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71f523979ced286fe544b672952956a5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 | What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$ | $2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3$
$=8-1+64-27+216-125+512-343+1000-729+1728-1331+2744-2197+4096-3375+5832-4913$
$= \boxed{3159}$ | D | 3159 |
71f523979ced286fe544b672952956a5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 | What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$ | Reduce all terms mod 9. This yields:
$2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3$ $\equiv -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 +0 - -1 (\mod 9)$ $\equiv 0 (\mod 9)$
The only answer choice which is also ≡0 mod 9 is $= \boxed{3159}$ | D | 3159 |
b26b50c49360088a9fc1f1d653723ffe | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_13 | 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?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$ | 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 $n-1$ triangular 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 $\boxed{36}$ | B | 36 |
b26b50c49360088a9fc1f1d653723ffe | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_13 | 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?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$ | First, we know that every player played every other player, so there's a total of $\dbinom{n}{2}$ games since each pair of players forms a bijection to a game. Therefore, that rules out D. Also, if we assume the right-handed players won a total of $x$ games, the left-handed players must have won a total of $\dfrac{7}{5}x$ games, meaning that the total number of games played was $\dfrac{12}{5}x$ . Thus, the total number of games must be divisible by $12$ . Therefore leaving only answer choices B and D. Since answer choice D doesn't satisfy the first condition, the only answer that satisfies both conditions is $\boxed{36}$ | B | 36 |
b26b50c49360088a9fc1f1d653723ffe | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_13 | 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?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$ | Let $r$ be the amount of games the right-handed won. Since the left-handed won $1.4r$ games, the total number of games played can be expressed as $(2.4)r$ , or $12/5r$ , meaning that the answer is divisible by 12. This brings us down to two answer choices, $B$ and $D$ .
We note that the answer is some number $x$ choose $2$ . This means the answer is in the form $x(x-1)/2$ . Since answer choice D gives $48 = x(x-1)/2$ , and $96 = x(x-1)$ has no integer solutions, we know that $\boxed{36}$ is the only possible choice. | B | 36 |
b26b50c49360088a9fc1f1d653723ffe | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_13 | 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?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$ | If there are $n$ players, the total number of games played must be $\binom{n}{2}$ , so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is $7:5$ , so the number of games played must also be divisible by $12$ . Finally, we notice that only $\boxed{36}$ satisfies both of these conditions. | B | 36 |
7133303c63722fce4fa2de69a16ce680 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_14 | How many complex numbers satisfy the equation $z^5=\overline{z}$ , where $\overline{z}$ is the conjugate of the complex number $z$
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~5\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~7$ | 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 $\boxed{7}$ | E | 7 |
7133303c63722fce4fa2de69a16ce680 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_14 | How many complex numbers satisfy the equation $z^5=\overline{z}$ , where $\overline{z}$ is the conjugate of the complex number $z$
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~5\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~7$ | Let $z = re^{i\theta}.$ We now have $\overline{z} = re^{-i\theta},$ and want to solve
\[r^5e^{5i\theta} = re^{-i\theta}.\]
From this, we have $r = 0$ as a solution, which gives $z = 0$ . If $r\neq 0$ , then we divide by it, yielding
\[r^4e^{5i\theta} = e^{-i\theta}.\]
Dividing both sides by $e^{-i\theta}$ yields $r^4e^{6i\theta} = 1$ .
Taking the magnitude of both sides tells us that $r^4 = 1$ , so $r^2 = \pm 1$ . However, if $r^2 = -1$ , then $r = \pm i$ , but $r$ must be real. Therefore, $r^2 = 1$
Multiplying both sides by $r^2$
\[r^6\cdot e^{6i\theta} = z^6 = 1.\]
Each of the $6$ th roots of unity is a solution to this, so there are $6 + 1 = \boxed{7}$ solutions. | D | 7 |
7133303c63722fce4fa2de69a16ce680 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_14 | How many complex numbers satisfy the equation $z^5=\overline{z}$ , where $\overline{z}$ is the conjugate of the complex number $z$
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~5\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~7$ | Let $z = a+bi$
Then, our equation becomes: $(a+bi)^5=a-bi$
Note that since every single term in the expansion contains either an $a$ or $b$ , simply setting $a=0$ and $b=0$ yields a solution.
Now, considering the other case that either $a$ or $b$ does not equal $0$
Multiplying both sides by $a+bi$ (or $z$ ), we get: $(a+bi)^6=a^2+b^2$ (since $i^2=-1$ ).
Substituting $z$ back into the left hand side, we get: $z^6=a^2+b^2$
Note that this will have 6 distinct, non-zero solutions since in this case, we consider that either $a$ or $b$ is not $0$ , and these are simply the sixth roots of a positive real number.
Adding up the solutions, we get $1+6=$ $\boxed{7}$ | E | 7 |
ba0f1b4cc50dec5c98db179914f42038 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_15 | 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 in a length that is $120$ meters? (This figure is not drawn to scale. Do not assume that he zigzag path has exactly four segments as shown; there could be more or fewer.)
[asy] import olympiad; draw((-50,15)--(50,15)); draw((50,15)--(50,-15)); draw((50,-15)--(-50,-15)); draw((-50,-15)--(-50,15)); draw((-50,-15)--(-22.5,15)); draw((-22.5,15)--(5,-15)); draw((5,-15)--(32.5,15)); draw((32.5,15)--(50,-4.090909090909)); label("$\theta$", (-41.5,-10.5)); label("$\theta$", (-13,10.5)); label("$\theta$", (15.5,-10.5)); label("$\theta$", (43,10.5)); dot((-50,15)); dot((-50,-15)); dot((50,15)); dot((50,-15)); dot((50,-4.09090909090909)); label("$D$",(-58,15)); label("$A$",(-58,-15)); label("$C$",(58,15)); label("$B$",(58,-15)); label("$S$",(58,-4.0909090909)); dot((-22.5,15)); dot((5,-15)); dot((32.5,15)); label("$P$",(-22.5,23)); label("$Q$",(5,-23)); label("$R$",(32.5,23)); [/asy]
$\textbf{(A)}~\arccos\frac{5}{6}\qquad\textbf{(B)}~\arccos\frac{4}{5}\qquad\textbf{(C)}~\arccos\frac{3}{10}\qquad\textbf{(D)}~\arcsin\frac{4}{5}\qquad\textbf{(E)}~\arcsin\frac{5}{6}$ | Drop an altitude from $P$ to $AB$ and let its base be $x$ . Note that if we repeat this for $Q$ and $R$ , all four right triangles (including $\triangle{RSC}$ ) will have the same trig ratios. By proportion, the hypotenuse $AP$ is $\frac{x}{100}(120) = \frac65 x$ , so $\cos\theta = \frac{x}{(\frac65x)} = \frac56 \Rightarrow \theta = \boxed{56}$ | A | 56 |
905b675d6b06136b95fe361d7e9fd4a6 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_16 | 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 $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$ | 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 \[m+n=\boxed{21}\] | B | 21 |
905b675d6b06136b95fe361d7e9fd4a6 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_16 | 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 $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$ | We are given that $1+z+z^2=c$ where $c$ is some complex number with magnitude $4$ . Rearranging the quadratic to standard form and applying the quadratic formula, we have \[z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{4c-3}}{2}.\]
The imaginary part of $z$ is maximized when $c=-4$ . (Why? See note below.)
Thus $z=i\sqrt{19}/2$ , and so the answer is $\boxed{21}$ | B | 21 |
905b675d6b06136b95fe361d7e9fd4a6 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_16 | 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 $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$ | [asy] size(250); import TrigMacros; rr_cartesian_axes(-6,5,-5,5,complexplane=true, usegrid = true); Label f; f.p=fontsize(6); xaxis(-6,5,Ticks(f, 1.0)); yaxis(-5,5,Ticks(f, 1.0)); dot((0,0)); draw(circle((-3/4, 0), 4), red + dashed); dot((-19/4, 0), blue); label("$\phi$", (-19/4, 0), NW); dot((0, 2.18), blue); label("$v'$", (0, 2.18), NE); draw(ellipse((0,0),1.8,2.18), green); [/asy] We can write the given condition as \[\left|\left(z+\frac{1}{2}\right)^2 + \frac{3}{4}\right| = 4.\] Letting $u = \left(z+\frac{1}{2}\right)^2$ , the equation $\left|u + \frac{3}{4}\right| = 4$ equates to the circle centered at $-\frac{3}{4}$ with radius $4$ in the complex plane, call it $\omega$ . Thus the locus of $\left(z+\frac{1}{2}\right)^2$ is $\omega$ . Let $v = z+\frac{1}{2}$ , and since the $+\frac{1}{2}$ does not change $z$ 's imaginary part, we now need to find $v$ with the largest imaginary part such that $v^2$ lies on $\omega$
Note that the point on $\omega$ with largest magnitude is $19/4$ and has argument $\pi$ , call it $\phi$ (The leftmost point on $\omega$ ). The value $v'$ with positive imaginary part such that $(v')^2 = \phi$ has an argument of $\frac{\pi}{2}$ and a magnitude of $\frac{\sqrt{19}}{2}$
Since across all values of $v$ the imaginary part is given by $r\sin{\theta}$ and $v'$ has the largest possible $r$ and the largest possible value of $\sin{\theta},$ it must have the largest imaginary part.
This can non-rigorously be seen by sketching the oval which is the locus of $v$
This gives $19 + 2 \implies \boxed{21}$ | B | 21 |
905b675d6b06136b95fe361d7e9fd4a6 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_16 | 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 $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$ | To start, we factor $1+z+z^2$ to get:
\[|(z-\frac{-1+\sqrt{3}i}{2})(z-\frac{-1-\sqrt{3}i}{2})|=4\]
Note that since the magnitude of a product of complex numbers is equal to the product of the magnitudes:
\[|(z+\frac{1-\sqrt{3}i}{2})||(z+\frac{1+\sqrt{3}i}{2})|=4\]
Now, we substitute $z=a+bi$ (Note that we are trying to maximize b):
\[|a+\frac{1}{2}+(b+\frac{\sqrt{3}}{2})i||a+\frac{1}{2}+(b-\frac{\sqrt{3}}{2})i|=4\]
Since we are trying to maximize $b$ , we want the real parts of the components to be as small as possible, which we can do by setting $a=-\frac{1}{2}$ . This leaves us with:
\[|(b+\frac{\sqrt{3}}{2})i||(b-\frac{\sqrt{3}}{2})i|=4\] \[(b+\frac{\sqrt{3}}{2})(b-\frac{\sqrt{3}}{2})=4\] \[b^2-\frac{3}{4}=4\] \[b^2=\frac{19}{4}\] \[b=\frac{\sqrt{19}}{2}\]
This gives us $19 + 2 \implies \boxed{21}$ | B | 21 |
3e0b2d49516628e5aa30ea0cf64c06b6 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_17 | 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 $\frac{1}{2^m}$
What is the probability that Flora will eventually land at 10?
