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Paul starts at $1$ and counts by threes: $1, 4, 7, 10, ... $. At the same time and at the same speed, Penny counts backwards from $2017$ by fi ves: $2017, 2012, 2007, 2002,...$ . Find the one number that both Paul and Penny count at the same time. | 1. Let's denote the sequence that Paul counts by as \( a_n \). Since Paul starts at 1 and counts by threes, the sequence can be written as:
\[
a_n = 1 + 3(n-1) = 3n - 2
\]
where \( n \) is the number of terms.
2. Similarly, let's denote the sequence that Penny counts by as \( b_m \). Since Penny starts at ... | 2017 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
When Phil and Shelley stand on a scale together, the scale reads $151$ pounds. When Shelley and Ryan stand on the same scale together, the scale reads $132$ pounds. When Phil and Ryan stand on the same scale together, the scale reads $115$ pounds. Find the number of pounds Shelley weighs. | 1. Define the variables for the weights of Phil, Shelley, and Ryan:
\[
p = \text{Phil's weight}, \quad s = \text{Shelley's weight}, \quad r = \text{Ryan's weight}
\]
2. Write down the given equations based on the problem statement:
\[
p + s = 151 \quad \text{(1)}
\]
\[
s + r = 132 \quad \text{(... | 84 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Find the least positive integer $m$ such that $lcm(15,m) = lcm(42,m)$. Here $lcm(a, b)$ is the least common multiple of $a$ and $b$. | To find the least positive integer \( m \) such that \( \text{lcm}(15, m) = \text{lcm}(42, m) \), we start by using the definition of the least common multiple (LCM) and the greatest common divisor (GCD). The LCM of two numbers \( a \) and \( b \) can be expressed as:
\[
\text{lcm}(a, b) = \frac{a \cdot b}{\text{gcd}(... | 70 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
A store had $376$ chocolate bars. Min bought some of the bars, and Max bought $41$ more of the bars than Min bought. After that, the store still had three times as many chocolate bars as Min bought. Find the number of chocolate bars that Min bought. | 1. Let \( a \) be the number of chocolate bars that Min bought.
2. Let \( b \) be the number of chocolate bars that Max bought.
3. According to the problem, Max bought 41 more bars than Min, so we can write the equation:
\[
b = a + 41
\]
4. The total number of chocolate bars initially was 376. After Min and Ma... | 67 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
For some constant $k$ the polynomial $p(x) = 3x^2 + kx + 117$ has the property that $p(1) = p(10)$. Evaluate $p(20)$. | 1. Given the polynomial \( p(x) = 3x^2 + kx + 117 \) and the condition \( p(1) = p(10) \), we need to find the value of \( k \) and then evaluate \( p(20) \).
2. First, we calculate \( p(1) \):
\[
p(1) = 3(1)^2 + k(1) + 117 = 3 + k + 117 = k + 120
\]
3. Next, we calculate \( p(10) \):
\[
p(10) = 3(10)^... | 657 | Algebra | other | Yes | Yes | aops_forum | false |
Consider an alphabetized list of all the arrangements of the letters in the word BETWEEN. Then BEEENTW would be in position $1$ in the list, BEEENWT would be in position $2$ in the list, and so forth. Find the position that BETWEEN would be in the list. | To find the position of the word "BETWEEN" in the alphabetized list of all its permutations, we need to consider the lexicographic order of the permutations. The word "BETWEEN" has 7 letters, with the letters B, E, E, E, N, T, W.
1. **Calculate the total number of permutations:**
Since the word "BETWEEN" has repeat... | 198 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of trailing zeros at the end of the base-$10$ representation of the integer $525^{25^2}
\cdot 252^{52^5}$ . | To find the number of trailing zeros at the end of the base-$10$ representation of the integer $525^{25^2} \cdot 252^{52^5}$, we need to determine the minimum of the powers of $2$ and $5$ in the prime factorization of the number. This is because a trailing zero is produced by a factor of $10$, which is $2 \times 5$.
1... | 1250 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of positive integers less than or equal to $2017$ that have at least one pair of adjacent digits that are both even. For example, count the numbers $24$, $1862$, and $2012$, but not $4$, $58$, or $1276$. | To find the number of positive integers less than or equal to $2017$ that have at least one pair of adjacent digits that are both even, we will use casework based on the number of digits and the positions of the even digits.
### Case 1: One-digit numbers
There are no one-digit numbers that satisfy the condition since ... | 738 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Dave has a pile of fair standard six-sided dice. In round one, Dave selects eight of the dice and rolls them. He calculates the sum of the numbers face up on those dice to get $r_1$. In round two, Dave selects $r_1$ dice and rolls them. He calculates the sum of the numbers face up on those dice to get $r_2$. In round t... | 1. **Expected Value of a Single Die Roll**:
Each face of a fair six-sided die has an equal probability of landing face up. The expected value \(E\) of a single die roll is calculated as follows:
\[
E = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5
\]
2. **Expected Value of \(r_1\)**:
In the first round, Da... | 343 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Find the sum of all integers $n$ for which $n - 3$ and $n^2 + 4$ are both perfect cubes. | 1. Let \( n - 3 = k^3 \) for some integer \( k \). Then \( n = k^3 + 3 \).
2. We need \( n^2 + 4 \) to be a perfect cube. Substituting \( n = k^3 + 3 \), we get:
\[
n^2 + 4 = (k^3 + 3)^2 + 4
\]
3. Expanding the square, we have:
\[
(k^3 + 3)^2 + 4 = k^6 + 6k^3 + 9 + 4 = k^6 + 6k^3 + 13
\]
4. We need \(... | 13 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
For real numbers $a, b$, and $c$ the polynomial $p(x) = 3x^7 - 291x^6 + ax^5 + bx^4 + cx^2 + 134x - 2$ has $7$ real roots whose sum is $97$. Find the sum of the reciprocals of those $7$ roots. | 1. **Identify the polynomial and its roots:**
The given polynomial is \( p(x) = 3x^7 - 291x^6 + ax^5 + bx^4 + cx^2 + 134x - 2 \). It has 7 real roots, say \( r_1, r_2, r_3, r_4, r_5, r_6, r_7 \).
2. **Sum of the roots:**
By Vieta's formulas, the sum of the roots of the polynomial \( p(x) = 3x^7 - 291x^6 + ax^5 +... | 67 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
In the $3$-dimensional coordinate space nd the distance from the point $(36, 36, 36)$ to the plane that passes
through the points $(336, 36, 36)$, $(36, 636, 36)$, and $(36, 36, 336)$. | 1. **Find the equation of the plane:**
We are given three points on the plane: \((336, 36, 36)\), \((36, 636, 36)\), and \((36, 36, 336)\). We need to find the equation of the plane in the form \(ax + by + cz = d\).
