AMC 12 2023 Test A
Instructions
- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
- No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
- Figures are not necessarily drawn to scale.
- You will have 75 minutes working time to complete the test.
Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$. Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet?
$\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
$\textbf{E}$
The relative speed of Alicia and Beth is $18+12=30$ miles per hour. It takes them $45/30=1.5$ hours to meet each other. The place where they meet is $18\times1.5=27$ miles away from City $A$.
The weight of $\dfrac{1}{3}$ of a large pizza together with $3 \dfrac{1}{2}$ cups of orange slices is the same weight of $\dfrac{3}{4}$ of a large pizza together with $\dfrac{1}{2}$ cups of orange slices. A cup of orange slices weigh $\dfrac{1}{4}$ of a pound. What is the weight, in pounds, of a large pizza?
$\textbf{(A) }1\dfrac{4}{5}\qquad\textbf{(B) }2\qquad\textbf{(C) }2\dfrac{2}{5}\qquad\textbf{(D) }3\qquad\textbf{(E) }3\dfrac{3}{5}$
$\textbf{A}$
Let $x$ and $y$ be the weight of a large pizza and a cup of orange slice in pound, respectively. Hence, we get $$\dfrac13x+3\dfrac12y=\dfrac34x+\dfrac12y\rightarrow x=\dfrac{36}{5}y$$ Given that $y=\dfrac14$, the weight of a large pizza is $x=\dfrac{36}{5}y=\dfrac95=1\dfrac45$ pounds.
How many positive perfect squares less than $2023$ are divisible by $5$?
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$
$\textbf{A}$
If a perfect square is divisible by 5, it is also divisible by $5^2=25$. Hence, the perfect squares are in the form of $5^2, 10^2, 15^2,\cdots$
We know that $45^2=2025$, which is slightly greater than 2023. Therefore, the greatest perfect square which satisfies the condition is $40^2$. The number of elements in the sequence $5^2,10^2,\cdots,40^2$ is $40/5=8$. So the answer is 8.
How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$?
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$
$\textbf{E}$
Prime factorization of the expression gives $8^5 \times 5^{10} \times 15^5=2^{15}\times3^{5}\times5^{15}=10^{15}\times3^5=243\times10^{15}$, which is 243 followed by 15 0s. So the number of digits is $3+15=18$.
Janet rolls a standard $6$-sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal $3?$
$\textbf{(A) }\dfrac{2}{9}\qquad\textbf{(B) }\dfrac{49}{216}\qquad\textbf{(C) }\dfrac{25}{108}\qquad\textbf{(D) }\dfrac{17}{72}\qquad\textbf{(E) }\dfrac{13}{54}$
$\textbf{B}$
There are $3$ cases where the running total will equal $3$: one roll, two rolls, or three rolls.
$\textbf{Case 1}$: The chance of rolling a running total of $3$ in exactly one roll, namely $(3)$, is $\dfrac{1}{6}$.
$\textbf{Case 2}$: The chance of rolling a running total of $3$ in exactly two rolls, namely $(1, 2)$ and $(2, 1)$, is $\dfrac{1}{6}\times\dfrac{1}{6}\times2=\dfrac{1}{18}$.
$\textbf{Case 3}$: The chance of rolling a running total of 3 in exactly three rolls, namely $(1, 1, 1)$, is $\dfrac{1}{6}\times\dfrac{1}{6}\times\dfrac{1}{6}=\dfrac{1}{216}$.
Hence, the probability of rolling a running total of 3 is $\dfrac{1}{6}+\dfrac{1}{18}+\dfrac{1}{216}=\dfrac{49}{216}$.
Points $A$ and $B$ lie on the graph of $y=\log_{2}x$. The midpoint of $\overline{AB}$ is $(6, 2)$. What is the positive difference between the $x$-coordinates of $A$ and $B$?
