id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
3,701 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19 | 2 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on... | Since the $13$ cards are picked up in two passes, the first pass must pick up the first $n$ cards and the second pass must pick up the remaining cards $m$ through $13$ .
Also note that if $m$ , which is the card that is numbered one more than $n$ , is placed before $n$ , then $m$ will not be picked up on the first pas... | 178 |
3,702 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19 | 3 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on... | To solve this problem, we can use recursion on $n$ . Let $A_n$ be the number of arrangements for $n$ numbers. Now, let's look at how these arrangements are formed by case work on the first number $a_1$
If $a_1 = 1$ , the remaining $n-1$ numbers from $2$ to $n$ are arranged in the same way just like number 1 to $n-1$ in... | 178 |
3,703 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19 | 4 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on... | When we have $3$ cards arranged in a row, after listing out all possible arrangements, we see that we have $4$ ones: $(1, 3, 2), (2, 1, 3), (2, 3, 1),$ and $(3, 1, 2)$ . When we have $4$ cards, we find $11$ possible arrangements: $(1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 3, 4), (2, 3, 1, 4), (2, 3... | 178 |
3,704 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19 | 5 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on... | Notice that for each card "position", we can choose for it to be picked up on the first or second pass, for a total of $2^{13}$ options. However, if all of the cards selected to be picked up first are before all of the cards to be picked up second, then this means that the list is in consecutive ascending order (and th... | 178 |
3,705 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24 | 1 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $... | For some $n$ , let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \ldots c_n$ each between $0$ and $n$ , and let $n$ cars come into this circle so that the $i$ th car tries to park at spot $c_i$ , but if it is already taken then it instead keeps going counterclockw... | 296 |
3,706 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24 | 2 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $... | Note that a valid string must have at least one $0.$
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:
Together, the answer is $1+75+500+600+120=\boxed{1296}.$ | 296 |
3,707 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24 | 3 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $... | Denote by $N \left( p, q \right)$ the number of $p$ -digit strings formed by using numbers $0, 1, \cdots, q$ , where for each $j \in \{1,2, \cdots , q\}$ , at least $j$ of the digits are less than $j$
We have the following recursive equation: \[N \left( p, q \right) = \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q... | 296 |
3,708 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24 | 4 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $... | The number of strings is $(n+1)^{(n-1)}$ as shown by Solution 1 (Parking Function), which is always equivalent to 1 (mod n). Thus you can choose $\boxed{1296}$ | 296 |
3,709 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24 | 5 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $... | Solution 4 tried to observe the answer modulo $5$ to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though:
Notice that for any valid set $\{a_1, a_2, a_3, a_4, a_5\}$ , if there is at least one element in the set that is unique (i.e. there is at least one digit in t... | 296 |
3,710 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_25 | 1 | A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}... | Suppose that with a pair $(a_i,b_i)$ the circle is an excircle. Then notice that the hypotenuse must be $(r-x)+(r-y)$ , so it must be the case that \[a_i^2+b_i^2=(2r-a_i-b_i)^2.\] Similarly, if with a pair $(a_i,b_i)$ the circle is an incircle, the hypotenuse must be $(x-r)+(y-r)$ , leading to the same equation.
Notice... | 17 |
3,711 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_25 | 2 | A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}... | Case 1: The tangent and the origin are on the opposite sides of the circle.