$\textbf{(A)}~\frac{5}{512}\qquad\textbf{(B)}~\frac{45}{1024}\qquad\textbf{(C)}~\frac{127}{1024}\qquad\textbf{(D)}~\frac{511}{1024}\qquad\textbf{(E)}~\frac{1}{2}$ | 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 $\boxed{12}$ | E | 12 |
5502e52f26febd5bff5305f2d44dc75e | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_19 | What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1$ $\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$ | 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 $\prod x=e^0=\boxed{1}$ | C | 1 |
5502e52f26febd5bff5305f2d44dc75e | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_19 | What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1$ $\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$ | \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
Rearranging it give us:
\[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x\]
\[(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)\]
let $\log_{2023}x$ be $a$ , we get
\[(\log_{2023}7+a)(\log_{2023}289+a)=1+a\]
\[a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a\]
\[a^2+\log_{2023}7 \cdot \log_{2023}289-1=0\]
by Vieta's Formulas,
\[a_1+a_2=0\]
\[\log_{2023}{x_1}+\log_{2023}{x_2}=0\]
\[\log_{2023}{x_1x_2}=0\]
\[x_1x_2=\boxed{1}\] | C | 1 |
5502e52f26febd5bff5305f2d44dc75e | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_19 | What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1$ $\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$ | Similar to solution 1, change the bases first \[\frac{\ln 289+\ln 7}{\ln7 + \ln{x}} \cdot \frac{\ln 289+\ln 7}{2\ln17 + \ln{x}} = \frac{\ln 289+\ln 7}{\ln7 + 2\ln17 + \ln{x}}\] Cancel and cross multiply to get \[(\ln7 + 2\ln17)(\ln7 + 2\ln17 + \ln{x}) = (\ln7 + \ln{x})(2\ln17 + \ln{x})\] 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 0, thus the product of all solutions of $x$ is $\boxed{1}$ | C | 1 |
5502e52f26febd5bff5305f2d44dc75e | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_19 | What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1$ $\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$ | We take the reciprocal of both sides: \[\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.\] Using logarithm properties, we have \[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.\] Simplify to obtain \[2023x^2=2023x,\] from which we have $x=\boxed{1}$ | C | 1 |
be405d0d4ddc50d7ee3148f37b9c9b76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
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 digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | 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 $8$ $8-3 = \boxed{5}$ | C | 5 |
be405d0d4ddc50d7ee3148f37b9c9b76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
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 digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | Let the sum of the numbers in row $2022$ be $S_{2022}$ . Let each number in row $2022$ be $x_i$ where $1 \leq i \leq 2022$
Then \begin{align*} S_{2023}&=1+(x_1+x_2+1)+(x_2+x_3+1)+...+(x_{2021}+x_{2022}+ 1)+1 \\ S_{2023}&=x_1+2(S_{2022}-x_1-x_{2022})+2023+x_{2022} \\ S_{2023} &= 2S_{2022} + 2021 \end{align*} From this we can establish: \begin{align*} S_n &= 2S_{n-1} + n-2 \\ S_{n-1} &= 2S_{n-2} + n-3 \\ S_n - S_{n-1} &= 2S_{n-1} - 2S_{n-2} + 1 \end{align*}
Let $B_{n} = S_n - S_{n-1}$
From this we have: \begin{align*} B_n + 1 &= 2(B_{n-1} + 1) \\ B_n &= 2^{n-1} - 1 \\ S_n - S_{n-1} &= 2^{n-1} - 1 \\ S_n &= 2^{n-1} + 2^{n-2} + ... + 2^1 - (n-1) + S_1 \\ S_n &= 2^n - n \\ S_{2023} &= 2^{2023} - 2023 \end{align*}
The problem requires us to find the last digit of $S_{2023}$ . We can use modular arithmetic. \begin{align*} S_{2023} &= 2^{2023} - 2023 \\ 2^{2023} - 2023 &\equiv 8 - 3\pmod{10} \\ &\equiv \boxed{5} luckuso | C | 5 |
be405d0d4ddc50d7ee3148f37b9c9b76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
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 digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | Let the sum of the $n^{th}$ row be $S_n$
For each of the $n-2$ non-1 entries in the $n^{th}$ row, they are equal to the sum of the $2$ numbers diagonally above it in the $n-1^{th}$ row plus $1$ . Therefore all $n-3$ non-1 entries in the $n-1^{th}$ row appear twice in the sum of the $n-2$ non-1 entries in the $n^{th}$ row, the two $1$ s on each end of the $n-1^{th}$ row only appear once in the sum of the $n-2$ non-1 entries in the $n^{th}$ row. Additionally, additional $1$ s are placed at each end of the $n^{th}$ row. Hence, $S_n = 2(S_{n-1} - 1) + n-2 + 1 + 1 = 2 S_{n-1} + n - 2$
$S_4 = 1+5+5+1 = 12$ $S_5 = 1+7+11+7+1 = 27$ . By using the recursive formula, $S_5 = 2 \cdot S_4 + 5-2 = 27$
\[S_n = 2 S_{n-1} + n - 2\] \[S_n + n = 2 S_{n-1} + 2n - 2\] \[S_n + n = 2( S_{n-1} + n - 1)\] $S_n + n$ is a geometric sequence by a ratio of $2$
$\because \quad S_1 = 1$
$\therefore \quad S_n + n = 2^n$ $S_n = 2^n - n$
$S_{2023} = 2^{2023} - 2023$
The unit digit of powers of $2$ is periodic by a cycle of $4$ digits: $2$ $4$ $8$ $6$ $2023 = 3 \mod 4$ , the unit digit of $2^{2023}$ is $8$
Therefore, the unit digit of $S_{2023}$ is $8-3 = \boxed{5}$ | C | 5 |
be405d0d4ddc50d7ee3148f37b9c9b76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
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 digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | Consider Pascal's triangle as the starting point. In the Pascal's triangle depicted below, the sum of the numbers in the $n$ th row is $2^{(n-1)}$ . For the 2023rd row in the Pascal's triangle, the sum of numbers is $2^{2022}$
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$2$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$3$", (-0.5,-2)); label("$3$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$4$", (-1,-8/3)); label("$6$", (0,-8/3)); label("$4$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
For the triangular array of integers in the problem, 1 is added to each interior entry, which propagates to two numbers diagonally below it in the next row, making the sum of numbers in the next row to increase by 2. And the addition of 1 continues to propagate to the next row, which makes the sum of numbers in the row below the next to increase by 4.
Examine the following triangular array of 1's, which indicates the 1's being added to each position of corresponding numbers in the Pascal's triangle.
[asy] size(4.5cm); label("$0$", (0,0)); label("$0$", (-0.5,-2/3)); label("$0$", (0.5,-2/3)); label("$0$", (-1,-4/3)); label("$1$", (0,-4/3)); label("$0$", (1,-4/3)); label("$0$", (-1.5,-2)); label("$1$", (-0.5,-2)); label("$1$", (0.5,-2)); label("$0$", (1.5,-2)); label("$0$", (-2,-8/3)); label("$1$", (-1,-8/3)); label("$1$", (0,-8/3)); label("$1$", (1,-8/3)); label("$0$", (2,-8/3)); [/asy]
For the 3rd row, 1 is added to the original number in the same position in the Pascal's triangle. And the addition of 1 in the 3rd row makes the sum of numbers in the 4th row to increase by 2, and makes the sum of numbers in the 5th row to increase by 4, and so forth. Therefore, the addition of a 1 in the 3rd row makes the sum of numbers in the 2023rd row to increase by $2^{2023-3}=2^{2020}$ . And similarly, the addition of each 1 in the 4th row makes the number of numbers in the 2023rd row to increase by $2^{2023-4}=2^{2019}$ . The 1's being added between the 3rd and 2022nd rows impact on the sum of numbers in the 2023rd row to increase by $2^{2020} + 2 \cdot 2^{2019} + 3 \cdot 2^{2018} + \dots + 2020 \cdot 2^1 = \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)})$
Therefore, the sum of numbers in the 2023rd row is the aggregation of the sum of numbers in the 2023rd row in the Pascal's triangle, the impact of addition of 1's between the 3rd row and the 2022nd row, and the addition of 1's on 2021 interior entries in the 2023rd row, which is $2^{2022} + \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)}) + 2021$
Because $2^{2022} = 2^{2020+2} = 16^{505} \cdot 2^2 = 16^{505} \cdot 4$ and $16^k$ will always ends with 6 as the unit digit, the unit digit of $2^{2022}$ is 4.
For $\sum_{n=1}^{2020}(n \cdot 2^{(2021-n)})$ , we can solve the sum of geometric sequence to be ${2^{2022} - 4 - 2020 \cdot 2}$ , which has 0 for the unit digit.
Therefore, for the sum of numbers in the 2023rd row, which is $2^{2022} + \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)}) + 2021$ , its unit digit is $4 + 0 + 1 = \boxed{5}$ | C | 5 |
be405d0d4ddc50d7ee3148f37b9c9b76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
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 digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | Observe that: $S_n = 2{S_{n-1}} + (n-2) \\ = 2{S_{n-1}} + D_{n} \\ = 2{S_{n-1}} + D_{n-1} + 1$ where $D_1=-1$ , and $D_n=D_{n-1}+1$ for $n>1$
Make a table of values, $(S_n, D_n) \mod 2$ and $(S_n, D_n) \mod 5$ , a until the cycle completes for each. ( $\mod 2$ has period $1$ $\mod 5$ has period $20$ ).
( In both cases, there is no "head" that is not part of the cycle.)
Mod 2: $S_{2023} \equiv S_{2023 \mod 2} \equiv S_1 = (1\cdot 2 +0)\cdot 0 + 1 \equiv 1 \pmod 2$
Mod 5: $S_{2023} \equiv S_{2023 \mod 20} \equiv S_3 \equiv (1\cdot 2 +0)\cdot 2 + 1 \equiv 5 \pmod 5$
Thus $S_{2023} \mod 10 = \boxed{5}$ | C | 5 |
11646ab7b2cc12aa78e4ed9da99561a5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_21 | 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 $d(Q, R) > d(R, S)$
$\textbf{(A) } \frac{7}{22} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{3}{8} \qquad \textbf{(D) } \frac{5}{12} \qquad \textbf{(E) } \frac{1}{2}$ | Since the icosahedron is symmetric polyhedron, we can rotate it so that R is on the topmost vertex. Since Q and
S basically the same, we can first count the probability that $d(Q,R) = d(R,S)$
$\mathfrak{Case} \ \mathfrak{1}: d(Q,R) = d(R,S) = 1$
There are 5 points $P$ such that $d(Q,P) = 1$ . There is $5 \times 4 = \boxed{20}$ ways to choose Q and S in this case. | null | 20 |
5b349718665c3da4a24a932d54990462 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_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 $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$ | 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 $p,q$ 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 $\boxed{96}$ | B | 96 |
5b349718665c3da4a24a932d54990462 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_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 $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$ | First, change the problem into an easier form. \[\sum_{d\mid n}d\cdot f(\frac{n}{d} )=\sum_{d\mid n}\frac{n}{d}f(d)=1\] So now we get \[\frac{1}{n}= \sum_{d\mid n}\frac{f(d)}{d}\] Also, notice that both $\frac{f(d)}{d}$ and $\frac{1}{n}$ are arithmetic functions. Applying Möbius inversion formula, we get \[\frac{f(n)}{n}=\sum_{d\mid n}\frac{ \mu (d) }{\frac{n}{d} }=\frac{1}{n} \sum_{d\mid n}d\cdot \mu (d)\] So \[f(n)=1-p_1-p_2-...+p_1p_2+...=(1-p_1)(1-p_2)...=\prod_{p\mid n}(1-p)\] So the answer should be $f(2023)=\prod_{p\mid 2023}(1-p)=(1-7)(1-17)=\boxed{96}$ | B | 96 |
5b349718665c3da4a24a932d54990462 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_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 $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$ | From the problem, we want to find $f(2023)$ . Using the problem, we get $f(2023)+7f(289)+17f(119)+119f(17)+289f(7)+2023f(1)=1$ . By plugging in factors of $2023$ , we get \begin{align} f(7)+7f(1)=1\\ f(17)+17f(1)=1\\ f(119)+7f(17)+17f(7)+119f(1)=1\\ f(289)+17f(17)+289f(1)=1 \end{align} Notice that $(4)-17(2)=f(289)$ , so $f(289)=-16$ . Similarly, notice that $(3)-17(1)=f(119)+7f(17)=-16$ . Now, substituting this all back into our equation to solve for $f(2023)$ , we get \begin{align*} f(2023)+7f(289)+17(f(119)+7f(17))+289(f(7)+7f(1))=1\\ f(2023)+7 \cdot (-16) + 17 \cdot (-16) + 289 \cdot (1) = 1\\ f(2023)=\boxed{96} -PhunsukhWangdu | B | 96 |
5b349718665c3da4a24a932d54990462 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_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 $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$ | Consider any $n \in \Bbb N$ with prime factorization $n = \Pi_{i=1}^k p_i^{\alpha_i}$ .
