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The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$. What is the missing number in the top row?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | Let the value in the empty box in the middle row be $x$, and the value in the empty box in the top row be $y$. $y$ is the answer we're looking for.
From the diagram, $600 = 30x$, making $x = 20$.
It follows that $20 = 5y$, so $y = \boxed{\textbf{(C)}\ 4}$. | 4 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Harry and Terry are each told to calculate $8-(2+5)$. Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$. If Harry's answer is $H$ and Terry's answer is $T$, what is $H-T$?
$\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10$ | We have $H=8-7=1$ and $T=8-2+5=11$. Clearly $1-11=-10$
, so our answer is $\boxed{\textbf{(A)}-10}$. | -10 | Algebra | 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 |
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 |
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 |
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 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 |
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 |
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 |
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 |
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 |
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 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$ | Draw the following four lines as shown:
We see these lines split the figure into five squares with side length $\sqrt2$. Thus, the area is $5\cdot\left(\sqrt2\right)^2=5\cdot 2 = \boxed{\textbf{(A) } 10}$.
~pog ~wamofan | 10 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many positive integers can fill the blank in the sentence below?
“One positive integer is _____ more than twice another, and the sum of the two numbers is $28$.”
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$ | Let $m$ and $n$ be positive integers such that $m>n$ and $m+n=28.$ It follows that $m=2n+d$ for some positive integer $d.$ We wish to find the number of possible values for $d.$
By substitution, we have $(2n+d)+n=28,$ from which $d=28-3n.$ Note that $n=1,2,3,\ldots,9$ each generate a positive integer for $d,$ so there ... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Mr. Ramos gave a test to his class of $20$ students. The dot plot below shows the distribution of test scores.
Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students $5$ extra points, which increased the median test score to $85$. What is... | We set up our cases as solution 1 showed, realizing that only the second case is possible.
We notice that $13$ students have scores under $85$ currently and only $5$ have scores over $85$. We find the median of these two numbers, getting:
\[13-5=8\]
\[\frac{8}{2}=4\]
\[13-4=9\]
Thus, we realize that $4$ students must ... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$?
$\text... | The sum of the numbers in each row is $12$. Consider the second row. In order for the sum of the numbers in this row to equal $12$, the two shaded numbers must add up to $13$:
If two numbers add up to $13$, one of them must be at least $7$: If both shaded numbers are no more than $6$, their sum can be at most $12$. Th... | 8 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the sam... | Let $x$ be the number of shots that Candace made in the first half, and let $y$ be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have $x+y=10+15=25.$ In addition, we have the following inequalities: \[\frac{x... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
When three positive integers $a$, $b$, and $c$ are multiplied together, their product is $100$. Suppose $a < b < c$. In how many ways can the numbers be chosen?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$ | The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\]
It is clear that $10\leq c\leq50,$ so we apply casework to $c:$
If $c=10,$ then $(a,b,c)=(2,5,10).$
If $c=20,$ then $(a,b,c)=(1,5,20).$
If $c=25,$ then $(a,b,c)=(1,4,25).$
If $c=50,$ then $(a,b,c)=(1,2,50).$
Together, the numbers $a,b,$ and $c$ can be cho... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned $6$ years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is $30$ years. How many years older than Bella is Anna?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qqu... | Five years ago, Bella was $6$ years old, and the kitten was $0$ years old.
Today, Bella is $11$ years old, and the kitten is $5$ years old. It follows that Anna is $30-11-5=14$ years old.
Therefore, Anna is $14-11=\boxed{\textbf{(C) } 3}$ years older than Bella.
~MRENTHUSIASM | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | Let the smallest number be $x.$ It follows that the largest number is $4x.$
Since $x,15,$ and $4x$ are equally spaced on a number line, we have
\begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{\textbf{(C) } 6}. \end{align*}
~MRENTHUSIASM | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Alina writes the numbers $1, 2, \dots , 9$ on separate cards, one number per card. She wishes to divide the cards into $3$ groups of $3$ cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(... | First, we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. $1 + 2 \cdots + 9 = \frac{9(10)}{2} = 45$. Then, dividing by $3$, we have $\frac{45}{3} = 15$, so each group of $3$ must have a sum of 15. To make the counting easier, we will just see the poss... | 2 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that
\[1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.\]
What is the sum of digits of $a_{14}?$
$\textbf{(A)}\ 8 \qqu... | We can find the possible values of the common difference by finding the numbers which satisfy the conditions. To do this, we find the minimum of the last two: $241-20=221$, and the maximum–$250-13=237$. There is a difference of $13$ between them, so only $17$ and $18$ work, as $17\cdot13=221$, so $17$ satisfies $221\le... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The numbers from $1$ to $49$ are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number $7.$ How many of these four numbers are prime?
