AMC8 2024

We can rewrite the expression as $222,222-(22,222+2,222+222+22+2)$. We note that the units digit of $22,222+2,222+222+22+2$ is $0$ because all the units digits of the five numbers are $2$ and $5\cdot2=10$, which has a units digit of $0$. Now, we have something with a units digit of $0$ subtracted from $222,222$, and so the units digit of this expression is $\boxed{\textbf{(B) } 2}$.

We see that $\frac{44}{11}$ is $4$; $\frac{110}{44}$ simplifies to $\frac{5}{2}$, which is $2.5$; and $\frac{44}{1100}$ simplifies to $\frac{1}{25}$, which is $0.04$; $4+2.5+0.04$ reveals \[\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}\] is $\boxed{\text{(C) 6.54}}$.

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 \[10^2 - 9^2 + 7^2 - 4^2 = 100 - 81 + 49 - 16 = 19 + 33 = \boxed{\textbf{(E)}\ 52}\] This problem appears multiple times in various math competitions including the AMC and MATHCOUNTS.

Adding numbers 1 through 9 gives us 45. This was her expected sum, but what she got was a perfect square. Since she got that perfect square sum by forgetting a number, that sum is less than 45. The square number right under 45 is 36. So 45 - 36 = 9. So the solution is $\boxed{\textbf{(E)}\ 9}$.

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: $3\times 9$ The only integer that cannot be the sum is $\boxed{\textbf{(B) } 6}$. We can prove that B is correct because the only ordered pairs that add to 6 is $(1,5)$ , $(5,1)$, $(2,4)$, $(4,2)$, and ,$(3,3)$, and these multiply to 5, 8, and 9. Slight changes ~ VersatileGopher38

You can measure the lengths of the paths until you find a couple of guaranteed true inferred statements as such: $Q$ is greater than $S$, $P$ is greater than $R$, and $R$ and $P$ are the smallest two, therefore the order is $R, P, S, Q.$ Thus we get the answer $\boxed{\textbf{(D)}~R, P, S, Q}$.

Let $x$ be the number of $1$ by $1$ tiles. There are $21$ squares and each $2$ by $2$ or $1$ by $4$ 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 each $2$ by $2$ and $1$ by $4$ shape takes up an equal number of blue and red squares, so there must be $3$ more $1$ by $1$ tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is (E) $5$, which can easily be confirmed to work.

How many dollar 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 $7$ dollars. Now, you have $2$ values for each of these. For $10$ dollars, you have $13$ dollars or $20$, for $8$ dollars, you have $16$ dollars or $11$, for $8$ dollars, you have $16$ dollars or $11$, and for $7$ dollars, you have $14$ dollars or $10$. On the final day, there are 11, 11, 16, and 16 repeating, leaving you with $8-2 = \boxed{\textbf{(D) 6}}$ different values.

Since Maria 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 $7$, as $x$ represents an integer, so the only possible answer is $\boxed{\textbf{(E) 28}}.$ Minor Edit - Runningmouse

This is a time period of $2030 - 1980 = 50$ years, so we can expect the ppm to increase by $50*1.515=75.75\approx 76$ ppm. Since we started with $338$ ppm, we have $76+338=\boxed{\textbf{(B) 414}}$.

Since the triangle has a base of $6$, we can plug in that value as the base. Then, we can solve the equation for the height. Doing so gives us, \[\dfrac{6h}{2}=3h=12.\] This means that $h=4$, so that means that we have to add 4 to the $y$-coordinate. So the answer is $7+4=\boxed{(D) 11}$

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 the second tank is $x+1$ - The number of guppies in the third tank is $x+1+2$ - The number of guppies in the fourth tank is $x+1+2+3$ 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 \[4x + 10 = 90.\] Solving the resulting equation gives $x = 20$, so the number of guppies in the fourth tank is $20+1+2+3 = \boxed{\textbf{(E)}\ 26}$.