Thus, the equation given in this problem can be equivalently written as \[ \sum_{\beta_1 = 0}^{\alpha_1} \sum_{\beta_2 = 0}^{\alpha_2} \cdots \sum_{\beta_k = 0}^{\alpha_k} \Pi_{i=1}^k p_i^{\alpha_i - \beta_i} \cdot f \left( \Pi_{i=1}^k p_i^{\beta_i} \right) = 1 . \]
$\noindent \textbf{Special case 1}$ $n = 1$
We have $f \left( 1 \right) = 1$
$\noindent \textbf{Special case 2}$ $n$ is a prime.
We have \[ 1 \cdot f \left( n \right) + n \cdot f \left( 1 \right) = 1 . \]
Thus, $f \left( n \right) = 1 - n$
$\noindent \textbf{Special case 3}$ $n$ is the square of a prime, $n = p_1^2$
We have \[ 1 \cdot f \left( p_1^2 \right) + p_1 \cdot f \left( p_1 \right) + p_1^2 \cdot f \left( 1 \right) = 1. \]
Thus, $f \left( p_1^2 \right) = 1 - p_1$
$\noindent \textbf{Special case 4}$ $n$ is the product of two distinct primes, $n = p_1 p_2$
We have \[ 1 \cdot f \left( p_1 p_2 \right) + p_1 \cdot f \left( p_2 \right) + p_2 \cdot f \left( p_1 \right) + p_1 p_2 \cdot f \left( 1 \right) = 1. \]
Thus, $f \left( p_1 p_2 \right) = 1 - p_1 - p_2 + p_1 p_2$
$\noindent \textbf{Special case 5}$ $n$ takes the form $n = p_1^2 p_2$ , where $p_1$ and $p_2$ are two distinct primes.
We have \[ 1 \cdot f \left( p_1^2 p_2 \right) + p_1 \cdot f \left( p_1 p_2 \right) + p_1^2 \cdot f \left( p_2 \right) + p_2 \cdot f \left( p_1^2 \right) + p_1 p_2 f \left( p_1 \right) + p_1^2 p_2 f \left( 1 \right) = 1. \]
Thus, $f \left( p_1^2 p_2 \right) = 1 - p_1 - p_2 + p_1 p_2$
The prime factorization of 2023 is $7 \cdot 17^2$ .
Therefore, \begin{align*} f \left( 2023 \right) & = 1 - 7 - 17 + 7 \cdot 17 \\ & = \boxed{96} | B | 96 |
33fa7ad3f4c10e63c94ce023046be9ca | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_23 | How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation \[(1+2a)(2+2b)(2a+b) = 32ab?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$ | Using AM-GM on the two terms in each factor on the left, we get \[(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,\] meaning the equality condition must be satisfied. This means $1 = 2a = b$ , so we only have $\boxed{1}$ solution. | null | 1 |
33fa7ad3f4c10e63c94ce023046be9ca | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_23 | How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation \[(1+2a)(2+2b)(2a+b) = 32ab?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$ | Equation $(1+2a)(2+2b)(2a+b)=32ab$ is equivalent to \[b(2a-1)^2+2a(b-1)^2+(2a-b)^2=0,\] where $a$ $b>0$ . Therefore $2a-1=b-1=2a-b=0$ , so $(a,b)=\left(\tfrac12,1\right)$ . Hence the answer is $\boxed{1}$ | B | 1 |
1f2ac522535b2962aa2bd710cbc6deca | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_24 | 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 $10$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | Consider any sequence with $n$ terms. Every 10 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 $\sum_{i=1}^{10}(i+1)^{10}\equiv \boxed{5}$ | C | 5 |
1f2ac522535b2962aa2bd710cbc6deca | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_24 | 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 $10$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | Let $f(x,\ell)$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_\ell$ such that each $A_i$ is a subset of $\{1, 2,\dots,x\}$ , and $A_i$ is a subset of $A_{i+1}$ for $i=1$ $2\dots$ $\ell-1$ . Then $f(x,1)=2^x$ and $f(0,\ell)=1$
If $\ell\ge2$ and $x\ge1$ , we need to get a recursive formula for $f(x,\ell)$ : If $|A_1|=i$ , then $A_1$ has $\text C_x^i$ possibilities, and the subsequence $\{A_i\}_{2\le i\le\ell}$ has $f(x-i,\ell-1)$ possibilities. Hence \[f(x,\ell)=\sum_{i=0}^x\text C_x^if(x-i,\ell-1).\] By applying this formula and only considering modulo $10$ , we get $f(1,2)=3$ $f(1,3)=4$ $f(1,4)=5$ $f(1,5)=6$ $f(1,6)=7$ $f(1,7)=8$ $f(1,8)=9$ $f(1,9)=0$ $f(1,10)=1$ $f(2,2)=9$ $f(2,3)=6$ $f(2,4)=5$ $f(2,5)=6$ $f(2,6)=9$ $f(2,7)=4$ $f(2,8)=1$ $f(2,9)=0$ $f(2,10)=1$ $f(3,2)=7$ $f(3,3)=4$ $f(3,4)=5$ $f(3,5)=6$ $f(3,6)=3$ $f(3,7)=2$ $f(3,8)=9$ $f(3,9)=0$ $f(3,10)=1$ $f(4,2)=1$ $f(4,3)=6$ $f(4,4)=5$ $f(4,5)=6$ $f(4,6)=1$ $f(4,7)=6$ $f(4,8)=1$ $f(4,9)=0$ $f(4,10)=1$ $f(5,2)=3$ $f(5,3)=4$ $f(5,4)=5$ $f(5,5)=6$ $f(5,6)=7$ $f(5,7)=8$ $f(5,8)=9$ $f(5,9)=0$ $f(5,10)=1$ $f(6,2)=9$ $f(6,3)=6$ $f(6,4)=5$ $f(6,5)=6$ $f(6,6)=9$ $f(6,7)=4$ $f(6,8)=1$ $f(6,9)=0$ $f(6,10)=1$ $f(7,2)=7$ $f(7,3)=4$ $f(7,4)=5$ $f(7,5)=6$ $f(7,6)=3$ $f(7,7)=2$ $f(7,8)=9$ $f(7,9)=0$ $f(7,10)=1$ $f(8,2)=1$ $f(8,3)=6$ $f(8,4)=5$ $f(8,5)=6$ $f(8,6)=1$ $f(8,7)=6$ $f(8,8)=1$ $f(8,9)=0$ $f(8,10)=1$ $f(9,2)=3$ $f(9,3)=4$ $f(9,4)=5$ $f(9,5)=6$ $f(9,6)=7$ $f(9,7)=8$ $f(9,8)=9$ $f(9,9)=0$ $f(9,10)=1$ $f(10,2)=9$ $f(10,3)=6$ $f(10,4)=5$ $f(10,5)=6$ $f(10,6)=9$ $f(10,7)=4$ $f(10,8)=1$ $f(10,9)=0$ $f(10,10)=1$
Lastly, we get $K\equiv\textstyle\sum\limits_{i=1}^{10}f(10,i)\equiv\boxed{5}$ .
~Quantum-Phantom | C | 5 |
1f2ac522535b2962aa2bd710cbc6deca | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_24 | 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 $10$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | Seeing that all the answers are different modulus 5, and that 10 is divisible by 5, we cheese this problem.
Let $A_1, A_2, \cdots, A_n$ be one sequence satisfying the constraints of the problem. Let $b_1, b_2, \cdots, b_n, b_{n+1}$ be the sequence of nonnegative integers such that $A_k$ has $b_k$ elements for all $1\leq{}k\leq{}n$ , and $b_{n+1}=10$ . Note that we can generate the number of valid sequences of $A$ by first generating all sequences of $b$ such that $b_i\leq{}b_{i+1}$ for all $1\leq{}i<n$ , then choosing the elements from $A_{k+1}$ that we keep in $A_k$ , given the sequence of $b$ as the restraint for the number of elements. For each sequence $b_1, b_2, \cdots{}, b_{n+1}$ , there are $\prod_{i=2}^{n+1}\binom{b_i}{b_{i-1}}$ corresponding sequences for $A$ . Now, consider two cases - either all terms in $b$ are either 0, 5, or 10, or there is at least one term in $b$ that is neither 0, 5 nor 10. In the second case, consider the last term in $b$ that is not 10 or 5, say $b_m$ . However, that implies $b_{m+1}=10$ , and so the number of corresponding sequences of $A$ is $\binom{10}{b_m}\cdot{}$ something or $\binom{5}{b_m}\cdot$ something, which is always a multiple of $5$ . Therefore, we only need to consider sequences of $b$ where each term is $0$ $5$ or $10$ . If all terms in $b$ are 0 or 10, then for each $1\leq{}n\leq{}10$ there are $n+1$ sequences of $b$ (since there are $n+1$ places to turn from $0$ to $10$ ), for a total of $2+3+\cdots+11=65\equiv0$ (mod 5). If there exists at least one term $b_k=5$ , then we use stars and bars to count the number of sequences of $b$ , and each sequence of $b$ corresponds to $\binom{10}{5}$ sequences of $A$ . For each $n$ , we must have at least one term of $5$ . After that, there are $n-1$ stars and $2$ bars (separating $0$ to $5$ and $5$ to $10$ ), so that is $\binom{n+1}{2}$ sequences of $b$ . So the sum is $\binom{2}{2}+\binom{3}{2}+\cdots+\binom{11}{2}=\binom{12}{3}\equiv0$ (mod 5). Therefore, the answer is 0 mod 5, and it must be $\boxed{5}$ | C | 5 |
1f2ac522535b2962aa2bd710cbc6deca | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_24 | 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 $10$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | We observe that in each sequence, if element $e \in A_i$ , then $e \in A_j$ for all $j \geq i$ .
Therefore, to determine a sequence with a fixed length $n$ , we only need to determine the first set $A_i$ that each element in $\left\{ 1, 2, \cdots , 10 \right\}$ is inserted into, or an element is never inserted into any subset.