$\textbf{(A)}~2\sqrt{11}\qquad\textbf{(B)}~4\sqrt{3}\qquad\textbf{(C)}~8\qquad\textbf{(D)}~4\sqrt{5}\qquad\textbf{(E)}~9$
$\textbf{D}$
Let the coordinates of points $A$ and $B$ be $A(x_A,\log_2x_A)$, $B(x_B,\log_2x_B)$. Since the midpoint of $\overline{AB}$ is $(6, 2)$, we have \begin{align*}
\frac{x_A + x_B}{2} = 6&\rightarrow x_A+x_B=12\\
\frac{\log_2 x_A + \log_2 x_B}{2} =\frac{\log_2x_Ax_B}2= 2&\rightarrow x_Ax_B=16
\end{align*} The positive difference between the $x$-coordinates of $A$ and $B$ is \begin{align*} \left| x_A - x_B \right| & = \sqrt{\left( x_A + x_B \right)^2 - 4 x_A x_B} \\ & = 4 \sqrt{5} \end{align*} The answer is D.
A digital display shows the current date as an $8$-digit integer consisting of a $4$-digit year, followed by a $2$-digit month, followed by a $2$-digit date within the month. For example, Arbor Day this year is displayed as $20230428$. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$
$\textbf{E}$
There is one $3$ in 2023, so we need one more 3 (three more 3s are impossible for month and day). For the same reason, we need one more $0$.
If $3$ is the units digit of the month, then $0$ must be the tens digit of the month. Therefore, the day must be 11 or 22. There are 2 valid 8-digit integers (20230311 and 20230322) in this case.
If $3$ is the tens digit of the day, then the units digit of the day must be 0 or 1. When the day is 30, the only possible number for month is 11. When the day is 31, the month could be 01 or 10. There are 3 valid 8-digit integers (20231130, 20230131 and 20231031) in this case.
If $3$ is the units digit of the day, then $0$ could go in any of the $3$ remaining slots. If 0 is the tens digit of the day, then the month must be $11$. If $0$ is the units digit of the month, then the other two slots must both be $1$. If $0$ is the tens digit of the month, then the other two slots can be either both $1$ or both $2$. There are $4$ valid 8-digit integers (20231103, 20231013, 20230113 and 20230223) in this case.
In total, we have $2+3+4=9$ valid 8-digit integers.
Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$. If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$. What is the mean of her quiz scores currently?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$
$\textbf{D}$
Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores currently.
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 $x=7$. The answer is 7.
A square of area $2$ is inscribed in a square of area $3$, creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?
$\textbf{(A) }\dfrac15\qquad\textbf{(B) }\dfrac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1$
$\textbf{C}$
Let $a,b$ be the lengths of the shorter leg and the longer leg of the shaded right triangle, respectively. By the Pythagorean Theorem, the length of the hypotenuse of the triangle is $\sqrt{a^2+b^2}$. Hence, the area of the smaller square is $$a^2+b^2=2\qquad(1)$$ The side length of the larger square is $a+b$, so its area is $$(a+b)^2=3\qquad(2)$$ Subtracting (1) from (2), we get $$2ab=1\qquad(3)$$ Dividing (1) by (3), we get $$\dfrac{a}{2b}+\dfrac{b}{2a}=2$$ where $\dfrac{a}{b}=r$ is the ratio we are looking for. Substituting and simplifying, we have $$r+\dfrac{1}{r}=4$$ Solving, we get the answer $r=2-\sqrt3$.
Positive real numbers $x$ and $y$ satisfy $y^3 = x^2$ and $(y-x)^2 = 4y^2$. What is $x+y$?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 42$
$\textbf{D}$
Taking the square root of both sides of the second equation, we get $$y-x = \pm2y$$ If $y-x = +2y$, then we have $x=-y$, which is impossible because both $x$ and $y$ are positive real numbers. Hence, we have $y-x=-2y$, which gives $x=3y$. Substituting $x = 3y$ into the first equation, we have $y^3=(3y)^2$. Solving, we get $y=9$. So the answer is $x+y=3y+y=4y=4\times9=36$.
What is the degree measure of the acute angle formed by lines with slopes $2$ and $\dfrac{1}{3}$?
