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The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$. In what year was Samantha born?
$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$ | The seventh AMC 8 would have been given in $1991$. If Samantha was 12 then, that means she was born 12 years ago, so she was born in $1991-12=1979$.
Our answer is $\boxed{(\text{A})1979}$
corrections made by DrDominic | 1979 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Jack wants to bike from his house to Jill's house, which is located three blocks east and two blocks north of Jack's house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill'... | We can apply complementary counting and count the paths that DO go through the blocked intersection, which is $\dbinom{2}{1}\dbinom{3}{1}=6$. There are a total of $\dbinom{5}{2}=10$ paths, so there are $10-6=4$ paths possible. $\boxed{(\text{A})4}$ is the correct answer. | 4 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$?
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$ | The area of $\bigtriangleup CDE$ is $\frac{DC\cdot CE}{2}$. The area of $ABCD$ is $AB\cdot AD=5\cdot 6=30$, which also must be equal to the area of $\bigtriangleup CDE$, which, since $DC=5$, must in turn equal $\frac{5\cdot CE}{2}$. Through transitivity, then, $\frac{5\cdot CE}{2}=30$, and $CE=12$. Then, using the Pyth... | 13 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The circumference of the circle with center $O$ is divided into $12$ equal arcs, marked the letters $A$ through $L$ as seen below. What is the number of degrees in the sum of the angles $x$ and $y$?
$\textbf{(A) }75\qquad\textbf{(B) }80\qquad\textbf{(C) }90\qquad\textbf{(D) }120\qquad\textbf{(E) }150$ | For this problem, it is useful to know that the measure of an inscribed angle is half the measure of its corresponding central angle. Since each unit arc is $\frac{1}{12}$ of the circle's circumference, each unit central angle measures $\left( \frac{360}{12} \right) ^{\circ}=30^{\circ}$. Then, we know that the central ... | 90 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?
$\textbf{(A) }60\q... | Within the conference, there are 8 teams, so there are $\dbinom{8}{2}=28$ pairings of teams, and each pair must play two games, for a total of $28\cdot 2=56$ games within the conference.
Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of $4\cdot 8 =32$ games outside the ... | 88 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must Geor... | Note that on a normal day, it takes him $1/3$ hour to get to school. However, today it took $\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4$ hour to walk the first $1/2$ mile. That means that he has $1/3 -1/4 = 1/12$ hours left to get to school, and $1/2$ mile left to go. Therefore, his speed must be $\frac{1/2 \text{ mile... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Paul owes Paula $35$ cents and has a pocket full of $5$-cent coins, $10$-cent coins, and $25$-cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textb... | The fewest amount of coins that can be used is $2$ (a quarter and a dime). The greatest amount is $7$, if he only uses nickels. Therefore we have $7-2=\boxed{\textbf{(E)}~5}$. | 5 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Rectangle $ABCD$ has sides $CD=3$ and $DA=5$. A circle of radius $1$ is centered at $A$, a circle of radius $2$ is centered at $B$, and a circle of radius $3$ is centered at $C$. Which of the following is closest to the area of the region inside the rectangle but outside all three circles?
$\text{(A) }3.5\qquad\text{(... | The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.
The area of the rectangle is $3\cdot5 =15$. The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$. T... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Three members of the Euclid Middle School girls' softball team had the following conversation.
Ashley: I just realized that our uniform numbers are all $2$-digit primes.
Bethany : And the sum of your two uniform numbers is the date of my birthday earlier this month.
Caitlin: That's funny. The sum of your two uniform nu... | The maximum amount of days any given month can have is $31$, and the smallest two-digit primes are $11, 13,$ and $17$. There are a few different sums that can be deduced from the following numbers, which are $24, 30,$ and $28$, all of which represent the three days. Therefore, since Bethany says that the other two peop... | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Isabella had a week to read a book for a school assignment. She read an average of $36$ pages per day for the first three days and an average of $44$ pages per day for the next three days. She then finished the book by reading $10$ pages on the last day. How many pages were in the book?
$\textbf{(A) }240\qquad\textbf{... | Isabella read $3\cdot 36+3\cdot 44$ pages in the first 6 days. Although this can be calculated directly, it is simpler to calculate it as $3\cdot (36+44)=3\cdot 80$, which gives that she read $240$ pages. On her last day, she read $10$ more pages for a total of $240+10=\boxed{\textbf{(B)}~250}$ pages. | 250 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The sum of two prime numbers is $85$. What is the product of these two prime numbers?
