AMC8 2026

$1+2-3+4+5-6+7+8-9+10+11-12$ = $\boxed{18}$

Notice that if we label the rows $A_1$, $A_2$, $A_3$, ..., $A_5$ top down, then, if we denote $S(A_n)$ to be the sum of the $A_n$ row, $A_1 = A_5$, $A_2 = A_4$, and $A_3 = A_3$ through symmetry. Thus the total sum is $2A_1 + 2A_2 + A_3$. Then, it is obvious that the sum of $A_1 = 7$, the sum of $A_2 = 2 + 2(5) = 12$, and the sum of $A_3 = 2(3) + 3(3) = 5(3) = 15$. The answer is then $2 \cdot 7 + 2 \cdot 12 + 15 = 14 + 15 + 24 = 29 + 24 = \boxed{\textbf{(C)}\ 53}$.

To solve this problem we can find the perimeter of each of the given spaces and check if it is less than 24. The first shape is a regular hexagon with side length $5$. The hexagon has a perimeter of $6 \cdot 5 = 30$ centimeters. The second shape is a square with area $36$ or a side length of $6$. Then the perimeter of all $4$ sides is $24$ centimeters. Finally, we need to find the perimeter of a right triangle with legs $6$ and $8$ centimeters. The hypotenuse of the right triangle is $10$ so the total perimeter is $6 + 8 + 10 = 24$ centimeters. Thus, only the square and the triangle can be made with 24 centimeters of wire so the answer is $\boxed{D}$.

We see that the change is $(1-0.2)(1+0.5)=0.8\cdot1.5=1.2$. That is an increase of $0.2$, or (E) $120\%$.

The average speed is $\frac{ \text{ Total distance } }{ \text{ Total time } }$, or $\frac{100}{3-t}$. Thus $\frac{100}{3-t} = 40 \Rightarrow 3-t = \frac{100}{40} \Rightarrow t = \frac{1}{2}$, which is $\boxed{ \textbf{(B) } 30}$ minutes.

We can calculate this reachable area to be $\text{Total area} - \text{Unreachable area}$. The total area is $8 \cdot 10 = 80$, and the unreachable area is $(8-2)(10-2) = 6\cdot 8 = 48$. Thus, our answer is $\frac{80-48}{80} = \frac{32}{80} = \boxed{ \textbf{(E) } \frac{2}{5} }$.

We can see that the total battery spent is $\frac{1}{2}+\frac{3}{10}=\frac{5+3}{10}=\frac{4}{5}$. We can thus write the proportion $\frac{40}{\frac{4}{5}}=\frac{x}{1}$. Since $\frac{40}{\frac{4}{5}}=40\cdot\frac{5}{4}=50$, $\frac{x}{1}=x=(C)\ 50$ miles.

We can see that $74 \% = \frac{74}{100}$ by definition. This fraction can be simplified to $\frac{37}{50}$, meaning $37$ out of $50$ people said "Yes". Since $\gcd(37,50) = 1$, if this fraction was divided any further, we would have fractional numerator and denominator, which is clearly impossible. Therefore, the minimum number of people surveyed was $\boxed{ \textbf{(D) } 50}$.

The inner square root cannot be negative. If the outer square root is negative, the denominator is also negative, resulting in a positive fraction, an identity. Thus, we only consider positive squares, in which the numerator simplifies to $\sqrt{16 \cdot 9} = \sqrt{16}\sqrt{9} = 4 \cdot 3 = 12$, and the denominator simplifies to $\sqrt{81 \cdot 4} = \sqrt{81}\sqrt{4} = 9 \cdot 2 = 18$. The answer is $\frac{12}{18} = \frac{4}{6} = \boxed{\frac{2}{3}}$. (Other solutions weren't explicit with logic).

Let the respective places be $L$, $M$, $N$, $O$, and $P$ respectively. Then, $P = 11+N$, $O-2 = M$, $O+3 = P$, and $O = 6+L$. Then, substitution for $P$ gives us $O+3 = N+11$, in which $O=N+8$. Then, $N+6 = M$, $N+8=6+L \Rightarrow N+2=L$. So $N$ finished m minutes behind everyone (sad), and thus was last. Now, $O=6+L$, in which $L+9 = P$, $L+7 = M$, and thus $\boxed{\text{Luke}}$ finished fourth.

We notice that we have multiple quarter-arcs (i.e. $\frac{1}{4}$ of a full circumference) of varying radii. However, since they all have the same formula, $\frac{1}{4} \cdot 2\pi r = \frac{\pi}{2}r$, we can take out $\frac{\pi}{2}$ by Distributive Property. Thus we have \[\frac{\pi}{2} (1 + 1 + 2 + 3 + 5) = \frac{\pi}{2} (12) = \boxed{ \textbf{(B) } 6\pi }\] as our total curve length.