$\tex... | We fill out the grid, as shown below:
From the four numbers that appear in the shaded squares, $\boxed{\textbf{(D)}\ 3}$ of them are prime: $19,23,$ and $47.$
~MathFun1000, MRENTHUSIASM | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The digits $2, 0, 2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?
$\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18$ | First, let us consider the case where $0$ is a base: This would result in the entire expression being $0.$ Contrastingly, if $0$ is an exponent, we will get a value greater than $0.$ $3^2\times2^0=9$ is greater than $2^3\times2^0=8$ and $2^2\times3^0=4.$ Therefore, the answer is $\boxed{\textbf{(C) }9}.$ | 9 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Malaika is skiing on a mountain. The graph below shows her elevation, in meters, above the base of the mountain as she skis along a trail. In total, how many seconds does she spend at an elevation between $4$ and $7$ meters?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad... | We mark the time intervals in which Malaika's elevation is between $4$ and $7$ meters in red, as shown below:
The requested time intervals are:
from the $2$nd to the $4$th seconds
from the $6$th to the $10$th seconds
from the $12$th to the $14$th seconds
In total, Malaika spends $(4-2) + (10-6) + (14-12) = \boxed{\te... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The coordinates of $\triangle ABC$ are $A(5,7)$, $B(11,7)$, and $C(3,y)$, with $y>7$. The area of $\triangle ABC$ is 12. What is the value of $y$?
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$ | The triangle has base $6,$ which means its height satisfies
\[\dfrac{6h}{2}=3h=12.\]
This means that $h=4,$ so the answer is $7+4=\boxed{(D) 11}$ | 11 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?
(For example, one sequence of hops is up-up-down-down-up-down.)
[2024-AMC8-q13.png](https://artofproblemsolving.com/wiki/index.php/File:2024-A... | Looking at the answer choices, you see that you can list them out.
Doing this gets you:
$UUDDUD$
$UDUDUD$
$UUUDDD$
$UDUUDD$
$UUDUDD$
Counting all the paths listed above gets you $\boxed{\textbf{(B)} \ 5}$.
~ALWAYSRIGHT11
~vockey(minor edits) | 5 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Minh enters the numbers $1$ through $81$ into the cells of a $9 \times 9$ grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by $3$?
$\textbf{(A) } 8\qquad\textbf{(B) } 9\qquad\textbf{(C) } 10\qquad\textb... | [asy] unitsize(0.2cm); draw((9,18)--(-9,18)); draw((9,16)--(-9,16)); draw((9,14)--(-9,14)); draw((9,12)--(-9,12)); draw((9,10)--(-9,10)); draw((9,8)--(-9,8)); draw((9,6)--(-9,6)); draw((9,4)--(-9,4)); draw((9,2)--(-9,2)); draw((9,0)--(-9,0)); draw((9,18)--(9,0)); draw((7,18)--(7,0)); draw((5,18)--(5,0)); draw((3,18)--... | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Any three vertices of the cube $PQRSTUVW$, shown in the figure below, can be connected to form a triangle. (For example, vertices $P$, $Q$, and $R$ can be connected to form isosceles $\triangle PQR$.) How many of these triangles are equilateral and contain $P$ as a vertex?
$\textbf{(A)}0 \qquad \textbf{(B) }1 \qquad \... | The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have $3$ possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are $3$ possible triangles. So the answer is $\boxed{\textbf{(D) }3}$
~Math645
~andliu766
~e___ | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high while the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides form a $45^\circ$ angle with the ground. The artwork has an area of $183$ square f... | Extend the "inner part" of the mountain so that the image is two right triangles that overlap in a third right triangle as shown.
The side length of the largest right triangle is $12\sqrt{2},$ which means its area is $144.$ Similarly, the area of the second largest right triangle is $64$ (the side length is $8\sqrt{2... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
When Yunji added all the integers from $1$ to $9$, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out? | The sum of digits 1-9 is 45, 45-9=36 36=6^2 Therefore, the solution is that the integer she forgot was 9. | 9 | Number Theory | math-word-problem | Yes | Yes | amc_aime | false |
Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of $6$. Which of the following integers cannot be the sum of the two numbers?
