AMC8 2023

By the order of operations, we have \[(8 \times 4 + 2) - (8 + 4 \times 2) = (32+2) - (8+8) = 34 - 16 = \boxed{\textbf{(D)}\ 18}.\]

Notice that when we unfold the paper along the vertical fold line, we get the following shape: Therefore the answer is $\boxed{\textbf{(E)}}$.

By substitution, we have \begin{align*} (\text{wind chill}) &= 36 - 0.7 \times 18 \\ &= 36 - 12.6 \\ &= 23.4 \\ &\approx \boxed{\textbf{(B)}\ 23}. \end{align*}

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 $47.$

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 $250\cdot6=\boxed{\textbf{(B)}\ 1500}$ fish in the lake.

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 $\boxed{\textbf{(C) }9}.$

If we extend the lines, we have the following diagram: Therefore, we see that the answer is $\boxed{\textbf{(B)}\ 1}.$

We can calculate the total number of wins ($1$'s) by seeing how many matches were players, which is $12$ matches played. Then, we can calculate the # of wins already on the table, which is $5 + 3 + 2 = 10$, so there are $12 - 10 = 2$ wins left in the mystery player. Now, we will make the key observation that there is only $2$ wins ($1$'s) per column as there are $2$ winners and $2$ losers in each round. Strategically looking through the columns counting the $1$'s and putting our own $2$ $1$'s when the column isn't already full yields $\boxed{\textbf{(A)}\ \texttt{000101}}$.

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 $(4-2) + (10-6) + (14-12) = \boxed{\textbf{(B)}\ 8}$ seconds at such elevation.

Note that: - Harold ate $\frac14$ of the pie. After that, $1-\frac14=\frac34$ of the pie was left behind. - The moose ate $\frac13\cdot\frac34 = \frac14$ of the pie. After that, $\frac34 - \frac14 = \frac12$ of the pie was left behind. - The porcupine ate $\frac13\cdot\frac12 = \frac16$ of the pie. After that, $\frac12 - \frac16 = \boxed{\textbf{(D)}\ \frac{1}{3}}$ of the pie was left behind. More simply, we can condense the solution above into the following equation: \[\left(1-\frac14\right)\left(1-\frac13\right)\left(1-\frac13\right) = \frac34\cdot\frac23\cdot\frac23 = \frac13.\]

Note that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Therefore, the speed (in miles per hour) is \[\frac{292{,}526{,}838}{6.5\cdot30\cdot24} \approx \frac{300{,}000{,}000}{6.5\cdot30\cdot24} = \frac{10{,}000{,}000}{6.5\cdot24} \approx \frac{10{,}000{,}000}{6.4\cdot25} = \frac{10{,}000{,}000}{160} = 62500 \approx \boxed{\textbf{(C)}\ 60{,}000}.\] As the answer choices are far apart from each other, we can ensure that the approximation is correct.

The large circle has radius 3 (in the standard diagram units), so its area is \[ \pi \cdot 3^2 = 9\pi. \] The shaded area consists of: - three small shaded circles, each with radius $\frac{1}{2}$, so each has area \[ \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}. \] Their total area is \[ 3 \times \frac{\pi}{4} = \frac{3\pi}{4}. \] - the equivalent of two full shaded circles of radius 1 (or the shaded regions that together contribute the same area as two such circles), giving area \[ 2 \times \pi \cdot 1^2 = 2\pi. \] The total shaded area is therefore \[ \frac{3\pi}{4} + 2\pi = \frac{3\pi}{4} + \frac{8\pi}{4} = \frac{11\pi}{4}. \] The fraction of the large circle that is shaded is \[ \frac{\frac{11\pi}{4}}{9\pi} = \frac{11}{4} \times \frac{1}{9} = \frac{11}{36}. \]

Suppose that it is $d$ miles long. The bathroom is located at \[\frac{d}{8}, \frac{2d}{8}, \ldots, \frac{7d}{8}\] miles from the start, and the water fountains are located at \[\frac{d}{3}, \frac{2d}{3}\] miles from the start. We are given that $\frac{3d}{8}=\frac{d}{3}+2,$ from which \begin{align*} \frac{9d}{24}&=\frac{8d}{24}+2 \\ \frac{d}{24}&=2 \\ d&=\boxed{\textbf{(D)}\ 48}. \end{align*}

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 cost to $7.10 - 3.00 = 4.10$. However, when we try to use quarters, the $25$ cents don’t fit evenly, so we have to give back $15$ cents to make the quarter amount $4.25$. The most efficient way to do this is to give back a $10$-cent (dime) stamp and a $5$-cent (nickel) stamp to have $38$ stamps used so far. Now, we just use $\frac{425}{25} = 17$ quarters to get a grand total of $38 + 17 = \boxed{\textbf{(E)}\ 55}$.

Note that Viswam walks at a constant speed of $60$ blocks per hour as he takes $1$ minute to walk each block. After walking $5$ blocks, he has taken $5$ minutes and has $5$ minutes remaining to walk $7$ blocks. Therefore, he must walk at a speed of $7 \cdot 60 \div 5 = 84$ blocks per hour to get to school on time, from the time he starts his detour. Since he normally walks $\frac{1}{2}$ mile, which is equal to $10$ blocks, $1$ mile is equal to $20$ blocks. Therefore, he must walk at $84 \div 20 = 4.2$ mph from the time he starts his detour to get to school on time, so the answer is $\boxed{\textbf{(B)}\ 4.2}$.

