AMC10 2022 B

We have \begin{align*} (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &= \boxed{\textbf{(A)}\ {-}2}. \end{align*}

Since $ABCD$ is a rhombus, all sides are equal, so \[ AB = AD = AP + PD = 3 + 2 = 5. \] Point $P$ lies on $\overline{AD}$ and $BP \perp AD$. Apply the Pythagorean theorem in right triangle $ABP$: \[ AP^2 + BP^2 = AB^2 \] \[ 3^2 + BP^2 = 5^2 \] \[ 9 + BP^2 = 25 \] \[ BP^2 = 16 \implies BP = 4. \] The area of a rhombus is base times height. Taking $AD$ as the base and $BP$ as the corresponding height, the area is \[ 5 \times 4 = 20. \] Thus, the area of rhombus $ABCD$ is $20$.

We use simple case work to solve this problem. Case 1: even, even, even = $4 \cdot 5 \cdot 5 = 100$ Case 2: even, odd, odd = $4 \cdot 5 \cdot 5 = 100$ Case 3: odd, even, odd = $5 \cdot 5 \cdot 5 = 125$ Case 4: odd, odd, even = $5 \cdot 5 \cdot 5 = 125$ Simply sum up the cases to get your answer, $100 + 100 + 125 + 125 = \boxed{\textbf{(D)~}450}$.

Since the donkey hiccupped the 1st hiccup at $4:00$, it hiccupped for $5 \cdot (700-1) = 3495$ seconds, which is $58$ minutes and $15$ seconds, so the answer is $\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}$.

We apply the difference of squares to the denominator, and then regroup factors: \begin{align*} \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}} &= \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}\cdot\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}}{\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\frac43\cdot\frac65\cdot\frac87}}{\sqrt{\frac23\cdot\frac45\cdot\frac67}} \\ &= \frac{\sqrt{4\cdot6\cdot8}}{\sqrt{2\cdot4\cdot6}} \\ &= \frac{\sqrt8}{\sqrt2} \\ &= \boxed{\textbf{(B)}\ 2}. \end{align*}

The $n$th term of this sequence is \[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\] It follows that the terms are \begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*} Therefore, there are $\boxed{\textbf{(A) } 0}$ prime numbers in this sequence.

Let $p$ and $q$ be the roots of $x^{2}+kx+36.$ By Vieta's Formulas, we have $p+q=-k$ and $pq=36.$ It follows that $p$ and $q$ must be distinct factors of $36.$ The possibilities of $\{p,q\}$ are \[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\] Each unordered pair gives a unique value of $k.$ Therefore, there are $\boxed{\textbf{(B) }8}$ values of $k,$ corresponding to $\mp37,\mp20,\mp15,\mp13,$ respectively.

There are \(\text{floor}\left(\frac{1000}{7}\right) = 142\) numbers divisible by 7. We split these into 100 sets containing 10 numbers each, giving us 1.42 multiples of 7 per set. After the first set, the numbers come evenly, and we multiply 100 by $1.42 - 1 = \boxed{\textbf{(B)}\ 42}.$

For all positive integers $n,$ we have \[\frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}.\] Note that the original sum is a telescoping series: \begin{align*} \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!} &= \left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\cdots+\left(\frac{1}{2021!}-\frac{1}{2022!}\right) \\ &= \frac{1}{1!}+\left(-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\cdots+\frac{1}{2021!}\right)-\frac{1}{2022!} \\ &= 1-\frac{1}{2022!}. \end{align*} Therefore, the answer is $1+2022=\boxed{\textbf{(D)}\ 2023}.$

Let $M$ be the median. It follows that the two largest integers are both $M+2.$ Let $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \[a,b,M,M+2,M+2.\] Since the median is $2$ greater than their arithmetic mean, we have $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \[a+b+14=2M.\] Note that $a+b$ must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let $a=1$ and $b=3,$ from which $M=9$ and $M+2=\boxed{\textbf{(D)}\ 11}.$

Let $B$ denote a school that is bigger than Euclid HS, and $M$ denote a school that sold more T-shirts than Euclid HS. It follows that $ eg B$ denotes a school that is not bigger than Euclid HS, and $ eg M$ denotes a school that did not sell more T-shirts than Euclid HS. Converting everything to conditional statements (if-then form), the given statement becomes \[B\implies eg M.\] Its contrapositive is $M\implies eg B,$ which is $\boxed{\textbf{(B)}}.$ Note that "not bigger than" does not mean "smaller than", and "not selling more" does not mean "selling fewer". There is an equality case. Therefore, none of the other answer choices is equivalent to $B\implies eg M.$

