AMC8 2025

Each of the unshaded triangles has base length $2$ and height $1$, so they all have area $\frac{2 \cdot 1}{2} = 1$. Each of the unshaded unit squares has area $1$. The area of the shaded region is equal to the area of the entire grid minus the area of the unshaded region, or $4^2 - 4 \cdot 1 - 4 \cdot 1 = 8$. The star is then $\frac{8}{16} = \frac{1}{2} = \frac{50}{100}$, or $\boxed{\textbf{(B)}~50}$ percent of the entire grid.

The first hieroglyph is worth $10,000$, the next 4 are worth $100 \cdot 4 = 400$, the next $2$ are worth $10 \cdot 2 = 20$, and the last $3$ are worth $1 \cdot 3 = 3$. Therefore, the answer is $10,000 + 400 + 20 + 3 = \boxed{\textbf{(B)}\ 10,423}$

We start with Annika and $3$ of her friends playing, meaning that there are $4$ players. This must mean that there is a total of $4 \cdot 15 = 60$ cards. If $2$ more players joined, there would be $6$ players, and since the cards need to be split evenly, this would mean that each player gets $\frac{60}{6}=\boxed{\text{(C) 10}}$ Buffalo Shuffle-O cards each meaning that the final answer is 10.

We plug $a=100, d=-7$ and $n=10$ into the formula $a+d(n-1)$ for the $n$th term of an arithmetic sequence whose first term is $a$ and common difference is $d$ to get $100-7(10-1) = \boxed{\text{(B) 37}}$.

Each shortest possible path from $A$ to $B$ follows the edges of the rectangle. The following path outlines a path of $\boxed{\textbf{(C)}\ 24}$ units:

Notice the $10$'s place as you erase one number; you get $40$, which is a multiple of $4$, so you can ignore the $10$'s place in the numbers provided altogether because it will always be a multiple of 40. This means that the only relevant numbers are the $1$'s digit ones. If you add the numbers now, you get $5+6+7+8+9=35.$ The closest multiple of $4$ underneath $35$ which can be achieved as a result of subtracting either $5$, $6$, $7$, $8$, or $9$ is $28$, which is achieved by subtracting $7$, therefore the answer is $\boxed{\textbf{(C)}~17}$.

$50$ people scored at least $80\%$, and out of these $50$ people, $13$ of them earned a score that was not less than $90\%$, so the number of people that scored in between at least $80\%$ and less than $90\%$ is $50-13 = \boxed{\text{(D) 37}}$.

Each of the $6$ faces of the cube have equal area, so the area of each face is equal to $\frac{18}{6} = 3$, making the side length $\sqrt3$. From this, we can see that the volume of the cube is $\sqrt{3}^3 = \boxed{\textbf{(A)}~3\sqrt{3}}$

This solution uses Gauss Summation. Our answer is

The area of each rectangle is $5 \cdot 3 = 15$. Then the sum of the areas of the two regions is the sum of the areas of the two rectangles, minus the area of their overlap. To find the area of the overlap, we note that the region of overlap is a square, each of whose sides have length $2.5$ (as they are formed by the midpoint of one of the long sides, the vertex, and also, since it is rotated 90 degrees). Then the answer is $15+15-2.5^2=\boxed{\textbf{(D)}~23.75}$.

The $3\times4$ rectangle allows for $7$ possible places to put the S piece, with each possible placement having an inverted version. One of the cases looks like this: As you can see, there is a hole in the top left corner of the board, which would be impossible to fill using the tetrominos. There are three cases in which a hole isn't created; the S lies flat in the bottom left corner, it lies flat in the top right corner, or it stands upright in the center. All three tilings are shown below. For each of the inverted cases, the L pieces can be inverted along with the S piece. Because the only cases that fill the rectangle after the S is placed are the ones that use two L pieces, the answer must be $\boxed{\textbf{(C)}~I \ and \ L}$.

The largest circle that can fit inside the figure has its center in the middle of the figure and will be tangent to the figure in $8$ points. By the Pythagorean Theorem, the distance from the center to one of these $8$ points is $\sqrt{2^2 + 1^2} = \sqrt5$, so the area of this circle is π52=(C) 5π.

Let's take the numbers 2 through 14 (evens). The remainders will be 2, 4, 6, 1, 3, 5, and 0. This sequence keeps repeating itself over and over. We can take floor(50/14) = 3, so after the number 42, every remainder has been achieved 3 times. However, since 44, 46, 48, and 50 are left, the remainders of those will be 2, 4, 6, and 1 respectively. The only histogram in which those 4 numbers are set at 4 is histogram $\boxed{\textbf{(A)}}$.