$\textbf{(A)}~30\qquad\textbf{(B)}~37.5\qquad\textbf{(C)}~45\qquad\textbf{(D)}~52.5\qquad\textbf{(E)}~60$
$\textbf{C}$
Let the angle formed by the line and the positive direction of $x$-axis be $\theta$. The slopes of the two lines can be expressed as $\tan\theta_1=2$, $\tan\theta_2=\dfrac13$. Hence, the acute angle formed by the two lines is $$\tan\left(\theta_1-\theta_2\right)=\dfrac{\tan\theta_1-\tan\theta_2}{1+\tan\theta_1\tan\theta_2}=1\rightarrow \theta_1-\theta_2=45^\circ$$
What is the value of\[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$
$\textbf{D}$
The original expression $S$ can be reorganized and simplified as \begin{align*}
S&=\left(2^3+4^3+\cdots+18^3\right)-\left(1^3+3^3+\cdots+17^3\right)\\
&=2\left(2^3+4^3+\cdots+18^3\right)-\left(1^3+2^3+3^3+4^3+\cdots+18^3\right)\\
&=2\times2^3\left(1^3+2^3+\cdots+9^3\right)-\left(1^3+2^3+3^3+4^3+\cdots+18^3\right)
\end{align*} Since $1^3+2^3+\cdots+n^3=\dfrac14n^2(n+1)^2$, it can be further simplified as \begin{align*}
S&=2\times2^3\times\left(\dfrac14\times9^2\times10^2\right)-\dfrac14\times18^2\times19^2\\
&=32400-29241\\
&=3159
\end{align*}
In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$
$\textbf{B}$
Let the number of left-handed players be $n$, so the number of right-handed players is $2n$.
The number of games won by the left-handed players comes in two ways:
$\bullet\quad$ The games played by two left-left pairs, which is $\dbinom{n}{2}=\dfrac{n(n-1)}{2}$, and$\newline$
$\bullet\quad$ The games played by left-right pairs, which we call it $x$.
Note that $x\leq 2n^2,$ which is the total number of games played by left-right pairs.
Using the same logic for the number of games won by the right-handed players, it also comes in two ways:
$\bullet\quad$ The games played by two right-right pairs, which is $\dbinom{2n}{2}=n(2n-1)$, and$\newline$
$\bullet\quad$ The games played by left-right pairs, which is $2n^2-x$.
Hence, we have \[\dfrac{\dfrac{n(n-1)}{2}+x}{n(2n-1)+2n^2-x}=1.4\]which gives\[x=\frac{17n^2}{8}-\frac{3n}{8}\]We know that $x\leq 2n^2$, which leads to \begin{align*} \frac{17n^2}{8}-\frac{3n}{8} &\le 2n^2 \\ 17n^2 - 3n &\le 16n^2 \\ n^2 - 3n &\le 0 \\ n &\le 3 \end{align*} So the total number of players $3n$ can only be $3$, $6$, and $9$.
The total number of games $\dbinom{3n}{2}=\dfrac{3n(3n-1)}{2}$ must be a multiple of $1+1.4=2.4$. Among $\{3,6,9\}$, only $3n = 9$ satisfies this condition. So the total number of games is $\dfrac{9\times8}{2} = 36.$
How many complex numbers satisfy the equation $z^{5}=\overline{z}$, where $\overline{z}$ is the conjugate of the complex number $z$?
$\textbf{(A)}~2\qquad\textbf{(B)}~3\qquad\textbf{(C)}~5\qquad\textbf{(D)}~6\qquad\textbf{(E)}~7$
$\textbf{E}$
Using the fact that $z\bar{z}=|z|^2$, we multiply both sides of the equation by $z$: $$z^6=\overline{z}z=|z|^2$$ Let $r$ be the magnitude of complex number $z$. Then we have $$r^6 = r^2$$ which can be rewrote as $$r^6-r^2=r^2(r^4-1)=0$$ From here, we have two cases: $r=0$ or $r=1$.
In the case that $r=0$, we have $z^6=0$, which leads to $z=0$. This gives one solution.
