AMC12 2023 A

This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$. The $18x$ is Alicia so $18\times1.5=\boxed{\textbf{(E) 27}}$

Use a system of equations. Let $x$ be the weight of a pizza and $y$ be the weight of a cup of orange slices. We have \[\frac{1}{3}x+\frac{7}{2}y=\frac{3}{4}x+\frac{1}{2}y.\] Rearranging, we get \begin{align*} \frac{5}{12}x&=3y, \\ x&=\frac{36}{5}y. \end{align*} Plugging in $\frac{1}{4}$ pounds for $y$ by the given gives $\frac{9}{5}=\boxed{\textbf{(A) }1\frac{4}{5}}.$

Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$, there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}$ perfect squares less than 2023 that are divisible by 5.

Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$. $10^{15}$ has $16$ digits and $243$ = $2.43*10^{2}$ gives us $3$ more digits. $16+2=\text{\boxed{\textbf{(E) }18}}$ $2.43*10^{17}$ has $18$ digits

There are $3$ cases where the running total will equal $3$: one roll; two rolls; or three rolls: Case 1: The chance of rolling a running total of $3$, namely $(3)$ in exactly one roll is $\frac{1}{6}$. Case 2: The chance of rolling a running total of $3$ in exactly two rolls, namely $(1, 2)$ and $(2, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$. Case 3: The chance of rolling a running total of 3 in exactly three rolls, namely $(1, 1, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$. Using the rule of sum we have $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$.

Let $A(6+m,2+n)$ and $B(6-m,2-n)$, since $(6,2)$ is their midpoint. Thus, we must find $2m$. We find two equations due to $A,B$ both lying on the function $y=\log_{2}x$. The two equations are then $\log_{2}(6+m)=2+n$ and $\log_{2}(6-m)=2-n$. Now add these two equations to obtain $\log_{2}(6+m)+\log_{2}(6-m)=4$. By logarithm rules, we get $\log_{2}((6+m)(6-m))=4$. By raising 2 to the power of both sides, we obtain $(6+m)(6-m)=16$. We then get \[36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}\]. Since we're looking for $2m$, we obtain $(2)(2\sqrt{5})=\boxed{\textbf{(D) }4\sqrt{5}}$

Do careful casework by each month. Make sure to start with 2023. In the month and the date, we need a $0$, a $3$, and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\boxed{\textbf{(E)}~9}$. For curious readers, the numbers (in chronological order) are:

Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously. We can write the following equations: \[\frac{ax+11}{a+1}=x+1\qquad (1)\] \[\frac{ax+33}{a+3}=x+2\qquad (2)\] Multiplying equation $(1)$ by $(a+1)$ and solving, we get: \[ax+11=ax+a+x+1\] \[11=a+x+1\] \[a+x=10\qquad (3)\] Multiplying equation $(2)$ by $(a+3)$ and solving, we get: \[ax+33=ax+2a+3x+6\] \[33=2a+3x+6\] \[2a+3x=27\qquad (4)\] Solving the system of equations for $(3)$ and $(4)$, we find that $a=3$ and $x=\boxed{\textbf{(D) }7}$.

The side lengths of the inner square and outer square are $\sqrt{2}$ and $\sqrt{3}$ respectively. Let the shorter side of our triangle be $x$, thus the longer leg is $\sqrt{3}-x$. Hence, by the Pythagorean Theorem, we have \begin{align*} (\sqrt{3}-x)^2+x^2&=(\sqrt{2})^2 \\ 3-2\sqrt{3}x+x^2+x^2&=2 \\ 2x^2-2\sqrt{3}x+1&=0. \end{align*} By the quadratic formula, we find that $x=\frac{\sqrt{3}\pm1}{2}$, so the answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.$

Because $y^3=x^2$, set $x=a^3$, $y=a^2$ ($a\neq 0$). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$. Solve the equation to get $a=3$ or $-1$. Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\boxed{\textbf{(D)} 36}$.

Remind that $\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta$ where $\theta$ is the angle between the slope and $x$-axis. $k_1=2=\tan \alpha$, $k_2=\dfrac{1}{3}=\tan \beta$. The angle formed by the two lines is $\alpha-\beta$. $\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1$. Therefore, $\alpha-\beta=\boxed{\textbf{(C)} 45^\circ}$.