In this case, $a, b > 2r$
We can easily prove that \[a + b - 2 r = c . \hspace{1cm} (1)\]
Recall that $c = \sqrt{a^2 + b^2}$
Taking square of (1) and reorganizing all terms, (1) is converted as \[\left( a - 2 r \right) \left( b - 2 r \right) = ... | 17 |
3,712 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_25 | 3 | A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}... | As $c_i$ is the length of the segment $(0,a_i)$ and $(b_i,0)$ $a_i^2+b_i^2=c_i^2$ . The equation for the line that passes through $(0,a_i)$ and $(b_i,0)$ is $a_ix+b_iy-a_ib_i=0$
By the point-line distance formula from point $(r,r)$ to line $a_ix+b_iy-a_ib_i=0$
\[r = \frac{ |a_ir+b_ir-a_ib_i| }{ \sqrt{a_i^2+b_i^2} }, \q... | 17 |
3,713 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_1 | 1 | Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$ | We have \begin{align*} (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &= \boxed{2} ~MRENTHUSIASM | 2 |
3,714 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_1 | 2 | Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$ | Observe that the $\diamond$ function is simply the positive difference between two numbers. Thus, we evaluate: the difference between $2$ and $3$ is $1;$ the difference between $1$ and $1$ is $0;$ the difference between $1$ and $2$ is $1;$ the difference between $1$ and $3$ is $2;$ and finally, $0-2=\boxed{2}.$ | 2 |
3,715 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_2 | 1 | In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10))... | [asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); label("$B$", B, NW); pair C = (6,3); label("$C$", C, NE); pair D = (3.75,0); label("$D$", D, SE); pair P = (2.25,0); label("$P$", P, S); draw(A--B--C--D--cycle); draw(P--B); draw(rightanglemark(B,P,D)); [/asy]
\[\textbf{Figure redrawn to scale.}\]
$AD = AP +... | 20 |
3,716 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_2 | 2 | In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10))... | [asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); label("$B$", B, NW); pair C = (6,3); label("$C$", C, NE); pair D = (3.75,0); label("$D$", D, SE); pair P = (2.25,0); label("$P$", P, S); draw(A--B--C--D--cycle); draw(D--B); draw(B--P); draw(rightanglemark(B,P,D)); [/asy]
The diagram is from as Solution 1, ... | 20 |
3,717 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_3 | 1 | How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$ | 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*} Therefor... | 0 |
3,718 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_3 | 2 | How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$ | Denote this sequence as $a_{n}$ , then we can find that \begin{align*} a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\ a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\ a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\ & \ \vdots \end{align*} So, we can induct that th... | 0 |
3,719 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_3 | 3 | How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$ | Observe how \begin{align*} 121 &= 110 + 11, \\ 11211 &= 11100 + 111, \\ 1112111 &= 1111000 + 1111, \\ & \ \vdots \end{align*} all take the form of \[\underbrace{111\ldots}_{n+1}\underbrace{00\ldots}_{n} + \underbrace{111\ldots}_{n+1} = \underbrace{111\ldots}_{n+1}(10^{n} + 1).\] Factoring each of the sums, we have \[11... | 0 |
3,720 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_3 | 4 | How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$ | Note that $121$ is divisible by $11$ and $11211$ is divisible by $3$ . Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is $\boxed{0}.$ | 0 |
3,721 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_4 | 2 | For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$ | Note that $k$ must be an integer. Using the quadratic formula $x=\frac{-k \pm \sqrt{k^2-144}}{2}.$ Since $4$ divides $144$ evenly, $k$ and $k^2-144$ have the same parity, so $x$ is an integer if and only if $k^2-144$ is a perfect square.
Let $k^2-144=n^2.$ Then, $(k+n)(k-n)=144.$ Since $k$ is an integer and $144$ is ev... | 8 |
3,722 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_4 | 3 | For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$ | Proceed similar to Solution 2 and deduce that the discriminant of $x^{2}+kx+36$ must be a perfect square greater than $0$ to satisfy all given conditions. Seeing something like $k^2-144$ might remind us of a right triangle, where $k$ is the hypotenuse, and $12$ is a leg. There are four ways we could have this: a $9$ $1... | 8 |
3,723 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_4 | 4 | For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$ | Since $36 = 2^2\cdot3^2$ , that means there are $(2+1)(2+1) = 9$ possible factors of $36$ . Since $6 \cdot 6$ violates the distinct root condition, subtract $1$ from $9$ to get $8$ . Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get $\boxed{8}$ | 8 |
3,724 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_6 | 1 | Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)... | We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
Each case has $\left\lfloor\frac{100}{7}\right\rfloor=14$ sets. Therefore, the answer is $14\cdot3=\boxed{42}.$ | 42 |
3,725 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_6 | 2 | Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)... | We find a pattern. \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} We can figure out that the first set has $1$ multiple of $7$ . The second set also has $1$ multiple of $7$ . The third set has $2$ multiples of $7$ . The ... | 42 |
3,726 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_7 | 1 | Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ ... | 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 $a+b$ mus... | 11 |
3,727 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_7 | 2 | Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ ... | We can also easily test all the answer choices. (This strategy is generally good to use for multiple-choice questions if you don't have a concrete method to proceed with!)