$\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$ | Since the two prime numbers sum to an odd number, one of them must be even. The only even prime number is $2$. The other prime number is $85-2=83$, and the product of these two numbers is $83\cdot2=\boxed{\textbf{(E)}~166}$. | 166 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Margie's car can go $32$ miles on a gallon of gas, and gas currently costs $$4$ per gallon. How many miles can Margie drive on $\textdollar 20$ worth of gas?
$\textbf{(A) }64\qquad\textbf{(B) }128\qquad\textbf{(C) }160\qquad\textbf{(D) }320\qquad \textbf{(E) }640$ | Margie can afford $20/4=5$ gallons of gas. She can go $32\cdot5=\boxed{\textbf{(C)}~160}$ miles on this amount of gas. | 160 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Six rectangles each with a common base width of $2$ have lengths of $1, 4, 9, 16, 25$, and $36$. What is the sum of the areas of the six rectangles?
$\textbf{(A) }91\qquad\textbf{(B) }93\qquad\textbf{(C) }162\qquad\textbf{(D) }182\qquad \textbf{(E) }202$ | The sum of the areas is equal to $2\cdot1+2\cdot4+2\cdot9+2\cdot16+2\cdot25+2\cdot36$. This is equal to $2(1+4+9+16+25+36)$, which is equal to $2\cdot91$. This is equal to our final answer of $\boxed{\textbf{(D)}~182}$. | 182 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$. What is the missing digit $A$ of this $3$-digit number?
$\textbf{(A) }0\qquad\textbf... | Since all the eleven members paid the same amount, that means that the total must be divisible by $11$. We can do some trial-and-error to get $A=3$, so our answer is $\boxed{\textbf{(D)}~3}$
~SparklyFlowers | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In $\bigtriangleup ABC$, $D$ is a point on side $\overline{AC}$ such that $BD=DC$ and $\angle BCD$ measures $70^\circ$. What is the degree measure of $\angle ADB$?
$\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150$ | Using angle chasing is a good way to solve this problem. $BD = DC$, so $\angle DBC = \angle DCB = 70$. Then $\angle CDB = 180-(70+70) = 40$. Since $\angle ADB$ and $\angle BDC$ are supplementary, $\angle ADB = 180 - 40 = \boxed{\textbf{(D)}~140}$. | 140 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Onkon wants to cover his room's floor with his favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)
$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }... | First, we multiply $12\cdot9$. To get that, we need $108$ square feet of carpet to cover the room's floor. Since there are $9$ square feet in a square yard, you divide $108$ by $9$ to get $12$ square yards, so our answer is $\bold{\boxed{\textbf{(A)}~12}}$. | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many integers between $1000$ and $9999$ have four distinct digits?
$\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561$ | There are $9$ choices for the first number, since it cannot be $0$, there are only $9$ choices left for the second number since it must differ from the first, $8$ choices for the third number, since it must differ from the first two, and $7$ choices for the fourth number, since it must differ from all three. This mean... | 4536 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG}$, does a cube have?
$\text{(A) }6 \quad\text{(B) }12 \quad\text{(C) } 18 \quad\text{(D) } 24 \quad \text{(E) } 36$ | Solution 1
We first count the number of pairs of parallel lines that are in the same direction as $\overline{AB}$. The pairs of parallel lines are $\overline{AB}\text{ and }\overline{EF}$, $\overline{CD}\text{ and }\overline{GH}$, $\overline{AB}\text{ and }\overline{CD}$, $\overline{EF}\text{ and }\overline{GH}$, $\ov... | 18 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is 6?
$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$ | Solution 1
Since there will be $9$ elements after removal, and their mean is $6$, we know their sum is $54$. We also know that the sum of the set pre-removal is $66$. Thus, the sum of the $2$ elements removed is $66-54=12$. There are only $\boxed{\textbf{(D)}~5}$ subsets of $2$ elements that sum to $12$: $\{1,11\}, \{2... | 5 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$. Each row and each column in this $5\times5$ array is an arith... | Solution 1
We begin filling in the table. The top row has a first term $1$ and a fifth term $25$, so we have the common difference is $\frac{25-1}4=6$. This means we can fill in the first row of the table:
The fifth row has a first term of $17$ and a fifth term of $81$, so the common difference is $\frac{81-17}4=16$... | 31 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Ralph went to the store and bought 12 pairs of socks for a total of $$24$. Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) }... | Solution 1
So, let there be $x$ pairs of $$1$ socks, $y$ pairs of $$3$ socks, and $z$ pairs of $$4$ socks.
We have $x+y+z=12$, $x+3y+4z=24$, and $x,y,z \ge 1$.
Now, we subtract to find $2y+3z=12$, and $y,z \ge 1$.