To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas. \[2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3\] $=(2-1)(2^2+1 \cdot 2+1^2)+(4-3)(4^2+4 \cdot 3+3^2)+(6-5)(6^2+6 \cdot 5+5^2)+...+(18-17)(18^2+18 \cdot 17+17^2)$ $=(2^2+1 \cdot 2+1^2)+(4^2+4 \cdot 3+3^2)+(6^2+6 \cdot 5+5^2)+...+(18^2+18 \cdot 17+17^2)$ $=1^2+2^2+3^2+4^2+5^2+6^2...+17^2+18^2+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$ $=\frac{18(18+1)(36+1)}{6}+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$ we could rewrite the second part as $\sum_{n=1}^{9}(2n-1)(2n)$ $(2n-1)(2n)=4n^2-2n$ $\sum_{n=1}^{9}4n^2=4(\frac{9(9+1)(18+1)}{6})$ $\sum_{n=1}^{9}-2n=-2(\frac{9(9+1)}{2})$ Hence, $1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18 = 4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$ Adding everything up: $2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$ $=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$ $=3(19)(37)+6(10)(19)-9(10)$ $=2109+1140-90$ $=\boxed{\textbf{(D) } 3159}$

We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$, and since $l = 1.4r$, $g = 2.4r$. Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is $n(n-1)/2$, the sum of the first $n-1$ numbers. Now, setting 36 and 48 equal to the equation will show that two consecutive numbers must have a product of 72 or 96. Clearly, $72=8*9$, so the answer is $\boxed{\textbf{(B) }36}$.

When $z^5=\overline{z}$, there are two conditions: either $z=0$ or $z\neq 0$. When $z\neq 0$, since $|z^5|=|\overline{z}|$, $|z|=1$. $z^5\cdot z=z^6=\overline{z}\cdot z=|z|^2=1$. Consider the $r(\cos \theta +i\sin \theta)$ form, when $z^6=1$, there are 6 different solutions for $z$. Therefore, the number of complex numbers satisfying $z^5=\bar{z}$ is $\boxed{\textbf{(E)} 7}$.

By "unfolding" $APQRS$ into a straight line, we get a right angled triangle $ABS'$. It follows that \begin{align*} \cos(\theta)&=\frac{100}{120} \\ \theta&=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}. \end{align*}

First, substitute in $z=a+bi$. \[|1+(a+bi)+(a+bi)^2|=4\] \[|(1+a+a^2-b^2)+ (b+2ab)i|=4\] \[(1+a+a^2-b^2)^2+ (b+2ab)^2=16\] \[(1+a+a^2-b^2)^2+ b^2(1+4a+4a^2)=16\] Let $p=b^2$ and $q=1+a+a^2$ \[(q-p)^2+ p(4q-3)=16\] \[p^2-2pq+q^2 + 4pq -3p=16\] We are trying to maximize $b=\sqrt p$, so we'll turn the equation into a quadratic to solve for $p$ in terms of $q$. \[p^2+(2q-3)p+(q^2-16)=0\] \[p=\frac{(-2q+3)\pm \sqrt{-12q+73}}{2}\] We want to maximize $p$. Since $q$ is always negatively contributing to $p$'s value, we want to minimize $q$. Due to the trivial inequality: $q=1+a+a^2=(a+\frac 12)^2+\frac{3}4 \geq \frac{3}4$ If we plug $q$'s minimum value in, we get that $p$'s maximum value is \[p=\frac{(-2(\frac 34)+3)+ \sqrt{-12(\frac 34)+73}}{2}=\frac{\frac 32+ 8}{2}=\frac{19}{4}\] Then \[b=\frac{\sqrt{19}}{2}\] and \[m+n=\boxed{\textbf{(B)}~21}\]

Initially, the probability of landing at $10$ and landing past $10$ (summing the infinte series) are exactly the same. Landing before 10 repeats this initial condition, with a different irrelevant scaling factor. Therefore, the probability must be $\boxed{\textbf{(E)}~\frac12}$.

Let $O$ be the center of the midpoint of the line segment connecting both the centers, say $A$ and $B$. Let the point of tangency with the inscribed circle and the right larger circles be $T$. Then $OT = BO + BT = BO + AT - \frac{1}{2} = \frac{1}{4} + 1 - \frac{1}{2} = \frac{3}{4}.$ Since $C_4$ is internally tangent to $C_1$, center of $C_4$, $C_1$ and their tangent point must be on the same line. Now, if we connect centers of $C_4$, $C_3$ and $C_1$/$C_2$, we get a right angled triangle. Let the radius of $C_4$ equal $r$. With the pythagorean theorem on our triangle, we have \[\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2\] Solving this equation gives us \[r = \boxed{\textbf{(D) } \frac{3}{28}}\]

For $\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$, transform it into $\dfrac{\ln 289+\ln 7}{\ln 7 + \ln x}\cdot \dfrac{\ln 289+\ln 7}{\ln 289 + \ln x}=\dfrac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}$. Replace $\ln x$ with $y$. Because we want to find the product of all solutions of $x$, it is equivalent to finding the exponential of the sum of all solutions of $y$. Change the equation to standard quadratic equation form, the term with 1 power of $y$ is canceled. By using Vieta, we see that since there does not exist a $by$ term, $\sum y=0$ and $\prod x=e^0=\boxed{\textbf{(C)} 1}$.