Looking at the answer choices, you see that you can list them out. Doing this gets you: $\mathit{UUDDUD}$ $\mathit{UDUDUD}$ $\mathit{UUUDDD}$ $\mathit{UDUUDD}$ $\mathit{UUDUDD}$ Counting all the paths listed above gets you $\boxed{\textbf{(B)} \ 5}$.

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. You can also simply brute-force it or sort of think ahead - for example, getting from A to M can be done $2$ ways; $A \rightarrow X \rightarrow M$ ($5+2$) or $A \rightarrow M (8)$, so you should take the shorter route ($5+2$). Another example is M to C, two ways - one is $6+5$ and the other is $14$. Take the shorter route. After this, you need to consider a few more times - consider if $5+10$ ($Y \rightarrow C \rightarrow Z$) is greater than $17 (Y \rightarrow Z$), which it is not, and consider if $25 (M \rightarrow Z$) is greater than $14+10$ ($M \rightarrow C \rightarrow Z$) or $6+5+10$ ($M \rightarrow Y \rightarrow C \rightarrow Z$) which it is not. TLDR: $5+2+6+5+10 = \boxed{28}$.

Let $\overline{FLY}$ be the $3$-digit number $100F+10L+Y$. Repeating a $3$-digit block twice gives \[\overline{FLYFLY}=1000\cdot \overline{FLY}+\overline{FLY}=1001\cdot \overline{FLY}.\] Similarly, \[\overline{BUGBUG}=1001\cdot \overline{BUG}.\] Given $8\cdot \overline{FLYFLY}=\overline{BUGBUG}$, substitute: \[8(1001\overline{FLY})=1001\overline{BUG}.\] Divide by $1001$: \[8\overline{FLY}=\overline{BUG}.\] Since $\overline{BUG}$ is a $3$-digit number, \[100\le 8\overline{FLY}\le 999 \quad\Rightarrow\quad 13\le \overline{FLY}\le 124.\] To maximize $\overline{FLYFLY}$, maximize $\overline{FLY}$. Try the largest possible value: \[\overline{FLY}=123 \quad\Rightarrow\quad \overline{BUG}=8\cdot 123=984.\] The digits $\{1,2,3,9,8,4\}$ are all distinct, so this works and is maximal. Therefore, \[\overline{FLY}+\overline{BUG}=123+984=1107.\] \[\boxed{1107}\]

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 minimum with $a+b=9+3=12$. If $ab=28$,our best is $a+b=7+4=11$. Note if $a+b=10$, $ab=25$. Because $25<27$, there is no smaller answer, and we get $\boxed{\textbf{(D)} 11}$.

If you place a king in any of the $4$ corners, the other king will have $5$ spots to go and there are $4$ corners, so $5 \times 4=20$. If you place a king in any of the $4$ edges, the other king will have $3$ spots to go and there are $4$ edges so $3 \times 4=12$. That gives us $20+12=32$ spots for the other king to go into in total. So $\boxed{\textbf{(E)} 32}$ is the answer.

Let $x=\angle{BOC}$. We see that the shaded region is the inner ring plus a sector $x^\circ$ of the outer ring. Using the formula for the area of a circle ($A = \pi r^2$), we find that the area of $x$ is $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$. This simplifies to $3 \pi + \frac{x}{360}(5 \pi)$. The unshaded portion is comprised of the smallest circle plus the sector $(360-x)^\circ$ of the outer ring, which evaluates to $\pi + \frac{360-x}{360}(5 \pi)$. We are told these are equal. Therefore, $3 \pi + \frac{x}{360}(5 \pi) = \pi + \frac{360-x}{360}(5 \pi)$. Solving for $x$ reveals $x=\boxed{\textbf{(A) } 108}$.