$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$ | First, figure out all pairs of numbers whose product is 6. Then, using the process of elimination, we can find the following:
$\textbf{(A)}$ is possible: $2\times 3$
$\textbf{(C)}$ is possible: $1\times 6$
$\textbf{(D)}$ is possible: $2\times 6$
$\textbf{(E)}$ is possible: $3\times 6$
The only integer that cannot be t... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
On Monday Taye has $2. Every day, he either gains $3 or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, 3 days later?
$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$ | How many values could be on the first day? Only $2$ dollars. The second day, you can either add $3$ dollars, or double, so you can have $5$ dollars, or $4$. For each of these values, you have $2$ values for each. For $5$ dollars, you have $10$ dollars or $8$, and for $4$ dollars, you have $8$ dollars or $$7$. Now, you ... | 6 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of ... | We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$.
It is easy to verify that player A can create a 5 in a row.
Let A place two counters anywhere, and B take off one of them. Then, A should create a "triangle of hexagons" by placing two adjacent counters also next to the unremoved one... | 6 | Combinatorics | proof | Yes | Yes | amc_aime | false |
Determine all the [roots](https://artofproblemsolving.com/wiki/index.php/Root), [real](https://artofproblemsolving.com/wiki/index.php/Real) or [complex](https://artofproblemsolving.com/wiki/index.php/Complex), of the system of simultaneous [equations](https://artofproblemsolving.com/wiki/index.php/Equation)
$x+y+z=3$,... | Let $x$, $y$, and $z$ be the [roots](https://artofproblemsolving.com/wiki/index.php/Root) of the [cubic polynomial](https://artofproblemsolving.com/wiki/index.php/Cubic_polynomial) $t^3+at^2+bt+c$. Let $S_1=x+y+z=3$, $S_2=x^2+y^2+z^2=3$, and $S_3=x^3+y^3+z^3=3$. From this, $S_1+a=0$, $S_2+aS_1+2b=0$, and $S_3+aS_2+bS_1... | 1 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|\] ends in the digit $9$. | Notice that $|a_i - b_i| \equiv 1 \pmod 5$, so $S=|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \equiv 1+1+\cdots + 1 \equiv 999 \equiv 4 \bmod{5}$.
Also, for integers $M, N$ we have $|M-N| \equiv M-N \equiv M+N \bmod{2}$.
Thus, we also have $S \equiv a_1+b_1+a_2+b_2+\cdots +a_{999}+b_{999} \equiv 1+2+ \cdots +1998 \eq... | 9 | Number Theory | proof | Yes | Yes | amc_aime | false |
Let $a, b, c$ be positive real numbers such that: $$ab - c = 3$$ $$abc = 18$$ Calculate the numerical value of $\frac{ab}{c}$ | 1. We start with the given equations:
\[
ab - c = 3
\]
\[
abc = 18
\]
2. From the first equation, we can express \( ab \) in terms of \( c \):
\[
ab = 3 + c
\]
3. Substitute \( ab = 3 + c \) into the second equation:
\[
abc = 18
\]
\[
c(3 + c) = 18
\]
4. Simplify the equa... | 2 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
If $x, y, z \in \mathbb{R}$ are solutions to the system of equations
$$\begin{cases}
x - y + z - 1 = 0\\
xy + 2z^2 - 6z + 1 = 0\\
\end{cases}$$
what is the greatest value of $(x - 1)^2 + (y + 1)^2$? | 1. **Rewrite the given system of equations:**
\[
\begin{cases}
x - y + z - 1 = 0 \\
xy + 2z^2 - 6z + 1 = 0
\end{cases}
\]
2. **Express \( y \) in terms of \( x \) and \( z \) from the first equation:**
\[
x - y + z - 1 = 0 \implies y = x + z - 1
\]
3. **Substitute \( y = x + z - 1 \) into t... | 11 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
A circle has radius $52$ and center $O$. Points $A$ is on the circle, and point $P$ on $\overline{OA}$ satisfies $OP = 28$. Point $Q$ is constructed such that $QA = QP = 15$, and point $B$ is constructed on the circle so that $Q$ is on $\overline{OB}$. Find $QB$.
[i]Proposed by Justin Hsieh[/i] | 1. **Identify the given information and draw the diagram:**
- Circle with center \( O \) and radius \( 52 \).
- Point \( A \) is on the circle, so \( OA = 52 \).