First, let $R(n)$ be the sum of the $n$th row. Now, with some observation and math instinct, we can guess that $R(n) = 2^n - n$. Now we try to prove it by induction, $R(1) = 2^n - n = 2^1 - 1 = 1$ (works for base case) $R(k) = 2^k - k$ $R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$ By definition from the question, the next row is always$:$ Double the sum of last row (Imagine each number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (minus leftmost and rightmost's 1) Which gives us $:$ $2(2^k - k) + (k + 1) - 2 = 2(2^k) - k - 1$ Hence, proven Last, simply substitute $n = 2023$, we get $R(2023) = 2^{2023} - 2023$ Last digit of $2^{2023}$ is $8$, $8-3 = \boxed{\textbf{(C) } 5}$

To find the total amount of vertices we first find the amount of edges, and that is $\frac{20 \times 3}{2}$. Next, to find the amount of vertices we can use Euler's characteristic, $V - E + F = 2$, and therefore the amount of vertices is $12$ So there are $\binom{12}{3} = 220$ ways to choose 3 distinct points. Now, the furthest distance we can get from one point to another point in an icosahedron is 3. Which gives us a range of $1 \leq d(Q, R), d(R, S) \leq 3$ With some case work, we get two cases: Case 1: $d(Q, R) = 3; d(R, S) = 1, 2$ Since we have only one way to choose Q, that is, the opposite point from R, we have one option for Q and any of the other points could work for S. Then, we get $12 \times 1 \times 10 = 120$ (ways to choose R × ways to choose Q × ways to choose S) Case 2: $d(Q, R) = 2; d(R, S) = 1$ We can visualize the icosahedron as 4 rows, first row with 1 vertex, second row with 5 vertices, third row with 5 vertices and fourth row with 1 vertex. We set R as the one vertex on the first row, and we have 12 options for R. Then, Q can be any of the 5 points on the third row and finally S can be one of the 5 points on the second row. Therefore, we have $12 \times 5 \times 5 = 300$ (ways to choose R × ways to choose Q × ways to choose S) In total, we have $420$ ways, but we must divide by $3!$ to account for permutations, giving us $70/220 = \boxed{\textbf{(A) } 7/22}.$

First, we note that $f(1) = 1$, since the only divisor of $1$ is itself. Then, let's look at $f(p)$ for $p$ a prime. We see that \[\sum_{d \mid p} d \cdot f\left(\frac{p}{d}\right) = 1\] \[1 \cdot f(p) + p \cdot f(1) = 1\] \[f(p) = 1 - p \cdot f(1)\] \[f(p) = 1-p\] Nice. Now consider $f(p^k)$, for $k \in \mathbb{N}$. \[\sum_{d \mid p^k} d \cdot f\left(\frac{p^k}{d}\right) = 1\] \[1 \cdot f(p^k) + p \cdot f(p^{k-1}) + p^2 \cdot f(p^{k-2}) + \dotsc + p^k f(1) = 1\]. It can be (strongly) inductively shown that $f(p^k) = f(p) = 1-p$. Here's how. We already showed $k=1$ works. Suppose it holds for $k = n$, then \[1 \cdot f(p^n) + p \cdot f(p^{n-1}) + p^2 \cdot f(p^{n-2}) + \dotsc + p^n f(1) = 1 \implies f(p^m) = 1-p \; \forall \; m \leqslant n\] For $k = n+1$, we have \[1 \cdot f(p^{n+1}) + p \cdot f(p^{n}) + p^2 \cdot f(p^{n-1}) + \dotsc + p^{n+1} f(1) = 1\], then using $f(p^m) = 1-p \; \forall \; m \leqslant n$, we simplify to \[1 \cdot f(p^{n+1}) + p \cdot (1-p) + p^2 \cdot (1-p) + \dotsc + p^n \cdot (1-p) + p^{n+1} f(1) = 1\] \[f(p^{n+1}) + \sum_{i=1}^n p^i (1-p) + p^{n+1} = 1\] \[f(p^{n+1}) + p(1 - p^n) + p^{n+1} = 1\] \[f(p^{n+1}) + p = 1 \implies f(p^{n+1}) = 1-p\]. Very nice! Now, we need to show that this function is multiplicative, i.e. $f(pq) = f(p) \cdot f(q)$ for $\textbf{distinct}$ $p,q$ prime. It's pretty standard, let's go through it quickly. \[\sum_{d \mid pq} d \cdot f\left(\frac{pq}{d}\right) = 1\] \[1 \cdot f(pq) + p \cdot f(q) + q \cdot f(p) + pq \cdot f(1) = 1\] Using our formulas from earlier, we have \[f(pq) + p(1-q) + q(1-p) + pq = 1 \implies f(pq) = 1 - p(1-q) - q(1-p) - pq = (1-p)(1-q) = f(p) \cdot f(q)\] Great! We're almost done now. Let's actually plug in $2023 = 7 \cdot 17^2$ into the original formula. \[\sum_{d \mid 2023} d \cdot f\left(\frac{2023}{d}\right) = 1\] \[1 \cdot f(2023) + 7 \cdot f(17^2) + 17 \cdot f(7 \cdot 17) + 7 \cdot 17 \cdot f(17) + 17^2 \cdot f(7) + 7 \cdot 17^2 \cdot f(1) = 1\] Let's use our formulas! We know \[f(7) = 1-7 = -6\] \[f(17) = 1-17 = -16\] \[f(7 \cdot 17) = f(7) \cdot f(17) = (-6) \cdot (-16) = 96\] \[f(17^2) = f(17) = -16\] So plugging ALL that in, we have \[f(2023) = 1 - \left(7 \cdot (-16) + 17 \cdot (-6) \cdot (-16) + 7 \cdot 17 \cdot (-16) + 17^2 \cdot (-6) + 7 \cdot 17^2\right)\] which, be my guest simplifying, is $\boxed{\textbf{(B)} \ 96}$

