and} \quad \angle CAE = \angle FAE$ $\therefore \quad \triangle AFE \cong \triangle ACE$ by $ASA$ Hence, $AF = AC = 20$ By the Law of Cosine $CF = \sqrt{20^2 + 20^2 - 2 \cdot 20 \cdot 20 (-\frac18)} = \sqrt{900} = \boxed{\textbf{C}~\text{30}}$ Note that $F$ is $C$'s reflection over line $AE$, quadrilateral $ACEF$ is a kite symmetrical by line $AE$, $AE \perp CF$.

Note that we can add $9$ to $R(n)$ to get $R(n+1)$, but must subtract $k$ for all $k|n+1$. Hence, we see that there are four ways to do that because $9=7+2=6+3=5+4=4+3+2$. Note that only $7+2$ is a plausible option, since $4+3+2$ indicates $n+1$ is divisible by $6$, $5+4$ indicates that $n+1$ is divisible by $2$, $6+3$ indicates $n+1$ is divisible by $2$, and $9$ itself indicates divisibility by $3$, too. So, $14|n+1$ and $n+1$ is not divisible by any positive integers from $2$ to $10$, inclusive, except $2$ and $7$. We check and get that only $n+1=14 \cdot 1$ and $n+1=14 \cdot 7$ give possible solutions so our answer is $\boxed{\textbf{(C) }2}$.