Rolling a pair of fair $6$-sided dice, the probability of getting a sum of $7$ is $\frac16:$ Regardless what the first die shows, the second die has exactly one outcome to make the sum $7.$ We consider the complement: The probability of not getting a sum of $7$ is $1-\frac16=\frac56.$ Rolling the pair of dice $n$ times, the probability of getting a sum of $7$ at least once is $1-\left(\frac56\right)^n.$ Therefore, we have $1-\left(\frac56\right)^n>\frac12,$ or \[\left(\frac56\right)^n<\frac12.\] Since $\left(\frac56\right)^4<\frac12<\left(\frac56\right)^3,$ the least integer $n$ satisfying the inequality is $\boxed{\textbf{(C) } 4}.$

Denote the two primes as \( a \) and \( b \). Then, \( a - b = 2 \) \( a^3 - b^3 = 31106 \) We see that \( a = 2 + b \) Now, we have \( (2+b)^3 - b^3 = 31106 \) We apply the binomial theorem (or just expand) to \( (2+b)^3 \), getting \( 8 + 12b + 6b^2 + b^3 - b^3 = 31106 \) \( \Rightarrow 8 + 12b + 6b^2 = 31106 \) Subtracting by 8 on both sides results in \( 6b^2 + 12b = 31098 \) \( b^2 + 2b - 5183 \) \( (b-71)(b+73) \) We see that \( b \in \{-73, 71\} \). We negate all negative values, and see that \( b = 71 \). Therefore \( a - b = 2 \), \( a = 2 + b \), \( a = 2 + 71 = 73 \). The next prime number greater than both of these is $79$, and therefore our answer is \( 7 + 9 = \) $\boxed{\textbf{(E) }16}$.

Let $M$ be the largest number in $S$. We categorize numbers $\left\{ 1, 2, \ldots , M-1 \right\}$ (except $\frac{M}{2}$ if $M$ is even) into $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups, such that the $i$th group contains two numbers $i$ and $M-i$. Recall that $M \in S$ and the sum of two numbers in $S$ cannot be equal to $M$, and the sum of numbers in each group above is equal to $S$. Thus, each of the above $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups can have at most one number in $S$. Therefore, \begin{align*} |S| & \leq 1 + \left\lfloor \frac{M-1}{2} \right\rfloor \\ & \leq 1 + \left\lfloor \frac{25}{2} \right\rfloor \\ & = 13. \end{align*} Next, we construct an instance of $S$ with $|S| = 13$. Let $S = \left\{ 13, 14, \ldots , 25 \right\}$. Thus, this set is feasible. Therefore, the most number of elements in $S$ is $\boxed{\textbf{(B) }13}$.

Let's say that our sequence is \[a, a+2, a+4, a+6, a+8, a+10, \ldots.\] Then, since the value of n doesn't matter in the quotient $\frac{S_{3n}}{S_n}$, we can say that \[\frac{S_{3}}{S_1} = \frac{S_{6}}{S_2}.\] Simplifying, we get $\frac{3a+6}{a}=\frac{6a+30}{2a+2}$, from which \[\frac{3a+6}{a}=\frac{3a+15}{a+1}.\] \[3a^2+9a+6=3a^2+15a\] \[6a=6\] Solving for $a$, we get that $a=1$. Since the sum of the first $n$ odd numbers is $n^2$, $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$. Note: you could also plug in the formulas for $\frac{S_{3n}}{S_{n}}$ and simplify, getting $3+ \frac{6n}{a+n-1}$ You would then find a=$1$

Let us label the points on the diagram. By doing some angle chasing using the fact that $\angle ACE$ and $\angle CEG$ are right angles, we find that $\angle BAC = \angle DCE = \angle FEG$. Similarly, $\angle ACB = \angle CED = \angle EGF$. Therefore, $\triangle ABC \sim \triangle CDE \sim \triangle EFG$. As we are given a rectangle and a square, $AB = 4$ and $AC = 5$. Therefore, $\triangle ABC$ is a $3$-$4$-$5$ right triangle and $BC = 3$. $CE$ is also $5$. So, using the similar triangles, $CD = 4$ and $DE = 3$. $EF = DF - DE = 4 - 3 = 1$. Using the similar triangles again, $EF$ is $\frac14$ of the corresponding $AB$. So, \begin{align*} [\triangle EFG] &= \left(\frac14\right)^2 \cdot [\triangle ABC] \\ &= \frac{1}{16} \cdot 6 \\ &= \frac38. \end{align*} Finally, we have \begin{align*} [ACEG] &= [ABDF] - [\triangle ABC] - [\triangle CDE] - [\triangle EFG] \\ &= 7 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac38 \\ &= 28 - 6 - 6 - \frac38 \\ &= \boxed{\textbf{(D) }15\dfrac{5}{8}}. \end{align*}