In the case that $r=1$, we have $z^6=1$, giving $6$ solutions for $z$.
Therefore, the answer is $6+1=7$.
Usain is walking for exercise by zigzagging across a $100$-meter by $30$-meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}$. He wants to increase the distance walked by zigzagging as shown in the figure below $(APQRS)$. What angle $\theta=\angle PAB=\angle QPC=\angle RQB=\cdots$ will produce a length that is $120$ meters? (This figure is not drawn to scale. Do not assume that the zigzag path has exactly four segments as shown; there could be more or fewer.)
$\textbf{(A)}~\arccos\dfrac{5}{6}\qquad\textbf{(B)}~\arccos\dfrac{4}{5}\qquad\textbf{(C)}~\arccos\dfrac{3}{10}\qquad\textbf{(D)}~\arcsin\dfrac{4}{5}\qquad\textbf{(E)}~\arcsin\dfrac{5}{6}$
$\textbf{A}$
The zigzag path can be connected to form segment $AS'$. Hence, we have $$\cos\theta=\dfrac{AB}{AS'}=\dfrac{100}{120}=\dfrac56$$ $$\theta=\arccos\dfrac56$$
Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$. The maximum value of the imaginary part of $z$ can be written in the form $\dfrac{\sqrt{m}}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$
$\textbf{B}$
Since the roots of $1+z+z^2=0$ are $z=\dfrac{-1\pm\sqrt3i}{2}$, the original expression can be rewrote as \[|(z-\frac{-1+\sqrt{3}i}{2})(z-\frac{-1-\sqrt{3}i}{2})|=4\] Note that the magnitude of a product of complex numbers is equal to the product of the magnitudes, we have \[|(z+\frac{1-\sqrt{3}i}{2})||(z+\frac{1+\sqrt{3}i}{2})|=4\] Substituting $z=a+bi$ into the equation, we have \[|a+\frac{1}{2}+(b+\frac{\sqrt{3}}{2})i||a+\frac{1}{2}+(b-\frac{\sqrt{3}}{2})i|=4\] Since we are trying to maximize $b$, we want the real parts of the components to be as small as possible, which can be done by setting $a=-\dfrac{1}{2}$. This leaves us with \[|(b+\frac{\sqrt{3}}{2})i||(b-\frac{\sqrt{3}}{2})i|=4\]\[(b+\frac{\sqrt{3}}{2})(b-\frac{\sqrt{3}}{2})=4\]\[b^2-\frac{3}{4}=4\]\[b^2=\frac{19}{4}\]\[b=\frac{\sqrt{19}}{2}\] So the answer is $19 + 2 =21$.
Flora the frog starts at $0$ on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\dfrac{1}{2^m}$. What is the probability that Flora will eventually land at $10$?
$\textbf{(A) } \dfrac{5}{512} \qquad \textbf{(B) } \dfrac{45}{1024} \qquad \textbf{(C) } \dfrac{127}{1024} \qquad \textbf{(D) } \dfrac{511}{1024} \qquad \textbf{(E) } \dfrac{1}{2}$
$\textbf{E}$
Let $P_n$ be the probability of landing on $n$. Clearly $P_0=1$, $P_1=\dfrac12$. For $n\geq1$, we have \[P_n = \frac{P_{n-1}}{2} + \frac{P_{n-2}}{4} + \dots + \frac{P_{n-m}}{2^m} \dots + \frac{P_0}{2^n}\] We see that \[P_{n-1} = \frac{P_{n-2}}{2} + \frac{P_{n-3}}{4} + \dots + \frac{P_0}{2^{n-1}}\] Hence, we get $$P_{n-1}=2\left(P_n-\frac{P_{n-1}}{2}\right)\rightarrow P_n=P_{n-1}$$ Thus, we have $$P_n=P_{n-1}=\cdots=P_1=\dfrac12$$ The answer is $P_{10}=\dfrac12$.
Circle $C_1$ and $C_2$ each have radius $1$, and the distance between their centers is $\dfrac{1}{2}$. Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2$. Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3$. What is the radius of $C_4$?