We have \begin{align*} K & = \sum_{n = 1}^{10} \left( n + 1 \right)^{10} \\ & = \sum_{m = 2}^{11} m^{10} . \end{align*}
Recalling or noticing 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 \\ & \equiv \sum_{m = 1}^{11} m^2 - 1^2 \\ & \equiv \frac{11 \cdot \left( 11 + 1 \right) \left( 2 \cdot 11 + 1 \right)}{6} - 1\\ & \equiv 505 \\ & \equiv \boxed{5} | C | 5 |
75ef21e032c6c8075dcb7863d30838d9 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_25 | 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 $a_{2023}?$
$\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023$ | \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^{2023} 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*}
\[a_{2023} = \boxed{1}\] | C | 1 |
75ef21e032c6c8075dcb7863d30838d9 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_25 | 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 $a_{2023}?$
$\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023$ | Note that $\tan{kx} = \frac{\binom{k}{1}\tan{x} - \binom{k}{3}\tan^{3}{x} + \cdots \pm \binom{k}{k}\tan^{k}{x}}{\binom{k}{0}\tan^{0}{x} - \binom{k}{2}\tan^{2}{x} + \cdots + \binom{k}{k-1}\tan^{k-1}{x}}$ , where k is odd and the sign of each term alternates between positive and negative. To realize this during the test, you should know the formulas of $\tan{2x}, \tan{3x},$ and $\tan{4x}$ , and can notice the pattern from that. The expression given essentially matches the formula of $\tan{kx}$ exactly. $a_{2023}$ is evidently equivalent to $\pm\binom{2023}{2023}$ , or 1. However, it could be positive or negative. Notice that in the numerator, whenever the exponent of the tangent term is congruent to 1 mod 4, the term is positive. Whenever the exponent of the tangent term is 3 mod 4, the term is negative. 2023, which is assigned to k, is congruent to 3 mod 4. This means that the term of $\binom{k}{k}\tan^{k}{x}$ is $\boxed{1}$ | C | 1 |
75ef21e032c6c8075dcb7863d30838d9 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_25 | 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 $a_{2023}?$
$\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023$ | For odd $n$ , 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} \\ & = \frac{ - 2 i \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}} {2 \sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \\ & = \frac{ \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}} {i \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \\ & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( i \tan x \right)^{2m + 1}} {i \sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( i \tan x \right)^{2m}} \\ & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1} i^{2m + 1}} {\sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \tan x \right)^{2m} i^{2m + 1}} \\ & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1} \left( -1 \right)^m} {\sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \tan x \right)^{2m} \left( -1 \right)^m} . \end{align*}
Thus, for $n = 2023$ , we have \begin{align*} a_{2023} & = \binom{2023}{2023} \left( -1 \right)^{(2023-1)/2} \\ & = \left( -1 \right)^{1011} \\ & = \boxed{1} | C | 1 |
8542a65522a2590dd85224c3708cc2f8 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_1 | Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?
$\textbf{(A) } \frac{1}{12} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{6} \qquad\textbf{(D) } \frac{1}{8} \qquad\textbf{(E) } \frac{2}{9}$ | Given that the first three glasses are full and the fourth is only $\frac{1}{3}$ full, let's represent their contents with a common denominator, which we'll set as 6. This makes the first three glasses $\dfrac{6}{6}$ full, and the fourth glass $\frac{2}{6}$ full.
To equalize the amounts, Mrs. Jones needs to pour juice from the first three glasses into the fourth. Pouring $\frac{1}{6}$ from each of the first three glasses will make them all $\dfrac{5}{6}$ full. Thus, all four glasses will have the same amount of juice. Therefore, the answer is $\boxed{16}.$ | C | 16 |
8542a65522a2590dd85224c3708cc2f8 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_1 | Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?
$\textbf{(A) } \frac{1}{12} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{6} \qquad\textbf{(D) } \frac{1}{8} \qquad\textbf{(E) } \frac{2}{9}$ | We let $x$ denote how much juice we take from each of the first $3$ children and give to the $4$ th child.
We can write the following equation: $1-x=\dfrac13+3x$ , since each value represents how much juice each child (equally) has in the end. (Each of the first three children now have $1-x$ juice, and the fourth child has $3x$ more juice on top of their initial $\dfrac13$ .)
Solving, we see that $x=\boxed{16}.$ | C | 16 |
2ca0547fbf70a2f4e406a3d263f87974 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$ | We can create the equation: \[0.8x \cdot 1.075 = 43\] using the information given. This is because x, the original price, got reduced by 20%, or multiplied by 0.8, and it also got multiplied by 1.075 on the discounted price. Solving that equation, we get \[\frac{4}{5} \cdot x \cdot \frac{43}{40} = 43\] \[\frac{4}{5} \cdot x \cdot \frac{1}{40} = 1\] \[\frac{1}{5} \cdot x \cdot \frac{1}{10} = 1\] \[x = \boxed{50}\] | null | 50 |
2ca0547fbf70a2f4e406a3d263f87974 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$ | The discounted shoe is $20\%$ off the original price. So that means $1 - 0.2 = 0.8$ . There is also a $7.5\%$ sales tax charge, so $0.8 * 1.075 = 0.86$ . Now we can set up the equation $0.86x = 43$ , and solving that we get $x=\boxed{50}$ ~ kabbybear | B | 50 |
2ca0547fbf70a2f4e406a3d263f87974 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$ | Let the original price be $x$ dollars.
After the discount, the price becomes $80\%x$ dollars.
After tax, the price becomes $80\% \times (1+7.5\%) = 86\% x$ dollars.
So, $43=86\%x$ $x=\boxed{50}.$ | B | 50 |
2ca0547fbf70a2f4e406a3d263f87974 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$ | We can assign a variable $c$ to represent the original cost of the shoes. Next, we set up the equation $80\%\cdot107.5\%\cdot c=43$ . We can solve this equation for $c$ and get $\boxed{50}$ | B | 50 |
2ca0547fbf70a2f4e406a3d263f87974 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
$\textbf{(A) }$46\qquad\textbf{(B) }$50\qquad\textbf{(C) }$48\qquad\textbf{(D) }$47\qquad\textbf{(E) }$49$ | We know the discount price will be 5/4, and 0.075 is equal to 3/40. So we look at answer choice $\textbf{(B) }$ , see that the discount price will be 40, and with sales tax applied it will be 43, so the answer choice is $\boxed{50}$ | B | 50 |
120d82221325acbc97c132bcf15a586a | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_4 | Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
$\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textbf{(E) } 16,250$ | $6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \times 2500$ , so the answer is $\boxed{1,625}$ | C | 1,625 |
120d82221325acbc97c132bcf15a586a | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_4 | Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
$\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textbf{(E) } 16,250$ | $6.5$ millimeters can be represented as $65 \times 10^{-2}$ centimeters. $25$ meters is $25 \times 10^{2}$ centimeters. Multiplying out these results in $(65 \times 10^{-2}) \times (25 \times 10^{2})$ , which is $65 \times 25$ making the answer $\boxed{1,625}$ | C | 1,625 |
48fece617762319902b04cd9594dc42f | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$ | First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5$ guesses, and one with $4$ . Since the problem is asking for the minimum number, the answer is $\boxed{4}$ | C | 4 |
48fece617762319902b04cd9594dc42f | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$ | Since the hidden rectangle can only hide two adjacent squares, we may think that we eliminate 8 squares and we're done, but think again. This is the AMC 10, so there must be a better solution (also note that every other solution choice is below 8 so we're probably not done) So, we think again, we notice that we haven't used the adjacent condition, and then it clicks. If we eliminate the four squares with only one edge on the boundary of the 9x9 square. We are left with 5 diagonal squares, since our rectangle cant be diagonal, we can ensure that we find it in 4 moves. So our answer is : $\boxed{4}$ | C | 4 |
48fece617762319902b04cd9594dc42f | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$ | The $3 \times 3$ grid can be colored like a checkerboard with alternating black and white squares.
Let the top left square be white, and the rest of the squares alternate colors.
Each $2 \times 1$ rectangle always covers $1$ white square and $1$ black square.
You can ensure that at least one of your guessed squares is covered by the rectangle by choosing either each of the white squares ( $5$ turns) or each of the black squares ( $4$ turns).
Since it is ideal to be the most efficient with our turns, by choosing all the black squares, we guarantee that one of the $\boxed{4}$ squares are of the $2 \times 1$ rectangle. | null | 4 |
48fece617762319902b04cd9594dc42f | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$ | We realize that every $2 \times 1$ rectangle must contain an edge and no more than one edge. There are a total of four edges so the answer is $\boxed{4.}$ .
~darrenn.cp | C | 4. |
f94ac51df8f5609c071ea4cd49f11383 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5$ | The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$ . We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are $\boxed{6}$ intervals. | C | 6 |
f94ac51df8f5609c071ea4cd49f11383 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5$ | The roots of the factorized polynomial are intervals from numbers 1 to 10. We take each interval as being defined as the number behind it. To make the function positive, we need to have an even number of negative expressions. Real numbers raised to even powers are always positive, so we only focus on $x-1$ $x-3$ $x-5$ $x-7$ , and $x-9$ . The intervals 1 and 2 leave 4 negative expressions, so they are counted. The same goes for intervals 5, 6, 9, and 10. Intervals 3 and 4 leave 3 negative expressions and intervals 7 and 8 leave 1 negative expression. The solution is the number of intervals which is $\boxed{6}$ | C | 6 |
f94ac51df8f5609c071ea4cd49f11383 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5$ | We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.
First, we evaluate any value on the interval $(-\infty, 1)$ . Since the degree of $P(x)$ is $1+2+...+9+10$ $\frac{10\times11}{2}$ $55$ , and every term in $P(x)$ is negative, multiplying $55$ negatives gives a negative value. So $(-\infty, 0)$ is a negative interval.
We know that the roots of $P(x)$ are at $1,2,...,10$ . When the degree of the term of each root is odd, the graph of $P(x)$ will pass through the graph and change signs, and vice versa. So at $x=1$ , the graph will change signs; at $x=2$ , the graph will not, and so on.
This tells us that the interval $(1,2)$ is positive, $(2,3)$ is also positive, $(3,4)$ is negative, $(4,5)$ is also negative, and so on, with the pattern being $+,+,-,-,+,+,-,-,...$
The positive intervals are therefore $(1,2)$ $(2,3)$ $(5,6)$ $(6,7)$ $(9,10)$ , and $(10,\infty)$ , for a total of $\boxed{6}$ | C | 6 |
f94ac51df8f5609c071ea4cd49f11383 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5$ | Denote by $I_k$ the interval $\left( k - 1 , k \right)$ for $k \in \left\{ 2, 3, \cdots , 10 \right\}$ and $I_1$ the interval $\left( - \infty, 1 \right)$
Therefore, the number of intervals that $P(x)$ is positive is \begin{align*} 1 + \sum_{i=1}^{10} \Bbb I \left\{ \sum_{j=i}^{10} j \mbox{ is even} \right\} & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{\left( i + 10 \right) \left( 11 - i \right)}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{- i^2 + i + 110}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{i^2 - i}{2} \mbox{ is odd} \right\} \\ & = \boxed{6} | C | 6 |
536444f8c922ff0dba205bd5e4ccbfd6 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_7 | For how many integers $n$ does the expression \[\sqrt{\frac{\log (n^2) - (\log n)^2}{\log n - 3}}\] represent a real number, where log denotes the base $10$ logarithm?
$\textbf{(A) }900 \qquad \textbf{(B) }2\qquad \textbf{(C) }902 \qquad \textbf{(D) } 2 \qquad \textbf{(E) }901$ | We have \begin{align*} \sqrt{\frac{\log \left( n^2 \right) - \left( \log n \right)^2}{\log n - 3}} & = \sqrt{\frac{2 \log n - \left( \log n \right)^2}{\log n - 3}} \\ & = \sqrt{\frac{\left( \log n \right) \left( 2 - \log n\right)}{\log n - 3}} . \end{align*}
Because $n$ is an integer and $\log n$ is well defined, $n$ must be a positive integer.
Case 1: $n = 1$ or $10^2$
The above expression is 0. So these are valid solutions.
Case 2: $n \neq 1, 10^2$
Thus, $\log n > 0$ and $2 - \log n \neq 0$ .
To make the above expression real, we must have $2 < \log n < 3$ .
Thus, $100 < n < 1000$ .
Thus, $101 \leq n \leq 999$ .
Hence, the number of solutions in this case is 899.