For answer choice $\textbf{(A)},$ the mode is $5,$ the median is $3,$ and the arithmetic mean is $1.$ However, we can quickly see this doesn't work,... | 11 |
3,728 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_9 | 1 | The sequence $a_0,a_1,a_2,\cdots$ is a strictly increasing arithmetic sequence of positive integers such that \[2^{a_7}=2^{27} \cdot a_7.\] What is the minimum possible value of $a_2$
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 22$ | We can rewrite the given equation as $2^{a_7-27}=a_7$ . Hence, $a_7$ must be a power of $2$ and larger than $27$ . The first power of 2 that is larger than $27$ , namely $32$ , does satisfy the equation: $2^{32 - 27} = 2^5 = 32$ . In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, ... | 12 |
3,729 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_11 | 3 | Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$ | We begin by recognizing this form looks similar to the definition of cosine: \[\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.\] We can convert our two terms into exponential form to find \[f(n) = \left( e^{\frac{2\pi i}{3}} \right ) ^n + \left ( e^{-\frac{2\pi i}{3}} \right ) ^n=e^{\frac{2 \pi i n}{3}} + e^{-\frac{2\pi i n}{3}}.\] ... | 2 |
3,730 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_11 | 4 | Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$ | Notice how this looks like the closed form of the Fibonacci sequence except different roots. This is motivation to turn this closed formula into a recurrence relation. The base of the exponents are the roots of the characteristic equation $r^3-1=0$ . So we have \begin{align*} a_0&=2\\ a_1&=-1\\ a_2&=-1\\ a_n&=a_{n-3} \... | 2 |
3,731 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_11 | 5 | Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$ | Converting the two terms into rectangular form,
\[f(2022)=\left(\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}\right)^{2022}+\left(\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}\right)^{2022}.\]
By DeMoivre's Theorem,
\[f(2022)=\left(\cos{\left(\frac{2\pi}{3}\cdot{2022}\right)}+i\sin{\left(\frac{2\pi}{3}\cdot{2022}\right)}\rig... | 2 |
3,732 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17 | 1 | How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{arr... | Note that the arrays and the sum configurations have one-to-one correspondence. Furthermore, the row sum configuration and the column sum configuration are independent of each other. Therefore, the answer is $(4!)^2=\boxed{576}.$ | 576 |
3,733 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17 | 2 | How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{arr... | In this problem, we call a matrix that satisfies all constraints given in the problem a feasible matrix.
First, we observe that if a matrix is feasible, and we swap two rows or two columns to get a new matrix, then this new matrix is still feasible.
Therefore, any feasible matrix can be obtained through a sequence of s... | 576 |
3,734 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17 | 3 | How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{arr... | Of the sixteen entries in one such array, there are six $0$ s and ten $1$ s. The rows and the columns each have zero, one, two, and three $0$ s, in some order. Once we decide the positions of the $0$ s, we form one such array.
We perform the following steps:
Together, the answer is $4\cdot4\cdot3\cdot3\cdot4=\boxed{576... | 576 |
3,735 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17 | 4 | How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{arr... | Since exactly $1$ row sum is $4$ and exactly $1$ column sum is $4$ , there is a unique entry in the array such that it, and every other entry in the same row or column, is a $1.$ Since there are $16$ total entries in the array, there are $16$ ways to choose the entry with only $1$ s in its row and column.
Without loss ... | 576 |
3,736 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17 | 5 | How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{arr... | Note that swapping any two rows or any two columns or both from the given example array, leads to a new array that satisfies the condition. There are $4$ rows, and you choose $2$ to swap, so $\frac{4!}{2!2!} = 6$ ; likewise for columns. Therefore, $6\cdot6 = 36$ . Clearly, the final answer must be divisible by $36$ , s... | 576 |
3,737 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_18 | 1 | Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
A sample transformation is shown in the figure below. [asy] import geometry; unitsize... | There are two cases for the initial configuration:
Together, the answer is $2+20=\boxed{22}.$ | 22 |
3,738 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_19 | 1 | In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$ , where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$
$\... | Let $AG = AE = GE = CE = 1$ . Since $G$ is the centroid, $DG = \frac12$ $BG = 2$
\[\angle BGD = \angle AGE = 60^{\circ}\]
By the Law of Cosine in $\triangle BGD$
\[BD^2 = BG^2 + DG^2 - 2 \cdot BG \cdot DG \cdot \cos \angle BGD\]
\[BD = \sqrt {2^2 + \left( \frac{1}{2} \right)^2 - 2 \cdot 2 \cdot \frac12 \cdot \cos \angl... | 44 |
3,739 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_19 | 2 | In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$ , where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$
$\... | Using reference triangle $\triangle AGE$ , we can let \[A=(1,0,0),G=(0,1,0),E=(0,0,1),C=(-1,0,2),D=(-\tfrac{1}{2},\tfrac{3}{2},0),B=(0,3,-2).\] If we move $A,B,C$ each over by $(1,0,-2)$ , leaving $\angle C$ unchanged, we have \[A=(2,0,-2),B=(1,3,-4),C=(0,0,0).\] The angle $\theta$ between vectors $\overrightarrow{CA}$... | 44 |