$\textbf{(A) } \dfrac{1}{14} \qquad \textbf{(B) } \dfrac{1}{12} \qquad \textbf{(C) } \dfrac{1}{10} \qquad \textbf{(D) } \dfrac{3}{28} \qquad \textbf{(E) } \dfrac{1}{9}$
$\textbf{D}$
Let $O_1,O_2$ be the centers of circle $C_1$ and $C_2$, respectively. Then we have $O_1O_2=\dfrac12$. By symmetry, the midpoint of line segment $O_1O_2$ is the center of circle $C_3$. Let's call it $O_3$.
Let the point of tangency of circle $O_1$ and $O_3$ be $P$.Then we have $$O_3P = O_1P-O_1O_3=1-\dfrac14=\dfrac{3}{4}$$ So the radius of circle $C_3$ is $\dfrac34$.
Let $O_4$ be the center of circle $C_4$, and $r$ be the radius of circle $C_4$. By symmetry, we see that $O_3O_4\perp O_1O_2$. In right triangle $O_1O_3O_4$, we have $O_1O_3=\dfrac14$, $O_3O_4=r+\dfrac34$, and $O_1O_4=1-r$. By the Pythagorean theorem, we have $$\left(\dfrac14\right)^2+\left(r+\dfrac34\right)^2=(1-r)^2$$ Solving, we get $r =\dfrac{3}{28}$.
What is the product of all the solutions to the equation\[\log_{7x}2023 \cdot \log_{289x} 2023 = \log_{2023x} 2023?\]
$\textbf{(A) }(\log_{2023}7 \cdot \log_{2023}289)^2 \qquad\newline \textbf{(B) }\log_{2023}7 \cdot \log_{2023}289\qquad\newline\textbf{(C) } 1 \newline\textbf{(D) }\log_{7}2023 \cdot \log_{289}2023\qquad\newline\textbf{(E) }(\log_{7}2023 \cdot \log_{289}2023)^2$
$\textbf{C}$
Since $2023=7\times289=7\times17^2$, the equation can be rewrote as \[\dfrac{\ln2023}{\ln(7x)}\cdot\dfrac{\ln2023}{\ln(289x)}=\dfrac{\ln2023}{\ln(2023x)}\]\[\frac{2\ln17+\ln 7}{\ln7 + \ln{x}} \cdot \frac{2\ln17+\ln 7}{2\ln17 + \ln{x}} = \frac{2\ln17+\ln 7}{2\ln17 +\ln7+ \ln{x}}\] Cancel and cross multiply to get\[(2\ln17+\ln7 )(2\ln17 +\ln7 + \ln{x}) = (\ln7 + \ln{x})(2\ln17 + \ln{x})\]Simplify to get \[(\ln{x})^2 = 4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2\]\[\ln{x} = \pm \sqrt{4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2}\]The sum of all possible $\ln{x}$ is $\ln x_1+\ln x_2=\ln(x_1x_2)=0$, so the product of all solutions of $x$ is $x_1x_2=1$.
Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below:
Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digit of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad\textbf{(E) }9$
$\textbf{C}$
Let the sum of all numbers in row $n$ be $S_n$. Then we have $S_1=1$, $S_2=2$. Let each number in row $2022$ be $a_i$, where $1 \leq i \leq 2022$. Then we have \begin{align*}
S_{2023}&=1+(a_1+a_2+1)+(a_2+a_3+1)+\cdots+(a_{2021}+a_{2022}+1)+1\\
&=a_1+a_{2022}+2(a_2+a_3+\cdots+a_{2021})+2023\\
&=1+1+2(S_{2022}-a_1-a_{2022})+2023\\
&=1+1+2(S_{2022}-1-1)+2023\\
&=2S_{2022}+2021
\end{align*} Hence, we have \begin{align*}
S_n &= 2S_{n-1} + n-2 \\
S_{n-1} &= 2S_{n-2} + n-3 \\
S_n - S_{n-1} &= 2\left(S_{n-1} - S_{n-2}\right) + 1
\end{align*} Let $A_{n} = S_n - S_{n-1}$, then we have $A_2=S_2-S_1=1$. The equation becomes \begin{align*}
A_n&=2A_{n-1}+1\\
A_n + 1 &= 2(A_{n-1} + 1)
\end{align*} Thus, we have $$A_n+1=2(A_{n-1}+1)=2^2(A_{n-2}+1)=\cdots=2^{n-2}(A_2+1)=2^{n-1}$$ $$A_n=2^{n-1}-1$$ Come back to $S_n$: $$S_n - S_{n-1}= 2^{n-1} - 1$$ \begin{align*}
S_n &= \left(2^{n-1}-1\right) + \left(2^{n-2}-1\right) + \cdots+ \left(2^1-1\right) + S_1 \\
&=\left(2^{n-1}+2^{n-2}+\cdots+2^1\right)-(n-1)+S_1\\
&=\left(2^n-2\right)-(n-1)+1\\
&= 2^n - n
\end{align*} Hence, we get $$S_{2023} = 2^{2023} - 2023 $$ The units digit of $2^n$ follows the pattern of $2,4,8,6,2,4,8,6\cdots$ So the units digit of $2^{2023}$ is 8. Therefore, the units digit of $S_{2023} = 2^{2023} - 2023 $ is $8-3=5$.
If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A, B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$. For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$, but if $\overline{AC}$ and $\overline{CB}$ are edges and $\overline{AB}$ is not an edge, then $d(A, B) = 2$. Let $Q$, $R$, and $S$ be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that $d(Q, R) > d(R, S)$?
$\textbf{(A)}~\dfrac{7}{22}\qquad\textbf{(B)}~\dfrac13\qquad\textbf{(C)}~\dfrac38\qquad\textbf{(D)}~\dfrac5{12}\qquad\textbf{(E)}~\dfrac12$
$\textbf{A}$
By symmetry, the probability $p$ that $d(Q, R) > d(R, S)$ is the same as $d(Q, R) < d(R, S)$. Therefore, we want to find the probability $p^\prime$ that $d(Q, R) = d(R, S)$, then we have $p=\dfrac{1-p^\prime}2$.
To find the total amount of vertices we first find the amount of edges. There are 20 faces, each with 3 edges, and each edge is shared by 2 faces. So the total amount of edge is $\dfrac{20 \times 3}{2}=30$. Next, to find the amount of vertices we can use Euler's formula, $\text{Vertices}+\text{Faces}-\text{Edges}= 2$, and therefore the amount of vertices is $12$.
Imagining the icosahedron as having 4 layers. 1 vertex at the top, 5 vertices below connected to the top vertex, 5 vertices below that are 2 edges away from the top vertex, and 1 vertex at the bottom that is 3 edges away from the top.
WLOG, supposing that R is the vertex at the top. If $d(Q, R) = d(R, S)$, there are 2 cases:
$\textbf{Case 1}$: Both $Q$ and $S$ are on the second layer. The probability of this case is $\dfrac{\dbinom52}{\dbinom{11}{2}}=\dfrac{2}{11}$.
$\textbf{Case 2}$: Both $Q$ and $S$ are on the third layer. The probability of this case is the same as Case 1.
Therefore, the probability that $d(Q, R) = d(R, S)$ is $p^\prime=\dfrac{2}{11}\times2=\dfrac{4}{11}$. So the probability that $d(Q, R) > d(R, S)$ is $p=\dfrac{1-p^\prime}2=\dfrac{7}{22}$.
Let $f$ be the unique function defined on the positive integers such that\[\sum_{d\mid n}d\cdot f\left(\frac{n}{d}\right)=1\]for all positive integers $n$, where the sum is taken over all positive divisors of $n$. What is $f(2023)$?
