Datasets:
id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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1 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_1 | 1 | What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$ | We can rewrite the expression as \[222,222-(22,222+2,222+222+22+2).\]
We note that the units digit of the addition is $0$ because all the units digits of the five numbers are $2$ and $5*2=10$ , which has a units digit of $0$
Now, we have something with a units digit of $0$ subtracted from $222,222$ . The units digit of... | 2 |
2 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_1 | 2 | What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$ | 222,222-22,222 = 200,000
200,000 - 2,222 = 197778
197778 - 222 = 197556
197556 - 22 = 197534
197534 - 2 = 1957532
So our answer is $\boxed{2}$ | 2 |
3 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_1 | 3 | What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$ | We only care about the unit's digits.
Thus, $2-2$ ends in $0$ $0-2$ ends in $8$ $8-2$ ends in $6$ $6-2$ ends in $4$ , and $4-2$ ends in $\boxed{2}$ | 2 |
4 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_1 | 4 | What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$ | We just take the units digit of each and subtract, or you can do it this way by adding an extra ten to the first number (so we don't get a negative number): \[12-2-(2+2+2+2)=10-8=2\] Thus, we get the answer $\boxed{2}$ | 2 |
5 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_3 | 1 | Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units? [asy] size(150); filldraw((0,0)... | We work inwards. The area of the outer shaded square is the area of the whole square minus the area of the second largest square. The area of the inner shaded region is the area of the third largest square minus the area of the smallest square. The sum of these areas is \[10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{52}\] | 52 |
6 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_3 | 2 | Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units? [asy] size(150); filldraw((0,0)... | We can calculate it as the sum of the areas of $2$ smaller trapezoids and $2$ larger trapezoids. \[2\left(\cfrac{(7+4)(7-4)}{2}\right)+2\left(\cfrac{(10+9)(10-9)}{2}\right)=10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{52}\] | 52 |
7 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_4 | 1 | 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?
$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$ | The sum of the digits from $1-9$ are $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$ . Note that one of the answer choices (that is equal to Yunju's digits) subtracted from her sum of $45$ must equal a square.
Note that $6^2 = 36$ is a very close square to the sum of 45. Checking, we see that $45 - 9 = 36 = 6^2$ works.
Theref... | 9 |
8 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_5 | 1 | 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 th... | 6 |
9 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_7 | 1 | $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?
$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$ | We can eliminate B, C, and D, because they are not $21$ subtracted by any multiple of $4$ . Finally, we see that there is no way to have A, so the solution is $\boxed{5}$ | 5 |
10 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_7 | 2 | $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?
$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$ | Let $x$ be the number of $1x1$ tiles. There are $21$ squares and each $2x2$ or $1x4$ tile takes up 4 squares, so $x \equiv 1 \pmod{4}$ , so it is either $1$ or $5$ . Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are $12$ red squares and $9$ blue squares, but ea... | 5 |
11 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_7 | 3 | $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?
$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$ | Suppose there are $a$ different $2\times 2$ tiles, $b$ different $4\times 1$ tiles and $c$ different $1\times 1$ tiles. Since the areas of these tiles must total up to $21$ (area of the whole grid), we have \[4a + 4b + c = 21.\] Reducing modulo $4$ gives $c\equiv 1\pmod{4}$ , or $c = 1$ or $c = 5$
If $c = 1$ , then $a ... | 5 |
12 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_8 | 1 | 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, ... | 6 |
13 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_8 | 2 | 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?
$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$ | Continue as in Solution 1 to get $7$ $8$ , or $10$ dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by $2$ or adding $3$ ) from here is if $7+3=10\cdot 2$ or $7+3=8\cdot 2$ which both aren't true. Hence our answer is $3\cdot2=\boxed{6}$ | 6 |
14 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_9 | 1 | All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
$\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textb... | Since she has half as many red marbles as green, we can call the number of red marbles $x$ , and the number of green marbles $2x$ .
Since she has half as many green marbles as blue, we can call the number of blue marbles $4x$ .
Adding them up, we have $7x$ marbles. The number of marbles therefore must be a multiple of... | 28 |
15 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_9 | 2 | All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
$\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textb... | Suppose Maria has $g$ green marbles and let $t$ be the total number of marbles. She then has $\frac{g}{2}$ red marbles and $2g$ blue marbles. Altogether, Maria has \[g + \frac{g}{2} + 2g = \frac{7g}{2} = t\] marbles, implying that $g = \dfrac{2t}{7},$ so $t$ must be a multiple of $7$ . The only multiple of $7$ is $\box... | 28 |
16 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_10 | 1 | In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.