It follows that $2y$ is a multiple of $3$ and $3z$ is a multiple of $3$. Since sum of 2 multiples of 3 = m... | 7 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$. What is the area of $\triangle KBC$?
$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$. | Clearly, since $\overline{FE}$ is a side of a square with area $32$, $\overline{FE} = \sqrt{32} = 4 \sqrt{2}$. Now, since $\overline{FE} = \overline{BC}$, we have $\overline{BC} = 4 \sqrt{2}$.
Now, $\overline{JB}$ is a side of a square with area $18$, so $\overline{JB} = \sqrt{18} = 3 \sqrt{2}$. Since $\Delta JBK$ is e... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process conti... | The text suggests that the number of students in the group has $12$ factors, since each arrangement is a factor. The smallest integer with $12$ factors is $2^2\cdot3\cdot5=\boxed{\textbf{(C) }60}$. | 60 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$. Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \te... | On one team they play $3N$ games in their division and $4M$ games in the other. This gives $3N+4M=76$.
Since $M>4$ we start by trying M=5. This doesn't work because $56$ is not divisible by $3$.
Next, $M=6$ does not work because $52$ is not divisible by $3$.
We try $M=7$ does work by giving $N=16$ ,$~M=7$ and thus $3\... | 48 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?
$\textbf{(A) }5\qq... | Using $d=rt$, we can set up an equation for when Jill arrives at the swimming pool:
$1=10t$
Solving for $t$, we get that Jill gets to the pool in $\frac{1}{10}$ of an hour, which is $6$ minutes. Doing the same for Jack, we get that
Jack arrives at the pool in $\frac{1}{4}$ of an hour, which in turn is $15$ minutes. ... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The Blue Bird High School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?
$\textbf{(A) }2\qquad\t... | There are $2! = 2$ ways to order the boys on the ends, and there are $3!=6$ ways to order the girls in the middle. We get the answer to be $2 \cdot 6 = \boxed{\textbf{(E) }12}$. | 12 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
In $\bigtriangleup ABC$, $AB=BC=29$, and $AC=42$. What is the area of $\bigtriangleup ABC$?
$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$ | Solution 1
We know the semi-perimeter of $\triangle ABC$ is $\frac{29+29+42}{2}=50$. Next, we use Heron's Formula to find that the area of the triangle is just $\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\boxed{\textbf{(B) }420}$.
Solution 2 (easier)
Splitting the isosceles triangle in half, we get a right... | 420 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the smallest whole number larger than the perimeter of any triangle with a side of length $5$ and a side of length $19$?
$\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57$ | We know from the [Triangle Inequality](https://artofproblemsolving.com/wiki/index.php/Triangle_Inequality) that the last side, $s$, fulfills $s<5+19=24$. Adding $5+19$ to both sides of the inequality, we get $s+5+19<48$, and because $s+5+19$ is the perimeter of our triangle, $\boxed{\textbf{(D)}\ 48}$ is our answer. | 48 | Geometry | MCQ | Yes | Yes | amc_aime | false |
On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }... | Solution 1
First, we have to find how many widgets she makes on Day $20$. We can write the linear equation $y=-1+2x$ to represent this situation. Then, we can plug in $20$ for $x$: $y=-1+2(20)$ -- $y=-1+40$ -- $y=39$. The sum of $1,3,5, ... 39$ is $\dfrac{(1 + 39)(20)}{2}= \boxed{\textbf{(D)}~400}$.
Solution 2
The su... | 400 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$ | Solution 1
It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is $60$ minutes in a hour. Therefore, there are $11 \cdot 60 = 660$ minutes in 11 hours. Adding the second part(the 5 minutes) we get $660 + 5 = \boxed{\textbf{(C)}\ 665}$.
Solution 2
The... | 665 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if
\[2 * (5 * x)=1\]
$\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$ | Let us plug in $(5 * x)=1$ into $3a-b$. Thus it would be $3(5)-x$. Now we have $2*(15-x)=1$. Plugging $2*(15-x)$ into $3a-b$, we have $6-15+x=1$. Solving for $x$ we have \[-9+x=1\]\[x=\boxed{\textbf{(D)} \, 10}\] | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$ | Solution 1 - kindlymath55532
We can write the two digit number in the form of $10a+b$; reverse of $10a+b$ is $10b+a$. The sum of those numbers is:
\[(10a+b)+(10b+a)=132\]\[11a+11b=132\]\[a+b=12\]
We can use brute force to find order pairs $(a,b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ h... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Karl's car uses a gallon of gas every $35$ miles, and his gas tank holds $14$ gallons when it is full. One day, Karl started with a full tank of gas, drove $350$ miles, bought $8$ gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that da... | Since he uses a gallon of gas every $35$ miles, he had used $\frac{350}{35} = 10$ gallons after $350$ miles. Therefore, after the first leg of his trip he had $14 - 10 = 4$ gallons of gas left. Then, he bought $8$ gallons of gas, which brought him up to $12$ gallons of gas in his gas tank. When he arrived, he had $\fra... | 525 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Annie and Bonnie are running laps around a $400$-meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(... | Solution 1
Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run $\boxed{\textbf{(D)}\ 5 }$ laps.