Putting all cases together, the total number of solutions is $\boxed{901}$ | E | 901 |
536444f8c922ff0dba205bd5e4ccbfd6 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_7 | For how many integers $n$ does the expression \[\sqrt{\frac{\log (n^2) - (\log n)^2}{\log n - 3}}\] represent a real number, where log denotes the base $10$ logarithm?
$\textbf{(A) }900 \qquad \textbf{(B) }2\qquad \textbf{(C) }902 \qquad \textbf{(D) } 2 \qquad \textbf{(E) }901$ | Notice $\log(n^2)$ can be written as $2\log(n)$ . Setting $a=\log(n)$ , the equation becomes $\sqrt{\frac{2a-a^2}{a-3}}$ which can be written as $\sqrt{\frac{a(2-a)}{a-3}}$
Case 1: $a \ge 3$ The expression is undefined when $a=3$ . For $a>3$ , it is trivial to see that the denominator is positive and the numerator is negative, thus resulting in no real solutions.
Case 2: $2 \le a<3$ For $a=2$ , the numerator is zero, giving us a valid solution. When $a>2$ , both the denominator and numerator are negative so all real values of a in this interval is a solution to the equation. All integers of n that makes this true are between $10^2$ and $10^3-1$ . There are 900 solutions here.
Case 3: $0<a<2$ The numerator will be positive but the denominator is negative, no real solutions exist.
Case 4: $a=0$ The expression evaluates to zero, $1$ valid solution exists.
Case 5: $a<0$ All values for $a<0$ requires $0<n<1$ , no integer solutions exist.
Adding up the cases: $900+1=\boxed{901}$ | E | 901 |
f643d2eb01b3ccbe875b1daaa9cb133a | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_8 | How many nonempty subsets $B$ of ${0, 1, 2, 3, \cdots, 12}$ have the property that the number of elements in $B$ is equal to the least element of $B$ ? For example, $B = {4, 6, 8, 11}$ satisfies the condition.
$\textbf{(A) } 256 \qquad\textbf{(B) } 136 \qquad\textbf{(C) } 108 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 156$ | There is no way to have a set with 0. If a set is to have its lowest element as 1, it must have only 1 element: 1. If a set is to have its lowest element as 2, it must have 2, and the other element will be chosen from the natural numbers between 3 and 12, inclusive. To calculate this, we do $\binom{10}{1}$ . If the set is the have its lowest element as 3, the other 2 elements will be chosen from the natural numbers between 4 and 12, inclusive. To calculate this, we do $\binom{9}{2}$ . We can see a pattern emerge, where the top is decreasing by 1 and the bottom is increasing by 1. In other words, we have to add $1 + \binom{10}{1} + \binom{9}{2} + \binom{8}{3} + \binom{7}{4} + \binom{6}{5}$ . This is $1+10+36+56+35+6 = \boxed{144}$ | D | 144 |
a620abdbf00b013c87a60f545ad1fe3b | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9 | What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$
$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$ | First consider, $|x-1|+|y-1| \le 1.$ We can see that it is a square with a radius of $1$ (diagonal $\sqrt{2}$ ). The area of the square is $\sqrt{2}^2 = 2.$
Next, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis.
So now we have $2$ squares.
Finally, we add one more absolute value and obtain $||x|-1|+||y|-1| \le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis.
Concluding, we have $4$ congruent squares. The total area is $4\cdot2 =$ $\boxed{8}$ | B | 8 |
a620abdbf00b013c87a60f545ad1fe3b | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9 | What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$
$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$ | We first consider the lattice points that satisfy $||x|-1| = 0$ and $||y|-1| = 1$ . The lattice points satisfying these equations
are $(1,0), (1,2), (1,-2), (-1,0), (-1,2),$ and $(-1,-2).$ By symmetry, we also have points $(0,1), (2,1), (-2,1), (0,-1), (2,-1),$ and $(-2,-1)$ when $||x|-1| = 1$ and $||y|-1| = 0$ . Graphing and connecting these points, we form 5 squares. However,
we can see that any point within the square in the middle does not satisfy the given inequality (take $(0,0)$ , for instance). As
noted in the above solution, each square has a diagonal $2$ for an area of $\frac{2^2}{2} = 2$ , so the total area is $4\cdot2 =$ $\boxed{8}.$ | B | 8 |
a620abdbf00b013c87a60f545ad1fe3b | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9 | What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$
$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$ | The value of $|x|$ and $|y|$ can be a maximum of 1 when the other is 0. Therefore the value of $x$ and $y$ range from -2 to 2. This forms a diamond shape which has area $4 \times \frac{2^2}{2}$ which is $\boxed{8}.$ | B | 8 |
a620abdbf00b013c87a60f545ad1fe3b | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9 | What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$
$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$ | We start by considering the graph of $|x|+|y|\leq 1$ . To get from this graph to $||x|-1|+||y|-1| \leq 1$ we have to translate it by $\pm 1$ on the $x$ axis and $\pm 1$ on the $y$ axis.
Graphing $|x|+|y|\leq 1$ we get a square with side length of $\sqrt{2}$ , so the area of one of these squares is just $2$
We have to multiply by $4$ since there are $4$ combinations of shifting the $x$ and $y$ axis.
So we have $2\times 4$ which is $\boxed{8}$ | B | 8 |
f266d07ceda6bbebf91d426246fd6e09 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_14 | For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 4$ | Denote three roots as $r_1 < r_2 < r_3$ .
Following from Vieta's formula, $r_1r_2r_3 = -6$
Case 1: All roots are negative.
We have the following solution: $\left( -3, -2, -1 \right)$
Case 2: One root is negative and two roots are positive.
We have the following solutions: $\left( -3, 1, 2 \right)$ $\left( -2, 1, 3 \right)$ $\left( -1, 2, 3 \right)$ $\left( -1, 1, 6 \right)$
Putting all cases together, the total number of solutions is $\boxed{5}$ | A | 5 |
50e54b7abfc96818f8df1db05b6483fc | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16 | In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\qquad\textbf{(E) }9$ | This problem asks to find largest $x$ that cannot be written as \[ 6 a + 10 b + 15 c = x, \hspace{1cm} (1) \] where $a, b, c \in \Bbb Z_+$
Denote by $r \in \left\{ 0, 1 \right\}$ the remainder of $x$ divided by 2.
Modulo 2 on Equation (1), we get
By using modulus $m \in \left\{ 2, 3, 5 \right\}$ on the equation above, we get $c \equiv r \pmod{2}$
Following from Chicken McNugget's theorem, we have that any number that is no less than $(3-1)(5-1) = 8$ can be expressed in the form of $3a + 5b$ with $a, b \in \Bbb Z_+$
Therefore, all even numbers that are at least equal to $2 \cdot 8 + 15 \cdot 0 = 16$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ .
All odd numbers that are at least equal to $2 \cdot 8 + 15 \cdot 1 = 31$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$
The above two cases jointly imply that all numbers that are at least 30 can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$
Next, we need to prove that 29 cannot be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$
Because 29 is odd, we must have $c \equiv 1 \pmod{2}$ .
Because $a, b, c \in \Bbb Z_+$ , we must have $c = 1$ .
Plugging this into Equation (1), we get $3 a + 5 b = 7$ .
However, this equation does not have non-negative integer solutions.
All analysis above jointly imply that the largest $x$ that has no non-negative integer solution to Equation (1) is 29.
Therefore, the answer is $2 + 9 = \boxed{11}$ | D | 11 |
50e54b7abfc96818f8df1db05b6483fc | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16 | In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\qquad\textbf{(E) }9$ | Let the number of $6$ cent coins be $a$ , the number of $10$ cent coins be $b$ , and the number of $15$ cent coins be $c$ . We get the Diophantine equation
\[6a + 10b + 15c = k\]
and we wish to find the largest possible value of $k$
Construct the following $\mod 6$ table of $6$ $10$ , and $15$
\[\begin{array}{c|ccc} & & & \\ \text{number of coins} & 6 & 10 & 15 \\ \hline & & & \\ 1 & 0 & 4 & 3\\ & & & \\ 2 & 0 & 2 & 0 \\ \end{array}\]
There are only $6$ possible residues for $6$ , they are: $0$ $1$ $2$ $3$ $4$ , and $5$
Hence, the largest value in cents we cannot obtain using $6$ $10$ , and $15$ cent coins is $29$ $2 + 9 = \boxed{11}$ | D | 11 |
50e54b7abfc96818f8df1db05b6483fc | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16 | In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\qquad\textbf{(E) }9$ | We claim that the largest number that cannot be obtained using $6$ $10$ , and $15$ cent coins is $29$
Let's first focus on the combination of $6$ $10$ . As both of them are even numbers, we cannot obtain any odd numbers from these two but requires $15$ to sum up to an odd number. Notice that by Chicken McNugget Theorem, the largest even number cannot be obtained by $6$ $10$ is $2(3\cdot 5-3-5)=14$ . Add this with $29$ , we can easily verify that $29$ cannot be obtained by $6$ $10$ , and $15$ as it needs at least one odd number, with the remaining part cannot be represented by $6$ and $10$
Let's show that any number greater than $29$ can be obtained. First, any even numbers greater than $29$ can be obtained by $6$ and $10$ by the Chicken McNugget Theorem. Next, any odd number greater than $29$ can be obtained by adding one $15$ with some $6$ s and $10$ s, which is also shown by the Chicken McNugget Theorem. This completes the proof. So the answer is $2+9 = \boxed{11}$ | D | 11 |
28ba9ce32fd109cb2a4c8284a55e162c | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18 | Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was $18$ points higher than her average for the first semester and was again $3$ points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?
$\textbf{(A)}$ Yolanda's quiz average for the academic year was $22$ points higher than Zelda's.
$\textbf{(B)}$ Zelda's quiz average for the academic year was higher than Yolanda's.
$\textbf{(C)}$ Yolanda's quiz average for the academic year was $3$ points higher than Zelda's.
$\textbf{(D)}$ Zelda's quiz average for the academic year equaled Yolanda's.
$\textbf{(E)}$ If Zelda had scored $3$ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda. | Denote by $A_i$ the average of person with initial $A$ in semester $i \in \left\{1, 2 \right\}$ Thus, $Y_1 = Z_1 + 3$ $Y_2 = Y_1 + 18$ $Y_2 = Z_2 + 3$
Denote by $A_{12}$ the average of person with initial $A$ in the full year.
Thus, $Y_{12}$ can be any number in $\left( Y_1 , Y_2 \right)$ and $Z_{12}$ can be any number in $\left( Z_1 , Z_2 \right)$
Therefore, the impossible solution is $\boxed{22}$ | A | 22 |
28ba9ce32fd109cb2a4c8284a55e162c | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18 | Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was $18$ points higher than her average for the first semester and was again $3$ points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?
$\textbf{(A)}$ Yolanda's quiz average for the academic year was $22$ points higher than Zelda's.
$\textbf{(B)}$ Zelda's quiz average for the academic year was higher than Yolanda's.
$\textbf{(C)}$ Yolanda's quiz average for the academic year was $3$ points higher than Zelda's.
$\textbf{(D)}$ Zelda's quiz average for the academic year equaled Yolanda's.
$\textbf{(E)}$ If Zelda had scored $3$ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda. | We can use process of elimination by finding possible solutions to answer choices. Let $y_1$ and $y_2$ be the number of quizzes Yolanda took in the first and second semesters, respectively. Define $z_1$ and $z_2$ similarly for Zelda.
Answer choice B is satisfied by $(y_1,y_2,z_1,z_2) = (289,1,1,289)$
Answer choice C and E are both satisfied by $(y_1,y_2,z_1,z_2) = (17,17,17,17)$
Answer choice D is satisfied by $(y_1,y_2,z_1,z_2) = (7,5,5,7)$
Therefore the answer is $\boxed{22}$ | A | 22 |
28ba9ce32fd109cb2a4c8284a55e162c | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18 | Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was $18$ points higher than her average for the first semester and was again $3$ points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?