- Point \( P \) is on \( \overline{OA} \) such that \( OP = 28 \).
- Point \( Q \) is constructed such that \( QA = QP = 15 \).
- Point \( B \)... | 11 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
In triangle $ABC$, let $I, O, H$ be the incenter, circumcenter and orthocenter, respectively. Suppose that $AI = 11$ and $AO = AH = 13$. Find $OH$.
[i]Proposed by Kevin You[/i] | 1. Given that \( AH = AO \), we start by using the known relationship between the orthocenter \( H \), circumcenter \( O \), and the circumradius \( R \) of a triangle. Specifically, we have:
\[
AH = 2R \cos A
\]
Since \( AH = AO = R \), we can set up the equation:
\[
2R \cos A = R
\]
Dividing b... | 10 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Starting with a $5 \times 5$ grid, choose a $4 \times 4$ square in it. Then, choose a $3 \times 3$ square in the $4 \times 4$ square, and a $2 \times 2$ square in the $3 \times 3$ square, and a $1 \times 1$ square in the $2 \times 2$ square. Assuming all squares chosen are made of unit squares inside the grid. In how m... | 1. **Choosing the $4 \times 4$ square:**
- The $5 \times 5$ grid has $2$ possible positions for the $4 \times 4$ square horizontally and $2$ possible positions vertically. Therefore, there are $2 \times 2 = 4$ ways to choose the $4 \times 4$ square.
2. **Choosing the $3 \times 3$ square within the $4 \times 4$ squa... | 1 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Let $ABCD$ be a square. Let $E, Z$ be points on the sides $AB, CD$ of the square respectively, such that $DE\parallel BZ$. Assume that the triangles $\triangle EAD, \triangle ZCB$ and the parallelogram $BEDZ$ have the same area.
If the distance between the parallel lines $DE$ and $BZ$ is equal to $1$, determine the a... | 1. **Identify the given conditions and relationships:**
- $ABCD$ is a square.
- $E$ and $Z$ are points on sides $AB$ and $CD$ respectively.
- $DE \parallel BZ$.
- The areas of $\triangle EAD$, $\triangle ZCB$, and parallelogram $BEDZ$ are equal.
- The distance between the parallel lines $DE$ and $BZ$ is ... | 4 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
An up-right path from lattice points $P$ and $Q$ on the $xy$-plane is a path in which every move is either one unit right or one unit up. The probability that a randomly chosen up-right path from $(0,0)$ to $(10,3)$ does not intersect the graph of $y=x^2+0.5$ can be written as $\tfrac mn$, where $m$ and $n$ are relativ... | To solve this problem, we need to calculate the probability that a randomly chosen up-right path from \((0,0)\) to \((10,3)\) does not intersect the graph of \(y = x^2 + 0.5\).
1. **Calculate the total number of up-right paths from \((0,0)\) to \((10,3)\):**
An up-right path from \((0,0)\) to \((10,3)\) consists o... | 2 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Let $P(x) = x^2-1$ be a polynomial, and let $a$ be a positive real number satisfying$$P(P(P(a))) = 99.$$ The value of $a^2$ can be written as $m+\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.
[i]Proposed by [b]HrishiP[/b][/i] | 1. First, we start with the polynomial \( P(x) = x^2 - 1 \). We need to find \( a \) such that \( P(P(P(a))) = 99 \).
2. Calculate \( P(P(a)) \):
\[
P(a) = a^2 - 1
\]
\[
P(P(a)) = P(a^2 - 1) = (a^2 - 1)^2 - 1 = a^4 - 2a^2 + 1 - 1 = a^4 - 2a^2
\]
3. Calculate \( P(P(P(a))) \):
\[
P(P(P(a))) = P... | 12 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Charlotte is playing the hit new web number game, Primle. In this game, the objective is to guess a two-digit positive prime integer between $10$ and $99$, called the [i]Primle[/i]. For each guess, a digit is highlighted blue if it is in the [i]Primle[/i], but not in the correct place. A digit is highlighted orange if ... | 1. We are given that Charlotte guesses the numbers $13$ and $47$ and receives the following feedback:
\[
\begin{array}{c}
\boxed{1} \,\, \boxed{3} [\smallskipamount]
\boxed{4}\,\, \fcolorbox{black}{blue}{\color{white}7}
\end{array}
\]
This feedback indicates that:
- The digits $1$, $3$... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
How many ways are there to fill in a $2\times 2$ square grid with the numbers $1,2,3,$ and $4$ such that the numbers in any two grid squares that share an edge have an absolute difference of at most $2$?