Let $O$ denote an odd number and $E$ denote an even. Then, we know that $O+E = O$, and thus the two neighboring circles that form sums 9 and 5 must be one of each odd and even. To obtain 5, we have $1+4$ or $2+3$. To obtain 9, we have $3+6$, $4+5$. Consider each case separately. Case 1: If $1+4$ makes the sum of 5, then the two numbers that make 9 must be $3+6$. Then, 3 is forced on top, and we require another 3. However, the problem doesn't allow this, so this is impossible. Case 2: $2+3$ is on 5, meaning that $4+5$ is on 9. Then, we have two subcases. Case A: 4 is on top. Then, we require 2 to complete the sum of 6, which is not possible as $2+3$ is on 5. Case B: 5 is on top, meaning that we require 1 right below, 3 on the bottom right, 2 to complete the sum of 5, and the rest are uniquely determined. The number is $\boxed{5}$.

Label the vertices $A$, $B$, $C$, $D$, in clockwise order. Then, we use coordinate geometry to easily solve this. Let point $A$ be on $(0, 0)$. Then, point $B$ is 1 unit up and 3 right of $(0, 0)$, or $(1, 3)$. The figure is said to be a square. and so we just calculate the distance from $(0, 0)$ to $(1, 3)$. Using the distance formula, we have $\sqrt{(1-0)^2 + (3-0)^2} = \sqrt{(1+9)} = \sqrt{10}$. The area is just this value squared, or $\boxed{10}$.

Since the integers are equally spaced, they form an arithmetic sequence, so we can label the middle term $a$, the smallest one $a-d$ and the largest one $a+d$. We are given that $(a-d)+a=2a-d=40$ and $a+(a+d)=2a+d=60$. Solving, $2d=60-40=20$ so $d=10$, which implies that $a=\dfrac{60-d}{2}=25$. Finally, the sum of all 3 numbers is $(a-d)+a+(a+d)=3a=3 \cdot 25 = \boxed{\textbf{(B)} 75}$.

We can have $4$ cubes with all their gray faces facing each other, and that will make no gray faces visible. Therefore, our answer is $\boxed{\textbf{(A)}\ 4}$.

The divisibility rule for $4$ is that the last 2 digits must be a multiple of $4$. Because of this, we can "discard" the first two digits, as the last two digits are the only ones that matter (this doesn't actually change the answer, since for every possible case that the first two digits can be, there are the same number of cases the last two digits can be, so the ratio of cases that work to the cases that don't work is still the same). We can also notice that the tens digit also doesn't matter, since every single digit in the number must be even. Therefore, we only need to care about the ones digit. We have the even digits $0$, $2$, $4$, $6$, and $8$, and there are 3 digits that are divisible by $4$, namely $0$, $4$, and $8$, out of $5$ total digits, so we have the final fraction $\boxed{\textbf{(D) }\frac{3}{5}}$.

Label the human beings A, B, C, and D in that order from left to right. It doesn't matter the original configuration. Now, the problem is: How many four letter strings ABCD exist such that A is not next to B, B not next to C, and C is not next to D. We can count the number of strings manually (or just do casework), to obtain $\boxed{2}$. This is incredibly similar to one AMC 10 problem, that asked how many four letter strings abcd have no two consecutive letters in the alphabet next to one another.

Note that the smallest consecutive sum of integers greater than 1 for 10 numbers itself is 55. Thus, we check $k$ consecutive numbers for $2 \le k \le 10$. Case 1: $k=2$ $n+n+1 = 2n+1 = 60$, not possible as $2n+1$ is odd. Case 2: $k=3$ $3n+3 = 60$, in which $n=29$. 1 case. Case 3: $k=4$ $4n+6 = 60$, not possible as 54 is not divisible by 4. Case 4: $k=5$ $5n+10 = 60$ in which $n = 5$, 1 case. Case 6: $k=6$ $6n+15 = 60$, in which 15 is not divisible by 6 and thus no case. Case 7: $k=7$ $7n+21 = 60$, in which 60 isn't divisible by 7 and there are 0 cases. Cases 8-10 continue the same way, with either $k(k+1)/2$ not divisible by $k$ or 60 not divisible by $k$. The answer is $\boxed{2}$.

Let Miguel's speed equal $s$. That makes Luna's speed $5s$. Let the distance between the park entrance and the tree be $d$. The total distance covered by Miguel and Luna is $s + 5s = 6s$, which is also equal to $2d$. Thus $6s = 2d \Rightarrow s = \boxed{ \textbf{(D) } \frac{1}{3}}$.

Let $x$ be the number of gold coins, and let $y$ be the number of silver coins. We know that $y$ must be $0$, $1$, and $2$, since if $y$ was $3$ or more, then the stack of coins would be more than $8$ mm tall. Now we can use casework to find all possible stacks. Case $1$: If $y=0$, then there is only one way to make the stack of coins, which is $8$ gold coins. Case $2$: If $y=1$, then there must be $5$ gold coins, since $8-3\cdot1=5$. There are $6$ coins in total, and we can put the silver coin anywhere in the stack, so there are $\binom{6}{1}=6$ ways to place the silver coin, giving $6$ different possible stacks with $1$ silver coin. (Note that when you place the silver coin anywhere, the rest of the stack is predetermined, because the rest of the stack must be gold, so we only have to account for one type of coin, in this case, the silver coin.) Case $3$: If $y=2$, then there must be $2$ gold coins, so we have 4 coins in total. We can distribute the $2$ silver coins in $\binom{4}{2}=6$ different ways, giving $6$ possible stacks with $2$ silver coins. (Again, note that wherever you place the silver coins, the rest of the stack is predetermined, because the rest of the stack must be gold, so we only have to account for one type of coin, in this case, the silver coin.) Adding all the possible number of cases in all $3$ cases gives $1+6+6=13$, thus our answer is $\boxed{\textbf{(D)}\ 13}$