Let's denote the points as \(A = (336, 36, 36)\), \(B = (36, 636, 36)\), and \(C = (36, 36, 336)\... | 188 | Geometry | other | Yes | Yes | aops_forum | false |
Find the greatest integer $n < 1000$ for which $4n^3 - 3n$ is the product of two consecutive odd integers. | 1. We start with the equation \(4n^3 - 3n\) and assume it is the product of two consecutive odd integers. Let these integers be \(2m - 1\) and \(2m + 1\). Therefore, we have:
\[
4n^3 - 3n = (2m - 1)(2m + 1)
\]
Simplifying the right-hand side, we get:
\[
(2m - 1)(2m + 1) = 4m^2 - 1
\]
Thus, the e... | 899 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
The familiar $3$-dimensional cube has $6$ $2$-dimensional faces, $12$ $1$-dimensional edges, and $8$ $0$-dimensional vertices. Find the number of $9$-dimensional sub-subfaces in a $12$-dimensional cube. | 1. **Understanding the problem**: We need to find the number of 9-dimensional sub-subfaces in a 12-dimensional cube. A sub-subface of a cube is a lower-dimensional face of the cube. For example, in a 3-dimensional cube (a regular cube), a 2-dimensional subface is a square face, a 1-dimensional subface is an edge, and a... | 1760 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Eight red boxes and eight blue boxes are randomly placed in four stacks of four boxes each. The probability that exactly one of the stacks consists of two red boxes and two blue boxes is $\frac{m}{n}$ , where m and n are relatively prime positive integers. Find $m + n$. | 1. **Total Number of Arrangements**:
We start by calculating the total number of ways to arrange 8 red boxes and 8 blue boxes in 16 positions. This is given by the binomial coefficient:
\[
\binom{16}{8} = \frac{16!}{8! \cdot 8!}
\]
Calculating this, we get:
\[
\binom{16}{8} = 12870
\]
2. **Spec... | 843 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Let $T_k = \frac{k(k+1)}{2}$ be the $k$-th triangular number. The in finite series
$$\sum_{k=4}^{\infty}\frac{1}{(T_{k-1} - 1)(Tk - 1)(T_{k+1} - 1)}$$
has the value $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 1. First, we need to simplify the expression for \( T_k - 1 \). Recall that \( T_k = \frac{k(k+1)}{2} \), so:
\[
T_k - 1 = \frac{k(k+1)}{2} - 1 = \frac{k^2 + k - 2}{2} = \frac{(k-1)(k+2)}{2}
\]
Therefore, for \( k-1 \), \( k \), and \( k+1 \), we have:
\[
T_{k-1} - 1 = \frac{(k-2)(k+1)}{2}, \quad T_k ... | 451 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
The gure below shows a large square divided into $9$ congruent smaller squares. A shaded square bounded by some of the diagonals of those smaller squares has area $14$. Find the area of the large square.
[img]https://cdn.artofproblemsolving.com/attachments/5/e/bad21be1b3993586c3860efa82ab27d340dbcb.png[/img] | 1. Let's denote the side length of each of the 9 congruent smaller squares as \( s \). Therefore, the side length of the large square is \( 3s \) since it is composed of a \( 3 \times 3 \) grid of smaller squares.
2. The shaded square is bounded by the diagonals of the smaller squares. Each side of the shaded square i... | 63 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The diagram below shows $\vartriangle ABC$ with point $D$ on side $\overline{BC}$. Three lines parallel to side $\overline{BC}$ divide segment $\overline{AD}$ into four equal segments. In the triangle, the ratio of the area of the shaded region to the area of the unshaded region is $\frac{49}{33}$ and $\frac{BD}{CD} = ... | 1. Let us denote the area of the smallest shaded triangle as \( x \) and the area of the smallest unshaded triangle as \( y \).
2. Since the segment \(\overline{AD}\) is divided into four equal segments by three lines parallel to \(\overline{BC}\), the areas of the triangles formed will be proportional to the square o... | 40 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The following figure is made up of many $2$ × $4$ tiles such that adjacent tiles always share an edge of length $2$. Find the perimeter of this figure. | 1. **Horizontal Edges Calculation:**
- We need to count the horizontal edges of the tiles.
- The problem states that adjacent tiles always share an edge of length 2.
- We go column by column from left to right.
- The calculation provided is:
\[
1 + 2 + 4 \cdot 10 + 2 = 3 + 40 + 2 = 45
\]
... | 158 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The positive integer $m$ is a multiple of $101$, and the positive integer $n$ is a multiple of $63$. Their sum is $2018$. Find $m - n$. | 1. Let \( m = 101x \) and \( n = 63y \), where \( x \) and \( y \) are positive integers.
2. Given that \( m + n = 2018 \), we can substitute the expressions for \( m \) and \( n \):
\[
101x + 63y = 2018
\]
3. We need to find integer solutions for \( x \) and \( y \). To do this, we can solve the equation modu... | 2 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Bradley is driving at a constant speed. When he passes his school, he notices that in $20$ minutes he will be exactly $\frac14$ of the way to his destination, and in $45$ minutes he will be exactly $\frac13$ of the way to his destination. Find the number of minutes it takes Bradley to reach his destination from the poi... | 1. Let \( T \) be the total time in minutes it takes Bradley to reach his destination from the point where he passes his school.
2. According to the problem, in 20 minutes, Bradley will be \(\frac{1}{4}\) of the way to his destination. Therefore, the time to reach \(\frac{1}{4}\) of the way is 20 minutes.
3. Similarly,... | 280 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
For some $k > 0$ the lines $50x + ky = 1240$ and $ky = 8x + 544$ intersect at right angles at the point $(m,n)$. Find $m + n$. | 1. **Determine the slopes of the lines:**
The given lines are:
\[
50x + ky = 1240
\]
and
\[
ky = 8x + 544
\]
To find the slopes, we rewrite each equation in slope-intercept form \(y = mx + b\).
For the first line:
\[
ky = -50x + 1240 \implies y = -\frac{50}{k}x + \frac{1240}{k}
\... | 44 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The triangle below is divided into nine stripes of equal width each parallel to the base of the triangle. The darkened stripes have a total area of $135$. Find the total area of the light colored stripes.
[img]https://cdn.artofproblemsolving.com/attachments/0/8/f34b86ccf50ef3944f5fbfd615a68607f4fadc.png[/img] | 1. Let the area of the uppermost strip be \( x \).
2. The areas of the darkened stripes are given as \( x, 5x, 9x, 13x, \) and \( 17x \). This is because the area of each strip increases quadratically as we move down the triangle, due to the similarity of the smaller triangles formed by the strips.
3. The total area of... | 60 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of positive integers less than $2018$ that are divisible by $6$ but are not divisible by at least one of the numbers $4$ or $9$. | 1. **Find the total number of integers less than $2018$ that are divisible by $6$:**
The number of integers less than $2018$ that are divisible by $6$ is given by:
\[
\left\lfloor \frac{2018}{6} \right\rfloor = 336
\]
2. **Find the number of integers less than $2018$ that are divisible by $6$ and also div... | 112 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Line segment $\overline{AB}$ has perpendicular bisector $\overline{CD}$, where $C$ is the midpoint of $\overline{AB}$. The segments have lengths $AB = 72$ and $CD = 60$. Let $R$ be the set of points $P$ that are midpoints of line segments $\overline{XY}$ , where $X$ lies on $\overline{AB}$ and $Y$ lies on $\overline{CD... | 1. **Identify the coordinates of points \(A\), \(B\), \(C\), and \(D\):**
- Let \(A = (-36, 0)\) and \(B = (36, 0)\) since \(AB = 72\) and \(C\) is the midpoint of \(\overline{AB}\).
- Therefore, \(C = (0, 0)\).
- Since \(\overline{CD}\) is the perpendicular bisector of \(\overline{AB}\) and \(CD = 60\), \(D\)... | 1080 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Suppose $x$ and $y$ are nonzero real numbers simultaneously satisfying the equations
$x + \frac{2018}{y}= 1000$ and $ \frac{9}{x}+ y = 1$.
Find the maximum possible value of $x + 1000y$. | 1. We start with the given equations:
\[
x + \frac{2018}{y} = 1000
\]
\[
\frac{9}{x} + y = 1
\]
2. We multiply these two equations:
\[
\left( x + \frac{2018}{y} \right) \left( \frac{9}{x} + y \right) = 1000 \cdot 1 = 1000
\]
3. Expanding the left-hand side:
\[
x \cdot \frac{9}{x} + x ... | 1991 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of ordered quadruples of positive integers $(a,b,c, d)$ such that $ab + cd = 10$. | We need to find the number of ordered quadruples of positive integers \((a, b, c, d)\) such that \(ab + cd = 10\). We will use casework based on the possible values of \(ab\) and \(cd\).