Solution 2
Call $x$ the distance... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
An ATM password at Fred's Bank is composed of four digits from $0$ to $9$, with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?
$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\tex... | Solution 1
For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{\textbf{(D)}\ 9990}$.
~CHECKMATE2021
Solution 2
Counting the prohibited cases, we find that there are 10 of them. This... | 9990 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
In an All-Area track meet, $216$ sprinters enter a $100-$meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion spri... | Solution 1
From any $n-$th race, only $\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.
Starting with the first race:
\[\frac{216}{6}=36\]
\[\frac{36}{6}=6\]
\[\frac{6}{6}=1\]
Adding all of the numbers in the second column yields $\boxed... | 43 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
The sum of $25$ consecutive even integers is $10,000$. What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | Let $n$ be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to $25n$ since $(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n$. Now, $25n=10000 \rightarrow n=400$. Remembering that this is the 13th integer, we wish to find the... | 424 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In rectangle $ABCD$, $AB=6$ and $AD=8$. Point $M$ is the midpoint of $\overline{AD}$. What is the area of $\triangle AMC$?
$\text{(A) }12\qquad\text{(B) }15\qquad\text{(C) }18\qquad\text{(D) }20\qquad \text{(E) }24$ | Solution 1
Using the triangle area formula for triangles: $A = \frac{bh}{2},$ where $A$ is the area, $b$ is the base, and $h$ is the height. This equation gives us $A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{\textbf{(A) } 12}$.
Solution 2
A triangle with the same height and base as a rectangle is half of the recta... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?
$\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180$ | We wish to find possible values of $a$, $b$, and $c$. By finding the greatest common factor of $12$ and $15$, we can find that $b$ is 3. Moving on to $a$ and $c$, in order to minimize them, we wish to find the least such that the least common multiple of $a$ and $3$ is $12$, $\rightarrow 4$. Similarly, with $3$ and $c$... | 20 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$. The circles intersect at two points, one of which is $E$. What is the degree measure of $\angle CED$?
$\textbf{(A) }90\qquad\... | Solution 1
Observe that $\triangle{EAB}$ is equilateral. Therefore, $m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}$. Since $CD$ is a straight line, we conclude that $m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}$. Since $BE=BD$ (both are radii of the same circle), $\triangle{BED}$ is isosceles, meaning that $... | 120 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The digits $1$, $2$, $3$, $4$, and $5$ are each used once to write a five-digit number $PQRST$. The three-digit number $PQR$ is divisible by $4$, the three-digit number $QRS$ is divisible by $5$, and the three-digit number $RST$ is divisible by $3$. What is $P$?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\q... | Solution 1 (Modular Arithmetic)
We see that since $QRS$ is divisible by $5$, $S$ must equal either $0$ or $5$, but it cannot equal $0$, so $S=5$. We notice that since $PQR$ must be even, $R$ must be either $2$ or $4$. However, when $R=2$, we see that $T \equiv 2 \pmod{3}$, which cannot happen because $2$ and $5$ are al... | 1 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Four students take an exam. Three of their scores are $70, 80,$ and $90$. If the average of their four scores is $70$, then what is the remaining score?
$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$ | Solution 1
Let $r$ be the remaining student's score. We know that the average, 70, is equal to $\frac{70 + 80 + 90 + r}{4}$. We can use basic algebra to solve for $r$: \[\frac{70 + 80 + 90 + r}{4} = 70\] \[\frac{240 + r}{4} = 70\] \[240 + r = 280\] \[r = 40\] giving us the answer of $\boxed{\textbf{(A)}\ 40}$.
Solut... | 40 | Algebra | MCQ | Yes | Yes | amc_aime | false |
When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes. As an old man, he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?
$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textb... | When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes $= 3\times60 + 30$ minutes $= 210$ minutes, thus running $\frac{210}{15} = 14$ minutes per mile. Now that he is an old man, he can walk $10$ miles in $4$ hours $= 4 \times 60$ minutes $= 240$ minutes, thus walking $\frac{240}{10} = 24$ minutes... | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The number $N$ is a two-digit number.