$\textbf{(A)}\... | This is a time period of $50$ years, so we can expect the ppm to increase by $50*1.515=75.75~76$ $76+338=\boxed{414}$ | 414 |
17 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_10 | 2 | In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.
$\textbf{(A)}\... | For each year that has passed, the ppm will increase by $1.515$ . In $2030$ , the CO2 would have increased by $50\cdot 1.515 \approx. 76,$ so the total ppm of CO2 will be $76 + 338 = \boxed{414}.$ | 414 |
18 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_10 | 3 | In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.
$\textbf{(A)}\... | 2030 - 1980 = 50 years.
338 + 50 * 1.515 = 338 + 75.75 = 413.75 for 2030 ppm level $=\boxed{414}$ | 414 |
19 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11 | 1 | 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 $y$
[asy] draw((3,11)--(11,7)--(5,7)--(3,11)); dot((5,7)); label("$A(5,7)$",(5,7),S); dot((11,7)); label("$B(11,7)$",(11,7),S); dot((3,11)); label("$C(3,y)$",(3,11),NW); [... | The triangle has base $6,$ which means its height satisfies \[\dfrac{6h}{2}=3h=12.\] This means that $h=4,$ so the answer is $7+4=\boxed{11}$ | 11 |
20 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11 | 2 | 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 $y$
[asy] draw((3,11)--(11,7)--(5,7)--(3,11)); dot((5,7)); label("$A(5,7)$",(5,7),S); dot((11,7)); label("$B(11,7)$",(11,7),S); dot((3,11)); label("$C(3,y)$",(3,11),NW); [... | [asy] size(10cm); draw((5,7)--(11,7)--(3,11)--cycle); draw((3,11)--(3,7)--(5,7),red); draw((3,7.5)--(3.5,7.5)--(3.5,7)); label("$A(5,7)$", (5,7),S); label("$B(11,7)$", (11,7),S); label("$C(3,y)$", (3,11),W); label("$D(3,7)$", (3,7),SW); [/asy] Label point $D(3,7)$ as the point at which $CD\perp DA$ . We now have $[\tri... | 11 |
21 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11 | 3 | 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 $y$
[asy] draw((3,11)--(11,7)--(5,7)--(3,11)); dot((5,7)); label("$A(5,7)$",(5,7),S); dot((11,7)); label("$B(11,7)$",(11,7),S); dot((3,11)); label("$C(3,y)$",(3,11),NW); [... | By the Shoelace Theorem, $\triangle ABC$ has area \[\frac{1}{2}|(y \cdot 11 + 7 \cdot 5 + 7 \cdot 3) - (3 \cdot 7 + 11 \cdot 7 + 5 \cdot y)| = \frac{1}{2}|(11y + 56) - (98 + 5y)| = \frac{1}{2}|6y - 42|\] . From the problem, this is equal to $12$ . We now solve for y.
$\frac{1}{2}|6y - 42| = 12$
$|6y-42| = 24$
$6y - 42 ... | 11 |
22 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11 | 4 | 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 $y$
[asy] draw((3,11)--(11,7)--(5,7)--(3,11)); dot((5,7)); label("$A(5,7)$",(5,7),S); dot((11,7)); label("$B(11,7)$",(11,7),S); dot((3,11)); label("$C(3,y)$",(3,11),NW); [... | As in the figure, the triangle is determined by the vectors $\begin{bmatrix}-2 \\ y-7\end{bmatrix}$ and $\begin{bmatrix}6\\0\end{bmatrix}$ . Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogr... | 11 |
23 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_12 | 1 | Rohan keeps a total of 90 guppies in 4 fish tanks.
How many guppies are in the 4th tank?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$ | Let $x$ denote the number of guppies in the first tank.
Then, we have the following for the number of guppies in the rest of the tanks:
The number of guppies in all of the tanks combined is 90, so we can write the equation
$x + x + 1 + x + 1 + 2 + x + 1 + 2 + 3 = 90$
Simplifying the equation gives
$4x + 10 = 90$
Solvin... | 26 |
24 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_12 | 2 | Rohan keeps a total of 90 guppies in 4 fish tanks.
How many guppies are in the 4th tank?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$ | Suppose there are no guppies in the first tank.
Then, the number of guppies in the other tanks are $1,3,$ and $6,$ or $10$ guppies in total.
We need to add $90 - 10 = 80$ guppies into $4$ tanks or $20$ guppies in each tank.