[i]Proposed by Andrew Wu[/i] | 1. **Identify the constraints**: We need to fill a $2 \times 2$ grid with the numbers $1, 2, 3,$ and $4$ such that the absolute difference between any two adjacent numbers is at most $2$.
2. **Analyze the placement of $1$ and $4$**: Since $1$ and $4$ have an absolute difference of $3$, they cannot be adjacent. Therefo... | 8 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Carissa is crossing a very, very, very wide street, and did not properly check both ways before doing so. (Don't be like Carissa!) She initially begins walking at $2$ feet per second. Suddenly, she hears a car approaching, and begins running, eventually making it safely to the other side, half a minute after she began ... | 1. Let \( a \) be the number of seconds Carissa spent walking. According to the problem, she spent \( n \) times as much time running as she did walking. Therefore, the time she spent running is \( an \) seconds.
2. The total time she spent crossing the street is given as 30 seconds. Thus, we have the equation:
\[
... | 10 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
[b]p1[/b] How many two-digit positive integers with distinct digits satisfy the conditions that
1) neither digit is $0$, and
2) the units digit is a multiple of the tens digit?
[b]p2[/b] Mirabel has $47$ candies to pass out to a class with $n$ students, where $10\le n < 20$. After distributing the candy as evenly as ... | To solve the problem, we need to find the smallest integer \( k \) such that when Mirabel distributes \( 47 - k \) candies to \( n \) students (where \( 10 \leq n < 20 \)), the candies can be distributed evenly.
1. **Identify the total number of candies after removing \( k \) candies:**
\[
47 - k
\]
We nee... | 2 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Let $n>4$ be a positive integer, which is divisible by $4$. We denote by $A_n$ the sum of the odd positive divisors of $n$. We also denote $B_n$ the sum of the even positive divisors of $n$, excluding the number $n$ itself. Find the least possible value of the expression $$f(n)=B_n-2A_n,$$
for all possible values of $n... | 1. **Define the problem and variables:**
Let \( n > 4 \) be a positive integer divisible by \( 4 \). We denote by \( A_n \) the sum of the odd positive divisors of \( n \) and by \( B_n \) the sum of the even positive divisors of \( n \), excluding \( n \) itself. We aim to find the least possible value of the expre... | 4 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, $(x_4, y_4)$, and $(x_5, y_5)$ be the vertices of a regular pentagon centered at $(0, 0)$. Compute the product of all positive integers k such that the equality $x_1^k+x_2^k+x_3^k+x_4^k+x_5^k=y_1^k+y_2^k+y_3^k+y_4^k+y_5^k$ must hold for all possible choices of the pentagon. | 1. **Vertices on the Unit Circle:**
Since the vertices of the pentagon lie on the unit circle centered at \((0,0)\), we can represent them using complex numbers. Let the vertices be \(z_1, z_2, z_3, z_4, z_5\) where \(z_i = \exp\left(\frac{2\pi i (i-1)}{5}\right)\) for \(i = 1, 2, 3, 4, 5\). Here, \(\exp\left(\frac{... | 5 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Suppose $P(x)$ is a monic polynomial of degree $2023$ such that $P(k) = k^{2023}P(1-\frac{1}{k})$ for every positive integer $1 \leq k \leq 2023$. Then $P(-1) = \frac{a}{b}$ where $a$ and $b$ are relatively prime integers. Compute the unique integer $0 \leq n < 2027$ such that $bn-a$ is divisible by the prime $2027$. | Given the monic polynomial \( P(x) \) of degree 2023 such that \( P(k) = k^{2023} P\left(1 - \frac{1}{k}\right) \) for every positive integer \( 1 \leq k \leq 2023 \), we need to find \( P(-1) \) in the form \( \frac{a}{b} \) where \( a \) and \( b \) are relatively prime integers. Then, we need to compute the unique i... | 0 | Algebra | math-word-problem | Yes | Yes | aops_forum | false |
Sets $A, B$, and $C$ satisfy $|A| = 92$, $|B| = 35$, $|C| = 63$, $|A\cap B| = 16$, $|A\cap C| = 51$, $|B\cap C| = 19$. Compute the number of possible values of$ |A \cap B \cap C|$. | 1. Let \( |A \cap B \cap C| = x \). We need to find the possible values of \( x \).