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$
$\textbf{B}$
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\[\sum_{d \mid p} d \cdot f\left(\frac{p}{d}\right) = 1\]\[1 \cdot f(p) + p \cdot f(1) = 1\]\[f(p) = 1 - p \cdot f(1)\]\[f(p) = 1-p\] Now consider $f(p^k)$, for $k \in \mathbb{N}$ \[\sum_{d \mid p^k} d \cdot f\left(\frac{p^k}{d}\right) = 1\]\[1 \cdot f(p^k) + p \cdot f(p^{k-1}) + p^2 \cdot f(p^{k-2}) + \dotsc + p^k f(1) = 1\qquad(1)\] We can find the pattern of $f(p^k)$ by considering $f(p^{k-1})$: \[\sum_{d \mid p^{k-1}} d \cdot f\left(\frac{p^{k-1}}{d}\right) = 1\]\[1 \cdot f(p^{k-1}) + p \cdot f(p^{k-2}) + p^2 \cdot f(p^{k-3}) + \dotsc + p^{k-1} f(1) = 1\qquad(2)\] We multiply equation $(2)$ by $p$, and then subtract it from equation $(1)$, which gives us $$f(p^k)=1-p$$ Next, we focus on the $f(pq)$ for distinct $p,q$ prime: \[\sum_{d \mid pq} d \cdot f\left(\frac{pq}{d}\right) = 1\]\[1 \cdot f(pq) + p \cdot f(q) + q \cdot f(p) + pq \cdot f(1) = 1\] where $f(p)=1-p$, $f(q)=1-q$, $f(1)=1$. Thus, we have \[f(pq) = 1 - p(1-q) - q(1-p) - pq = (1-p)(1-q) = f(p) \cdot f(q)\] Now all the preliminary preparations have been completed.
For $2023 = 7 \times 17^2$, we have \[\sum_{d \mid 2023} d \cdot f\left(\frac{2023}{d}\right) = 1\]\[1 \cdot f(2023) + 7 \cdot f(17^2) + 17 \cdot f(7 \cdot 17) + 7 \cdot 17 \cdot f(17) + 17^2 \cdot f(7) + 7 \cdot 17^2 \cdot f(1) = 1\] where \[f(7) = 1-7 = -6\]\[f(17) = 1-17 = -16\]\[f(7 \cdot 17) = f(7) \cdot f(17) = (-6) \cdot (-16) = 96\]\[f(17^2) = 1-17 = -16\] Substituting the values, we get\[f(2023) = 1 - \left[7 \cdot (-16) + 17 \cdot (-6) \cdot (-16) + 7 \cdot 17 \cdot (-16) + 17^2 \cdot (-6) + 7 \cdot 17^2\right]=96\]
How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation\[(1+2a)(2+2b)(2a+b) = 32ab?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$
$\textbf{B}$
Applying the AM-GM inequality on each term on the left side of the equation, we have \begin{align*}
1+2a&\geq2\sqrt{1\cdot2a}\\
2+2b&\geq2\sqrt{2\cdot2b}\\
2a+b&\geq2\sqrt{2a\cdot b}
\end{align*} which gives \[(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab\] This means that all inequalities have reached their minimum values. Therefore, we must have \begin{align*}
1&=2a\\
2&=2b\\
2a&=b
\end{align*} Hence, the only possible solution is $(a,b)=(\dfrac12,1)$.
Let $K$ be the number of sequences $A_1$, $A_2$, $\dots$, $A_n$ such that $n$ is a positive integer less than or equal to $10$, each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$, and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$, inclusive. For example, $\{\}$, $\{5, 7\}$, $\{2, 5, 7\}$, $\{2, 5, 7\}$, $\{2, 5, 6, 7, 9\}$ is one such sequence, with $n = 5$. What is the remainder when $K$ is divided by $10$?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$
$\textbf{C}$
Consider any sequence $A_1$, $A_2$, $\dots$, $A_n$ with $n$ terms. Each number in $\{1, 2, 3, \dots, 10\}$ has such choices: never appear, appear the first time in $A_1$, appear the first time in $A_2$,$\cdots$, and appear the first time in $A_n$, 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 total number of sequences is $$K=\sum_{n=1}^{10}(n+1)^{10}=\sum_{m=2}^{11}m^{10}$$ Now we want to find the units digits of $2^{10}, 3^{10} \cdots 11^{10}$. We can find the pattern that $$x^n \equiv x^{n \mod 4} \pmod {10}$$ Then, Modulo 10, we have\begin{align*}
K & \equiv \sum_{m = 2}^{11} m^2 \pmod{10}\\
& \equiv\left( \sum_{m = 1}^{11} m^2\right) - 1^2 \pmod{10}\\
& \equiv \dfrac16\times11\times(11+1)\times(2\times11+1) - 1\pmod{10}\\
& \equiv 505 \pmod{10}\\
& \equiv 5\pmod{10}
\end{align*} The answer is C.
