AMC8 2020

We have $\text{water} : \text{sugar} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1,$ so Luka needs $3 \cdot 8 = \boxed{\textbf{(E) }24}$ cups.

The friends earn $\$\left(15+20+25+40\right)=\$100$ in total. Since they decided to split their earnings equally, it follows that each person will get $\$\left(\frac{100}{4}\right)=\$25$. Since the friend who earned $\$40$ will need to leave with $\$25$, he will have to give $\$\left(40-25\right)=\boxed{\textbf{(C) }\$15}$ to the others.

Note that the unit of the answer is strawberries, which is the product of By conversion factors, we have \[\left(6 \ \color{red}\cancel{\mathrm{ft}}\color{black}\cdot8 \ \color{red}\cancel{\mathrm{ft}}\color{black}\right)\cdot\left(4 \ \frac{\color{green}\cancel{\mathrm{plants}}}{\color{red}\cancel{\mathrm{ft}^2}}\right)\cdot\left(10 \ \frac{\mathrm{strawberries}}{\color{green}\cancel{\mathrm{plant}}}\right)=6\cdot8\cdot4\cdot10 \ \mathrm{strawberries}=\boxed{\textbf{(D) }1920} \ \mathrm{strawberries}.\]

Looking at the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots, and the third has $3+4+5+4+3$ dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has $4+5+6+7+6+5+4=\boxed{\textbf{(B) }37}$ dots.

Each cup is filled with $\frac{3}{4} \cdot \frac{1}{5} = \frac{3}{20}$ of the amount of juice in the pitcher, so the percentage is $\frac{3}{20} \cdot 100 = \boxed{\textbf{(C) }15}$.

Write the order of the cars as $\square\square\square\square\square$, where the left end of the row represents the back of the train and the right end represents the front. Call the people $A$, $D$, $K$, $M$, and $S$ respectively. The first condition gives $M\square\square\square\square$, so we try $MAS\square\square$, $M\square AS\square$, and $M\square\square AS$. In the first case, as $D$ sat in front of $A$, we must have $MASDK$ or $MASKD$, which do not comply with the last condition. In the second case, we obtain $MKASD$, which works, while the third case is obviously impossible since it results in there being no way for $D$ to sit in front of $A$. It follows that, with the only possible arrangement being $MKASD$, the person sitting in the middle car is $\boxed{\textbf{(A) }\text{Aaron}}$.

Firstly, we can observe that the second digit of such a number cannot be $1$ or $2$ because the digits must be distinct and in increasing order. The second digit also cannot be $4$ as the number must be less than $2400$, so the second digit must be $3$. It remains to choose the latter two digits, which must be $2$ distinct digits from $\left\{4,5,6,7,8,9\right\}$. That can be done in $\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15$ ways; there is then only $1$ way to order the digits, namely in increasing order. This means the answer is $\boxed{\textbf{(C) }15}$.

Clearly, the amount of money Ricardo has will be maximized when he has the maximum number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is $2019$, giving a total of $(2019\cdot 5 + 1)$ cents. Analogously, the amount of money he has will be least when he has the greatest number of pennies; as he must have at least one nickel, the greatest number of pennies he can have is also $2019$, giving him a total of $(2019\cdot 1 + 5)$ cents. Hence the required difference is \[(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4\cdot 2018=\boxed{\textbf{(C) }8072}\]

Notice that, for a small cube which does not form part of the bottom face, it will have exactly $2$ faces with icing on them only if it is one of the $2$ center cubes of an edge of the larger cube. There are $12-4 = 8$ such edges (as we exclude the $4$ edges of the bottom face), so this case yields $2 \cdot 8 = 16$ small cubes. As for the bottom face, we can see that only the $4$ corner cubes have exactly $2$ faces with icing, so the total is $16+4 = \boxed{\textbf{(D) }20}$.

Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$, respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each other. If we place $S$ and $T$ next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. $ST\square\square, \square ST\square, \square\square ST$). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS\square\square, \square TS\square, \square\square TS$). Thus there are 6 ways of placing $S$ and $T$ directly next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$. Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ next to each other. Subtracting this from the total number of arrangements gives us $24-12=12$ total arrangements $\implies\boxed{\textbf{(C) }12}$. We can also solve this problem directly by looking at the number of ways that we can place $S$ and $T$ such that they are not directly next to each other. Observe that there are three ways to place $S$ and $T$ (in that order) into the four slots so they are not next to each other (i.e. $S\square T\square, \square S\square T, S\square\square T$). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $T\square S\square, \square T\square S, T\square\square S$). Thus there are 6 ways of placing $S$ and $T$ so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$. Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ such that they are not next to each other $\implies\boxed{\textbf{(C) }12}$.

Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since $\text{speed} = \frac{\text{distance}}{\text{time}}$, her speed is $\frac{6}{\left(\frac{1}{6}\right)} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph, so the answer is $36-12 = \boxed{\textbf{(E) }24}$.

We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$, and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$. Therefore, the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$, and so $12 \cdot 10! = 12 \cdot N!$. Cancelling the $12$s, it is clear that $N=\boxed{\textbf{(A) }10}$.

After Jamal adds $x$ purple socks, he has $(18+x)$ purple socks and $6+18+12+x=(36+x)$ total socks. This means the probability of drawing a purple sock is $\frac{18+x}{36+x}$, so we obtain \[\frac{18+x}{36+x}=\frac{3}{5}\] Since $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}$, the answer is $\boxed{\textbf{(B) }9}$.

We can see that the dotted line is exactly halfway between $4{,}500$ and $5{,}000$, so it is at $4{,}750$. As this is the average population of all $20$ cities, the total population is simply $4{,}750 \cdot 20 = \boxed{\textbf{(D) }95{,}000}$.

Since $20\% = \frac{1}{5}$, multiplying the given condition by $5$ shows that $y$ is $15 \cdot 5 = \boxed{\textbf{(C) }75}$ percent of $x$.

We can form the following expressions for the sum along each line: \[ \begin{cases} A+B+C\\ A+E+F\\ C+D+E\\ B+D\\ B+F \end{cases} \] Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$, i.e. $2(A+B+C+D+E+F)+B=47$. Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$, we obtain $A+B+C+D+E+F=1+2+3+4+5+6=21$ (where the order doesn't matter as addition is commutative), so our equation simplifies to $42 + B = 47$. This means $B = \boxed{\textbf{(E) }5}$.

Since $2020 = 2^2 \cdot 5 \cdot 101$, we can simply list its factors: \[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.\] There are $12$ factors; only $1, 2, 4, 5, 101$ don't have over $3$ factors, so the remaining $12-5 = \boxed{\textbf{(B) }7}$ factors have more than $3$ factors.

Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$, so $OC = 17$. By symmetry, $O$ is the midpoint of $DA$, so $OD=OA=\frac{16}{2}= 8$. By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$), we have that $CD$ (or $AB$) is $\sqrt{17^2-8^2}=15$. Accordingly, the area of $ABCD$ is $16\cdot 15=\boxed{\textbf{(A) }240}$.

A number is divisible by $15$ precisely if it is divisible by $3$ and $5$. The latter means the last digit must be either $5$ or $0$, and the former means the sum of the digits must be divisible by $3$. If the last digit is $0$, the first digit would be $0$ (because the digits alternate), which is not possible. Hence the last digit must be $5$, and the number is of the form $5\square 5\square 5$. If the unknown digit is $x$, we deduce $5+x+5+x+5 \equiv 0 \pmod{3} \Rightarrow 2x \equiv 0 \pmod{3}$. We know $2^{-1}$ exists modulo $3$ because 2 is relatively prime to 3, so we conclude that $x$ (i.e. the second and fourth digit of the number) must be a multiple of $3$. It can be $0$, $3$, $6$, or $9$, so there are $\boxed{\textbf{(B) }4}$ options: $50505$, $53535$, $56565$, and $59595$.