So the number of guppies in the fourth tank is $20 + 6 = \boxed{26}.$ | 26 |
25 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13 | 1 | 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
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\... | 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 $\boxed{5}$ | 5 |
26 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13 | 2 | 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
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\... | Any combination can be written as some re-arrangement of $UUUDDD$ . Clearly we must end going down, and start going up, so we need the number of ways to insert 2 $U$ 's and 2 $D$ 's into $U\, \_ \, \_ \, \_ \, \_ \, D$ . There are ${4\choose 2}=6$ ways, but we have to remove the case $UDDUUD$ , giving us $\boxed{5}$ | 5 |
27 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13 | 4 | 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
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\... | First step is U, last step is D.
After third step we can get only positions 3 or 1.
In the first case there is only one way UUUDDD.
In the second case we have two way to get this position UDU and UUD.
Similarly, we have two way return to position 0 (UDD and DUD).
Therefore, we have $1 + 2 \cdot 2 = \boxed{5}$ | 5 |
28 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13 | 5 | 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
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\... | We can find the total cases then deduct the ones that don't work.
Let $U$ represent "Up" and $D$ represent "Down". We know that in order to land back at the bottom of the stairs, we must have an equal number of $U$ 's and $D$ 's, therefore six hops means $3$ of each.
The number of ways to arrange $3$ $U$ 's and $3$ $D$... | 5 |
29 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14 | 1 | The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
2024-AMC8-q14.png
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \t... | We can simply see that path $A \rightarrow X \rightarrow M \rightarrow Y \rightarrow C \rightarrow Z$ will give us the smallest value. Adding, $5+2+6+5+10 = \boxed{28}$ . This is nice as it’s also the smallest value, solidifying our answer. | 28 |
30 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14 | 2 | The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
2024-AMC8-q14.png
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \t... | We can execute Dijkstra's algorithm by hand to find the shortest path from $A$ to every other town, including $Z$ . This effectively proves that, assuming we execute the algorithm correctly, that we will have found the shortest distance. The distance estimates for each step of the algorithm (from $A$ to each node) are ... | 28 |
31 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14 | 3 | The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
2024-AMC8-q14.png
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \t... | From $A$ , we want to find the shortest route to $Z$ , so we want to try to find the shortest path through each node (not necessarily all of them). We should follow the arrows, since all of them lead to $Z$ . From $A$ , there are $2$ paths we can take, to $M$ $(8)$ , or to $X$ $(5)$ . We travel to $X$ , since $5 < 8$ .... | 28 |
32 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14 | 4 | The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
2024-AMC8-q14.png
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \t... | We can cross off a few routes:
Finally, we are left with a single path AXMYCZ from A to Z which adding it up gives $28 = \boxed{28}.$ | 28 |
33 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_15 | 1 | Let the letters $F$ $L$ $Y$ $B$ $U$ $G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation
\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\u... | The highest that $FLYFLY$ can be would have to be $124124$ , and it cannot be higher than that because then it would exceed the $6$ -digit limit set on $BUGBUG$
So, if we start at $124124\cdot8$ , we get $992992$ , which would be wrong because both $B \& U$ would be $9$ , and the numbers cannot be repeated between diff... | 107 |
34 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_15 | 2 | Let the letters $F$ $L$ $Y$ $B$ $U$ $G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation
\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\u... | Notice that $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y} = 1000(\underline{F}~\underline{L}~\underline{Y}) + \underline{F}~\underline{L}~\underline{Y}$
Likewise, $\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G} = 1000(\underline{B}~\underline{U... | 107 |
35 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_15 | 3 | Let the letters $F$ $L$ $Y$ $B$ $U$ $G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation
\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\u... | Note that $FLY+BUG = 9 \cdot FLY$ . Thus, we can check the answer choices and find $FLY$ through each of the answer choices, we find the 1107 works, so the answer is $\boxed{1107}$ | 107 |
36 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_16 | 2 | 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\textbf... | Note you can swap/rotate any configuration of rows, such that all the rows and columns that have a product of 3 are in the top left. Hence the points are bounded by a $a \times b$ rectangle. This has $ab$ area and $a+b$ rows and columns divisible by $3$ . We want $ab\ge 27$ and $a+b$ minimized.
If $ab=27$ , we achieve ... | 11 |
37 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_16 | 3 | 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\textbf... | For a row or column to have a product divisible by $3$ , there must be a multiple of $3$ in the row or column. To create the least amount of rows and columns with multiples of $3$ , we must find a way to keep them all together, to minimize the total number of rows and columns. From $1$ to $81$ , there are $27$ multiple... | 11 |
38 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_16 | 4 | 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\textbf... | In the numbers $1$ to $81$ , there are 27 multiples of three. In order to minimize the rows and columns, the best way is to make a square. However, the closest square is $25$ , meaning there are two multiples of three remaining. However, you can place these multiples right above the 5x5 square, meaning the answer is $\... | 11 |
39 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18 | 1 | Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded ... | Let $x=\angle{BOC}$
We see that the shaded region is the inner ring plus a sector $x^\circ$ of the outer ring. The area of this in terms of $x$ is $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$ . This simplifies to $3 \pi + \frac{x}{360}(5 \pi)$
Also, the unshaded portion is comprised of the sm... | 108 |
40 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18 | 2 | Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded ... | Notice for the 3rd most outer ring of the circle, the ratio of the shaded region to non-shaded region is the ratio of $\angle{BOC}$ to $360-\angle{BOC}$ . With that, all we need to do is solve for the shaded region.