1. **Case 1: \(ab = 1\) and \(cd = 9\)**
- The only way to have \(ab = 1\) is \(a = 1\) and \(b = 1\).
- For \(cd = 9\), the p... | 58 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
On $\vartriangle ABC$ let $D$ be a point on side $\overline{AB}$, $F$ be a point on side $\overline{AC}$, and $E$ be a point inside the triangle so that $\overline{DE}\parallel \overline{AC}$ and $\overline{EF} \parallel \overline{AB}$. Given that $AF = 6, AC = 33, AD = 7, AB = 26$, and the area of quadrilateral $ADEF... | 1. **Identify the given information and draw the diagram:**
- Let $\triangle ABC$ be given with points $D$ on $\overline{AB}$, $F$ on $\overline{AC}$, and $E$ inside the triangle such that $\overline{DE} \parallel \overline{AC}$ and $\overline{EF} \parallel \overline{AB}$.
- Given: $AF = 6$, $AC = 33$, $AD = 7$, ... | 286 | Geometry | other | Yes | Yes | aops_forum | false |
Let $a, b, c$, and $d$ be real numbers such that $a^2 + b^2 + c^2 + d^2 = 3a + 8b + 24c + 37d = 2018$. Evaluate $3b + 8c + 24d + 37a$. | 1. We are given the equations:
\[
a^2 + b^2 + c^2 + d^2 = 2018
\]
and
\[
3a + 8b + 24c + 37d = 2018
\]
2. We need to evaluate the expression \(3b + 8c + 24d + 37a\).
3. Let's assume \(a, b, c, d\) are such that \(a^2 = 3a\), \(b^2 = 8b\), \(c^2 = 24c\), and \(d^2 = 37d\). This implies:
\[
a... | 1215 | Algebra | other | Yes | Yes | aops_forum | false |
Rectangle $ABCD$ has side lengths $AB = 6\sqrt3$ and $BC = 8\sqrt3$. The probability that a randomly chosen point inside the rectangle is closer to the diagonal $\overline{AC}$ than to the outside of the rectangle is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 1. **Simplify the Problem:**
Given the rectangle \(ABCD\) with side lengths \(AB = 6\sqrt{3}\) and \(BC = 8\sqrt{3}\), we need to find the probability that a randomly chosen point inside the rectangle is closer to the diagonal \(\overline{AC}\) than to the sides of the rectangle.
2. **Normalize the Rectangle:**
... | 17 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let $ABCD$ be a square with side length $6$. Circles $X, Y$ , and $Z$ are congruent circles with centers inside the square such that $X$ is tangent to both sides $\overline{AB}$ and $\overline{AD}$, $Y$ is tangent to both sides $\overline{AB}$ and $\overline{BC}$, and $Z$ is tangent to side $\overline{CD}$ and both cir... | 1. Let us denote the centers of circles \(X, Y, Z\) as \(X', Y', Z'\) respectively, and let the radius of each circle be \(r\). Since circle \(X\) is tangent to both sides \(\overline{AB}\) and \(\overline{AD}\), the center \(X'\) is at a distance \(r\) from both these sides. Similarly, circle \(Y\) is tangent to both ... | 195 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Find the positive integer $n$ such that $\frac12 \cdot \frac34 + \frac56 \cdot \frac78 + \frac{9}{10}\cdot \frac{11}{12 }= \frac{n}{1200}$ . | 1. First, we need to simplify each term in the given expression:
\[
\frac{1}{2} \cdot \frac{3}{4} = \frac{1 \cdot 3}{2 \cdot 4} = \frac{3}{8}
\]
\[
\frac{5}{6} \cdot \frac{7}{8} = \frac{5 \cdot 7}{6 \cdot 8} = \frac{35}{48}
\]
\[
\frac{9}{10} \cdot \frac{11}{12} = \frac{9 \cdot 11}{10 \cdot 12} ... | 2315 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Find $x$ so that the arithmetic mean of $x, 3x, 1000$, and $3000$ is $2018$. | 1. Start with the given arithmetic mean equation:
\[
\frac{x + 3x + 1000 + 3000}{4} = 2018
\]
2. Combine like terms in the numerator:
\[
\frac{4x + 4000}{4} = 2018
\]
3. Simplify the left-hand side by dividing each term in the numerator by 4:
\[
x + 1000 = 2018
\]
4. Solve for \( x \) by i... | 1018 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
The following diagram shows a grid of $36$ cells. Find the number of rectangles pictured in the diagram that contain at least three cells of the grid.
[img]https://cdn.artofproblemsolving.com/attachments/a/4/e9ba3a35204ec68c17a364ebf92cc107eb4d7a.png[/img] | 1. **Calculate the total number of rectangles in the grid:**
The grid is a $6 \times 6$ grid, which means it has 7 horizontal and 7 vertical lines. The number of ways to choose 2 horizontal lines from 7 is $\binom{7}{2}$, and the number of ways to choose 2 vertical lines from 7 is also $\binom{7}{2}$. Therefore, th... | 345 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
One afternoon at the park there were twice as many dogs as there were people, and there were twice as many people as there were snakes. The sum of the number of eyes plus the number of legs on all of these dogs, people, and snakes was $510$. Find the number of dogs that were at the park. | 1. Let \( D \) be the number of dogs, \( P \) be the number of people, and \( S \) be the number of snakes.
2. According to the problem, we have the following relationships:
\[
D = 2P \quad \text{(twice as many dogs as people)}
\]
\[
P = 2S \quad \text{(twice as many people as snakes)}
\]
3. Each dog ... | 60 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
In $10$ years the product of Melanie's age and Phil's age will be $400$ more than it is now. Find what the sum of Melanie's age and Phil's age will be $6$ years from now. | 1. Let \( A \) be Melanie's age now.
2. Let \( B \) be Phil's age now.
3. According to the problem, in 10 years, the product of their ages will be 400 more than it is now. This can be written as:
\[
(A + 10)(B + 10) = AB + 400
\]
4. Expanding the left-hand side of the equation:
\[
AB + 10A + 10B + 100 = ... | 42 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Let $a$ and $b$ be positive integers such that $2a - 9b + 18ab = 2018$. Find $b - a$. | 1. We start with the given equation:
\[
2a - 9b + 18ab = 2018
\]
2. Move \(2018\) to the left-hand side:
\[
18ab + 2a - 9b - 2018 = 0
\]
3. To factor this expression, we use Simon's Favorite Factoring Trick. We add and subtract 1 to the left-hand side:
\[
18ab + 2a - 9b - 2018 + 1 - 1 = 0
\... | 223 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of positive integers $k \le 2018$ for which there exist integers $m$ and $n$ so that $k = 2^m + 2^n$.
For example, $64 = 2^5 + 2^5$, $65 = 2^0 + 2^6$, and $66 = 2^1 + 2^6$. | To find the number of positive integers \( k \leq 2018 \) for which there exist integers \( m \) and \( n \) such that \( k = 2^m + 2^n \), we need to consider the binary representation of \( k \).