$\textbf{(A)}$ Yolanda's quiz average for the academic year was $22$ points higher than Zelda's.
$\textbf{(B)}$ Zelda's quiz average for the academic year was higher than Yolanda's.
$\textbf{(C)}$ Yolanda's quiz average for the academic year was $3$ points higher than Zelda's.
$\textbf{(D)}$ Zelda's quiz average for the academic year equaled Yolanda's.
$\textbf{(E)}$ If Zelda had scored $3$ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda. | Let Yolanda's average for semester $1$ be $y_1$ , the number of quizzes Yolanda took in semester $1$ be $n_1$ , Zelda's average for semester $1$ be $z_1$ , the number of quizzes Zelda took in semester $1$ be $k_1$ , Yolanda's average for semester $2$ be $y_2$ , the number of quizzes Yolanda took in semester $2$ be $n_2$ , Zelda's average for semester $2$ be $z_2$ , the number of quizzes Zelda took in semester $2$ be $k_2$ , Yolanda's average for the entire year be $y$ , Zelda's average for the entire year be $z$
From the problem we know that
\[y_1 = z_1 + 3, \quad y_2 = z_2 + 3\]
\[y_2 = y_1 + 18, \quad z_2 = z_1 + 18\]
\[y = \frac{ y_1 n_1 + y_2 n_2 }{ n_1 + n_2} = \frac{ y_1 n_1 + (y_1 + 18) n_2 }{ n_1 + n_2} = y_1 + \frac{18 y_2 }{ n_1 + n_2}\]
\[z = \frac{ z_1 k_1 + z_2 k_2 }{ k_1 + k_2} = \frac{ z_1 k_1 + (z_1 + 18) k_2 }{ k_1 + k_2} = z_1 = \frac{ 18 k_2 }{ k_1 + k_2}\]
\[y - z = y_1 + \frac{18 y_2 }{ n_1 + n_2} - z_1 - \frac{ 18 k_2 }{ k_1 + k_2} = y_1 - z_1 + 18 \left( \frac{y_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right) = 3 + 18 \left( \frac{y_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right)\]
$\frac{y_2 }{ n_1 + n_2}$ at most is $1$ $\frac{k_2 }{ k_1 + k_2}$ is at least $0$ , meaning that $\frac{y_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2}$ is at most $1$
Therefore, \[y - z \le 3 + 18 = 21\]
Hence, $\boxed{22}$ is not possible. | A | 22 |
28ba9ce32fd109cb2a4c8284a55e162c | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18 | Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was $18$ points higher than her average for the first semester and was again $3$ points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?
$\textbf{(A)}$ Yolanda's quiz average for the academic year was $22$ points higher than Zelda's.
$\textbf{(B)}$ Zelda's quiz average for the academic year was higher than Yolanda's.
$\textbf{(C)}$ Yolanda's quiz average for the academic year was $3$ points higher than Zelda's.
$\textbf{(D)}$ Zelda's quiz average for the academic year equaled Yolanda's.
$\textbf{(E)}$ If Zelda had scored $3$ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda. | Denote $y_1$ and $y_2$ as the quiz averages of Yolanda in the $1$ st and $2$ nd semesters, respectively. Similarly, denote $z_1$ and $z_2$ as the quiz averages of Zelda in the $1$ st and $2$ nd semesters.
We have $y_1 = z_1 + 3$ , so $y_1 - 3 = z_1$ . We also know that $y_2 = 18 +y_1 = 3 + z_2$ , implying $z_2 = 15 + y_1$
The average quiz scores for both students must lie between the averages of each semester, i.e $\mathrm{Avg_{y}}\in[y_1, 18+y_1],$ and $\mathrm{Avg_{z}}\in[y_1-3, y_1+15].$ Since $z_2>z_1$ $y_2 > y_1$ , and $y_1>z_1$ , we have $\min\{\mathrm{Avg}\} = \min\{\mathrm{Avg}_z\} = z_1$ and $\max\{\mathrm{Avg}\} = \max\{\mathrm{Avg}_y\} = y_2$ . Therefore the maximum difference between the two yearly averages is
\[|y_2 - z_1| = |(y_1 + 18)-(y_1-3)| = 21 < 22.\]
Therefore, $\boxed{22}$ is not possible. | A | 22 |
66d18e7d966a9c114b3b52a1226deef5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_19 | Each of $2023$ balls is randomly placed into one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?
$\textbf{(A) } \frac{2}{3} \qquad\textbf{(B) } \frac{3}{10} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{1}{3} \qquad\textbf{(E) } \frac{1}{4}$ | We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.
If a $E$ denotes an even number and a $O$ denotes an odd number, then the distribution of balls for $2022$ balls could be $EEE,EOO,OEO,$ or $OOE$ . With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
From $EEE$ , it is not possible to get to all odd by adding one ball; we could either get $OEE,EOE,$ or $EEO$ . For the other $3$ cases, though, if we add a ball to the exact right place, then it'll work.
For each of the working cases, we have $1$ possible slot the ball can go into (for $OEO$ , for example, the new ball must go in the center slot to make $OOO$ ) out of the $3$ slots, so there's a $\dfrac13$ chance. We have a $\dfrac34$ chance of getting one of these working cases, so our answer is $\dfrac34\cdot\dfrac13=\boxed{14.}$ | E | 14. |
188ab6f20a1737b9212f31d99b8063db | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_22 | A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b) + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$ | Substituting $a = b$ we get \[f(2a) + f(0) = 2f(a)^2\] Substituting $a= 0$ we find \[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\] This gives \[f(2a) = 2f(a)^2 - f(0) \geq 0-1\] Plugging in $a = \frac{1}{2}$ implies $f(1) \geq -1$ , so answer choice $\boxed{2}$ is impossible. | E | 2 |
188ab6f20a1737b9212f31d99b8063db | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_22 | A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b) + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$ | First, we set $a \leftarrow 0$ and $b \leftarrow 0$ .
Thus, the equation given in the problem becomes $[ f(0) + f(0) = 2 f(0) \times f(0) . ]$
Thus, $f(0) = 0$ or 1.
Case 1: $f(0) = 0$
We set $b \leftarrow 0$ .
Thus, the equation given in the problem becomes $[ 2 f(a) = 0 . ]$
Thus, $f(a) = 0$ for all $a$
Case 2: $f(0) = 1$
We set $b \leftarrow a$ .
Thus, the equation given in the problem becomes \[[ f(2a) + 1 = 2 \left( f(a) \right)^2. ]\]
Thus, for any $a$ \begin{align*} f(2a) & = -1 + 2 \left( f(a) \right)^2 \\ & \geq -1 . \end{align*}
Therefore, an infeasible value of $f(1)$ is $\boxed{2}.$ | E | 2 |
188ab6f20a1737b9212f31d99b8063db | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_22 | A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b) + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$ | The relationship looks suspiciously like a product-to-sum identity. In fact, \[\cos(\alpha)\cos(\beta) = \frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))\] which is basically the relation. So we know that $f(x) = \cos(x)$ is a valid solution to the function. However, if we define $x=ay,$ where $a$ is arbitrary, the above relation should still hold for $f(x) = \cos(ay) = \cos(a(1))$ so any value in $[-1,1]$ can be reached, so choices $A,B,$ and $C$ are incorrect.
In addition, using the similar formula for hyperbolic cosine, we know \[\cosh(\alpha)\cosh(\beta) = \frac{1}{2}(\cosh(\alpha-\beta)+\cosh(\alpha+\beta))\] The range of $\cosh(ay)$ is $[1,\infty)$ so choice $D$ is incorrect.
Therefore, the remaining answer is choice $\boxed{2}.$ | E | 2 |
cc5239b25b6f929cf833dce40fb6377f | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_23 | When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$
$\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9$ | We start by trying to prove a function of $n$ , and then we can apply the function and equate it to $936$ to find the value of $n$
It is helpful to think of this problem in the format $(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots$ . Note that if we represent the scenario in this manner, we can think of picking a $1$ for one factor and then a $5$ for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for $n=2$ $4$ can be reached by picking $1$ and $4$ or $2$ and $2$ . However, this form gives us insights that will be useful later in the problem.
Note that there are only $3$ primes in the set $\{1,2,3,4,5,6\}$ $2,3,$ and $5$ . Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form $2^h \cdot 3^i \cdot 5^j$ (the choice of variables will become clear later), for integer nonnegative values $h,i,j$ . So now the problem boils down to how many distinct triplets $(h,i,j)$ can be formed by taking the product of $n$ dice values.
We start our work on representing $j$ : the powers of $5$ , because it is the simplest in this scenario because there is only one factor of $5$ in the set. Because of this, having $j$ fives in our prime factorization of the product is equivalent to picking $j$ factors from the polynomial $(1+\dots + 6) \cdots$ and choosing each factor to be a $5$ . Now that we've selected $j$ factors, there are $n-j$ factors remaining to choose our powers of $3$ and $2$
Suppose our prime factorization of this product contains $i$ powers of $3$ . These powers of $3$ can either come from a $3$ factor or a $6$ factor, but since both $3$ and $6$ contain only one power of $3$ , this means that a product with $i$ powers of $3$ corresponds directly to picking $i$ factors from the polynomial, each of which is either $3$ or $6$ (but this distinction doesn't matter when we consider only the powers of $3$
Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair $(i,j)$ that match the requirements, corresponding to the number of $3$ 's and the number of $5$ 's our product will have. Then how many different $h$ values for the powers of $2$ are possible?
In the $i+j$ factors we have already chosen, we obviously can't have any factors of $2$ in the $j$ factors with $5$ . However, we can have a factor of $2$ pairing with factors of $3$ , if we choose a $6$ . The maximal possible power of $2$ in these $i$ factors is thus $2^i$ , which occurs when we pick every factor to be $6$
We now have $n-i-j$ factors remaining, and we want to allocate these to solely powers of $2$ . For each of these factors, we can choose either a $1,2,$ or $4$ . Therefore the maximal power of $2$ achieved in these factors is when we pick $4$ for all of them, which is equivalent to $2^{2\cdot (n-i-j)}$
Now if we multiply this across the total $n$ factors (or $n$ dice) we have a total of $2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}$ , which is the maximal power of $2$ attainable in the product for a pair $(i,j)$ . Now note that every power of $2$ below this power is attainable: we can simply just take away a power of $2$ from an existing factor by dividing by $2$ . Therefore the powers of $2$ , and thus the $h$ value ranges from $h=0$ to $h=2n-i-2j$ , so there are a total of $2n+1-i-2j$ distinct values for $h$ for a given pair $(i,j)$
Now to find the total number of distinct triplets, we must sum this across all possible $i$ s and $j$ s. Lets take note of our restrictions on $i,j$ : the only restriction is that $i+j \leq n$ , since we're picking factors from $n$ dice.
\[\sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j\]
We start by calculating the first term. $2n+1$ is constant, so we just need to find out how many pairs there are such that $i+j \leq n$ . Set $i$ to $0$ $j$ can range from $0$ to $n$ , then set $i$ to $1$ $j$ can range from $0$ to $n-1$ , etc. The total number of pairs is thus $n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}$ . Therefore the left summation evaluates to \[\frac{(2n+1)(n+1)(n+2)}{2}\]
Now we calculate $\sum_{i+j \leq n}^{} i+2j$ . This simplifies to $\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j$ . Note that because $i+j = n$ is symmetric with respect to $i,j$ , the sum of $i$ in all of the pairs will be equal to the sum of $j$ in all of the pairs. Thus this is equal to calculating $3 \cdot \sum_{i+j \leq n}^{} i$
In the pairs, $i=1$ appears for $j$ ranging between $0$ and $n-1$ so the sum here is $1 \cdot (n)$ . Similarly $i=2$ appears for $j$ ranging from $0$ to $n-2$ , so the sum is $2 \cdot (n-1)$ . If we continue the pattern, the sum overall is $(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1$ . We can rearrange this as $((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)$
\[= \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1\]
We can write this in easier terms as $\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k$ \[=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)\] \[= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})\] \[= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}\] \[= \frac{n(n+1)(n+2)}{6}\]
We multiply this by $3$ to obtain that \[\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}\]
Thus our final answer for the number of distinct triplets $(h,i,j)$ is: \[\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}\] \[= \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)\] \[= \frac{(n+1)^2(n+2)}{2}\]
Now most of the work is done. We set this equal to $936$ and prime factorize. $936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13$ , so $(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13$ . Clearly $13$ cannot be anything squared and $2^4 \cdot 3^2$ is a perfect square, so $n+2 = 13$ and $n = 11 = \boxed{11}$ | A | 11 |
cc5239b25b6f929cf833dce40fb6377f | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_23 | When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$
$\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9$ | The product can be written as \begin{align*} 2^a 3^b 4^c 5^d 6^e & = 2^{a + 2c + e} 3^{b + e} 5^d . \end{align*}
Therefore, we need to find the number of ordered tuples $\left( a + 2c + e, b+e, d \right)$ where $a$ $b$ $c$ $d$ $e$ are non-negative integers satisfying $a+b+c+d+e \leq n$ .