The inner most circle has radius $1$ , and the second circle has radius 2. Therefore, the first shaded a... | 108 |
41 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18 | 3 | Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded ... | The AMC 8s allow a ruler tool that you can rotate and drag. You can use the tool to make a straight segment (which we know is $180$ degrees), and we let the angle of desire be $x$ . We can estimate that $180-x$ is just about $30$ degrees short of $x$ itself, so $x-30=180-x$ , solving gives $x=105$ , therefore the close... | 108 |
42 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18 | 4 | Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded ... | Suppose the desired angle is some fraction $x$ of the total degree measure of the circle. We now compile a list of the shaded and unshaded areas. The inner circle of radius $1$ is completely unshaded, so it contributes $1$ to the unshaded area. (Everything will be a multiple of $\pi$ , so we omit it.) The inner annulus... | 108 |
43 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_20 | 2 | 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?
[asy] unitsize(4); pair P,Q,R,S,T,U,V,W; P=(... | Each other compatible point must be an even number of edges away from P, so the compatible points are R, V, and T. Therefore, we must choose two of the three points, because P must be a point in the triangle. So, the answer is ${3 \choose 2} = \boxed{3}$ | 3 |
44 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_20 | 3 | 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?
[asy] unitsize(4); pair P,Q,R,S,T,U,V,W; P=(... | List them out- you get $PRV$ $PRT$ , and $PVT$ . Therefore, the answer is $\boxed{3}$ | 3 |
45 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_21 | 1 | A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow
when in the sun. Initially, the ratio of green to yellow frogs was $3 : 1$ . Then $3$ green frogs moved to the
sunny side and $5$ yellow frogs moved to the shady side. Now the ratio is $4 : 1$ . What is the diff... | Since the original ratio is $3:1$ and the new ratio is $4:1$ , the number of frogs must be a multiple of $12$ , the only solutions left are $(B)$ and $(E)$
Let's start with $12$ frogs:
We must have $9$ frogs in the shade and $3$ frogs in the sun. After the change, there would be $11$ frogs in the shade and $1$ frog in ... | 24 |
46 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_22 | 1 | A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.
$\textb... | The roll of tape is $1/0.015=$ 66 layers thick. In order to find the total length, we have to find the average of each concentric circle and multiply it by $66$ . Since the diameter of the small circle is $2$ inches and the diameter of the large one is $4$ inches, the "middle value" is $3$ . Therefore, the average circ... | 600 |
47 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_22 | 2 | A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.
$\textb... | There are about $\dfrac{1}{0.015}=\dfrac{200}{3}$ "full circles" of tape, and with average circumference of $\dfrac{4+2}{2}\pi=3\pi.$ $\dfrac{200}{3} \cdot 3\pi=200\pi,$ which means the answer is $\boxed{600}$ | 600 |
48 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_22 | 3 | A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.
$\textb... | The volume of the tape is always the same, but we can either calculate it when the tape is unrolled as a really long, thin rectangular prism, or we can calculate it as a cylinder with a hole cut out of it. When we calculate it as a long rectangular prism, we can say that the length is $X$ (this is what the problem want... | 600 |
49 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_22 | 4 | A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.
$\textb... | If you cannot notice that the average diameter is $3$ , you can still solve this problem by the following method.