• When $N$ is divided by $9$, the remainder is $1$.
• When $N$ is divided by $10$, the remainder is $3$.
What is the remainder when $N$ is divided by $11$?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$ | From the second bullet point, we know that the second digit must be $3$, for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$, the multiple of $9$ must end in a $2$ for it to have the desired remainder$\pmod {10}.$ We now look for this one:
$9(1)=9\\ 9(2)=18\\ 9(3)=... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$ | Solution 1
We first notice that the median name will be the $(19+1)/2=10^{\mbox{th}}$ name. The $10^{\mbox{th}}$ name is $\boxed{\textbf{(B)}\ 4}$.
Solution 2
To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us $7 + 3 + 1 + 4 + 4 = 19$. Thus the index of the medi... | 4 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Find the value of the expression
\[100-98+96-94+92-90+\cdots+8-6+4-2.\]$\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$ | Solution 1
We can group each subtracting pair together:
\[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).\]
After subtracting, we have:
\[2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).\]
There are $50$ even numbers, therefore there are $\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{\textbf{(C) }50}$
Solu... | 50 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the sum of the distinct prime integer divisors of $2016$?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$ | Solution 1
The prime factorization is $2016=2^5\times3^2\times7$. Since the problem is only asking us for the distinct prime factors, we have $2,3,7$. Their desired sum is then $\boxed{\textbf{(B) }12}$.
Solution 2
We notice that $9 \mid 2016$, since $2+0+1+6 = 9$, and $9 \mid 9$. We can divide $2016$ by $9$ to get ... | 12 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$ | Since the number of tiles lying on both diagonals is $37$, counting one tile twice, there are $37=2x-1\implies x=19$ tiles on each side. Therefore, our answer is $19^2=361=\boxed{\textbf{(C)}\ 361}$.
~AllezW | 361 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
$\textbf{(A) }8\qquad... | Notice that the upper-most section contains a 3 by 3 square that looks like:
It has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply ${6 \cdot 4}$ to get $\boxed{\textbf{(D)}\ 24}$ total paths. | 24 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty. So instead I put $6$ gold coins in each treasure chest, but then I had $3$ gold coins left over. How many gold coins did I have?
$\textbf{(A) }9\qquad\textbf{(B)... | We can represent the amount of gold with $g$ and the amount of chests with $c$. We can use the problem to make the following equations:
\[9c-18 = g\]
\[6c+3 = g\]
We do this because for 9 chests there are 2 empty and if 9 were in each 9 multiplied by 2 is 18 left.
Therefore, $6c+3 = 9c-18.$ This implies that $c = 7.$ W... | 45 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Factoring out $98!+99!+100!$, we have $98! (1+99+99*100)$, which is $98! (10000)$. Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$. The $19$ is because of all the multiples of $5$.The $3$ is because of all the multiples of $25$. Now, $10,000$... | 26 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?
$\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qqu... | Let $x$ be the total amount of votes casted. From the chart, Brenda received $30\%$ of the votes and had $36$ votes. We can express this relationship as $\frac{30}{100}x=36$. Solving for $x$, we get $x=\boxed{\textbf{(E)}\ 120}.$ | 120 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Suppose $a$, $b$, and $c$ are nonzero real numbers, and $a+b+c=0$. What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$?
$\text{(A) }0\qquad\text{(B) }1\text{ and }-1\qquad\text{(C) }2\text{ and }-2\qquad\text{(D) }0,2,\text{ and }-2\qquad\text{(E) }0,1,\text{ and }-1$ | There are $2$ cases to consider:
Case $1$: $2$ of $a$, $b$, and $c$ are positive and the other is negative. Without loss of generality (WLOG), we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.\]
Case $2$: ... | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance trav... | It is well known that $\text{Distance}=\text{Speed} \cdot \text{Time}$. In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour. By the equation derived from $\text{Distance}=\text{Speed} \cdot \text{Time}$, we have $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$, so ... | 25 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }7... | We use Principle of Inclusion-Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is \[\left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor.\]
To this result we add the number... | 146 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the value of the expression $\sqrt{16\sqrt{8\sqrt{4}}}$?
$\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16$ | $\sqrt{16\sqrt{8\sqrt{4}}}$ = $\sqrt{16\sqrt{8\cdot 2}}$ = $\sqrt{16\sqrt{16}}$ = $\sqrt{16\cdot 4}$ = $\sqrt{64}$ = $\boxed{\textbf{(C)}\ 8}$. | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
When $0.000315$ is multiplied by $7,928,564$ the product is closest to which of the following?