We denote this number as $f(n)$
Denote by $g \left( k \right)$ the number of ordered tuples $\left( a + 2c + e, b+e \right)$ where $\left( a, b, c, e \right) \in \Delta_k$ with $\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}$
Thus, \begin{align*} f \left( n \right) & = \sum_{d = 0}^n g \left( n - d \right) \\ & = \sum_{k = 0}^n g \left( k \right) . \end{align*}
Next, we compute $g \left( k \right)$
Denote $i = b + e$ . Thus, for each given $i$ , the range of $a + 2c + e$ is from 0 to $2 k - i$ .
Thus, the number of $\left( a + 2c + e, b + e \right)$ is \begin{align*} g \left( k \right) & = \sum_{i=0}^k \left( 2 k - i + 1 \right) \\ & = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) . \end{align*}
Therefore, \begin{align*} f \left( n \right) & = \sum_{k = 0}^n g \left( k \right) \\ & = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \\ & = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2 - \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \\ & = \frac{3}{2} \cdot \frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right) - \frac{1}{2} \cdot \frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \\ & = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) . \end{align*}
By solving $f \left( n \right) = 936$ , we get $n = \boxed{11}$ | A | 11 |
cc5239b25b6f929cf833dce40fb6377f | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_23 | When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$
$\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9$ | The product can be written as \begin{align*} 2^x 3^y 5^z \end{align*}
Letting $n=1$ , we get $(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)$ , 6 possible values. But if the only restriction of the product if that $2x\le n,y\le n,z\le n$ , we can get $(2+1)(1+1)(1+1)=12$ possible values. We calculate the ratio \[r = \frac{\text{possible values of real situation}}{\text{possible values of ideal situation}} = \frac{6}{12}=0.5.\]
Letting $n=2$ , we get $(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),$ $(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)$ , 17 possible values.
The number of possibilities in the ideal situation is $5*3*3=45$ , making $r = 17/45 \approx 0.378$
Now we can predict the trend of $r$ : as $n$ increases, $r$ decreases.
Letting $n=3$ , you get possible values of ideal situation= $7*4*4=112$ $n=4$ , the number= $9*5*5=225$ $n=5$ , the number= $11*6*6=396$ $n=6$ , the number= $13*7*7=637,637<936$ so 6 is not the answer. $n=7$ , the number= $15*8*8=960$ $n=8$ , the number= $17*9*9=1377$ ,but $1377*0.378$ $521$ still much smaller than 936. $n=9$ , the number= $19*10*10=1900$ ,but $1900*0.378$ $718$ still smaller than 936. $n=10$ , the number= $21*11*11=2541$ $2541*0.378$ $960$ is a little bigger 936, but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is $\boxed{11}$ | A | 11 |
0729f0c94b88d88f4d4f6f99e514fbd0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_24 | Suppose that $a$ $b$ $c$ and $d$ are positive integers satisfying all of the following relations.
\[abcd=2^6\cdot 3^9\cdot 5^7\] \[\text{lcm}(a,b)=2^3\cdot 3^2\cdot 5^3\] \[\text{lcm}(a,c)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(a,d)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(b,c)=2^1\cdot 3^3\cdot 5^2\] \[\text{lcm}(b,d)=2^2\cdot 3^3\cdot 5^2\] \[\text{lcm}(c,d)=2^2\cdot 3^3\cdot 5^2\]
What is $\text{gcd}(a,b,c,d)$
$\textbf{(A)}~30\qquad\textbf{(B)}~45\qquad\textbf{(C)}~3\qquad\textbf{(D)}~15\qquad\textbf{(E)}~6$ | Denote by $\nu_p (x)$ the number of prime factor $p$ in number $x$
We index Equations given in this problem from (1) to (7).
First, we compute $\nu_2 (x)$ for $x \in \left\{ a, b, c, d \right\}$
Equation (5) implies $\max \left\{ \nu_2 (b), \nu_2 (c) \right\} = 1$ .
Equation (2) implies $\max \left\{ \nu_2 (a), \nu_2 (b) \right\} = 3$ .
Equation (6) implies $\max \left\{ \nu_2 (b), \nu_2 (d) \right\} = 2$ .
Equation (1) implies $\nu_2 (a) + \nu_2 (b) + \nu_2 (c) + \nu_2 (d) = 6$
Therefore, all above jointly imply $\nu_2 (a) = 3$ $\nu_2 (d) = 2$ , and $\left( \nu_2 (b), \nu_2 (c) \right) = \left( 0 , 1 \right)$ or $\left( 1, 0 \right)$
Second, we compute $\nu_3 (x)$ for $x \in \left\{ a, b, c, d \right\}$
Equation (2) implies $\max \left\{ \nu_3 (a), \nu_3 (b) \right\} = 2$ .
Equation (3) implies $\max \left\{ \nu_3 (a), \nu_3 (c) \right\} = 3$ .
Equation (4) implies $\max \left\{ \nu_3 (a), \nu_3 (d) \right\} = 3$ .
Equation (1) implies $\nu_3 (a) + \nu_3 (b) + \nu_3 (c) + \nu_3 (d) = 9$
Therefore, all above jointly imply $\nu_3 (c) = 3$ $\nu_3 (d) = 3$ , and $\left( \nu_3 (a), \nu_3 (b) \right) = \left( 1 , 2 \right)$ or $\left( 2, 1 \right)$
Third, we compute $\nu_5 (x)$ for $x \in \left\{ a, b, c, d \right\}$
Equation (5) implies $\max \left\{ \nu_5 (b), \nu_5 (c) \right\} = 2$ .
Equation (2) implies $\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3$ .
Thus, $\nu_5 (a) = 3$
From Equations (5)-(7), we have either $\nu_5 (b) \leq 1$ and $\nu_5 (c) = \nu_5 (d) = 2$ , or $\nu_5 (b) = 2$ and $\max \left\{ \nu_5 (c), \nu_5 (d) \right\} = 2$
Equation (1) implies $\nu_5 (a) + \nu_5 (b) + \nu_5 (c) + \nu_5 (d) = 7$ .
Thus, for $\nu_5 (b)$ $\nu_5 (c)$ $\nu_5 (d)$ , there must be two 2s and one 0.
Therefore, \begin{align*} {\rm gcd} (a,b,c,d) & = \Pi_{p \in \{ 2, 3, 5\}} p^{\min\{ \nu_p (a), \nu_p(b) , \nu_p (c), \nu_p(d) \}} \\ & = 2^0 \cdot 3^1 \cdot 5^0 \\ & = \boxed{3} | C | 3 |
1b1f489744d81ebb17c8bd3e16b38e80 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_2 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$ | Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$
We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$
Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{5}.$ | E | 5 |
1b1f489744d81ebb17c8bd3e16b38e80 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_2 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$ | Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: \begin{align*} x+y+z&=96, \\ x&=6z, \\ z&=y-40. \end{align*} Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$
Therefore, we get \begin{align*} 6z+(z+40)+z&=96 \\ 8z+40&=96 \\ 8z&=56 \\ z&=7. \end{align*} Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47.$
So, the answer is $|x-y|=|42-47|=\boxed{5}.$ | E | 5 |
1b1f489744d81ebb17c8bd3e16b38e80 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_2 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$ | In accordance with Solution 2, \[y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{5}.\] [email protected], vvsss | E | 5 |
198c83a8aae2415bd0e7046faef3a4aa | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_4 | The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | Note that \begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*} Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that:
Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{6}.$ | B | 6 |
198c83a8aae2415bd0e7046faef3a4aa | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_4 | The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | The options for $\text{lcm}(x, 18)=180$ are $20$ $60$ , and $180$ . The options for $\text{gcd}(y, 45)=15$ are $15$ $30$ $60$ $75$ , etc. We see that $60$ appears in both lists; therefore, $6+0=\boxed{6}$ | B | 6 |
4eb6bd114745345c53cd0e62e1c26831 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924$ | Let us consider the number of points for a certain $x$ -coordinate. For any $x$ , the viable points are in the range $[-20 + |x|, 20 - |x|]$ . This means that our total sum is equal to \begin{align*} 1 + 3 + 5 + \cdots + 41 + 39 + 37 + \cdots + 1 &= (1 + 3 + 5 + \cdots + 39) + (1 + 3 + 5 + \cdots + 41) \\ & = 20^2 + 21^2 \\ & = 29^2 \\ &= \boxed{841} ~mathboy100 | C | 841 |
4eb6bd114745345c53cd0e62e1c26831 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924$ | Since the second point is the origin, this is equivalent to finding all points $(x, y)$ such that $|x| + |y| \leq 20$ . Due to the absolute values, the set of all such points will be symmetric about the origin meaning we can focus on the first quadrant and multiply by $4$
To avoid overcounts, ignore points on the axes. This means $x, y > 0$ . If $x = 1$ , there are $19$ solutions for $y$ $y = 1, 2, 3, \ldots, 19$ ). If $x = 2$ , there are $18$ solutions. This pattern repeats until $x = 19$ , at which point there is $1$ solution for $y$
So we get $19 + 18 + 17 + \cdots + 1 = \frac{19(20)}{2} = 190$ points in the first quadrant. Multiplying by $4$ gives $760$ . Now, the $x$ axis has $y = 0$ which gives $|x| \leq 20$ , meaning there are $41$ solutions. This is the same with the $y$ axis, but we overcounted the origin by $1$
Our final answer is $760 + 41 + 41 - 1 = \boxed{841}$ | C | 841 |
4eb6bd114745345c53cd0e62e1c26831 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924$ | This solution refers to the Diagram section. [asy] /* Made by MRENTHUSIASM */ size(350); for (int y = 20; y >= 1; --y) { for (int x = 0; x <= 20-y; ++x) { dot((x,y),green+linewidth(4)); } } for (int y = 0; y >= -20; --y) { for (int x = 1; x <= y+20; ++x) { dot((x,y),blue+linewidth(4)); } } for (int y = 20; y >= 0; --y) { for (int x = y-20; x <= -1; ++x) { dot((x,y),purple+linewidth(4)); } } for (int y = -1; y >= -20; --y) { for (int x = -y-20; x <= 0; ++x) { dot((x,y),red+linewidth(4)); } } dot(origin,black+linewidth(4)); [/asy] The problem can be visualized as depicted on the right split equally into four "triangular" parts excluding the origin. The "triangular" parts are identical the ones that would be used in a visual proof of the formula for triangular numbers. Becuase of this the number of points in each part is equal to $\frac{n(n+1)}{2}$ where $n$ is the length of a "leg" of the "triangle" which is $20$ for this problem. Substituting and computing, we get $210.$ Multiplying by $4$ and adding $1$ to account for all parts and the origin, we get $210\cdot4 + 1 = \boxed{841}.$ | C | 841 |
4eb6bd114745345c53cd0e62e1c26831 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924$ | This solution refers to the Diagram section.