1. **Binary Representation Analysis**:
- If \( k = 2^m + 2^n \), then \( k \) has exactly two 1's in its binary repres... | 65 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
A jeweler can get an alloy that is $40\%$ gold for $200$ dollars per ounce, an alloy that is $60\%$ gold for $300$ dollar per ounce, and an alloy that is $90\%$ gold for $400$ dollars per ounce. The jeweler will purchase some of these gold alloy products, melt them down, and combine them to get an alloy that is $50\%$ ... | 1. Let \( x \) be the number of ounces of the 40% gold alloy, and \( y \) be the number of ounces of the 90% gold alloy. We want to find the minimum cost per ounce for an alloy that is 50% gold.
2. The total amount of gold in the mixture is given by:
\[
0.4x + 0.9y
\]
3. The total weight of the mixture is:
... | 240 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Five lighthouses are located, in order, at points $A, B, C, D$, and $E$ along the shore of a circular lake with a diameter of $10$ miles. Segments $AD$ and $BE$ are diameters of the circle. At night, when sitting at $A$, the lights from $B, C, D$, and $E$ appear to be equally spaced along the horizon. The perimeter in ... | 1. **Understanding the Geometry**:
- The lake is circular with a diameter of 10 miles, so the radius \( r \) is \( \frac{10}{2} = 5 \) miles.
- Points \( A, B, C, D, \) and \( E \) are located on the circumference of the circle.
- Segments \( AD \) and \( BE \) are diameters, so they each span 180 degrees.
2.... | 95 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
If you roll four standard, fair six-sided dice, the top faces of the dice can show just one value (for example, $3333$), two values (for example, $2666$), three values (for example, $5215$), or four values (for example, $4236$). The mean number of values that show is $\frac{m}{n}$ , where $m$ and $n$ are relatively pri... | To solve this problem, we need to calculate the expected number of distinct values that appear when rolling four six-sided dice. We will use the concept of expected value in probability.
1. **Calculate the total number of possible outcomes:**
Each die has 6 faces, so the total number of outcomes when rolling four d... | 887 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Find the positive integer $k$ such that the roots of $x^3 - 15x^2 + kx -1105$ are three distinct collinear points in the complex plane. | 1. **Identify the roots**: Given that the roots of the polynomial \(x^3 - 15x^2 + kx - 1105\) are three distinct collinear points in the complex plane, we can assume the roots are of the form \(a+bi\), \(a\), and \(a-bi\). This is because collinear points in the complex plane that are symmetric about the real axis will... | 271 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Suppose that $a$ and $b$ are positive real numbers such that $3\log_{101}\left(\frac{1,030,301-a-b}{3ab}\right) = 3 - 2 \log_{101}(ab)$. Find $101 - \sqrt[3]{a}- \sqrt[3]{b}$. | 1. Start with the given equation:
\[
3\log_{101}\left(\frac{1,030,301 - a - b}{3ab}\right) = 3 - 2 \log_{101}(ab)
\]
2. Divide both sides by 3:
\[
\log_{101}\left(\frac{1,030,301 - a - b}{3ab}\right) = 1 - \frac{2}{3} \log_{101}(ab)
\]
3. Let \( x = \log_{101}(ab) \). Then the equation becomes:
\... | 0 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Aileen plays badminton where she and her opponent stand on opposite sides of a net and attempt to bat a birdie back and forth over the net. A player wins a point if their opponent fails to bat the birdie over the net. When Aileen is the server (the first player to try to hit the birdie over the net), she wins a point ... | 1. We start by noting that Aileen wins a point if her opponent fails to return the birdie. The probability of Aileen winning a point when she is the server is given as $\frac{9}{10}$.
2. Let $p$ be the probability that Aileen successfully returns the birdie. We are given that her opponent returns the birdie with proba... | 73 | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
Let $x$ be in the interval $\left(0, \frac{\pi}{2}\right)$ such that $\sin x - \cos x = \frac12$ . Then $\sin^3 x + \cos^3 x = \frac{m\sqrt{p}}{n}$ , where $m, n$, and $p$ are relatively prime positive integers, and $p$ is not divisible by the square of any prime. Find $m + n + p$. | 1. Let $\sin x = a$ and $\cos x = b$. Given that $a - b = \frac{1}{2}$ and $a^2 + b^2 = 1$.
2. We need to find $\sin^3 x + \cos^3 x = a^3 + b^3$.
3. Recall the identity for the sum of cubes:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
4. First, solve for $a$ and $b$ using the given equations:
\[
a - b = \... | 28 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Positive integers $a$ and $b$ satisfy $a^3 + 32b + 2c = 2018$ and $b^3 + 32a + 2c = 1115$. Find $a^2 + b^2 + c^2$. | 1. We start with the given equations:
\[
a^3 + 32b + 2c = 2018
\]
\[
b^3 + 32a + 2c = 1115
\]
2. Subtract the second equation from the first:
\[
(a^3 + 32b + 2c) - (b^3 + 32a + 2c) = 2018 - 1115
\]
Simplifying, we get:
\[
a^3 - b^3 + 32b - 32a = 903
\]
3. Factor the left-hand si... | 226 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Suppose $p < q < r < s$ are prime numbers such that $pqrs + 1 = 4^{p+q}$. Find $r + s$. | 1. **Identify the properties of the primes:**
Given that \( p < q < r < s \) are prime numbers and \( pqrs + 1 = 4^{p+q} \), we first note that \( p, q, r, s \) cannot be 2 because if any of them were 2, the left-hand side (LHS) would be odd, while the right-hand side (RHS) would be even, which is a contradiction. T... | 274 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
In $\vartriangle ABC$ points $D, E$, and $F$ lie on side $\overline{BC}$ such that $\overline{AD}$ is an angle bisector of $\angle BAC$, $\overline{AE}$ is a median, and $\overline{AF}$ is an altitude. Given that $AB = 154$ and $AC = 128$, and $9 \times DE = EF,$ fi nd the side length $BC$. | 1. **Set up the coordinate system:**
Let \( A = (a, b) \), \( B = (0, 0) \), and \( C = (x, 0) \). Given \( AB = 154 \) and \( AC = 128 \), we can use the distance formula to express \( a \) and \( b \) in terms of \( x \).
2. **Use the distance formula:**
\[
AB = \sqrt{a^2 + b^2} = 154 \quad \text{and} \quad... | 94 | Geometry | other | Yes | Yes | aops_forum | false |
Find the number of positive integers less than or equal to $2019$ that are no more than $10$ away from a perfect square. | 1. **Identify the range of perfect squares:**
We need to find the perfect squares less than or equal to \(2019\). The largest integer \(n\) such that \(n^2 \leq 2019\) is \(n = 44\) because \(44^2 = 1936\) and \(45^2 = 2025\).
2. **Count the numbers close to each perfect square:**
For each perfect square \(k^2\)... | 905 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Suppose that $a, b, c$, and $d$ are real numbers simultaneously satisfying
$a + b - c - d = 3$
$ab - 3bc + cd - 3da = 4$
$3ab - bc + 3cd - da = 5$
Find $11(a - c)^2 + 17(b -d)^2$. | 1. We start with the given equations:
\[
a + b - c - d = 3
\]
\[
ab - 3bc + cd - 3da = 4
\]
\[
3ab - bc + 3cd - da = 5
\]
2. From the first equation, we can rewrite it as:
\[
(a - c) + (b - d) = 3
\]
Let \( x = a - c \) and \( y = b - d \). Thus, we have:
\[
x + y = 3
\]... | 63 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Squares $ABCD$ and $AEFG$ each with side length $12$ overlap so that $\vartriangle AED$ is an equilateral triangle as shown. The area of the region that is in the interior of both squares which is shaded in the diagram is $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of an... | 1. **Identify the given information and the goal:**
- We have two overlapping squares \(ABCD\) and \(AEFG\), each with side length 12.
- \(\triangle AED\) is an equilateral triangle.