$\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000$ | We can approximate $7,928,564$ to $8,000,000$ and $0.000315$ to $0.0003.$ Multiplying the two yields $2400.$ Thus, it shows our answer is $\boxed{\textbf{(D)}\ 2400}.$ | 2400 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$?
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$ | Directly calculating:
We evaluate both the top and bottom: $\frac{40320}{36}$. This simplifies to $\boxed{\textbf{(B)}\ 1120}$. | 1120 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If the degree measures of the angles of a triangle are in the ratio $3:3:4$, what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$ | The sum of the ratios is $10$. Since the sum of the angles of a triangle is $180^{\circ}$, the ratio can be scaled up to $54:54:72$ $(3\cdot 18:3\cdot 18:4\cdot 18).$ The numbers in the ratio $54:54:72$ represent the angles of the triangle. The question asks for the largest, so the answer is $\boxed{\textbf{(D) }72}... | 72 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$?
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$ | To check, if a number is divisible by 19, take its unit digit and multiply it by 2, then add the result to the rest of the number, and repeat this step until the number is reduced to two digits. If the result is divisible by 19, then the original number is also divisible by 19. Or we could just try to divide the exampl... | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."
(1) It is prime.
(2) It is even.
(3) It is divisible by 7.
(4) One of its digi... | Notice that (1) cannot be true. Otherwise, the number would have to be prime and be either even or divisible by 7. This only happens if the number is 2 or 7, neither of which are two-digit numbers, so we run into a contradiction. Thus, we must have (2), (3), and (4) be true. By (2), the $2$-digit number is even, and th... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$ | The $6$ green marbles and yellow marbles form $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles. Now, suppose the total number of marbles is $x$. We know the number of yellow marbles is $\frac{5}{12}x - 6$ and a positive integer. Therefore, $12$ must divide $x$. Trying the smallest multiples of $12$ f... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
An amusement park has a collection of scale models, with a ratio of $1: 20$, of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\q... | You can see that since the ratio of real building's heights to the model building's height is $1:20$. We also know that the U.S Capitol is $289$ feet in real life, so to find the height of the model, we divide by 20. That gives us $14.45$ which rounds to 14. Therefore, to the nearest whole number, the duplicate is $\bo... | 14 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for La... | Say Laila gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=410.$
The value $y$ has to be greater than $82$, because otherwise she would receive the same score on her last test. Additionally, the greatest value for $y$ is $98$ (as $y=100$ would make $x$ as a decimal), so therefo... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $N$ be the greatest five-digit number whose digits have a product of $120$. What is the sum of the digits of $N$?
$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$ | If we start off with the first digit, we know that it can't be $9$ since $9$ is not a factor of $120$. We go down to the digit $8$, which does work since it is a factor of $120$. Now, we have to know what digits will take up the remaining four spots. To find this result, just divide $\frac{120}{8}=15$. The next place c... | 18 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760... | Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $5$ factorial. Now, we multiply this product by $2!$ because there are $... | 5760 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \t... | Every 10 feet Bella goes, Ella goes 50 feet, which means a total of 60 feet. They need to travel that 60 feet $10560\div60=176$ times to travel the entire 2 miles. Since Bella goes 10 feet 176 times, this means that she travels a total of 1760 feet. And since she walks 2.5 feet each step, $1760\div2.5=\boxed{\textbf{(A... | 704 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many positive factors does $23,232$ have?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | We can first find the prime factorization of $23,232$, which is $2^6\cdot3^1\cdot11^2$. Now, we add one to our powers and multiply. Therefore, the answer is $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{\textbf{(E) }42}$
Note:
23232 is a large number, so we can look for shortcuts to factor it.
One way to factor it q... | 42 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the ... | You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:
+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is $\boxed{\textbf{(C) } 8}$.
-NinjaBoi2000 | 8 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
What is the value of the product
\[\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?\]
$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(... | By adding up the numbers in each of the $6$ parentheses, we get:
$\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}$.
Using telescoping, most of the terms cancel out diagonally. We are left with $\frac{7}{1}$ which is equivalent to $7$. Thus, the answer would be $\bo... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$ | Let the area of $\triangle CEF$ be $x$. Thus, the area of triangle $\triangle ACD$ is $45+x$ and the area of the square is $2(45+x) = 90+2x$.
By AA similarity, $\triangle CEF \sim \triangle ABF$ with a 1:2 ratio, so the area of triangle $\triangle ABF$ is $4x$. Now, consider trapezoid $ABED$. Its area is $45+4x$, which... | 108 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$ | We compute $2^8+1=257$. We're all familiar with what $6^3$ is, namely $216$, which is too small. The smallest cube greater than it is $7^3=343$. $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^6)^3=64^3$, which therefore will clearly be the largest cube less than $2^{18}+1$. So, the required number ... | 58 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$?