3,740 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20 | 1 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
th... | 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 res... | 23 |
3,741 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20 | 2 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
th... | Let $P(x) = Q(x)(x^2+x+1) + x + 2$ , then $P(x) = Q(x)(x^2+1) + xQ(x) + x + 2$ , therefore $xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}$ , or $xQ(x) \equiv x-1 \pmod{x^2+1}$ . Clearly the minimum is when $Q(x) = x+1$ , and expanding gives $P(x) = x^3+2x^2+3x+3$ . Summing the squares of coefficients gives $\boxed{23}$ | 23 |
3,742 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20 | 3 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
th... | Let $P(x) = (x^2+x+1)Q_1(x) + x + 2$ ,
then $P(x) = (x^2+1)Q_1(x) + xQ_1(x) + x + 2$
Also $P(x) = (x^2+1)Q_2(x) + 2x + 1$
We infer that $Q_1(x)$ and $Q_2(x)$ have same degree, we can assume $Q_1(x) = x + a$ , and $Q_2(x) = x + b$ , since $P(x)$ has least degree. If this cannot work, we will try quadratic, etc.
Then we... | 23 |
3,743 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20 | 4 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
th... | Notice that we cannot have the quotients equal to some constants, since the same constant will yield different constant terms for $P(x)$ (which is bad) and different constants will yield different first coefficients (also bad). Thus, we try setting the quotients equal to linear terms (for minimizing degree).
Let $P(x)=... | 23 |
3,744 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20 | 5 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
th... | We construct the following equations in terms of $P(x)$ and the information given by the problem: \[\textbf{(1) } P(x)=(x^2+x+1)\cdot Q(x)+x+2\] \[\textbf{(2) } P(x)=(x^2+1)\cdot R(x)+2x+1\] Upon inspection, $Q(x)$ and $R(x)$ cannot be constant, so the smallest possible degree of $P(x)$ is $3,$ and both $Q(x)$ and $R(x... | 23 |
3,745 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20 | 6 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
th... | By remainder theorem, the polynomial can be written as follows.
\[P(x) = (x^2+x+1)Q_{1}(x)+x+2 = (x^2+1)Q_{2}(x)+2x+1\] This is a timed exam, we can use the information given by answer choices. The answer choices tell us this is the polynomial with integer coefficients, and we need to find the polynomial with the least... | 23 |
3,746 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_21 | 1 | Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$ $x^{2}+y^{2}=64$ , and $(x-5)^{2}+y^{2}=3$ . What is the sum of the areas of all circles in $S$
$\textbf{(A)}~48\pi\qquad\textbf{(B)}~68\pi\qquad\textbf{(C)}~96\pi\qquad\textbf{(D)}~102\pi\... | We denote by $C_1$ the circle that has the equation $x^2 + y^2 = 4$ .
We denote by $C_2$ the circle that has the equation $x^2 + y^2 = 64$ .
We denote by $C_3$ the circle that has the equation $(x-5)^2 + y^2 = 3$
We denote by $C_0$ a circle that is tangent to $C_1$ $C_2$ and $C_3$ .
We denote by $\left( u, v \right)$ t... | 136 |
3,747 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_23 | 1 | Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qq... | 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 ... | 6 |
3,748 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_23 | 2 | Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qq... | First, notice that \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Then since $S_n$ is the modular inverse of $7$ in $\mathbb{Z}_{2^n}$ , we can perform the Euclidean Algorithm to find it for $n = 2019,2023$
Starting with $2019$ \begin{align*} 7S_{2019} &\equiv 1 \pmod{2^{2019}}... | 6 |
3,749 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_23 | 4 | Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qq... | Note that, as in Solution 2, we have \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] This is because \[S_{2023} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2019}2^{2019} + \cdots + x_{2022}2^{2022}\] and \[S_{2019} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2018}2^{2018}.\] Note that \... | 6 |
3,750 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | 1 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$
Therefore, the sum of the $4$ th powers of these lengths is \begin{align*} 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} ... | 147 |
3,751 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | 2 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$
Therefore, the sum of the $4$ th powers of these lengths is \begin{align*} & 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7... | 147 |
3,752 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | 3 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | As explained in Solutions 1 and 2, what we are trying to find is $7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7}$ . Using trig we get \begin{align*} & \sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7} \\ = & \sin^2 \frac{\pi}{7} \left(1 -... | 147 |
3,753 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | 4 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:
\[\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}\]
\begin{align*} S &= \cos \frac{2\pi}{7} + \cos \frac{4\pi... | 147 |
3,754 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | 5 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | Let $x$ $y$ , and $z$ be the lengths of the chords with arcs $\frac{2\pi}{7}$ $\frac{4\pi}{7}$ and $\frac{6\pi}{7}$ respectively.