- We need to find the area of the region that is in the interior of both squares, expressed as \(m\sqrt{n}\), and then find \(m ... | 111 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The diagram shows a polygon made by removing six $2\times 2$ squares from the sides of an $8\times 12$ rectangle. Find the perimeter of this polygon.
[img]https://cdn.artofproblemsolving.com/attachments/6/3/c23510c821c159d31aff0e6688edebc81e2737.png[/img] | 1. **Calculate the perimeter of the original $8 \times 12$ rectangle:**
The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2 \times (\text{length} + \text{width})
\]
Here, the length is 12 units and the width is 8 units. Therefore,
\[
P = 2 \times (12 + 8) = 2 \times 20 = 40 \text... | 52 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The diagram below shows four congruent squares and some of their diagonals. Let $T$ be the number of triangles and $R$ be the number of rectangles that appear in the diagram. Find $T + R$.
[img]https://cdn.artofproblemsolving.com/attachments/1/5/f756bbe67c09c19e811011cb6b18d0ff44be8b.png[/img] | 1. **Counting the Triangles:**
- Each square is divided into 4 triangles by its diagonals.
- There are 4 squares, so the total number of triangles formed by the diagonals within the squares is \(4 \times 4 = 16\).
- However, some triangles are counted multiple times. Specifically, the triangles formed by the i... | 25 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
A semicircle has diameter $\overline{AD}$ with $AD = 30$. Points $B$ and $C$ lie on $\overline{AD}$, and points $E$ and $F$ lie on the arc of the semicircle. The two right triangles $\vartriangle BCF$ and $\vartriangle CDE$ are congruent. The area of $\vartriangle BCF$ is $m\sqrt{n}$, where $m$ and $n$ are positive in... | 1. **Identify the given information and setup the problem:**
- The diameter \( \overline{AD} \) of the semicircle is \( AD = 30 \).
- Points \( B \) and \( C \) lie on \( \overline{AD} \).
- Points \( E \) and \( F \) lie on the arc of the semicircle.
- The right triangles \( \triangle BCF \) and \( \triang... | 52 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The mean of $\frac12 , \frac34$ , and $\frac56$ differs from the mean of $\frac78$ and $\frac{9}{10}$ by $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 1. **Calculate the mean of $\frac{1}{2}$, $\frac{3}{4}$, and $\frac{5}{6}$:**
- First, find the sum of the fractions:
\[
\frac{1}{2} + \frac{3}{4} + \frac{5}{6}
\]
- To add these fractions, find a common denominator. The least common multiple (LCM) of 2, 4, and 6 is 12.
\[
\frac{1}{2} = \... | 859 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Evaluate
$$\frac{(2 + 2)^2}{2^2} \cdot \frac{(3 + 3 + 3 + 3)^3}{(3 + 3 + 3)^3} \cdot \frac{(6 + 6 + 6 + 6 + 6 + 6)^6}{(6 + 6 + 6 + 6)^6}$$ | 1. Start by simplifying each term in the given expression:
\[
\frac{(2 + 2)^2}{2^2} \cdot \frac{(3 + 3 + 3 + 3)^3}{(3 + 3 + 3)^3} \cdot \frac{(6 + 6 + 6 + 6 + 6 + 6)^6}{(6 + 6 + 6 + 6)^6}
\]
2. Simplify the first fraction:
\[
\frac{(2 + 2)^2}{2^2} = \frac{4^2}{2^2} = \frac{16}{4} = 4
\]
3. Simplify ... | 108 | Algebra | other | Yes | Yes | aops_forum | false |
Find the positive integer $n$ such that $32$ is the product of the real number solutions of $x^{\log_2(x^3)-n} = 13$ | 1. Start with the given equation:
\[
x^{\log_2(x^3) - n} = 13
\]
2. Take the logarithm base \( x \) of both sides:
\[
\log_x \left( x^{\log_2(x^3) - n} \right) = \log_x 13
\]
3. Simplify the left-hand side using the property of logarithms \(\log_b(a^c) = c \log_b(a)\):
\[
\log_2(x^3) - n = \lo... | 15 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Let $m > n$ be positive integers such that $3(3mn - 2)^2 - 2(3m -3n)^2 = 2019$. Find $3m + n$.
| 1. We start with the given equation:
\[
3(3mn - 2)^2 - 2(3m - 3n)^2 = 2019
\]
2. Expand the squares:
\[
3(9m^2n^2 - 12mn + 4) - 2(9m^2 - 18mn + 9n^2)
\]
3. Distribute the constants:
\[
27m^2n^2 - 36mn + 12 - 18m^2 + 36mn - 18n^2
\]
4. Combine like terms:
\[
27m^2n^2 - 18m^2 - 18n^2 +... | 46 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
There are relatively prime positive integers $m$ and $n$ so that the parabola with equation $y = 4x^2$ is tangent to the parabola with equation $x = y^2 + \frac{m}{n}$ . Find $m + n$. | 1. Let \(\frac{m}{n} = e\). We substitute \(y = 4x^2\) into \(x = y^2 + e\) to obtain:
\[
x = (4x^2)^2 + e \implies x = 16x^4 + e
\]
2. We want the quartic polynomial \(16x^4 - x + e\) to have a repeated real root, which implies that the polynomial must have a discriminant of zero. The discriminant \(\Delta\) ... | 19 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Suppose $a$ is a real number such that $\sin(\pi \cdot \cos a) = \cos(\pi \cdot \sin a)$. Evaluate $35 \sin^2(2a) + 84 \cos^2(4a)$. | 1. Given the equation $\sin(\pi \cdot \cos a) = \cos(\pi \cdot \sin a)$, we need to find the value of $35 \sin^2(2a) + 84 \cos^2(4a)$.
2. First, let's analyze the given trigonometric equation. We know that $\sin(x) = \cos(\pi/2 - x)$. Therefore, we can rewrite the given equation as:
\[
\sin(\pi \cdot \cos a) = \... | 21 | Algebra | other | Yes | Yes | aops_forum | false |
Find the number of ordered triples of sets $(T_1, T_2, T_3)$ such that
1. each of $T_1, T_2$, and $T_3$ is a subset of $\{1, 2, 3, 4\}$,
2. $T_1 \subseteq T_2 \cup T_3$,
3. $T_2 \subseteq T_1 \cup T_3$, and
4. $T_3\subseteq T_1 \cup T_2$. | 1. **Understanding the problem**: We need to find the number of ordered triples of sets \((T_1, T_2, T_3)\) such that each of \(T_1, T_2\), and \(T_3\) is a subset of \(\{1, 2, 3, 4\}\), and they satisfy the conditions \(T_1 \subseteq T_2 \cup T_3\), \(T_2 \subseteq T_1 \cup T_3\), and \(T_3 \subseteq T_1 \cup T_2\).
... | 625 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of ordered pairs of integers $(x, y)$ such that $$\frac{x^2}{y}- \frac{y^2}{x}= 3 \left( 2+ \frac{1}{xy}\right)$$ | To find the number of ordered pairs of integers \((x, y)\) that satisfy the equation
\[
\frac{x^2}{y} - \frac{y^2}{x} = 3 \left( 2 + \frac{1}{xy} \right),
\]
we start by simplifying and manipulating the given equation.
1. **Multiply both sides by \(xy\) to clear the denominators:**
\[
xy \left( \frac{x^2}{y} - \... | 0 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
There are positive integers $m$ and $n$ such that $m^2 -n = 32$ and $\sqrt[5]{m +\sqrt{n}}+ \sqrt[5]{m -\sqrt{n}}$ is a real root of the polynomial $x^5 - 10x^3 + 20x - 40$. Find $m + n$. | 1. Let \( p = \sqrt[5]{m + \sqrt{n}} \) and \( q = \sqrt[5]{m - \sqrt{n}} \). We are given that \( x = p + q \) is a real root of the polynomial \( x^5 - 10x^3 + 20x - 40 \).