As shown below, the taxicab distance between each red point and the origin is even, and the taxicab distance between each blue point and the origin is odd. [asy] /* Made by MRENTHUSIASM */ size(350); for (int y = 20; y >= 0; --y) { for (int x = y-20; x <= 20-y; x+=2) { dot((x,y),red+linewidth(4)); } } for (int y = -1; y >= -20; --y) { for (int x = -y-20; x <= y+20; x+=2) { dot((x,y),red+linewidth(4)); } } for (int y = 19; y >= 0; --y) { for (int x = y-19; x <= 19-y; x+=2) { dot((x,y),blue+linewidth(4)); } } for (int y = -1; y >= -19; --y) { for (int x = -y-19; x <= y+19; x+=2) { dot((x,y),blue+linewidth(4)); } } draw((20,0)--(0,20)--(-20,0)--(0,-20)--cycle,red+linewidth(1.25)); draw((19,0)--(0,19)--(-19,0)--(0,-19)--cycle,blue+linewidth(1.25)); [/asy] Note that the red array consists of $21^2=441$ points, and the blue array consists of $20^2=400$ points.
Together, the answer is $441+400=\boxed{841}.$ | C | 841 |
4eb6bd114745345c53cd0e62e1c26831 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924$ | Let $P = (x, y)$ . Since the problem asks for taxicab distances from the origin, we want $|x| + |y| \le 20$ . The graph of all solutions to this equation on the $xy$ -plane is a square with vertices at $(0, \pm 20)$ and $(\pm 20, 0)$ (In order to prove this, one can divide the sections of this graph into casework on the four quadrants, and tie together the resulting branches.) We want the number of lattice points on the border of the square and inside the square.
Each side of the square goes through an equal number of lattice points, so if we focus on one side going from $(0,20)$ to $(20, 0)$ , we can see that it goes through $21$ points in total. In addition, each of the vertices gets counted twice, so the total number of border points is $21\cdot4 - 4 = 80$ . Also, the area of the square is $800$ , so when we plug this information inside Pick's theorem, we get $800 = i + \frac{80}{2} - 1 \implies i = 761$ . Then our answer is $761+80 = \boxed{841}.$ | C | 841 |
4eb6bd114745345c53cd0e62e1c26831 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924$ | Instead of considering all points with integer coordinates, first consider points with nonnegative coordinates only. Then, we want $x + y \le 20$ where $x$ and $y$ are nonnegative integers. We can introduce a third variable, say $z$ , such that $z = 20 - (x + y)$ . Note that counting the number of ways to have $x + y + z = 20$ is the same as counting the number of ways to have $x + y \le 20$ . Therefore, by stars and bars, there are $\dbinom{20 + 3 - 1}{3 - 1} = 231$ solutions with nonnegative integer coordinates.
Then, we can copy our solutions over to the other four quadrants. First, so as not to overcount, we remove all points on the axes. There are $20 + 20 + 1 = 41$ such points with nonnegative integer coordinates. We multiply the $190$ remaining points by $4$ to get $760$ points that are not on the axes. Then, we can add back the $41$ nonnegative points on the axes, as well as the $40$ other points on the negative axes to get $760 + 41 + 40 = \boxed{841}.$ | C | 841 |
306ad56644222e4293d34fe57f1b9ca3 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_6 | A data set consists of $6$ (not distinct) positive integers: $1$ $7$ $5$ $2$ $5$ , and $X$ . The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all possible values of $X$
$\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40$ | First, note that $1+7+5+2+5=20$ . There are $3$ possible cases:
Case 1: the mean is $5$
$X = 5 \cdot 6 - 20 = 10$
Case 2: the mean is $7$
$X = 7 \cdot 6 - 20 = 22$
Case 3: the mean is $X$
$X= \frac{20+X}{6} \Rightarrow X=4$
Therefore, the answer is $10+22+4=\boxed{36}$ | D | 36 |
958b419d57c566a5ed2c3527b6a7a84a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_7 | A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
[asy] size(5.5cm); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((2,0)--(8,0)--(8,2)--(2,2)--cycle); draw((8,0)--(12,0)--(12,2)--(8,2)--cycle); draw((0,2)--(6,2)--(6,4)--(0,4)--cycle); draw((6,2)--(12,2)--(12,4)--(6,4)--cycle); [/asy]
$\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720$ | The top left rectangle can be $5$ possible colors. Then the bottom left region can only be $4$ possible colors, and the bottom middle can only be $3$ colors since it is next to the top left and bottom left. Similarly, we have $3$ choices for the top right and $3$ choices for the bottom right, which gives us a total of $5\cdot4\cdot3\cdot3\cdot3=\boxed{540}$ | D | 540 |
958b419d57c566a5ed2c3527b6a7a84a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_7 | A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
[asy] size(5.5cm); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((2,0)--(8,0)--(8,2)--(2,2)--cycle); draw((8,0)--(12,0)--(12,2)--(8,2)--cycle); draw((0,2)--(6,2)--(6,4)--(0,4)--cycle); draw((6,2)--(12,2)--(12,4)--(6,4)--cycle); [/asy]
$\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720$ | Case 1: All the rectangles are different colors. It would be $5! = 120$ choices.
Case 2: Two rectangles that are the same color. Grouping these two rectangles as one gives us $5\cdot4\cdot3\cdot2 = 120$ . But, you need to multiply this number by three because the same-colored rectangles can be chosen at the top left and bottom right, the top right and bottom left, or the bottom right and bottom left, which gives us a grand total of $360$
Case 3: We have two sets of rectangles chosen from these choices (top right & bottom left, top left & bottom right) that have the same color. However, the choice of the bottom left and bottom right does not work for this case, as the second pair would be chosen from two touching rectangles. Again, grouping the same-colored rectangles gives us $5\cdot4\cdot3 = 60$
Therefore, we have $120 + 360 + 60 = \boxed{540}$ | D | 540 |
73f2bab466470a9d11e6bc33c39cc117 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_8 | The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?
$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$ | We can write $\sqrt[3]{10}$ as $10 ^ \frac{1}{3}$ . Similarly, $\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$
By continuing this, we get the form \[10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,\] which is \[10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}.\] Using the formula for an infinite geometric series $S = \frac{a}{1-r}$ , we get \[\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}.\] Thus, our answer is $10 ^ \frac{1}{2} = \boxed{10}$ | A | 10 |
73f2bab466470a9d11e6bc33c39cc117 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_8 | The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?
$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$ | We can write this infinite product as $L$ (we know from the answer choices that the product must converge): \[L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\] If we raise everything to the third power, we get: \[L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \cdots = 10L \implies L^3 - 10L = 0 \implies L \in \left\{0, \pm \sqrt{10}\right\}.\] Since $L$ is positive (as it is an infinite product of positive numbers), it must be that $L = \boxed{10}.$ | A | 10 |
73f2bab466470a9d11e6bc33c39cc117 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_8 | The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?
$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$ | Move the first term inside the second radical. We get \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\] Do this for the third radical as well: \[\sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10}\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10\sqrt[3]{10\cdots}}}.\] It is clear what the pattern is. Setting the answer as $P,$ we have \[P = \sqrt[3]{10P},\] from which $P = \boxed{10}.$ | null | 10 |
591e95d2b1d2b47c854d366a8684f9e5 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_9 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$ children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the $15$ children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who
answered yes.
How many pieces of candy in all did the principal give to the children who always
tell the truth?
$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$ | Note that:
Suppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.
The conditions of the first two questions imply that \begin{align*} T+L+A&=22, \\ L+A&=15. \end{align*} Subtracting the second equation from the first, we have $T=22-15=\boxed{7}.$ | A | 7 |
591e95d2b1d2b47c854d366a8684f9e5 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_9 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$ children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the $15$ children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who
answered yes.
How many pieces of candy in all did the principal give to the children who always
tell the truth?
$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$ | Consider when the principal asks "Are you a liar?": The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all $9$ children that answered yes are alternaters that falsely answer Questions 1 and 3, and truthfully answer Question 2. The rest of the alternaters, however many there are, have the opposite behavior.
Consider the second question, "Are you an alternater?": The truth tellers again answer no, the liars falsely answer yes, and alternaters that truthfully answer also say yes. From the previous part, we know that $9$ alternaters truthfully answer here. Because only liars and $9$ alternaters answer yes, we can deduce that there are $15-9=6$ liars.
Consider the first question, "Are you a truth teller?": Truth tellers say yes, liars also say yes, and alternaters that lie on this question also say yes. From the first part, we know that $9$ alternaters lie here. From the previous part, we know that there are $6$ liars. Because only the number of truth tellers is unknown here, we can deduce that there are $22-9-6=7$ truth tellers.
The final question is how many pieces of candy did the principal give to truth tellers. Because truth tellers answer yes on only the first question, we know that all $7$ of them said yes once, resulting in $\boxed{7}$ pieces of candy. | A | 7 |
591e95d2b1d2b47c854d366a8684f9e5 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_9 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$ children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the $15$ children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who
answered yes.
How many pieces of candy in all did the principal give to the children who always
tell the truth?
$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$ | Note that we have $4$ types of people:
Given this information, it is reasonable to ignore the fourth type, because they will never answer yes to any question. Hence, we only consider people of type $1, 2,$ and $3.$
The principal's first question implies that \[T + L + A = 22.\] The second question implies that \[L + A = 15.\] The third question implies that \[A = 9.\]
Solving, we find that $T = 7,$ so $\boxed{7}$ is our answer. We can also note that $T = 7, L = 6, A = 9,$ and there are $9$ alternators who answer no to every question. | A | 7 |
15422b10cd2f7a57f4d8a412de7ac36b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_10 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | Clearly, the integers from $8$ through $14$ must be in different pairs, and $7$ must pair with $14.$
Note that $6$ can pair with either $12$ or $13.$ From here, we consider casework:
Together, the answer is $72+72=\boxed{144}.$ | E | 144 |
15422b10cd2f7a57f4d8a412de7ac36b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_10 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | As said in Solution 1, clearly, the integers from $8$ through $14$ must be in different pairs.
We know that $8$ or $9$ can pair with any integer from $1$ to $4$ $10$ or $11$ can pair with any integer from $1$ to $5$ , and $12$ or $13$ can pair with any integer from $1$ to $6$ . Thus, $8$ will have $4$ choices to pair with, $9$ will then have $3$ choices to pair with ( $9$ cannot pair with the same number as the one $8$ pairs with). $10$ cannot pair with the numbers $8$ and $9$ has paired with but can also now pair with $5$ , so there are $3$ choices. $11$ cannot pair with $8$ 's, $9$ 's, or $10$ 's paired numbers, so there will be $2$ choices for $11$ $12$ can pair with an integer from $1$ to $5$ that hasn't been paired with already, or it can pair with $6$ $13$ will only have one choice left, and $7$ must pair with $14$
So, the answer is $4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{144}.$ | E | 144 |
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