The median of the list is $7$, so the mean of the new list will be $7 \cdot 2 = 14$. Since there are $6$ numbers in the new list, the sum of the $6$ numbers will be $14 \cdot 6 = 84$. Therefore, $2+6+7+7+28+N = 84 \implies N = \boxed{\text{(E) 34}}$

We can see that the least number of gold-on-gold pairs will be obtained when the $13$ silver squares are placed on the two sides so that they don't overlap when folded over (because then it will minimize the number of gold-on-golds). We can see that if we split them up $6$ and $7$ on both sides, and then fold it, the number of gold-on-golds will be $18-13 = 5$. The maximum number of gold-on-golds will be achieved when the silver squares overlap when folded over, which will increase the number of gold-on-golds. If we align 6 silver squares with each other on each side, and put the last one somewhere else, we get the maximum is $18 - 7 = 11$. Therefore, the answer is $11+5=\boxed{\textbf{(C)}~16}$.

Note that for no two numbers to differ by $10$, every number chosen must have a different units digit. To make computations easier, we can choose $(1, 2, 3, 4, 5)$ from the first group and $(16, 17, 18, 19, 20)$ from the second group. Then the sum evaluates to $1+2+3+4+5+16+17+18+19+20 = \boxed{\text{(C) 105}}$.

There are $100 \cdot (\frac{1}{4} + \frac{1}{5}) = 100 \cdot \frac{9}{20} = 45$ people who do not work in city $A$ that live in city $A$, meaning that $100 - 45 = 55$ people who live in city $A$ work in city $A$. There are $\frac{1}{3} \cdot 120 = 40$ people who live in city $B$ and work in $A$, as well as $\frac{1}{8} \cdot 160 = 20$ people who live in city $C$ that work in city $A$. Therefore, the answer is $55 + 40 + 20 = \boxed{\textbf{(D)}\ 115}$.

The area of the shaded region in the circle on the left is the area of the circle minus the area of the square, or $\big(\pi-2)$. The shaded area in the circle on the right is $\dfrac{1}{4}$ of the area of the circle minus the area of the square, or $\dfrac{\pi R^2-2R^2}{4}$, which can be factored as $\dfrac{R^2(\pi-2)}{4}$. Since the shaded areas are equal to each other, we have $\pi-2=\dfrac{R^2(\pi-2)}{4}$, which simplifies to $R^2=4$. Taking the square root, we have $R=\boxed{\text{(B) 2}}$

The first car, moving from town $A$ at $25$ miles per hour, takes $\frac{5}{25} = \frac{1}{5} \text{hours} = 12$ minutes. The second car, traveling another $5$ miles from town $B$, takes $\frac{5}{20} = \frac{1}{4} \text{hours} = 15$ minutes. The first car has traveled for 3 minutes or $\frac{1}{20}$th of an hour at $40$ miles per hour when the second car has traveled 5 miles. The first car has traveled $40 \cdot \frac{1}{20} = 2$ miles from the previous $5$ miles it traveled at $25$ miles per hour. They have $3$ miles left, and they travel at the same speed, so they meet $1.5$ miles through, so they are $5 + 2 + 1.5 = \boxed{\textbf{(D) }8.5}$ miles from town $A$.

Let the total amount of cheese be $1$. We will track the amount of cheese Sarika eats throughout the process. First Round: Sarika eats half of the total cheese, so she eats: \[\frac{1}{2}.\] Second Round: Dev eats half of what remains after Sarika's turn, which is: \[\frac{1}{4}.\] Third Round: Rajiv eats half of the remaining cheese after Dev’s turn, which is: \[\frac{1}{8}.\] At the end of the first round, the total cheese eaten is: \[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}.\] We observe that Sarika’s consumption follows a geometric sequence. In the first round, she eats $\frac{1}{2}$, and in subsequent rounds, she eats half of what remains from the previous round. This gives the following series for Sarika’s total consumption: \[\frac{1}{2} + \frac{1}{16} + \frac{1}{128} + \cdots\] This is a geometric series with first term $\frac{1}{2}$ and common ratio $\frac{1}{8}$. The sum $S$ of this infinite geometric series is given by the formula: \[S = \frac{a}{1 - r},\] where $a$ is the first term and $r$ is the common ratio. Substituting $a = \frac{1}{2}$ and $r = \frac{1}{8}$: \[S = \frac{\frac{1}{2}}{1 - \frac{1}{8}} = \frac{\frac{1}{2}}{\frac{7}{8}} = \frac{1}{2} \times \frac{8}{7} = \frac{4}{7}.\] Thus, Sarika eats $\frac{4}{7}$ of the original block of cheese. The correct answer is: \[\boxed{\textbf{(A) } \frac{4}{7}}.\]