2. Using the principle of inclusion-exclusion for three sets, we have:
\[
|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|
\]
Substituting the given values:
\[
|A \cup B \... | 10 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Consider the set $\mathcal{T}$ of all triangles whose sides are distinct prime numbers which are also in arithmetic progression. Let $\triangle \in \mathcal{T}$ be the triangle with least perimeter. If $a^{\circ}$ is the largest angle of $\triangle$ and $L$ is its perimeter, determine the value of $\frac{a}{L}$. | 1. **Identify the triangle with the least perimeter:**
- We need to find a triangle with sides that are distinct prime numbers in arithmetic progression.
- The smallest such set of primes is \(3, 5, 7\).
- Therefore, the triangle with sides \(3, 5, 7\) has the least perimeter.
2. **Calculate the perimeter \(L... | 8 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Let $P_0 = (3,1)$ and define $P_{n+1} = (x_n, y_n)$ for $n \ge 0$ by $$x_{n+1} = - \frac{3x_n - y_n}{2}, y_{n+1} = - \frac{x_n + y_n}{2}$$Find the area of the quadrilateral formed by the points $P_{96}, P_{97}, P_{98}, P_{99}$. | 1. **Define the sequence and transformation matrix:**
Let \( P_0 = (3,1) \) and define \( P_{n+1} = (x_n, y_n) \) for \( n \ge 0 \) by:
\[
x_{n+1} = - \frac{3x_n - y_n}{2}, \quad y_{n+1} = - \frac{x_n + y_n}{2}
\]
We can represent this transformation using a matrix \( M \):
\[
\vec{v}_n = \begin{pm... | 8 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Suppose $a,b,c,x,y,z$ are pairwisely different real numbers. How many terms in the following can be $1$ at most:
$$\begin{aligned}
&ax+by+cz,&&&&ax+bz+cy,&&&&ay+bx+cz,\\
&ay+bz+cx,&&&&az+bx+cy,&&&&az+by+cx?
\end{aligned}$$ | 1. **Claim and Example:**
We claim that the maximum number of terms that can be equal to 1 is 2. To illustrate this, consider the example where \(a = 1\), \(b = 2\), \(c = 3\), \(x = 4\), \(y = 27\), and \(z = -19\). In this case:
\[
ax + by + cz = 1 \quad \text{and} \quad ay + bz + cx = 1
\]
Therefore, ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
A finite set $M$ of real numbers has the following properties: $M$ has at least $4$ elements, and there exists a bijective function $f:M\to M$, different from the identity, such that $ab\leq f(a)f(b)$ for all $a\neq b\in M.$ Prove that the sum of the elements of $M$ is $0.$ | 1. Let \( M = \{a_1, a_2, \ldots, a_n\} \) be the set of real numbers, where \( n \geq 4 \). There exists a bijective function \( f: M \to M \) such that \( ab \leq f(a)f(b) \) for all \( a \neq b \in M \).
2. Since \( f \) is bijective, we can denote \( f(a_i) = b_i \) for \( i = 1, 2, \ldots, n \). Thus, \( M = \{b_... | 0 | Logic and Puzzles | proof | Yes | Yes | aops_forum | false |
For every positive integer $N\geq 2$ with prime factorisation $N=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ we define \[f(N):=1+p_1a_1+p_2a_2+\cdots+p_ka_k.\] Let $x_0\geq 2$ be a positive integer. We define the sequence $x_{n+1}=f(x_n)$ for all $n\geq 0.$ Prove that this sequence is eventually periodic and determine its fund... | 1. **Define the function \( f(N) \):**
Given a positive integer \( N \geq 2 \) with prime factorization \( N = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \), the function \( f(N) \) is defined as:
\[
f(N) = 1 + p_1 a_1 + p_2 a_2 + \cdots + p_k a_k
\]
2. **Behavior of \( f(N) \) for composite numbers:**
For a ... | 2 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $a$ be a non-negative real number and a sequence $(u_n)$ defined as:
$u_1=6,u_{n+1} = \frac{2n+a}{n} + \sqrt{\frac{n+a}{n}u_n+4}, \forall n \ge 1$
a) With $a=0$, prove that there exist a finite limit of $(u_n)$ and find that limit
b) With $a \ge 0$, prove that there exist a finite limit of $(u_n)$ | ### Part (a)
Given \( a = 0 \), we need to prove that there exists a finite limit of \( (u_n) \) and find that limit.