We will show that $22$, $11$, $22$, $44$, and $22$ meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height $11$ meters, we can deduce that Trees 1 and 3 both have a height of $22$ meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of $11$ and $22$, $44$ and $88$, or $44$ and $22$. Checking each of these, in the first case, the average is $17.6$ meters, which doesn't end in $.2$ as the problem requires. Therefore, we consider the other cases. With $44$ and $88$, the average is $37.4$ meters, which again does not end in $.2$, but with $44$ and $22$, the average is $24.2$ meters, which does. Consequently, the answer is $\boxed{\textbf{(B) }24.2}$.

Notice that, from one of the $1$ or $2$ white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from $P$ to that square is the sum of the number of ways to move from $P$ to each of the white squares immediately beneath it (also called the Waterfall Method). To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). The answer is therefore $\boxed{\textbf{(A) }28}$. Note: This is a classic example of a problem involving Pascal's triangle.

We start with final output of $1$ and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage: \[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}\] where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$), but $16$ could come from $32$ or $5$ (as $\frac{32}{2} = 3 \cdot 5 + 1 = 16$, and $32$ is even while $5$ is odd). By construction, last set in this sequence contains all the numbers which will lead to number $1$ to end of the $6$-step process, and sum is $1+8+10+64=\boxed{\textbf{(E) }83}$.

Firstly, observe that a single student can't receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$, $2$, or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ award; if a student receives $2$ awards, then another student must also receive $2$ awards and the remaining student must receive $1$ award. We consider each of these two cases in turn. If a student receives three awards, there are $3$ ways to choose which student this is, and $\binom{5}{3}$ ways to give that student $3$ out of the $5$ awards. Next, there are $2$ students left and $2$ awards to give out, with each student getting one award. There are clearly just $2$ ways to distribute these two awards out, giving $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case. In the other case, two students receive $2$ awards and one student receives $1$ award. We know there are $3$ choices for which student gets $1$ award. There are $\binom{3}{1}$ ways to do this. Then, there are $\binom{5}{2}$ ways to give the first student his two awards, leaving $3$ awards yet to be distributed. There are then $\binom{3}{2}$ ways to give the second student his $2$ awards. Finally, there is only $1$ student and $1$ award left, so there is only $1$ way to distribute this award. This results in $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ ways to distribute the awards in this case. Adding the results of these two cases, we get $60+90=\boxed{\textbf{(B) }150}$.

The area of the shaded region is $(24s)^2$. To find the area of the large square, we note that there is a $d$-inch border between each of the $23$ pairs of consecutive squares, as well as from between the first/last squares and the large square, for a total of $23+2 = 25$ times the length of the border, i.e. $25d$. Adding this to the total length of the consecutive squares, which is $24s$, the side length of the large square is $(24s+25d)$, yielding the equation $\frac{(24s)^2}{(24s+25d)^2}=\frac{64}{100}$. Taking the square root of both sides (and using the fact that lengths are non-negative) gives $\frac{24s}{24s+25d}=\frac{8}{10} = \frac{4}{5}$, and cross-multiplying now gives $120s = 96s + 100d \Rightarrow 24s = 100d \Rightarrow \frac{d}{s} = \frac{24}{100} = \boxed{\textbf{(A) }\frac{6}{25}}$. Note: Once we obtain $\tfrac{24s}{24s+25d} = \tfrac{4}{5},$ to ease computation, we may take the reciprocal of both sides to yield $\tfrac{24s+25d}{24s} = 1 + \tfrac{25d}{24s} = \tfrac{5}{4},$ so $\tfrac{25d}{24s} = \tfrac{1}{4}.$ Multiplying both sides by $\tfrac{24}{25}$ yields the same answer as before.

Let the side length of each square $S_k$ be $s_k$. Then, from the diagram, we can line up the top horizontal lengths of $S_1$, $S_2$, and $S_3$ to cover the top side of the large rectangle, so $s_{1}+s_{2}+s_{3}=3322$. Similarly, the short side of $R_2$ will be $s_1-s_2$, and lining this up with the left side of $S_3$ to cover the vertical side of the large rectangle gives $s_{1}-s_{2}+s_{3}=2020$. We subtract the second equation from the first to obtain $2s_{2}=1302$, and thus $s_{2}=\boxed{\textbf{(A) }651}$.