The same with solution 1, we have $\frac{1000}{0.015}$ layers of tape. If we consider every layers with the diameter $2$ , the length should be $\frac{1000}{0.015}2\pi\approx 400$ . If the diameter is seem ... | 600 |
50 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_23 | 2 | Rodrigo has a very large sheet of graph paper. First he draws a line segment connecting point $(0,4)$ to point $(2,0)$ and colors the $4$ cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point $(2000,3000)$ to point $(5000,8000)$ . How many cells will he color th... | Draw a line in the lattice which from $(2,3)$ to $(5,8)$ , notice that the line crossed 7 blocks in this pattern. Such a pattern is repeated 1000 times between $(2000,3000)$ and $(5000,8000)$ , then the answer is $\boxed{7000}$ | 0 |
51 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_24 | 1 | 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. [asy] unitsize(.2cm); draw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,linewidth(1)); draw((11,5)--(6, 0)--(16, 0)--cycle,linewidth(0.5)); label("$8\sqrt{2}$",(4,4),NW); label("$12\sqrt{2}$... | 5 |
52 | https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_24 | 2 | 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... | You can measure $h$ with a ruler(rulers are allowed on the AMC 8), and see that $h$ is closest to $\boxed{5}.$ | 5 |
53 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_1 | 2 | What is the value of $(8 \times 4 + 2) - (8 + 4 \times 2)$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 24$ | We can simplify the expression above in another way: \[(8 \times 4 + 2) - (8 + 4 \times 2)=8\times4+2-8-4\times2=32+2-8-8=34-16=\boxed{18}.\] | 18 |
54 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_3 | 2 | Wind chill is a measure of how cold people feel when exposed to wind outside. A good estimate for wind chill can be found using this calculation \[(\text{wind chill}) = (\text{air temperature}) - 0.7 \times (\text{wind speed}),\] where temperature is measured in degrees Fahrenheit $(^{\circ}\text{F})$ and the wind spee... | $0.7$ is very close to $\frac{2}{3}$ - therefore, we can substitute $\frac{2}{3}$ into the equation to get $36 - \frac{2}{3} * 18$ , which is $36 - 12 = 24$ . As $\frac{2}{3}$ is slightly less than $0.7$ , the correct answer is slightly less than $24$ . Therefore, the answer is $\boxed{23}$ | 23 |
55 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_5 | 1 | A lake contains $250$ trout, along with a variety of other fish. When a marine biologist catches and releases a sample of $180$ fish from the lake, $30$ are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?
$... | Note that \[\frac{\text{number of trout}}{\text{total number of fish}} = \frac{30}{180} = \frac16.\] So, the total number of fish is $6$ times the number of trout. Since the lake contains $250$ trout, there are $250\cdot6=\boxed{1500}$ fish in the lake. | 500 |
56 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_6 | 1 | 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?
[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path b... | 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 $\boxed{9}.$ | 9 |
57 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_6 | 2 | 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?
[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path b... | The maximum possible value of using the digits $2,0,2,$ and $3$ : We can maximize our value by keeping the $3$ and $2$ together in one power (the biggest with the biggest and the smallest with the smallest). This shows $3^{2}\times2^{0}=9\times1=9.$ (We don't want $0^{2}$ because that is $0$ .) It is going to be $\boxe... | 9 |
58 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_6 | 3 | 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?
[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path b... | Trying all $12$ distinct orderings, we see that the only possible values are $0,4,8,$ and $9,$ the greatest of which is $\boxed{9}.$ | 9 |
59 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_7 | 1 | A rectangle, with sides parallel to the $x$ -axis and $y$ -axis, has opposite vertices located at $(15, 3)$ and $(16, 5)$ . A line is drawn through points $A(0, 0)$ and $B(3, 1)$ . Another line is drawn through points $C(0, 10)$ and $D(2, 9)$ . How many points on the rectangle lie on at least one of the two lines? [asy... | If we extend the lines, we have the following diagram: [asy] usepackage("mathptmx"); size(9cm); draw((0,-.5)--(0,11),EndArrow(size=.15cm)); draw((1,0)--(1,11),mediumgray); draw((2,0)--(2,11),mediumgray); draw((3,0)--(3,11),mediumgray); draw((4,0)--(4,11),mediumgray); draw((5,0)--(5,11),mediumgray); draw((6,0)--(6,11),m... | 1 |
60 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_7 | 2 | A rectangle, with sides parallel to the $x$ -axis and $y$ -axis, has opposite vertices located at $(15, 3)$ and $(16, 5)$ . A line is drawn through points $A(0, 0)$ and $B(3, 1)$ . Another line is drawn through points $C(0, 10)$ and $D(2, 9)$ . How many points on the rectangle lie on at least one of the two lines? [asy... | Note that the $y$ -intercepts of line $AB$ and line $CD$ are $0$ and $10$ . If the analytic expression for line $AB$ is $y=k_{1}x$ , and the analytic expression for line $CD$ is $y=k_{2}x+10$ , we have equations: $3k_{1} = 1$ and $2k_{2} + 10 = 9$ . Solving these equations, we can find out that $k_{1} = \frac{1}{3}$ an... | 1 |
61 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_9 | 1 | 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? [asy] // Diagram by TheMathGuyd. Found cubic, so graph is perfect. import graph; size(8cm); int... | We mark the time intervals in which Malaika's elevation is between $4$ and $7$ meters in red, as shown below: [asy] // Diagram by TheMathGuyd. Found cubic, so graph is perfect. import graph; size(8cm); int i; for(i=1; i<9; i=i+1) { draw((-0.2,2i-1)--(16.2,2i-1), mediumgrey); draw((2i-1,-0.2)--(2i-1,16.2), mediumgrey); ... | 8 |
62 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_9 | 2 | 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? [asy] // Diagram by TheMathGuyd. Found cubic, so graph is perfect. import graph; size(8cm); int... | Notice that the entire section between the $2$ second mark and the $14$ second mark is between the $4$ and $7$ feet elevation level except the $2$ seconds where she skis just under the $4$ feet mark and when she skis just above the $7$ feet mark, making the answer $14-2-2-2=\boxed{8}.$ | 8 |
63 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_11 | 1 | NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292526838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?