Then by the law of cosines we get: \begin{align*} x^2 &= 2\left(1-\cos\frac{2\pi}{7}\right), \\ y^2 &= 2\left(1-\cos\frac{4\pi}{7}\right), \\ z^2 &= 2\left(1-\cos\frac{6\pi}{7}\right). \end{... | 147 |
3,755 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | 6 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.
First, measuring the radius of the circle obtains $2.9$ cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by $2.9$
Measuring the sides of the circle gets $2.5$ cm. The ... | 147 |
3,756 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | 7 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle $\frac{2\pi}{7},$ which are \[\left(\cos\dfrac{2\pi n}{7}, \sin\dfrac{2\pi n}{7}\right)\] ... | 147 |
3,757 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | 8 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | Place the figure on the complex plane and let $w = e^{\frac{2\pi}{7}}$ (a $7$ th root of unity). The vertices of the $7$ -gon are $w^0,w^1,w^2,\dots,w^6$ . We wish to find \[\sum_{i=0}^5\sum_{j=i+1}^6\lvert w^i-w^j\rvert^4 = \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4.\] The second expression is more conv... | 147 |
3,758 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | 9 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | This is how I solve this problem:
It's easy to solve for $3$ -gon, $4$ -gon, and $6$ -gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon)
For $3$ -gon, the sum is equal to $3$ times the $4$ th power of an edge. Thus, \[S_3=3\,\cdot\,\left(\sqrt{3}\right)^4=27.\]
For $4$ -... | 147 |
3,759 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25 | 1 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
Wh... | [asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*... | 4 |
3,760 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25 | 2 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
Wh... | [asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(... | 4 |
3,761 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25 | 3 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
Wh... | [asy] import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(... | 4 |
3,762 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25 | 4 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
Wh... | We calculate the area as the area of the red octagon minus the four purple congruent triangles: [asy] import geometry; unitsize(3cm); draw((1-sqrt(3),1-sqrt(3))--(1-sqrt(3),sqrt(3))--(sqrt(3),sqrt(3))--(sqrt(3),1-sqrt(3))--cycle,dashed); filldraw((0,1-sqrt(3))--(1,1-sqrt(3))--(sqrt(3),0)--(sqr... | 4 |
3,763 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25 | 5 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
Wh... | Note that each of the green sections is a rectangle, so its interior angles are all $90^{\circ}.$ Since $\angle{ABC}=120^{\circ}$ , every one of the orange sections is a $30-60-90$ right triangle.
Define $x$ to be the distance from the corner of the square with side length $1$ to the corner of the larger blue square. D... | 4 |
3,764 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_1 | 1 | What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$ | We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\left(\frac{91}{13}\right)^2=7^2=\boxed{49}.\] ~MRENTHUSIASM | 49 |
3,765 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_1 | 2 | What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$ | We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{13^2}=\frac{((10+3)(10-3))^2}{13^2}=\frac{(13\cdot7)^2}{13^2}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{49}.\] | 49 |
3,766 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_1 | 3 | What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$ | We know that $2112-2021 = 91$ . Approximate this as $100$ as it is pretty close to it. Also, approximate $169$ to $170$ . We then have \[\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.\] Now check the answer choices. The two closest answers are $49$ and $64$ . As the numerator ... | 49 |
3,767 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_2 | 1 | Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?
$\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textb... | We construct the following table: \[\begin{array}{c||c|c||c} & & & \\ [-2.5ex] \textbf{Scenario} & \textbf{Length} & \textbf{Width} & \textbf{Area} \\ [0.5ex] \hline & & & \\ [-2ex] \text{Initial} & 4 & 6 & 24 \\ \text{Menkara shortens one side.} & 3 & 6 & 18 \\ \text{Menkara shortens other side instead.} & 4 & 5 & 20 ... | 20 |
3,768 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_4 | 1 | The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$ | First, modulo $2$ or $5$ $\underline{20210A} \equiv A$ .