On the second turn, she cannot be on an outer point. Thus, for the first move she can do from an outer point to an inner, she has 2 ways to go. Then, from said point, she again has 2 ways to go. Finally, once more she has 2 ways to go that will end her up on an outer point. Then, there is just 8 configurations for her that are favorable. There are a total of $2 \times 4 \times 4 = 32$ possible movements. The probability is $\frac{8}{32} = \boxed{\frac{1}{4}}$.

Consider using a five by five square table to solve this problem. The rows represent the five groups, the center column is the median of each group and the center square cell is the least possible value of $M$. The only numbers that really matter is the ones in the shaded area, the rest can be ignored. In order to find the least possible value of $M$ the smallest numbers, $1\ldots 9$, need to be in these nine square cells. $9$ is the smallest number that is able to satisfy both median of a group and median of five medians. Final answer, the least possible value of $M$ is (A) $9$.

In order to find the length of the elastic band, perpendicular lines are added to the figure to divide the band into sections for easier calculation. After the additions, the length of band consist of five identical lines and a full circle, the four shaded areas. The total length is five lines, the diameter of the coins, plus the circumference of a coin, so 5d+4π. Finally, the length of the elastic band is $\boxed{\textbf{(C)}4\pi+20}.$

The superfactorial $51!^1$ is defined as \[ 51!^1 = 1! \times 2! \times 3! \times \cdots \times 51!. \] The exponent of 7 in the prime factorization of $51!^1$ is \[ \sum_{k=1}^{51} v_7(k!), \] where $v_7(k!)$ is the exponent of 7 in $k!$. We know that \[ v_7(k!) = \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor + \left\lfloor \frac{k}{343} \right\rfloor + \cdots. \] Since $343 > 51$, only the first two terms matter. Thus, the total exponent is \[ \sum_{k=1}^{51} \left( \left\lfloor \frac{k}{7} \right\rfloor + \left\lfloor \frac{k}{49} \right\rfloor \right) = \sum_{k=1}^{51} \left\lfloor \frac{k}{7} \right\rfloor + \sum_{k=1}^{51} \left\lfloor \frac{k}{49} \right\rfloor. \] **First sum:** $\sum_{k=1}^{51} \left\lfloor k/7 \right\rfloor$ - $k=7$ to $13$: $\left\lfloor k/7 \right\rfloor = 1$ (7 terms) $\to 7 \times 1 = 7$ - $k=14$ to $20$: $=2$ (7 terms) $\to 14$ - $k=21$ to $27$: $=3$ (7 terms) $\to 21$ - $k=28$ to $34$: $=4$ (7 terms) $\to 28$ - $k=35$ to $41$: $=5$ (7 terms) $\to 35$ - $k=42$ to $48$: $=6$ (7 terms) $\to 42$ - $k=49$ to $51$: $=7$ (3 terms) $\to 21$ Total: $7 + 14 + 21 + 28 + 35 + 42 + 21 = 168$. **Second sum:** $\sum_{k=1}^{51} \left\lfloor k/49 \right\rfloor$ Only $k=49,50,51$ give 1, so the sum is $3$. **Overall exponent:** $168 + 3 = 171$. Therefore, the answer is $\boxed{\textbf{(E) 171}}$.

Let the side lengths of the hexagon be $a, b, c, d, e, f$. Obviously, each one of these is to be in between 1, 2, or 3 . Every 2 can be followed by a 2 or a 3. Every 3 must be followed by a 1 or 2, and every 1 must be followed by a 3. The question is now how many strings of length 6 satisfy these conditions, in which the first and last numbers also satisfy these conditions. To count that, we use recursion. Let $a_n$ be a string of length $n$ ending in 1, $b_n$ is a string of length $n$ ending in 2, and $c_n$ is a string that ends in 3. For $a_{n+1}$, we have that it must come from $c_n$, so $a_{n+1} = c_n$. Additionally, $b_{n+1} = b_n + c_n$, and $c_{n+1} = a_n$. Then, $a_1 = 1$, $b_1 = 1$, and $c_1 = 1$. Thus, $a_2 = 1$, $b_2 = 2$, and $c_2 = 1$; $a_3 = 1$, $b_3 = 3$, and $c_3 = 1$; $a_4 = 1$, $b_4 = 4$, $c_4 = 1$; $a_5 = 1$, $b_5 = 5$, $c_5 = 1$; and finally $a_6 = 1$, $b_6 = 6$, and $c_6 = 1$. Our answer is $a_6 + b_6 + c_6$, or $1+6+1 = \boxed{8}$.