Jordan has $10$ high top sneakers, and $6$ white sneakers. We would want as many white high-top sneakers as possible, so we set $6$ high-top sneakers to be white. Then, we have $10-6=4$ red high-top sneakers, so the answer is $\boxed{\dfrac{4}{15}}.$

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 $\boxed{\textbf{(D) }3}$

Let the initial number of green frogs be $g$ and the initial number of yellow frogs be $y$. Since the ratio of the number of green frogs to yellow frogs is initially $3 : 1$, $g = 3y$. Now, $3$ green frogs move to the sunny side and $5$ yellow frogs move to the shade side, thus the new number of green frogs is $g + 2$ and the new number of yellow frogs is $y - 2$. We are given that $\frac{g + 2}{y - 2} = \frac{4}{1}$, so $g + 2 = 4y - 8$, since $g = 3y$, we have $3y + 2 = 4y - 8$, so $y = 10$ and $g = 30$. Thus the answer is $(g + 2) - (y - 2) = 32 - 8 = \boxed{(E) \hspace{1 mm} 24}.$

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" (or mean) is $3$. Therefore, the average circumference is $3\pi$. Multiplying $3\pi \cdot 66$ gives approximately $(B) \boxed{600}$.

Let $f(x, y)$ be the number of cells the line segment from $(0, 0)$ to $(x, y)$ passes through. The problem is then equivalent to finding \[f(5000-2000, 8000-3000)=f(3000, 5000).\] Sometimes the segment passes through lattice points in between the endpoints, which happens $\text{gcd}(3000, 5000)-1=999$ times. This partitions the segment into $1000$ congruent pieces that each pass through $f(3, 5)$ cells, which means the answer is \[1000f(3, 5).\] Note that a new square is entered when the lines pass through one of the lines in the coordinate grid, which for $f(3, 5)$ happens $3-1+5-1=6$ times. Because $3$ and $5$ are relatively prime, no lattice point except for the endpoints intersects the line segment from $(0, 0)$ to $(3, 5).$ This means that including the first cell closest to $(0, 0),$ The segment passes through $f(3, 5)=6+1=7$ cells. Thus, the answer is $\boxed{\textbf{(C)}7000}.$ Alternatively, $f(3, 5)$ can be found by drawing an accurate diagram, leaving you with the same answer. Note: A general form for finding $f(x, y)$ is $x+y-\text{gcd}(x, y).$ We subtract $\text{gcd}(x, y)$ to account for overlapping, when the line segment goes through a lattice point.

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 $8\sqrt{2}$), and the area of the overlap is $h^2$ (the side length is $h\sqrt{2}$). Because the right triangles have a side ratio of 1:1:$\sqrt{2}$.Thus, \[144+64-h^2=183,\] which means that the answer is $\boxed{\mathbf{(B)}\text{ 5}}.$

Suppose the passengers are indistinguishable. There are $\binom{12}{8} = 495$ total ways to distribute the passengers. We proceed with complementary counting, and instead, will count the number of passenger arrangements such that the couple cannot sit anywhere. Consider the partitions of $8$ among the rows of $3$ seats, to make our lives easier, assuming they are non-increasing. We have $(3, 3, 2, 0), (3, 3, 1, 1), (3, 2, 2, 1), (2, 2, 2, 2)$. For the first partition, clearly, the couple will always be able to sit in the row with $0$ occupied seats, so we have $0$ cases here. For the second partition, there are $\frac{4!}{2!2!} = 6$ ways to permute the partition. Now the rows with exactly $1$ passenger must be in the middle, so this case generates $6$ cases. For the third partition, there are $\frac{4!}{2!} = 12$ ways to permute the partition. For rows with $2$ passengers, there are $\binom{3}{2} = 3$ ways to arrange them in the row so that the couple cannot sit there. The row with $1$ passenger must be in the middle. We obtain $12 \cdot 3^2 = 108$ cases. For the fourth partition, there is $1$ way to permute the partition. As said before, rows with $2$ passengers can be arranged in $3$ ways, so we obtain $3^4 = 81$ cases. Collectively, we obtain a total of $6 + 108 + 81 = 195$ cases. The final probability is $1 - \frac{195}{495} = \boxed{\textbf{(C)}~\frac{20}{33}}$.