1. **Initial Condition and Recurrence Relation:**
\[
u_1 = 6, \quad u_{n+1} = \frac{2n}{n} + \sqrt{\frac{n}{n}u_n + 4} = 2 + \sqrt{u_n + 4}
\]
2. **Establishing a Lower Bound:**
\[
u_n ... | 5 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Consider an integrable function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $af(a)+bf(b)=0$ when $ab=1$. Find the value of the following integration:
$$ \int_{0}^{\infty} f(x) \,dx $$ | 1. Given the condition \( af(a) + bf(b) = 0 \) when \( ab = 1 \), we can rewrite this as:
\[
af(a) + \frac{1}{a} f\left(\frac{1}{a}\right) = 0
\]
This implies:
\[
a f(a) = -\frac{1}{a} f\left(\frac{1}{a}\right)
\]
Multiplying both sides by \( a \), we get:
\[
a^2 f(a) = -f\left(\frac{1}{a}... | 0 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Determine the least positive integer $n{}$ for which the following statement is true: the product of any $n{}$ odd consecutive positive integers is divisible by $45$. | To determine the least positive integer \( n \) for which the product of any \( n \) odd consecutive positive integers is divisible by \( 45 \), we need to ensure that the product is divisible by both \( 3^2 = 9 \) and \( 5 \).
1. **Divisibility by \( 3^2 = 9 \):**
- For any set of \( n \) consecutive odd integers... | 6 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
3. We have a $n \times n$ board. We color the unit square $(i,j)$ black if $i=j$, red if $i<j$ and green if $i>j$. Let $a_{i,j}$ be the color of the unit square $(i,j)$. In each move we switch two rows and write down the $n$-tuple $(a_{1,1},a_{2,2},\cdots,a_{n,n})$. How many $n$-tuples can we obtain by repeating this p... | 1. **Understanding the Problem:**
We have an \( n \times n \) board where each unit square \((i,j)\) is colored based on the following rules:
- Black if \( i = j \)
- Red if \( i < j \)
- Green if \( i > j \)
We are interested in the \( n \)-tuple \((a_{1,1}, a_{2,2}, \ldots, a_{n,n})\) obtained by swit... | 0 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
A regular polygon with $20$ vertices is given. Alice colors each vertex in one of two colors. Bob then draws a diagonal connecting two opposite vertices. Now Bob draws perpendicular segments to this diagonal, each segment having vertices of the same color as endpoints. He gets a fish from Alice for each such segment he... | 1. **Initial Setup and Assumptions:**
- We have a regular polygon with 20 vertices.
- Alice colors each vertex in one of two colors.
- Bob draws a diagonal connecting two opposite vertices and then draws perpendicular segments to this diagonal, each segment having vertices of the same color as endpoints.
- ... | 4 | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Suppose $\{a_1,a_2,\ldots,a_7\}$ is a set of pair-wisely different positive integers. If $a_1,2a_2,\ldots,7a_7$ can form an arithmetic series (in this order), find the smallest positive value of $|a_7-a_1|$. | Given that $\{a_1, a_2, \ldots, a_7\}$ is a set of pairwise different positive integers, and $a_1, 2a_2, \ldots, 7a_7$ can form an arithmetic series, we need to find the smallest positive value of $|a_7 - a_1|$.
1. **Define the Arithmetic Series:**
Let the common difference of the arithmetic series be $d$. Then the... | 1 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Determine the smallest natural number $n$ for which there exist distinct nonzero naturals $a, b, c$, such that $n=a+b+c$ and $(a + b)(b + c)(c + a)$ is a perfect cube. | To determine the smallest natural number \( n \) for which there exist distinct nonzero naturals \( a, b, c \) such that \( n = a + b + c \) and \((a + b)(b + c)(c + a)\) is a perfect cube, we can proceed as follows:
1. **Assume \( \gcd(a, b, c, n) = d \)**:
Let \( a = dx \), \( b = dy \), \( c = dz \), and \( n = ... | 10 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Find all positive integers $b$ with the following property: there exists positive integers $a,k,l$ such that $a^k + b^l$ and $a^l + b^k$ are divisible by $b^{k+l}$ where $k \neq l$. | 1. We need to find all positive integers \( b \) such that there exist positive integers \( a, k, l \) with \( k \neq l \) and both \( a^k + b^l \) and \( a^l + b^k \) are divisible by \( b^{k+l} \).