2. Consider the polynomial \( x^5 - 10x^3 + 20x - 40 \). We need to express \( x^5 \) in terms of \( p \) and \( q \).
3. First, note that:
... | 388 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
The diagram below shows a $12$ by $20$ rectangle split into four strips of equal widths all surrounding an isosceles triangle. Find the area of the shaded region.
[img]https://cdn.artofproblemsolving.com/attachments/9/e/ed6be5110d923965c64887a2ca8e858c977700.png[/img] | 1. **Understanding the Problem:**
The problem involves a $12$ by $20$ rectangle split into four strips of equal widths, surrounding an isosceles triangle. We need to find the area of the shaded region.
2. **Identifying the Shapes:**
The rectangle is divided into:
- Two right triangles
- One isosceles trian... | 150 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The following diagram shows four adjacent $2\times 2$ squares labeled $1, 2, 3$, and $4$. A line passing through the lower left vertex of square $1$ divides the combined areas of squares $1, 3$, and $4$ in half so that the shaded region has area $6$. The difference between the areas of the shaded region within square $... | 1. **Understanding the Problem:**
We have four adjacent \(2 \times 2\) squares labeled \(1, 2, 3,\) and \(4\). A line passing through the lower left vertex of square \(1\) divides the combined areas of squares \(1, 3,\) and \(4\) in half, resulting in a shaded region with an area of \(6\). We need to find the differ... | 49 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The circle centered at point $A$ with radius $19$ and the circle centered at point $B$ with radius $32$ are both internally tangent to a circle centered at point $C$ with radius $100$ such that point $C$ lies on segment $\overline{AB}$. Point $M$ is on the circle centered at $A$ and point $N$ is on the circle centered ... | 1. **Identify the given information and set up the problem:**
- Circle centered at \( A \) with radius \( 19 \).
- Circle centered at \( B \) with radius \( 32 \).
- Both circles are internally tangent to a larger circle centered at \( C \) with radius \( 100 \).
- Point \( C \) lies on segment \( \overline... | 140 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The following diagram shows equilateral triangle $\vartriangle ABC$ and three other triangles congruent to it. The other three triangles are obtained by sliding copies of $\vartriangle ABC$ a distance $\frac18 AB$ along a side of $\vartriangle ABC$ in the directions from $A$ to $B$, from $B$ to $C$, and from $C$ to $A... | 1. **Understanding the Problem:**
We are given an equilateral triangle $\triangle ABC$ and three other triangles congruent to it. These triangles are obtained by sliding copies of $\triangle ABC$ a distance $\frac{1}{8} AB$ along a side of $\triangle ABC$ in the directions from $A$ to $B$, from $B$ to $C$, and from ... | 768 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
In the diagram below, points $D, E$, and $F$ are on the inside of equilateral $\vartriangle ABC$ such that $D$ is on $\overline{AE}, E$ is on $\overline{CF}, F$ is on $\overline{BD}$, and the triangles $\vartriangle AEC, \vartriangle BDA$, and $\vartriangle CFB$ are congruent. Given that $AB = 10$ and $DE = 6$, the per... | 1. **Identify the given information and the goal:**
- We have an equilateral triangle \( \triangle ABC \) with side length \( AB = 10 \).
- Points \( D, E, \) and \( F \) are inside the triangle such that \( D \) is on \( \overline{AE} \), \( E \) is on \( \overline{CF} \), and \( F \) is on \( \overline{BD} \).
... | 308 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of integers $n$ for which $\sqrt{\frac{(2020 - n)^2}{2020 - n^2}}$ is a real number. | 1. We need to find the number of integers \( n \) for which the expression \(\sqrt{\frac{(2020 - n)^2}{2020 - n^2}}\) is a real number. For the expression under the square root to be real, the radicand (the expression inside the square root) must be non-negative.
2. The radicand is \(\frac{(2020 - n)^2}{2020 - n^2}\).... | 90 | Inequalities | math-word-problem | Yes | Yes | aops_forum | false |
Camilla drove $20$ miles in the city at a constant speed and $40$ miles in the country at a constant speed that was $20$ miles per hour greater than her speed in the city. Her entire trip took one hour. Find the number of minutes that Camilla drove in the country rounded to the nearest minute. | 1. Let \( x \) be Camilla's speed in the city in miles per hour. Then her speed in the country is \( x + 20 \) miles per hour.
2. The time taken to drive 20 miles in the city is \( \frac{20}{x} \) hours.
3. The time taken to drive 40 miles in the country is \( \frac{40}{x+20} \) hours.
4. According to the problem, the ... | 38 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
There is a complex number $K$ such that the quadratic polynomial $7x^2 +Kx + 12 - 5i$ has exactly one root, where $i =\sqrt{-1}$. Find $|K|^2$. | 1. **Determine the condition for a double root:**
For the quadratic polynomial \(7x^2 + Kx + (12 - 5i)\) to have exactly one root, the discriminant must be zero. The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[
\Delta = b^2 - 4ac
\]
Here, \(a = 7\), \(b = K\), and... | 364 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
There are two distinct pairs of positive integers $a_1 < b_1$ and $a_2 < b_2$ such that both $(a_1 + ib_1)(b_1 - ia_1) $ and $(a_2 + ib_2)(b_2 - ia_2)$ equal $2020$, where $i =\sqrt{-1}$. Find $a_1 + b_1 + a_2 + b_2$. | 1. We start by expanding the given expressions. For the first pair \((a_1, b_1)\):
\[
(a_1 + ib_1)(b_1 - ia_1) = a_1b_1 + a_1(-ia_1) + ib_1b_1 - i^2a_1b_1
\]
Since \(i^2 = -1\), this simplifies to:
\[
a_1b_1 - ia_1^2 + ib_1^2 + a_1b_1 = 2a_1b_1 + i(b_1^2 - a_1^2)
\]
Given that this expression eq... | 714 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
There are relatively prime positive integers $s$ and $t$ such that $$\sum_{n=2}^{100}\left(\frac{n}{n^2-1}- \frac{1}{n}\right)=\frac{s}{t}$$ Find $s + t$. | 1. We start by simplifying the given sum:
\[
\sum_{n=2}^{100}\left(\frac{n}{n^2-1}- \frac{1}{n}\right)
\]
Notice that \( \frac{n}{n^2-1} \) can be decomposed using partial fractions:
\[
\frac{n}{n^2-1} = \frac{n}{(n-1)(n+1)} = \frac{A}{n-1} + \frac{B}{n+1}
\]
Solving for \( A \) and \( B \):
... | 127 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
In isosceles $\vartriangle ABC, AB = AC, \angle BAC$ is obtuse, and points $E$ and $F$ lie on sides $AB$ and $AC$, respectively, so that $AE = 10, AF = 15$. The area of $\vartriangle AEF$ is $60$, and the area of quadrilateral $BEFC$ is $102$. Find $BC$. | 1. Given that $\triangle ABC$ is isosceles with $AB = AC$ and $\angle BAC$ is obtuse, we need to find the length of $BC$.
2. Points $E$ and $F$ lie on sides $AB$ and $AC$ respectively, such that $AE = 10$ and $AF = 15$. The area of $\triangle AEF$ is $60$.
3. Using the area formula for a triangle, we have:
\[
\fr... | 36 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The mean number of days per month in $2020$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 1. **Determine the total number of days in the year 2020:**
Since 2020 is a leap year (divisible by 4 and not a century year unless divisible by 400), it has 366 days.