The key observation for this solution is to observe that pods $C$ and $F$ both have degree 5; that is, they are connected to five other pods. This implies that $C$ and $F$ must contain grades 1 and 7 in either order (if $C$ or $F$ contained any of grades 2, 3, 4, 5, or 6, then there would only be four possible grades for five pods, a contradiction). It does not matter which grade $C$ and $F$ gets (see Solution 2 below), so we will assume that pod $C$ is assigned grade 1, and pod $F$ is assigned grade 7. Next, pod $D$ is the only pod which is not adjacent to pod $F$, so pod $D$ must be assigned grade 6. Similarly, pod $G$ must be assigned grade 2. Lastly, we need to assign grades 3, 4, and 5 to pods $A$, $B$, and $E$. Note that $A$ and $B$ are adjacent; therefore, pods $A$ and $B$ must contain grades 3 and 5. We conclude that $E$ must be assigned grade 4. The requested sum is $1+4+7 = \mathbf{(A)\,} 12$.

Suppose there are $c$ coats on the rack. Notice that there are $c+1$ "gaps" formed by these coats, each of which must have the same number of empty spaces (say, $k$). Then the values $k$ and $c$ must satisfy $c+k(c+1)=35 \implies kc+k+c=35$. We now use Simon's Favorite Factoring Trick as follows: \[kc+k+c=35\] \[\implies kc+k+c+1=36\] \[\implies (k+1)(c+1)=36\] Our only restrictions now are that $k>0 \implies k+1 > 1$ and $c>0 \implies c+1>1$. Other than that, each factor pair of $36$ produces a valid solution $(k,c)$, which in turn uniquely determines an arrangement. Since $36$ has $9$ factors, our answer is $9-2=\boxed{\textbf{(D)}~7}$. ~cxsmi

The "Condition (II)" perfect square must end in "$00$" because $...99+1=...00$ Condition (I). Four-digit perfect squares ending in "$00$" are ${40, 50, 60, 70, 80, 90}$. Condition (II) also says the number is in the form $n^2-1$. By the Difference of Squares, $n^2-1 = (n+1)(n-1)$. Hence: - $40^2-1 = (39)(41)$ - $50^2-1 = (49)(51)$ - $60^2-1 = (59)(61)$ - $70^2-1 = (69)(71)$ - $80^2-1 = (79)(81)$ - $90^2-1 = (89)(91)$ On this list, the only number that is the product of $2$ prime numbers (condition $3$) is $60^2-1 = (59)(61)$, so the answer is $\boxed{\text{(B) 1}}$.

Let $a$ be the length of the shorter base, and let $b$ be the length of the longer one. Note that these two parameters, along with the angle measures and the fact that the trapezoid is isosceles, uniquely determine a trapezoid. We drop perpendiculars down from the endpoints of the top base. Then the length from the foot of this perpendicular to either vertex will be half the difference between the lengths of the two bases, or $\frac{b-a}{2}$. Now, since we have a 30-60-90 triangle and this side length corresponds to the "30" part, the length of the hypotenuse (one of the legs) is $2 \cdot \frac{b-a}{2} = b-a$. Then the perimeter of the trapezoid is $2(b-a)+a+b=3b-a=30$. The only other stipulation for this trapezoid to be valid is that $b>a$ (which was our assumption). We can now easily count the valid pairs $(a,b)$, yielding $(3,11),(6,12),(9,13),(12,14)$. It is clear that proceeding further would cause $a \geq b$, so we have $\boxed{\textbf{(E)}~4}$ valid trapezoids.

Step 1: To find the total number of paths, observe that all paths will have $10$ total steps. We have to choose which $5$ of these steps will be NE (the rest will be NW). So the total number of paths is $\binom{10}{5}$. The formula for combinations is: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ and $\binom{10}{5} = \frac{10!}{5!\times5!}=252$. Step 2: Each path splits the total area of $25$ in two parts. So, for any path that gives area = $A$, you can find a unique "sister" path that has an area = $25-A$ (in other words, the pair of paths have a combined area of 25). Possible ways to define the "sister" path are: - Rotate the entire grid $180^{\circ}$ - Swap each step of the original paths (for example, each NW becomes NE) (this is a reflection over the diagonal) Step 3: There are a few ways to get from this observation to the total area: - There are $252/2 = 126$ pairs of such paths, and the total area of each pair is $25$. So the total area given by all paths is $126 \times 25$. - Each of the $252$ paths gives an area of $25$ if you also count the "sister" paths. Since each "sister" path is also one of the $252$, you have to divide by $2$ to avoid double counting. So the total area given by all paths is $\frac{252 \times 25}{2}$. - Note that the average area of two "sister" paths is $\frac{25}{2}$, so you can think about every path having this area on average. So the total area given by all paths is $252 \times \frac{25}{2}$. The final answer is $\boxed{\textbf{(B)}~3150}.$ Note: This problem has a bijection (or 1-1 correspondence) , check out Intermediate Counting & Probability, Chapter 4, and Introduction to Counting & Probability, Chapter 5