There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that\[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\]whenever $\tan 2023x$ is defined. What is $a_{2023}?$
$\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023$
$\textbf{C}$
For positive integer $n$, we have \begin{align*}
\tan nx & = \frac{\sin nx}{\cos nx} \\
& = \frac{\frac{1}{2i} \left( e^{i n x} - e^{-i n x} \right)} {\frac{1}{2} \left( e^{i n x} + e^{-i n x} \right)} \\
& = - i \frac{e^{i n x} - e^{-i n x}}{e^{i n x} + e^{-i n x}} \\
& = - i \frac{\left( \cos x + i \sin x \right)^n - \left( \cos x - i \sin x \right)^n} {\left( \cos x + i \sin x \right)^n + \left( \cos x - i \sin x \right)^n} \\
&=-i\dfrac{\sum_{m=0}^n\binom{n}{m}\cos^{n-m}x\left(i\sin x\right)^{m}-\sum_{m=0}^n\binom{n}{m}\cos^{n-m}x\left(-i\sin x\right)^{m}}{\sum_{m=0}^n\binom{n}{m}\cos^{n-m}x\left(i\sin x\right)^{m}+\sum_{m=0}^n\binom{n}{m}\cos^{n-m}x\left(-i\sin x\right)^{m}}
\end{align*} When $n$ is odd, the numerator $\left( \cos x + i \sin x \right)^n - \left( \cos x - i \sin x \right)^n$ becomes $$2\left[\binom{n}1\cos^{n-1}x\left(i\sin x\right)^1+\binom{n}{3}\cos^{n-3}x\left(i\sin x\right)^3+\cdots+\binom{n}{n}\left(i\sin x\right)^n\right]$$ and the denominator $\left( \cos x + i \sin x \right)^n + \left( \cos x - i \sin x \right)^n$ becomes $$2\left[\binom{n}0\cos^nx+\binom{n}2\cos^{n-2}x\left(i\sin x\right)^2+\cdots+\binom{n}{n-1}\cos x\left(i\sin x\right)^{n-1}\right]$$ Now we have \begin{align*}
\tan nx&=-i\dfrac{\binom{n}1\cos^{n-1}x\left(i\sin x\right)^1+\binom{n}{3}\cos^{n-3}x\left(i\sin x\right)^3+\cdots+\binom{n}{n}\left(i\sin x\right)^n}{\binom{n}0\cos^nx+\binom{n}2\cos^{n-2}x\left(i\sin x\right)^2+\cdots+\binom{n}{n-1}\cos x\left(i\sin x\right)^{n-1}}\\
&=-i\dfrac{i\binom{n}{1}\tan x+i^3\binom{n}{3}\tan^3x+\cdots+i^n\binom{n}{n}\tan^nx}{\binom{n}{0}+i^2\binom{n}{2}\tan^2x+\cdots+i^{n-1}\tan^{n-1}x}\\
&=\frac{\binom{n}{1}\tan{x} - \binom{n}{3}\tan^{3}{x} + \binom{n}{5}\tan^{5}{x} - \binom{n}{7}\tan^{7}{x} + \cdots}{1 - \binom{n}{2}\tan^{2}{x} + \binom{n}{4}\tan^{4}{x} - \binom{n}{6}\tan^{6}{x} + \cdots}
\end{align*} For $n=2023$, the coefficient of $\tan^{2023}x$ in the numerator is $$a_{2023}=-\binom{2023}{2023}=-1$$