Hence, $A \neq 0, 2, 4, 5, 6, 8$
Second modulo $3$ $\underline{20210A} \equiv 2 + 0 + 2 + 1 + 0 + A \equiv 5 + A$ .
Hence, $A \neq 1, 4, 7$
Third, modulo $11$ $\underline{20210A} \equiv A + 1 + 0 - 0 - 2 - 2 \equiv A - 3$ .
Hence, $A \neq 3$
Therefore, the answer... | 9 |
3,769 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_4 | 2 | The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$ | Any number ending in $5$ is divisible by $5$ . So we can eliminate option $\textbf{(C)}$
If the sum of the digits of a number is divisible by $3$ , the number is divisible by $3$ . The sum of the digits of this number is $2 + 0 + 2 + 1 + 0 + A = 5 + A$ . If $5 + A$ is divisible by $3$ , the number is divisible by $3$ .... | 9 |
3,770 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_4 | 3 | The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$ | $202100 \implies$ divisible by $2$
$202101 \implies$ divisible by $3$
$202102 \implies$ divisible by $2$
$202103 \implies$ divisible by $11$
$202104 \implies$ divisible by $2$
$202105 \implies$ divisible by $5$
$202106 \implies$ divisible by $2$
$202107 \implies$ divisible by $3$
$202108 \implies$ divisible by $2$
This... | 9 |
3,771 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_5 | 1 | Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in f... | There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet.
Each of Oscar's leaps covers $132\div12=11$ feet, and each of Elmer's strides covers $132\div44=3$ feet.
Therefore, Oscar's leap is $11-3=\boxed{8}$ feet longer than Elmer's stride. | 8 |
3,772 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_5 | 2 | Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in f... | There are $41-1=40$ gaps between the $41$ telephone poles, so Elmer takes $44 \cdot 40 = 1760$ strides in total, and Oscar takes $12 \cdot 40 = 480$ leaps in total. Therefore, the answer is $(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{8}$ | 8 |
3,773 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_6 | 1 | As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); p... | By angle subtraction, we have $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.$ Note that $\triangle DEF$ is isosceles, so $\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.$ Finally, we get $\angle AFE = 180^\circ - \angle DFE = \boxed{170}$ degrees. | 170 |
3,774 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_6 | 2 | As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); p... | We can extend $\overline{AD}$ to $G$ , making $\angle CDG$ a right angle. It follows that $\angle GDE$ is $110^\circ - 90^\circ = 20^\circ$ , as shown below. [asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, S... | 170 |
3,775 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_8 | 1 | Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40.$ What is the value of $\frac{N}{M}?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 37 \qquad\textbf{(D)}\ 74 \qquad\textbf{(E)}\ 2886$ | By the definition of least common mutiple, we take the greatest powers of the prime numbers of the prime factorization of all the numbers, that we are taking the $\text{lcm}$ of. In this case, \[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\] Now, using the same logic, we fi... | 74 |
3,776 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_9 | 1 | A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_{2}x, \log_{3}x,$ and $\log_{4}x.$ What is $x?$
$\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)}\ 6\sqrt{6} \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 576$ | The surface area of this right rectangular prism is $2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x).$
The volume of this right rectangular prism is $\log_{2}x\log_{3}x\log_{4}x.$
Equating the numerical values of the surface area and the volume, we have \[2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\lo... | 576 |
3,777 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_15 | 1 | Recall that the conjugate of the complex number $w = a + bi$ , where $a$ and $b$ are real numbers and $i = \sqrt{-1}$ , is the complex number $\overline{w} = a - bi$ . For any complex number $z$ , let $f(z) = 4i\hspace{1pt}\overline{z}$ . The polynomial \[P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1\] has four complex roots: $z_1... | By Vieta's formulas, $z_1z_2+z_1z_3+\dots+z_3z_4=3$ , and $B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).$
Since $\overline{a}\cdot\overline{b}=\overline{ab},$ \[B=(4i)^2\left(\overline{z_1z_2}+\overline{z_1z_3}+\overline{z_1z_4}+\overline{z_2z_3... | 208 |
3,778 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_16 | 1 | An organization has $30$ employees, $20$ of whom have a brand A computer while the other $10$ have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if th... | We claim that to maximize the number of cables used, we isolate one computer and connect all cables for the remaining $29$ computers, then connect one more cable for the isolated computer.