2. Consider the \( p \)-adic valuation \( \nu_p \) for a prime \( p \). Recall that for any integers \( x \) and \( y \... | 1 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Given the sequence $(u_n)$ satisfying:$$\left\{ \begin{array}{l}
1 \le {u_1} \le 3\\
{u_{n + 1}} = 4 - \dfrac{{2({u_n} + 1)}}{{{2^{{u_n}}}}},\forall n \in \mathbb{Z^+}.
\end{array} \right.$$
Prove that: $1\le u_n\le 3,\forall n\in \mathbb{Z^+}$ and find the limit of $(u_n).$ | 1. **Define the functions and initial conditions:**
Given the sequence \((u_n)\) defined by:
\[
\begin{cases}
1 \le u_1 \le 3 \\
u_{n+1} = 4 - \frac{2(u_n + 1)}{2^{u_n}}, \forall n \in \mathbb{Z^+}
\end{cases}
\]
We need to prove that \(1 \le u_n \le 3\) for all \(n \in \mathbb{Z^+}\) and find t... | 3 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Determine the largest natural number $k$ such that there exists a natural number $n$ satisfying:
\[
\sin(n + 1) < \sin(n + 2) < \sin(n + 3) < \ldots < \sin(n + k).
\] | 1. We start with the inequality given in the problem:
\[
\sin(n + 1) < \sin(n + 2) < \sin(n + 3) < \ldots < \sin(n + k).
\]
2. To analyze this, we need to understand the behavior of the sine function. The sine function is periodic with period \(2\pi\) and oscillates between -1 and 1. The function \(\sin(x)... | 3 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Determine the smallest possible value of the expression $$\frac{ab+1}{a+b}+\frac{bc+1}{b+c}+\frac{ca+1}{c+a}$$ where $a,b,c \in \mathbb{R}$ satisfy $a+b+c = -1$ and $abc \leqslant -3$
| To determine the smallest possible value of the expression
\[ S = \frac{ab+1}{a+b} + \frac{bc+1}{b+c} + \frac{ca+1}{c+a} \]
where \(a, b, c \in \mathbb{R}\) satisfy \(a+b+c = -1\) and \(abc \leq -3\), we proceed as follows:
1. **Substitute \(1\) with \((a+b+c)^2\):**
Since \(a+b+c = -1\), we have:
\[ 1 = (a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | aops_forum | false |
There are $10$ cups, each having $10$ pebbles in them. Two players $A$ and $B$ play a game, repeating the following in order each move:
$\bullet$ $B$ takes one pebble from each cup and redistributes them as $A$ wishes.
$\bullet$ After $B$ distributes the pebbles, he tells how many pebbles are in each cup to $A$. The... | 1. **Initial Setup:**
- There are 10 cups, each containing 10 pebbles.
- Players A and B take turns according to the rules specified.
2. **Strategy of Player A:**
- Player A instructs Player B to distribute the pebbles such that the cup with the most pebbles gets $\left\lfloor \frac{k}{2} \right\rfloor$ pebbl... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
Set $M$ contains $n \ge 2$ positive integers. It's known that for any two different $a, b \in M$, $a^2+1$ is divisible by $b$. What is the largest possible value of $n$?
[i]Proposed by Oleksiy Masalitin[/i] | 1. **Assume $M$ contains $n$ positive integers $a_1, a_2, \ldots, a_n$.**
We are given that for any two different $a, b \in M$, $a^2 + 1$ is divisible by $b$. This implies that for any $a_i, a_j \in M$ with $i \neq j$, $a_i^2 + 1$ is divisible by $a_j$.
2. **Consider the smallest element $a_1$ in $M$.**
Withou... | 2 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $ABC$ be an equilateral triangle with side length $1$. Points $A_1$ and $A_2$ are chosen on side $BC$, points $B_1$ and $B_2$ are chosen on side $CA$, and points $C_1$ and $C_2$ are chosen on side $AB$ such that $BA_1<BA_2$, $CB_1<CB_2$, and $AC_1<AC_2$.
Suppose that the three line segments $B_1C_2$, $C_1A_2$, $A_... | 1. **Setup and Initial Conditions:**
- Let $ABC$ be an equilateral triangle with side length $1$.
- Points $A_1$ and $A_2$ are chosen on side $BC$ such that $BA_1 < BA_2$.
- Points $B_1$ and $B_2$ are chosen on side $CA$ such that $CB_1 < CB_2$.
- Points $C_1$ and $C_2$ are chosen on side $AB$ such that $AC... | 1 | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
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