2. **Calculate the mean number of days per month:**
The mean number of days per month is given by the total number of days divided by the number ... | 63 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Alex launches his boat into a river and heads upstream at a constant speed. At the same time at a point $8$ miles upstream from Alex, Alice launches her boat and heads downstream at a constant speed. Both boats move at $6$ miles per hour in still water, but the river is owing downstream at $2\frac{3}{10}$ miles per ho... | 1. **Determine the effective speeds of Alex and Alice:**
- The river flows downstream at \(2\frac{3}{10} = \frac{23}{10}\) miles per hour.
- Alex's speed in still water is \(6\) miles per hour. Since he is going upstream, his effective speed is:
\[
6 - \frac{23}{10} = 6 - 2.3 = 3.7 = \frac{37}{10} \text... | 52 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Find a positive integer $n$ such that there is a polygon with $n$ sides where each of its interior angles measures $177^o$ | 1. We start with the formula for the sum of the interior angles of a polygon with \( n \) sides, which is given by:
\[
(n-2) \times 180^\circ
\]
2. Since each interior angle of the polygon measures \( 177^\circ \), the total sum of the interior angles can also be expressed as:
\[
177^\circ \times n
\]... | 120 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Patrick started walking at a constant rate along a straight road from school to the park. One hour after Patrick left, Tanya started running along the same road from school to the park. One hour after Tanya left, Jose started bicycling along the same road from school to the park. Tanya ran at a constant rate of $2$ mil... | 1. Let \( v \) be Patrick's speed in miles per hour (mph) and \( t \) be the total time in hours he took to get to the park. Therefore, the distance \( d \) from the school to the park is given by:
\[
d = vt
\]
2. Tanya started one hour after Patrick and ran at a speed of \( v + 2 \) mph. She took \( t - 1 \)... | 277 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Mary mixes $2$ gallons of a solution that is $40$ percent alcohol with $3$ gallons of a solution that is $60$ percent alcohol. Sandra mixes $4$ gallons of a solution that is $30$ percent alcohol with $\frac{m}{n}$ gallons of a solution that is $80$ percent alcohol, where $m$ and $n$ are relatively prime positive intege... | 1. **Calculate the total amount of alcohol in Mary's solution:**
- Mary mixes 2 gallons of a 40% alcohol solution and 3 gallons of a 60% alcohol solution.
- The amount of alcohol in the 2 gallons of 40% solution is:
\[
2 \times 0.40 = 0.8 \text{ gallons}
\]
- The amount of alcohol in the 3 gallo... | 29 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Let $a$ and $b$ be positive integers such that $(a^3 - a^2 + 1)(b^3 - b^2 + 2) = 2020$. Find $10a + b$. | 1. We start with the given equation:
\[
(a^3 - a^2 + 1)(b^3 - b^2 + 2) = 2020
\]
2. Factorize 2020:
\[
2020 = 2^2 \times 5 \times 101
\]
3. We need to find values of \(a\) and \(b\) such that:
\[
(a^3 - a^2 + 1)(b^3 - b^2 + 2) = 4 \times 5 \times 101
\]
4. Let's consider the factors of 202... | 53 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Six different small books and three different large books sit on a shelf. Three children may each take either two small books or one large book. Find the number of ways the three children can select their books. | To solve this problem, we need to consider the different ways the three children can select their books. Each child can either take two small books or one large book. We will use casework to count the number of ways based on how many children take small books.
1. **Case 1: 0 children take small books (all take large b... | 1041 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Daniel had a string that formed the perimeter of a square with area $98$. Daniel cut the string into two pieces. With one piece he formed the perimeter of a rectangle whose width and length are in the ratio $2 : 3$. With the other piece he formed the perimeter of a rectangle whose width and length are in the ratio $3 :... | 1. **Determine the perimeter of the original square:**
- Given the area of the square is \(98\), we can find the side length \(s\) of the square using the formula for the area of a square:
\[
s^2 = 98 \implies s = \sqrt{98} = 7\sqrt{2}
\]
- The perimeter \(P\) of the square is:
\[
P = 4s ... | 67 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Find the number of permutations of the letters $ABCDE$ where the letters $A$ and $B$ are not adjacent and the letters $C$ and $D$ are not adjacent. For example, count the permutations $ACBDE$ and $DEBCA$ but not $ABCED$ or $EDCBA$. | 1. **Calculate the total number of permutations of the letters \(ABCDE\):**
\[
5! = 120
\]
There are \(120\) permutations of the letters \(ABCDE\).
2. **Calculate the number of permutations where \(A\) and \(B\) are adjacent:**
- Treat \(A\) and \(B\) as a single unit. This gives us the units \(\{AB, C,... | 48 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Wendy randomly chooses a positive integer less than or equal to $2020$. The probability that the digits in Wendy's number add up to $10$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 1. **Case 1: 2-digit numbers**
We need to find all 2-digit numbers whose digits sum to 10. Listing them:
\[
19, 28, 37, 46, 55, 64, 73, 82, 91
\]
This gives us a total of 9 numbers.
2. **Case 2: 3-digit numbers**
We need to find all 3-digit numbers whose digits sum to 10. We can break this down by ... | 107 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Right $\vartriangle ABC$ has side lengths $6, 8$, and $10$. Find the positive integer $n$ such that the area of the region inside the circumcircle but outside the incircle of $\vartriangle ABC$ is $n\pi$.
[img]https://cdn.artofproblemsolving.com/attachments/d/1/cb112332069c09a3b370343ca8a2ef21102fe2.png[/img] | 1. **Identify the side lengths and the type of triangle:**
Given a right triangle $\vartriangle ABC$ with side lengths $6$, $8$, and $10$. Since $10^2 = 6^2 + 8^2$, it confirms that $\vartriangle ABC$ is a right triangle with the hypotenuse $10$.
2. **Calculate the circumradius $R$:**
For a right triangle, the c... | 21 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Construct a geometric gure in a sequence of steps. In step $1$, begin with a $4\times 4$ square. In step $2$, attach a $1\times 1$ square onto the each side of the original square so that the new squares are on the outside of the original square, have a side along the side of the original square, and the midpoints of ... | 1. **Step 1: Calculate the area of the initial square.**
- The initial square is a \(4 \times 4\) square.
- The area of the initial square is:
\[
4 \times 4 = 16
\]
2. **Step 2: Calculate the area of the squares added in step 2.**
- In step 2, four \(1 \times 1\) squares are added, one on each ... | 285 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The diagram shows two intersecting line segments that form some of the sides of two squares with side lengths $3$ and $6$. Two line segments join vertices of these squares. Find the area of the region enclosed by the squares and segments.