Using the AM-GM inequality on the two terms in each factor on the left-hand side, we get \[(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,\] This means the equality condition must be satisfied. Therefore, we must have $1 = 2a = b$, so the only solution is $\boxed{\textbf{(B) }1}$.

Consider any sequence with $n$ terms. Every number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot ..., and appear the first time in the $n$th spot, which means every number has $(n+1)$ choices to show up in the sequence. Consequently, for each sequence with length $n$, there are $(n+1)^{10}$ possible ways. Thus, the desired value is $\sum_{n=1}^{10}(n+1)^{10}\equiv \boxed{\textbf{(C) } 5}\pmod{10}$.

\begin{align*} \cos 2023 x + i \sin 2023 x &= (\cos x + i \sin x)^{2023}\\ &= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2023}{3} \cos^{2020} x (i \sin x)^{3}\\ &+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\\ &= \cos^{2023} x + i \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{2} \cos^{2021} x \sin^{2} x - i\binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots\\ &- \binom{2023}{2022} \cos x \sin^{2022} x - i \sin^{2023} x\\ \end{align*} By equating real and imaginary parts: \[\cos 2023 x = \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x\] \[\sin 2023 x = \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x\] \begin{align*} \tan2023x &= \frac{ \sin2023x }{ \cos2023x } = \frac{ \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x }{ \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x }\\ &= \frac{ \binom{2023}{1} \frac{\cos^{2022} x \sin x}{\cos^{2023} x} - \binom{2023}{3} \frac{\cos^{2020} x \sin^{3} x}{\cos^{2023} x} + \dots - \frac{\sin^{2023} x}{\cos^{2023} x} }{ \frac{\cos^{2023} x}{\cos^{2023} x} - \binom{2023}{2} \frac{\cos^{2021} x \sin^{2} x}{\cos^{2023} x} + \dots - \binom{2023}{2022} \frac{\cos x \sin^{2022} x}{\cos^{2023} x} }\\ &= \frac{ \binom{2023}{1} \tan x - \binom{2023}{3} \tan^{3}x + \dots - \tan^{2023}x }{ 1 - \binom{2023}{2} \tan^{2}x + \dots - \binom{2023}{2022} \tan^{2022} x }\\ \end{align*} \[a_{2023} = \boxed{\textbf{(C)}-1}\] This problem is the same as problem 7.64 in the Art of Problem Solving textbook Precalculus chapter 7 that asks to prove $\tan{nx} = \frac{\binom{n}{1}\tan{x} - \binom{n}{3}\tan^{3}{x} + \binom{n}{5}\tan^{5}{x} - \binom{n}{7}\tan^{7}{x} + \dots}{1 - \binom{n}{2}\tan^{2}{x} + \binom{n}{4}\tan^{4}{x} - \binom{n}{6}\tan^{6}{x} + \dots}$