$\textbf{(A)}\ 6000 \qquad \textbf{(B)}\ 12000 \qquad \textbf{(C)}... | Note that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Therefore, the speed (in miles per hour) is \[\frac{292526838}{6.5\cdot30\cdot24} \approx \frac{300000000}{6.5\cdot30\cdot24} = \frac{10000000}{6.5\cdot24} \approx \frac{10000000}{6.4\cdot25} = \frac{10000000}{160} = 62500 \approx \boxed{60000}.\] As th... | 0 |
64 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_11 | 2 | NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292526838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?
$\textbf{(A)}\ 6000 \qquad \textbf{(B)}\ 12000 \qquad \textbf{(C)}... | Note that $292526838 \approx 300000000$ miles. We also know that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Now, we can calculate the speed in miles per hour, which we find is about \[\dfrac{300000000}{6.5\cdot30\cdot24}=\dfrac{10000000}{6.5\cdot24}=\dfrac{10000000}{13\cdot12}=\dfrac{10000000}{156}\approx... | 0 |
65 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_14 | 1 | Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$ -cent, $10$ -cent, and $25$ -cent stamps, with exactly $20$ of each type. What is the greatest number of ... | Let's use the most stamps to make $7.10.$ We have $20$ of each stamp, $5$ -cent (nickels), $10$ -cent (dimes), and $25$ -cent (quarters).
If we want the highest number of stamps, we must have the highest number of the smaller value stamps (like the coins above). We can use $20$ nickels and $20$ dimes to bring our total... | 55 |
66 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_14 | 2 | Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$ -cent, $10$ -cent, and $25$ -cent stamps, with exactly $20$ of each type. What is the greatest number of ... | The value of his entire stamp collection is $8$ dollars. To make $$7.10$ with stamps, he should remove $90$ cents worth of stamps with as few stamps as possible. To do this, he should start by removing as many $25$ cent stamps as possible as they have the greatest denomination. He can remove at most $3$ of these stamps... | 55 |
67 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_17 | 1 | regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$
[asy] // Diagram by TheMathGuyd import graph; // The Solid // To sa... | We color face $6$ red and face $5$ yellow. Note that from the octahedron, face $5$ and face $?$ do not share anything in common. From the net, face $5$ shares at least one vertex with all other faces except face $1,$ which is shown in green: [asy] /* Diagram by TheMathGuyd Edited by MRENTHUSIASM */ import graph; // The... | 1 |
68 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_17 | 2 | regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$
[asy] // Diagram by TheMathGuyd import graph; // The Solid // To sa... | We label the octohedron going triangle by triangle until we reach the $?$ triangle. The triangle to the left of the $Q$ should be labeled with a $6$ . Underneath triangle $6$ is triangle $5$ . The triangle to the right of triangle $5$ is triangle $4$ and further to the right is triangle $3$ . Finally, the side of trian... | 1 |
69 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_17 | 3 | regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$
[asy] // Diagram by TheMathGuyd import graph; // The Solid // To sa... | Notice that the triangles labeled $2, 3, 4,$ and $5$ make the bottom half of the octahedron, as shown below: [asy] /* Diagram by TheMathGuyd Edited by MRENTHUSIASM */ import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk;... | 1 |
70 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_17 | 4 | regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$
[asy] // Diagram by TheMathGuyd import graph; // The Solid // To sa... | The first half of the octahedron will need $4$ triangles connected to one another to form it. We can choose the triangles $4$ $5$ $6$ , and $7$ and form the half around the vertex they all share. That leaves triangles $1$ $3$ $2$ , and $Q$ to form the second half. Triangle $3$ will definitely share its sides with trian... | 1 |
71 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_18 | 1 | Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?