If a brand A computer is isolated, then the technician can use at most $19\cdot10+1=191$ cables. If a brand B computer is isolated ... | 191 |
3,779 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17 | 2 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | Similar to Solution 1, use the discriminant to get $b^2\leq 4c$ and $c^2\leq 4b$ . These can be rearranged to $c\geq \frac{1}{4}b^2$ and $b\geq \frac{1}{4}c^2$ . Now, we can roughly graph these two inequalities, letting one of them be the $x$ axis and the other be $y$ .
The graph of solutions should be above the parab... | 6 |
3,780 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17 | 3 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | We need to solve the following system of inequalities: \[ \left\{ \begin{array}{ll} b^2 - 4 c \leq 0 \\ c^2 - 4 b \leq 0 \end{array} \right.. \] Feasible solutions are in the region formed between two parabolas $b^2 - 4 c = 0$ and $c^2 - 4 b = 0$
Define $f \left( b \right) = \frac{b^2}{4}$ and $g \left( b \right) = 2 \... | 6 |
3,781 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17 | 4 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | A quadratic equation $Ax^2+Bx+C=0$ has one real solution if and only if $\sqrt{B^2-4AC}=0.$ Similarly, it has imaginary solutions if and only if $\sqrt{B^2-4AC}<0.$ We proceed as following:
We want both $x^2+bx+c$ to be $1$ value or imaginary and $x^2+cx+b$ to be $1$ value or imaginary. $x^2+4x+4$ is one such case sinc... | 6 |
3,782 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17 | 5 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | We see that $b^2 \leq 4c$ and $c^2 \leq 4b.$ WLOG, assume that $b \geq c.$ Then we have that $b^2 \leq 4c \leq 4b$ , so $b^2 \leq 4b$ and therefore $b \leq 4$ , also meaning that $c \leq 4.$ This means that we only need to try 16 cases. Now we can get rid of the assumption that $b \geq c$ , because we want ordered pair... | 6 |
3,783 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17 | 6 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | We need both $b^2\leq 4c$ and $c^2\leq 4b$
If $b=c$ then the above become $b^2\leq 4b\iff b\leq 4$ , so we have four solutions $(k,k)$ , where $k=1$ $2$ $3$ $4$
If $b<c$ then we only need $c^2\leq 4b$ since it implies $b^2< 4c$ . Now $c^2\leq 4b\leq 4(c-1) \implies (c-2)^2\leq 0 \implies c=2$ , so $b=1$ . We plug $b=1$... | 6 |
3,784 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_17 | 7 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | Since $b^{2} - 4c \le 0$ and $c^{2} - 4b \le 0$ , adding the two together yields $b^{2} + c^{2} \le 4(c+b)$ . Obviously, this is not true if either $b$ or $c$ get too large, and they are equal when $b = c = 4$ , so the greatest pair is $(4,4)$ and both numbers must be lesser for further pairs. For there to be two disti... | 6 |
3,785 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_18 | 1 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ ... | For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.\] Therefore, the answer is \[\frac pq=\fr... | 16 |
3,786 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_18 | 2 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ ... | For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Let $q=\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.
We can take $1$ ball from one bin and place it in another bin so that some bin ends up with $... | 16 |
3,787 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_18 | 3 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ ... | Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \cdot \binom {5}{2} = 20$ ways to choose where to place the $3$ and the $5$ . After that, there are $\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the bins. For the $4,4,4,4,4$ case, after we c... | 16 |
3,788 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_18 | 4 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ ... | Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} =... | 16 |
3,789 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_19 | 1 | Let $x$ be the least real number greater than $1$ such that $\sin(x)= \sin(x^2)$ , where the arguments are in degrees. What is $x$ rounded up to the closest integer?
$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$ | The smallest $x$ to make $\sin(x) = \sin(x^2)$ would require $x=x^2$ , but since $x$ needs to be greater than $1$ , these solutions are not valid.
The next smallest $x$ would require $x=180-x^2$ , or $x^2+x=180$
After a bit of guessing and checking, we find that $12^2+12=156$ , and $13^2+13=182$ , so the solution lies ... | 13 |
3,790 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_20 | 2 | For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | First, take note that the maximum possible value of $f_1(n)$ for $1 \le n \le k$ increases as $k$ increases (it is a step function), i.e. it is increasing. Likewise, as $k$ decreases, the maximum possible value of $f_1(n)$ decreases as well. Also, let $f_1(n) = 2d(n)$ where $d(n)$ is the number of divisors of n.