For $\textbf{(A)}$ modulo $3,$ \begin{align*} 2^{606} - 1 & \equiv (-1)^{606} - 1 \\ & \equiv 1 - 1 \\ & \equiv 0 . \end{align*} Thus, $2^{606} - 1$ is divisible by $3.$ For $\textbf{(B)}$ modulo $5,$ \begin{align*} 2^{606} + 1 & \equiv 2^{{\rm Rem} ( 606, \phi(5) )} + 1 \\ & \equiv 2^{{\rm Rem} ( 606, 4 )} + 1 \\ & \equiv 2^2 + 1 \\ & \equiv 0 . \end{align*} Thus, $2^{606} + 1$ is divisible by $5.$ For $\textbf{(D)}$ modulo $3,$ \begin{align*} 2^{607} + 1 & \equiv (-1)^{607} + 1 \\ & \equiv - 1 + 1 \\ & \equiv 0 . \end{align*} Thus, $2^{607} + 1$ is divisible by $3.$ For $\textbf{(E)}$ modulo $5,$ \begin{align*} 2^{607} + 3^{607} & \equiv 2^{607} + (-2)^{607} \\ & \equiv 2^{607} - 2^{607} \\ & \equiv 0 . \end{align*} Thus, $2^{607} + 3^{607}$ is divisible by $5.$ Therefore, the answer is $\boxed{\textbf{(C) }2^{607} - 1}.$

Let $M_1=\begin{bmatrix}a_1 & b_1 & c_1\end{bmatrix}, M_2=\begin{bmatrix}a_2 & b_2 & c_2\end{bmatrix},$ and $M_3=\begin{bmatrix}a_3 & b_3 & c_3\end{bmatrix}.$ We wish to count the ordered triples $(M_1,M_2,M_3)$ of row matrices. We perform casework: 1. $M_1=M_2=M_3.$ 2. Exactly two of $M_1,M_2,$ and $M_3$ are equal. 3. All of $M_1,M_2,$ and $M_3$ are different. Together, the answer is $8+168+126+36=\boxed{\textbf{(B)}\ 338}.$

There are two cases for the initial configuration: 1. The center square is filled. 2. The center square is empty. Together, the answer is $2+20=\boxed{\textbf{(C)}\ 22}.$

Extend segments $\overline{AD}$ and $\overline{BE}$ until they meet at point $G$. Because $\overline{AB} \parallel \overline{ED}$, we have $\angle ABG = \angle DEG$ and $\angle GDE = \angle GAB$, so $\triangle ABG \sim \triangle DEG$ by AA. Because $ABCD$ is a rhombus, $AB = CD = 2DE$, so $AG = 2GD$, meaning that $D$ is a midpoint of segment $\overline{AG}$. Now, $\overline{AF} \perp \overline{BE}$, so $\triangle GFA$ is right and median $FD = AD$. So now, because $ABCD$ is a rhombus, $FD = AD = CD$. This means that there exists a circle from $D$ with radius $AD$ that passes through $F$, $A$, and $C$. AG is a diameter of this circle because $\angle AFG=90^\circ$. This means that $\angle GFC = \angle GAC = \frac{1}{2} \angle GDC$, so $\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ$, which means that $\angle BFC = \boxed{\textbf{(D)} \ 113}$

Given that all the answer choices and coefficients are integers, we hope that $P(x)$ has positive integer coefficients. Throughout this solution, we will express all polynomials in base $x$. E.g. $x^2 + x + 1 = 111_{x}$. We are given: \[111a + 12 = 101b + 21 = P(x).\] We add $111$ and $101$ to each side and balance respectively: \[111(a - 1) + 123 = 101(b - 1) + 122 = P(x).\] We make the unit's digits equal: \[111(a - 1) + 123 = 101(b - 2) + 223 = P(x).\] We now notice that: \[111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).\] Therefore $a = 11_{x} = x + 1$, $b = 12_{x} = x + 2$, and $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$. $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$‘s digit in $101b + 21$ when $b$ is an integer. The desired sum is $1^2 + 2^2 +3^2+ 3^2 = \boxed{\textbf{(E)} \ 23}$ P.S. The four computational steps can be deduced through quick experimentation.