$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \... | We have $2$ directions going $5$ right or $3$ left. We can assign a variable to each of these directions. We can call going right $1$ direction $\text{X}$ and we can call going $1$ left $\text{Y}$ . We can build a equation of $5\text{X}-3\text{Y}=2023$ , where we have to limit the number of moves we do. We can do this ... | 411 |
72 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_18 | 2 | Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?
$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \... | Notice that $2023 \equiv 3\pmod{5}$ , and jumping to the left increases the value of Greta's position $\pmod{5}$ by $2$ . Therefore, the number of jumps to the left must be $4 \pmod{5}$ . As the number of jumps to the left increases, so does the number of jumps to the right, we must minimize both, which occurs when we ... | 411 |
73 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_18 | 3 | Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?
$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \... | $5y - 2023$ must be divisible by 3. The smallest value of $y$ that will achieve this is $407$ , which lands it at $2035$ . After that, it takes $4$ jumps back, making a total of $\boxed{411}$ | 411 |
74 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_20 | 1 | Two integers are inserted into the list $3, 3, 8, 11, 28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?
$\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61$ | To double the range, we must find the current range, which is $28 - 3 = 25$ , to then double to: $2(25) = 50$ . Since we do not want to change the median, we need to get a value less than $8$ (as $8$ would change the mode) for the smaller, making $53$ fixed for the larger. Remember, anything less than $3$ is not benefi... | 60 |
75 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_21 | 2 | 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{(... | The group with $5$ must have the two other numbers adding up to $10$ , since the sum of all the numbers is $(1 + 2 \cdots + 9) = \frac{9(10)}{2} = 45$ . The sum of the numbers in each group must therefore be $\frac{45}{3}=15$ . We can have $(1, 5, 9)$ $(2, 5, 8)$ $(3, 5, 7)$ , or $(4, 5, 6)$ . With the first group, we ... | 2 |
76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_22 | 1 | In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is $4000$ . What is the first term?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$ | In this solution, we will use trial and error to solve. $4000$ can be expressed as $200 \times 20$ . We divide $200$ by $20$ and get $10$ , divide $20$ by $10$ and get $2$ , and divide $10$ by $2$ to get $\boxed{5}$ . No one said that they have to be in ascending order! | 5 |
77 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_25 | 1 | 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 $22... | 8 |
78 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_25 | 2 | 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... | Let the common difference between consecutive $a_i$ be $d$ .
Since $a_{15} - a_1 = 14d$ , we find from the first and last inequalities that $231 \le 14d \le 249$ . As $d$ must be an integer, this means $d = 17$ . Substituting this into all of the given inequalities so we may extract information about $a_1$ gives \[1 \... | 8 |
79 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | 1 | 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?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | Draw the following four lines as shown: [asy] usepackage("mathptmx"); defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)... | 10 |
80 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | 2 | 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?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | There are $5$ lattice points in the interior of the logo and $12$ lattice points on the boundary of the logo. Because of Pick's Theorem, the area of the logo is $5+\frac{12}{2}-1=\boxed{10}$ | 10 |
81 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | 3 | 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?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | Notice that the area of the figure is equal to the area of the $4 \times 4$ square subtracted by the $12$ triangles that are half the area of each square, which is $1$ . The total area of the triangles not in the figure is $12 \cdot \frac{1}{2} = 6$ , so the answer is $16-6 = \boxed{10}$ | 10 |
82 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | 4 | 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?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | Draw the following four lines as shown:
[asy] usepackage("mathptmx"); defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)... | 10 |
83 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | 5 | 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?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | The coordinates are $(1,2), (2,1), (3,2), (4,1), (5,2), (4,3), (5,4), (4,5), (3,4), (2,5), (1,4), (2,3)$ Use the Shoelace Theorem to get $\boxed{10}$ | 10 |
84 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1 | 6 | 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?