In our $5\times5$ grid, there are $8,9$ and $8$ of the letters $\text{P}, \text{Q},$ and $\text{R}$, respectively, and in a $2\times2$ grid, there are $1,2$ and $1$ of the letters $\text{P}, \text{Q},$ and $\text{R}$, respectively. We see that in both grids, there are $x, x+1,$ and $x$ of the $\text{P}, \text{Q},$ and $\text{R}$, respectively. This is because in any $n\times n$ grid with $n\equiv2\pmod3$, there are $x, x+1,$ and $x$ of the $\text{P}, \text{Q},$ and $\text{R}$, respectively. We can see that the only answer choice which satisfies this condition is $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.$

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: Therefore, the answer is $\boxed{\textbf{(A)}\ 1}.$

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 an equation of $5\text{X}-3\text{Y}=2023$, where we have to limit the number of moves we do. We can do this by making more of our moves the $5$ move turn than the $3$ move turn. The first obvious step is to go some amount of moves in the right direction then subtract off in the left direction to land on $2023$. The least amount of $3$’s added to $2023$ to make a multiple of $5$ is $4$ as $2023 + 4(3) = 2035$. So now, we have solved the problem as we just go $\frac{2035}{5} = 407$ hops right, and just do 4 more hops left. Yielding $407 + 4 = \boxed{\textbf{(D)}\ 411}$ as our answer.

All equilateral triangles are similar. For the outer equilateral triangle to the inner equilateral triangle, since their side-length ratio is $\frac32,$ their area ratio is $\left(\frac32\right)^2=\frac94.$ It follows that the area ratio of three trapezoids to the inner equilateral triangle is $\frac94-1=\frac54,$ so the area ratio of one trapezoid to the inner equilateral triangle is \[\frac54\cdot\frac13=\frac{5}{12}=\boxed{\textbf{(C) } 5 : 12}.\]

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, which will be $7$. This fixes $53$ for the larger value. Anything less than $3$ is not beneficial to the optimization because you want to get the largest range without changing the mode. So, taking our optimal values of $7$ and $53$, we have an answer of $7 + 53 = \boxed{\textbf{(D)}\ 60}$.

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 have $(2, 3, 4, 6, 7, 8)$ left over. The only way to form a group of $3$ numbers that add up to $15$ is with $(3, 4, 8)$ or $(2, 6, 7)$. One of the possible arrangements is therefore $(1, 5, 9) (3, 4, 8) (2, 6, 7)$. Then, with the second group, we have $(1, 3, 4, 6, 7, 9)$ left over. With these numbers, there is no way to form a group of $3$ numbers adding to $15$. Similarly, with the third group there is $(1, 2, 4, 6, 8, 9)$ left over and we can make a group of $3$ numbers adding to $15$ with $(1, 6, 8)$ or $(2, 4, 9)$. Another arrangement is $(3, 5, 7) (1, 6, 8) (2, 4, 9)$. Finally, the last group has $(1, 2, 3, 7, 8, 9)$ left over. There is no way to make a group of $3$ numbers adding to $15$ with this, so the arrangements are $(1, 5, 9) (3, 4, 8) (2, 6, 7)$ and $(3, 5, 7) (1, 6, 8) (2, 4, 9)$. So，there are $\boxed{\textbf{(C)}\ 2}$ sets that can be formed.

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 $\boxed{\textbf{(D)}\ 5}$. No one said that they have to be in ascending order! Solution by ILoveMath31415926535 and clarification edits by apex304 Anonymous question here, if 5 is first term, how do you get 10? (answer to anonymous question) "No one said they had to be in ascending order!" means after 5, the term doesnt need to increase each time.

There are $4$ cases that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids, as shown below: There are $4^5$ ways to decide the $5$ white squares for each case, and the cases do not have any overlap. So, the requested probability is \[\frac{4\cdot4^5}{4^9} = \frac{4^6}{4^9} = \frac{1}{4^3} = \boxed{\textbf{(C) } \frac{1}{64}}.\]

First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$. Similarly, we can find that the area of the gray part in the second triangle is $[ABC]\cdot\left(\tfrac{h-5}{h}\right)^2$. These areas are equal, so $1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2$. Simplifying yields $10h=146$ so $h=\boxed{\textbf{(A) }14.6}$.

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 $221\leq 13x\leq237$. The number $18$ is similarly found. $19$, however, is too much. Now, we check with the first and last equations using the same method. We know $241-10\leq 14x\leq250-1$. Therefore, $231\leq 14x\leq249$. We test both values we just got, and we can realize that $18$ is too large to satisfy this inequality. On the other hand, we can now find that the difference will be $17$, which satisfies this inequality. The last step is to find the first term. We know that the first term can only be from $1$ to $3$ since any larger value would render the second inequality invalid. Testing these three, we find that only $a_1=3$ will satisfy all the inequalities. Therefore, $a_{14}=13\cdot17+3=224$. The sum of the digits is therefore $\boxed{\textbf{(A)}\ 8}$.