$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$ | Rearranging the terms, we get $(1-2)+(3-4)+(5-6)+...(2017-2018)+2019$, and our answer is $-1009+2019=\boxed{\textbf{(E) }1010}$.
If you were stuck on this problem, refer to AOPS arithmetic lessons.
~Nivaar | 1010 | Algebra | MCQ | Yes | Yes | amc_aime | false |
On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
$\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }... | Since Anh spends half an hour to drive 10 miles on the coastal road, his speed is $r=\frac dt=\frac{10}{0.5}=20$ mph. His speed on the highway then is $60$ mph. He drives $50$ miles, so he drives for $\frac{5}{6}$ hours, which is equal to $50$ minutes (Note that $60$ miles per hour is the same as $1$ mile per minute). ... | 80 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The $5$-digit number $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9$. What is the remainder when this number is divided by $8$?
$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | We use the property that the digits of a number must sum to a multiple of $9$ if it are divisible by $9$. This means $2+0+1+8+U$ must be divisible by $9$. The only possible value for $U$ then must be $7$. Since we are looking for the remainder when divided by $8$, we can ignore the thousands. The remainder when $187$ i... | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,
and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how
many items will they buy?
$\textb... | We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$. Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign l... | Let $x$ be the number of students taking both a math and a foreign language class.
By P-I-E, we get $70 + 54 - x$ = $93$.
Solving gives us $x = 31$.
But we want the number of students taking only a math class,
which is $70 - 31 = 39$.
$\boxed{\textbf{(D)}\ 39}$
~phoenixfire | 39 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$?
$\textbf{(A) }... | Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \ldots, 99$. Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be writ... | 2 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$ | The only option that is easily divisible by $55$ is $110$, which gives 2 hours of travel. And, the formula is $\frac{15}{30} + \frac{110}{55} = \frac{5}{2}$.
And, $\text{Average Speed}$ = $\frac{\text{Total Distance}}{\text{Total Time}}$.
Thus, $\frac{125}{50} = \frac{5}{2}$.
Both are equal and thus our answer is $\box... | 110 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for ... | This isn't finished
to another. This gives equality, as each team wins once and loses once as well. For a win, we have $3$ points, so a team gets $3\times2=6$ points if they each win a game and lose a game. This case brings a total of $18+6=24$ points.
Therefore, we use Case 2 since it brings the greater amount of po... | 24 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Three identical rectangles are put together to form rectangle $ABCD$, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$?
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }... | Solution 1
We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$, so the bigger side is $10$, if we do $5 \cdot 2 = 10$. Now we get the sides of the big rectangle being $15$ and $10$, so the area is $\boxed{\textb... | 150 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$ | We have that $(x^2-5)^2 = 16$ if and only if $x^2-5 = \pm 4$. If $x^2-5 = 4$, then $x^2 = 9 \implies x = \pm 3$, giving 2 solutions. If $x^2-5 = -4$, then $x^2 = 1 \implies x = \pm 1$, giving 2 more solutions. All four of these solutions work, so the answer is $\boxed{\textbf{(D) }4}$. Further, the equation is a [quart... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$?
$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$ | First, we need to find the coordinates where the graphs intersect.
We want the points x and y to be the same. Thus, we set $5=x+1,$ and get $x=4.$ Plugging this into the equation, $y=1-x,$
$y=5$, and $y=1+x$ intersect at $(4,5)$, we call this line x.
Doing the same thing, we get $x=-4.$ Thus, $y=5$. Also,
$y=5$ and $y... | 16 | Geometry | MCQ | Yes | Yes | amc_aime | false |
A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased$?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }3... | Suppose the fraction of discount is $x$. That means $(1-x)(1+x)=0.84$; so, $1-x^{2}=0.84$, and $(x^{2})=0.16$, obtaining $x=0.4$. Therefore, the price was increased and decreased by $40$%, or $\boxed{\textbf{(E)}\ 40}$. | 40 | Algebra | MCQ | Yes | Yes | amc_aime | false |
After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $... | Given the information above, we start with the equation $\frac{t}{4}+\frac{2t}{7} + 15 + x = t$,where $t$ is the total number of points scored and $x\le 14$ is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation $x+15 = \frac{13}{28}t$, or $28x+28\cdot 15=13t$... | 11 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In triangle $\triangle ABC$, point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$. Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$. Given that the area of $\triangle ABC$ is $360$, what is the area of $\triangle EBF$?