Since ... | 10 |
3,791 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_20 | 3 | For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | $\textbf{Observation 1}$ $f_1 \left( 12 \right) = 12$
Hence, if $n$ has the property that $f_j \left( n \right) = 12$ for some $j$ , then $f_k \left( n \right) = 12$ for all $k > j$
$\textbf{Observation 2}$ $f_1 \left( 8 \right) = 8$
Hence, if $n$ has the property that $f_j \left( n \right) = 8$ for some $j$ , then $f_... | 10 |
3,792 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_22 | 1 | Azar and Carl play a game of tic-tac-toe. Azar places an in $X$ one of the boxes in a $3$ -by- $3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all boxes are filled or one of the players has of their symbols in a row... | We need to find out the number of configurations with 3 $O$ and 3 $X$ with 3 $O$ in a row, and 3 $X$ not in a row.
$\textbf{Case 1}$ : 3 $O$ are in a horizontal row or a vertical row.
Step 1: We determine the row that 3 $O$ occupy.
The number of ways is 6.
Step 2: We determine the configuration of 3 $X$
The number of w... | 148 |
3,793 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_24 | 1 | Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a... | Let $E$ be a point on $\overline{AB}$ such that $BCDE$ is a parallelogram. Suppose that $BC=ED=b, CD=BE=c,$ and $DA=d,$ so $AE=18-c,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E; A = (0,0); B = (18,0); D = A+9*dir(60); C = D+(7,0); E = D+(B-C); dot("$A$",A,1.5*SW,linewidth(4)); dot("$... | 84 |
3,794 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_24 | 2 | Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a... | Let $b, c$ , and $d$ denote the sides $BC, CD$ , and $AD$ respectively. [asy] size(250); pair A, B, C, D, E; A = (0,0); B = (18,0); D = A+9*dir(60); C = D+(7,0); E = D+(B-C); pen pdot=linewidth(3)+fontsize(12); dot("$A$",A,SW,pdot); dot("$B$",B,SE,pdot); dot("$C$",C,NE,pdot); dot("$D$",D,NW,pdot); draw(A--B--C--D--... | 84 |
3,795 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_24 | 3 | Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a... | Denote $x = AD$ $\theta = \angle B$ .
Hence, $BC = \frac{\sqrt{3}}{2} \cdot \frac{x}{\sin \theta}$ $DC = 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta$
$\textbf{Case 1}$ $DC = AD = BC = AB$
This is a rhombus. So each side has length $18$
For the following cases, we consider four sides that have distinct lengths.
... | 84 |
3,796 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_25 | 1 | Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $... | For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4).$
Let $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\frac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$
Since $\gcd(m,4)=1,$ we conclude that \[\{k+4(0),k+4(1),k+4(2),\ldots... | 11 |
3,797 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_25 | 2 | Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $... | Define \[ b_i = \left\{ \begin{array}{ll} a_i & \mbox{ if } 1 \leq a_i \leq \frac{m-1}{2} \\ a_i - m & \mbox{ if } \frac{m-1}{2} + 1 \leq a_i \leq m - 1 \\ 0 & \mbox{ if } a_i = m \end{array} \right.. \]
Hence, $b_i$ is a one-to-one and onto function of $a_i$ , and the range of $b_i$ is $\left\{- \frac{m-1}{2} , \cdot... | 11 |
3,798 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_25 | 3 | Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $... | As before, note that we have $m(m-1)(m-2)$ numbers we can choose as $a,b,c.$ From here, there is exactly one possible value of $1 \leq d \leq m$ that could make $S=a+b+c+d$ divisible by $m.$ However, there is a $\frac{3}{m}$ chance that this value of $d$ has already been chosen as $a,b$ or $c$ . Thus our polynomial is ... | 11 |
3,799 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_1 | 1 | What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$ | We see that $1, 2, 3,$ and $4$ each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$ , we find that the sum is equal to \[10\cdot(1+10+100+1000)=\boxed{11110}.\] Note that it is equally valid to manually add all four numbers together to get the answer. | 110 |
3,800 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_1 | 2 | What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$ | We have \[1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{11110}.\] ~Steven Chen (www.professorchenedu.com) | 110 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.