The circles match up as follows: Case $1$ is brown, Case $2$ is blue, Case $3$ is green, and Case 4 is red. Let $x^2 + y^2 = 64$ be circle $O$, $x^2 + y^2 = 4$ be circle $P$, and $(x-5)^2 + y^2 = 3$ be circle $Q$. All the circles in S are internally tangent to circle $O$. There are four cases with two circles belonging to each: $*$ $P$ and $Q$ are internally tangent to $S$. $*$ $P$ and $Q$ are externally tangent to $S$. $*$ $P$ is externally and Circle $Q$ is internally tangent to $S$. $*$ $P$ is internally and Circle $Q$ is externally tangent to $S$. Consider Cases $1$ and $4$ together. Since circles $O$ and $P$ have the same center, the line connecting the center of $S$ and the center of $O$ will pass through the tangency point of both $S$ and $O$ and the tangency point of $S$ and $P$. This line will be the diameter of $S$ and have length $r_P + r_O = 10$. Therefore the radius of $S$ in these cases is $5$. Consider Cases $2$ and $3$ together. Similarly to Cases $1$ and $4$, the line connecting the center of $S$ to the center of $O$ will pass through the tangency points. This time, however, the diameter of $S$ will have length $r_P-r_O=6$. Therefore, the radius of $S$ in these cases is $3$. The set of circles $S$ consists of $8$ circles - $4$ of which have radius $5$ and $4$ of which have radius $3$. The total area of all circles in $S$ is $4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}$.

Let $x$ and $y$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,\] as shown below: Let $x,y,$ and $z$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,\] as shown below: We have two cases: 1. Amelia takes exactly $2$ steps. 2. Amelia takes exactly $3$ steps. Together, the answer is $\frac14 + \frac{5}{12} = \boxed{\textbf{(C) }\frac{2}{3}}.$

We have \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| &&(\bigstar) \\ &\leq \frac12\left|\frac12|800-400|\right| \\ &= 100, \end{align*} from which we eliminate answer choices $\textbf{(D)}$ and $\textbf{(E)}.$ Note that \begin{alignat*}{8} |f(800)-f(300)| &\leq \frac12|800-300| &&= 250, \\ |f(800)-f(900)| &\leq \frac12|800-900| &&= 50, \\ |f(400)-f(300)| &\leq \frac12|400-300| &&= 50, \\ |f(400)-f(900)| &\leq \frac12|400-900| &&= 250. \\ \end{alignat*} Let $a=f(300)=f(900).$ Together, it follows that \begin{align*} |f(800)-a|&\leq 50, \\ |f(400)-a|&\leq 50. \\ \end{align*} We rewrite $(\bigstar)$ as \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\ &= \frac12|(f(800)-a)-(f(400)-a)| \\ &\leq \frac12|50-(-50)| \\ &=\boxed{\textbf{(B)}\ 50}. \end{align*}

In binary numbers, we have \[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0})_2.\] It follows that \[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0}000)_2.\] We obtain $7S_n$ by subtracting the equations: \[\begin{array}{clccrccccccr} & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_2 & x_1 & x_0)_2 \\ \hline & & & & & & & & & & & \\ [-2.5ex] & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}\] We work from right to left: \begin{alignat*}{6} x_0=x_1=x_2=1 \quad &\implies \quad &x_3 &= 0& \\ \quad &\implies \quad &x_4 &= 1& \\ \quad &\implies \quad &x_5 &= 1& \\ \quad &\implies \quad &x_6 &= 0& \\ \quad &\implies \quad &x_7 &= 1& \\ \quad &\implies \quad &x_8 &= 1& \\ \quad &\quad \vdots & & & \end{alignat*} For all $n\geq3,$ we conclude that - $x_n=0$ if and only if $n\equiv 0\pmod{3}.$ - $x_n=1$ if and only if $n\not\equiv 0\pmod{3}.$ Finally, we get $(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),$ from which \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \boxed{\textbf{(A) } 6}.\]