$\textbf... | We use the line-segment ratios to infer area ratios and height ratios.
Areas:
$AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120$.
$BE:BD = 1:2 \text{ (midpoint)} \implies [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60$.
Heights:
Let $h_A$ = height (of altitude) from $\overline{BC}$ to $A$.
$AD:DC ... | 30 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?
$\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$ | Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).
This method uses the counting method of stars and bars (non-negative version). Since each person must have at least $2$ apples, we can ... | 190 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$?
$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$ | A rhombus has sides of equal length. Because the perimeter of the rhombus is $52$, each side is $\frac{52}{4}=13$. In a rhombus, diagonals are perpendicular and bisect each other, which means $\overline{AE}$ = $12$ = $\overline{EC}$.
Consider one of the right triangles:
$\overline{AB}$ = $13$, and $\overline{AE}$ = $1... | 120 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ , $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) ... | We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] We can now cross multiply to get rid of the denominator. \[257... | 48 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
$\textbf{(A) }20\q... | After Gilda gives $20$% of the marbles to Pedro, she has $80$% of the marbles left. If she then gives $10$% of what's left to Ebony, she has $(0.8*0.9)$ = $72$% of what she had at the beginning. Finally, she gives $25$% of what's left to her brother, so she has $(0.75*0.72)$ $\boxed{\textbf{(E)}\ 54}$ of what she had i... | 54 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E... | We have $\text{water} : \text{sugar} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1,$ so Luka needs $3 \cdot 8 = \boxed{\textbf{(E) }24}$ cups. | 24 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }1... | Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$, respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each o... | 12 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?
$\textbf{(A) }6 \qquad \textbf{(B) }12 \... | Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since $\text{speed} = \frac{\text{distance}}{\text{time}}$, her speed is $\frac{6}{\left(\frac{1}{6}\right)} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph, so the answer is $36-12 = \boxed{\textbf{(E) }2... | 24 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For a positive integer $n$, the factorial notation $n!$ represents the product of the integers from $n$ to $1$. What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$ | We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$, and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$. Therefore, the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$, and so $12 \cdot 10! = 12 \cdot N!$. Cancelling the $12$s, it is clear that $N=\boxed{\textbf{(A) }10}$. | 10 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$ | Since $20\% = \frac{1}{5}$, multiplying the given condition by $5$ shows that $y$ is $15 \cdot 5 = \boxed{\textbf{(C) }75}$ percent of $x$. | 75 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $... | We can form the following expressions for the sum along each line:
\[\begin{dcases}A+B+C\\A+E+F\\C+D+E\\B+D\\B+F\end{dcases}\]
Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$, i.e. $2(A+B+C+D+E+F)+B=47$. Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$, we obtain $A+B+C+D+E+F=1+2+3+4+5+6=21$ (wher... | 5 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$)
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$ | Since $2020 = 2^2 \cdot 5 \cdot 101$, we can simply list its factors: \[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.\] There are $12$ factors; only $1, 2, 4, 5, 101$ don't have over $3$ factors, so the remaining $12-5 = \boxed{\textbf{(B) }7}$ factors have more than $3$ factors. | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$
$\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$ | Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$, so $OC = 17$. By symmetry, $O$ is the midpoint of $DA$, so $OD=OA=\frac{16}{2}= 8$. By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$), we have that $CD$ (or $AB$) is $\sqrt{17^2-8^2}=15$. Accordingly, the area of ... | 240 | Geometry | MCQ | Yes | Yes | amc_aime | false |
A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | A number is divisible by $15$ precisely if it is divisible by $3$ and $5$. The latter means the last digit must be either $5$ or $0$, and the former means the sum of the digits must be divisible by $3$. If the last digit is $0$, the first digit would be $0$ (because the digits alternate), which is not possible. Hence t... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Four friends do yardwork for their neighbors over the weekend, earning $$15, $20, $25,$ and $$40,$ respectively. They decide to split their earnings equally among themselves. In total, how much will the friend who earned $$40$ give to the others?
$\textbf{(A) }$5 \qquad \textbf{(B) }$10 \qquad \textbf{(C) }$15 \qquad \... | The friends earn $$\left(15+20+25+40\right)=$100$ in total. Since they decided to split their earnings equally, it follows that each person will get $$\left(\frac{100}{4}\right)=$25$. Since the friend who earned $$40$ will need to leave with $$25$, he will have to give $$\left(40-25\right)=\boxed{\textbf{(C) }$15}$ to ... | 15 | Algebra | MCQ | Yes | Yes | amc_aime | false |
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