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); ... | If the triangles are rearranged such that the gaps are filled, there would be a $4$ by $2$ rectangle, and two $1$ by $1$ squares are present. Thus, the answer is $\boxed{10}$ | 10 |
85 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_3 | 1 | 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 $c:$
Together, the numbers $a,b,$ and $c$ can be chosen in $\boxed{4}$ ways. | 4 |
86 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_3 | 2 | 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 \[1,2,4,5,10,20,25,50,100.\] We apply casework to $a$
If $a=1$ , then there are $3$ cases:
If $a=2$ , then there is only $1$ case:
In total, there are $3+1=\boxed{4}$ ways to choose distinct positive integer values of $a,b,c$ | 4 |
87 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_5 | 1 | 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 $14-11=\boxed{3}$ years older than Bella. | 3 |
88 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | 1 | 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{6} ~MRENTHUSIASM | 6 |
89 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | 2 | 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 common difference of the arithmetic sequence be $d$ . Consequently, the smallest number is $15-d$ and the largest number is $15+d$ . As the largest number is $4$ times the smallest number, $15+d=60-4d\implies d=9$ . Finally, we find that the smallest number is $15-9=\boxed{6}$ | 6 |
90 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | 3 | 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 $x$ . Since the integers are equally spaced, and there are three of them, the middle number ( $15$ ) is the arithmetic mean of the other two numbers ( $x$ and $4x$ ). Thus, we set up the equation $(4x + x)/3 = 15$ , and, solving for $x$ , get $x = 6$ . Since $6$ is the smallest number out of ... | 6 |
91 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6 | 4 | 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 $x$ . Because $x$ and $4x$ are equally spaced from $15$ $15$ must be the average. By adding $x$ and $4x$ and dividing by $2$ , we get that the mean is also $2.5x$ . We get that $2.5x=15$ , and solving gets $x=\boxed{6}$ | 6 |
92 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_7 | 1 | When the World Wide Web first became popular in the $1990$ s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)
$\textbf{(A) } 0.6 \qquad \textbf{... | Notice that the number of kilobits in this song is $4.2 \cdot 8000 = 8 \cdot 7 \cdot 6 \cdot 100.$
We must divide this by $56$ in order to find out how many seconds this song would take to download: $\frac{\cancel{8}\cdot\cancel{7}\cdot6\cdot100}{\cancel{56}} = 600.$
Finally, we divide this number by $60$ because this... | 10 |
93 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_7 | 2 | When the World Wide Web first became popular in the $1990$ s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)
$\textbf{(A) } 0.6 \qquad \textbf{... | We seek a value of $x$ that makes the following equation true, since every other quantity equals $1$
\[\frac{x\ \text{min}}{4.2\ \text{mb}} \cdot \frac{56\ \text{kb}}{1\ \text{sec}} \cdot \frac{1\ \text{mb}}{8000\ \text{kb}} \cdot \frac{60\ \text{sec}}{1\ \text{min}} = 1.\] Solving yields $x=\boxed{10}$ | 10 |
94 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_9 | 1 | A cup of boiling water ( $212^{\circ}\text{F}$ ) is placed to cool in a room whose temperature remains constant at $68^{\circ}\text{F}$ . Suppose the difference between the water temperature and the room temperature is halved every $5$ minutes. What is the water temperature, in degrees Fahrenheit, after $15$ minutes?
$... | Initially, the difference between the water temperature and the room temperature is $212-68=144$ degrees Fahrenheit.
After $5$ minutes, the difference between the temperatures is $144\div2=72$ degrees Fahrenheit.
After $10$ minutes, the difference between the temperatures is $72\div2=36$ degrees Fahrenheit.
After $15$ ... | 86 |
95 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_11 | 1 | Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating $3$ inches of pasta from the middle of one piece. In the end, he has $10$ pieces of pasta whose total length is $17$ inches. How long, in inches, was the piece of pasta he started with?
$\textbf{(A) } 34\qquad\textbf{... | If there are $10$ pieces of pasta, Henry took $10-1=9$ bites. Each of these $9$ bites took $3$ inches of pasta out, and thus his bites in total took away $9\cdot 3 = 27$ inches of pasta. Thus, the original piece of pasta was $27+17=\boxed{44}$ inches long. | 44 |
96 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_13 | 1 | 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 |
97 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_14 | 1 | In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $\textbf{E}$ s do not appear together?
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$ | All valid arrangements of the letters must be of the form \[\textbf{E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E}.\] The problem is equivalent to counting the arrangements of $\textbf{B},\textbf{K},\textbf{P},$ and $\textbf{R}$ into the four blanks, in which ther... | 24 |
98 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_15 | 1 | Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*... | [asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*"$1$", (1,-0.3), black); label(scale(0.7)*"$2$", (2,-0.3), black); label(scale(0.7)*"$3$", (3,-0.3), black); label(scale(0.7)*"$4$", (4,-0.3), black); label(scale(0.7)*"$5$",... | 3 |
99 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_15 | 2 | Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*... | By the answer choices, we can disregard the points that do not have integer weights. As a result, we obtain the following diagram:
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*"$1$", (1,-0.3), black); label(scale(0.7)*"... | 3 |
100 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_15 | 3 | Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*... | We can find the lowest point in each line ( $1$ $2$ $3$ $4$ , or $5$ ) and find the price per pound. (Note that we don't need to find the points higher than the points below since we are finding the lowest price per pound.)
[asy] //diagram by pog size(5.5cm); usepackage("mathptmx"); defaultpen(mediumgray*0.5+gray*0.5+l... | 3 |
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