| 1. **Calculate the areas of the squares:**
- The side length of the first square is \(3\). Therefore, its area is:
\[
\text{Area of the first square} = 3^2 = 9
\]
- The side length of the second square is \(6\). Therefore, its area is:
\[
\text{Area of the second square} = 6^2 = 36
\... | 63 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
A building contractor needs to pay his $108$ workers $\$200$ each. He is carrying $122$ one hundred dollar bills and $188$ fifty dollar bills. Only $45$ workers get paid with two $\$100$ bills. Find the number of workers who get paid with four $\$50$ bills. | 1. **Determine the number of workers paid with two $100 bills:**
- Given that 45 workers are paid with two $100 bills each, the total number of $100 bills used is:
\[
45 \times 2 = 90 \text{ hundred dollar bills}
\]
2. **Calculate the remaining $100 bills:**
- The contractor initially has 122 one ... | 31 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
There were three times as many red candies as blue candies on a table. After Darrel took the same number of red candies and blue candies, there were four times as many red candies as blue candies left on the table. Then after Cloe took $12$ red candies and $12$ blue candies, there were five times as many red candies as... | 1. Let \( r \) be the number of red candies and \( b \) be the number of blue candies initially. According to the problem, we have:
\[
r = 3b
\]
2. Let \( x \) be the number of candies (both red and blue) that Darrel took. After Darrel took \( x \) red candies and \( x \) blue candies, the number of red candi... | 48 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
A rectangular wooden block has a square top and bottom, its volume is $576$, and the surface area of its vertical sides is $384$. Find the sum of the lengths of all twelve of the edges of the block. | 1. Let the side length of the square base be \( s \) and the height of the block be \( h \). We are given two pieces of information:
\[
s^2 \cdot h = 576
\]
\[
4sh = 384
\]
2. From the second equation, solve for \( sh \):
\[
4sh = 384 \implies sh = \frac{384}{4} = 96
\]
3. Substitute \( sh ... | 112 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Pam lists the four smallest positive prime numbers in increasing order. When she divides the positive integer $N$ by the first prime, the remainder is $1$. When she divides $N$ by the second prime, the remainder is $2$. When she divides $N$ by the third prime, the remainder is $3$. When she divides $N$ by the fourth pr... | 1. Identify the four smallest positive prime numbers: 2, 3, 5, and 7.
2. Set up the system of congruences based on the problem statement:
\[
\begin{cases}
N \equiv 1 \pmod{2} \\
N \equiv 2 \pmod{3} \\
N \equiv 3 \pmod{5} \\
N \equiv 4 \pmod{7}
\end{cases}
\]
3. Solve the first congruence:
\[
... | 53 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
A semicircle has diameter $AB$ with $AB = 100$. Points $C$ and $D$ lie on the semicircle such that $AC = 28$ and $BD = 60$. Find $CD$. | 1. Let the center of the semicircle be \( O \). The radius of the semicircle is \( \frac{AB}{2} = \frac{100}{2} = 50 \). Therefore, \( OA = OB = 50 \).
2. Draw \( OC \) and \( OD \). Since \( O \) is the center, \( OC = OD = 50 \).
3. We are given \( AC = 28 \) and \( BD = 60 \). We need to find \( CD \).
4. Use the... | 60 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let $L_1$ and $L_2$ be perpendicular lines, and let $F$ be a point at a distance $18$ from line $L_1$ and a distance $25$ from line $L_2$. There are two distinct points, $P$ and $Q$, that are each equidistant from $F$, from line $L_1$, and from line $L_2$. Find the area of $\triangle{FPQ}$. | 1. **Identify the coordinates of point \( F \)**:
- Let \( L_1 \) be the line \( x = 0 \) and \( L_2 \) be the line \( y = 0 \).
- Point \( F \) is at a distance of 18 units from \( L_1 \) and 25 units from \( L_2 \).
- Therefore, the coordinates of \( F \) are \( (18, 25) \).
2. **Determine the coordinates o... | 210 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Each of the cells of a $7 \times 7$ grid is painted with a color chosen randomly and independently from a set of $N$ fixed colors. Call an edge hidden if it is shared by two adjacent cells in the grid that are painted the same color. Determine the least $N$ such that the expected number of hidden edges is less than $3$... | 1. **Define the problem and the probability of a hidden edge:**
- An edge is considered hidden if two adjacent cells sharing that edge are painted the same color.
- The probability that two adjacent cells are painted the same color is $\frac{1}{N}$, where $N$ is the number of colors.
2. **Count the total number ... | 29 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
The side lengths of a scalene triangle are roots of the polynomial $$x^3-20x^2+131x-281.3.$$ Find the square of the area of the triangle. | 1. Let the side lengths of the scalene triangle be \(a\), \(b\), and \(c\). These are the roots of the polynomial \(x^3 - 20x^2 + 131x - 281.3\).
2. By Vieta's formulas, we know:
\[
a + b + c = 20,
\]
\[
ab + bc + ca = 131,
\]
\[
abc = 281.3.
\]
3. To find the square of the area of the tria... | 287 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let $a, b, c, d$ be an increasing arithmetic sequence of positive real numbers with common difference $\sqrt2$. Given that the product $abcd = 2021$, $d$ can be written as $\frac{m+\sqrt{n}}{\sqrt{p}}$ , where $m, n,$ and $p$ are positive integers not divisible by the square of any prime. Find $m + n + p$. | 1. Given that \(a, b, c, d\) are in an increasing arithmetic sequence with a common difference of \(\sqrt{2}\), we can express \(a, b, c, d\) in terms of \(d\):
\[
a = d - 3\sqrt{2}, \quad b = d - 2\sqrt{2}, \quad c = d - \sqrt{2}, \quad d
\]
2. The product of these terms is given as:
\[
abcd = 2021
... | 100 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Let $ABCD$ be a convex quadrilateral with positive integer side lengths, $\angle{A} = \angle{B} = 120^{\circ}, |AD - BC| = 42,$ and $CD = 98$. Find the maximum possible value of $AB$. | 1. **Extend lines \(AD\) and \(BC\) to form an equilateral triangle:**
- Let \(P\) be the point where the extensions of \(AD\) and \(BC\) intersect.
- Since \(\angle A = \angle B = 120^\circ\), \(\triangle APD\) and \(\triangle BPC\) are equilateral triangles.
2. **Assign variables:**
- Let \(s = PC\).
- S... | 69 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let $x$ be a real number such that $$4^{2x}+2^{-x}+1=(129+8\sqrt2)(4^{x}+2^{-x}-2^{x}).$$ Find $10x$. | 1. Let \( u = 2^x \). Then the given equation becomes:
\[
4^{2x} + 2^{-x} + 1 = (129 + 8\sqrt{2})(4^x + 2^{-x} - 2^x)
\]
Since \( 4^{2x} = (2^x)^4 = u^4 \) and \( 4^x = (2^x)^2 = u^2 \), we can rewrite the equation as:
\[
u^4 + \frac{1}{u} + 1 = (129 + 8\sqrt{2})\left(u^2 + \frac{1}{u} - u\right)
\... | 35 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
The area of the triangle whose altitudes have lengths $36.4$, $39$, and $42$ can be written as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 1. Let the sides of the triangle be \(a\), \(b\), and \(c\). Given the altitudes corresponding to these sides are \(36.4\), \(39\), and \(42\) respectively, we can write the relationships:
\[
42a = 39b = 36.4c
\]
To simplify, we multiply each term by 10 to eliminate the decimal:
\[
420a = 390b = 364c
... | 3553 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
The product $$\left(\frac{1}{2^3-1}+\frac12\right)\left(\frac{1}{3^3-1}+\frac12\right)\left(\frac{1}{4^3-1}+\frac12\right)\cdots\left(\frac{1}{100^3-1}+\frac12\right)$$ can be written as $\frac{r}{s2^t}$ where $r$, $s$, and $t$ are positive integers and $r$ and $s$ are odd and relatively prime. Find $r+s+t$. | 1. **Simplify the given product:**
\[
\left(\frac{1}{2^3-1}+\frac{1}{2}\right)\left(\frac{1}{3^3-1}+\frac{1}{2}\right)\left(\frac{1}{4^3-1}+\frac{1}{2}\right)\cdots\left(\frac{1}{100^3-1}+\frac{1}{2}\right)
\]
2. **Rewrite each term:**
\[
\frac{1}{k^3-1} + \frac{1}{2} = \frac{1}{k^3-1} + \frac{1}{2} = \... | 3769 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
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