2025 AMC 8

Solution 1

Each of the unshaded triangles has base length and height , so they all have area . Each of the unshaded unit squares has area . The area of the shaded region is equal to the area of the entire grid minus the area of the unshaded region, or . The star is then , or percent of the entire grid.

Solution 2

There are total squares in the diagram and each square has triangles whose areas are half the area of a unit square. Thus, the total number of triangles in the diagram is triangles. There are shaded triangles in the diagram, so the area of the star is , or percent.

Solution 1

The first hieroglyph is worth , the next 4 are worth , the next are worth , and the last are worth . Therefore, the answer is

Solution 2

Simply notice that the first hieroglyph represents , and the next ones represent hundreds. The only answer choice with a in the thousands place and a in the hundreds place is answer choice

We start with Annika and of her friends playing, meaning that there are players. This must mean that there is a total of cards. If more players joined, there would be players, and since the cards need to be split evenly, this would mean that each player gets Buffalo Shuffle-O cards each meaning that the final answer is 10.

Solution 1

We plug and into the formula for the th term of an arithmetic sequence whose first term is and common difference is to get .

Solution 2

Since we want to find the th number Lucius says after he says , is subtracted from his number times, so our answer is

Solution 1

Each shortest possible path from to follows the edges of the rectangle. The following path outlines a path of units:

Solution 2

We can find the shortest distance using a line diagonally from one point to the other, creating a sort of slope, then find the sum of rise and run of the slope, which happens to be the shortest distance, repeat this process until you get back to Point , and you should get , which is equal to .

Solution 1

The sum of all five numbers is . Since is more than a multiple of , the number being subtracted must be more than a multiple of . Thus, the answer is .

Solution 2

The sum of the residues of these numbers modulo is . Hence, the number being subtracted must be congruent to modulo . The only such answer is .

Solution 1

people scored at least , and out of these people, of them earned a score that was not less than , so the number of people that scored in between at least and less than is .

Solution 2

Let denote the number of people who had a score of at least , but less than , and let denote the number of people who had a score of at least but less than . Our answer is equal to . We find , while . Thus, the answer is .

Solution 1

Each of the faces of the cube have equal area, so the area of each face is equal to , making the side length . From this, we can see that the volume of the cube is

Solution 2

Let the side length of the cube shown be equal to centimeters. The surface area of this cube is equal to the area of the net of the cube, which is equal to square centimeters. The surface area of this cube is also square centimeters, so we have

However, we aren't done. We have found that the side length of the cube is , but the question asks for the volume of the cube, which is equal to cubic centimeters.

Solution 1

Our answer is

Solution 2

We proceed with brute force. All of the pairs of opposite numbers on the clock are , , , , , , where order doesn't matter. The averages of each of these pairs are and respectively, and the average these numbers is

The area of each rectangle is . 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 (as they are formed by the midpoint of one of the long sides and a vertex). Then the answer is .

The rectangle allows for 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 .

Solution 1

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 points. By the Pythagorean Theorem, the distance from the center to one of these points is , so the area of this circle is .

Solution 2

Draw the circle in the grid and analyze the radius. Its radius is a little more than 2 but a lot less than 2.5, so the area is a little more than . So, the area of the circle is with a radius of approximately 2.23.

Solution 1

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 .

Solution 2

Writing down all the remainders gives us

In this list, there are numbers with remainder , numbers with remainder , numbers with remainder , numbers with remainder , numbers with remainder , numbers with remainder , and numbers with remainder . Manually computation of every single term can be avoided by recognizing the pattern alternates from to and there are terms. The only histogram that matches this is .

Solution 1

The median of the list is , so the mean of the new list will be . Since there are numbers in the new list, the sum of the numbers will be . Therefore,

Solution 2

Since the average right now is 10, and the median is 7, we see that N must be larger than 10, which means that the median of the 6 resulting numbers should be 7, making the mean of these 14. We can do 2 + 6 + 7 + 7 + 28 + N = 14 * 6 = 84. 50 + N = 84, so N =

Solution 1

First, we note that there are "pairs" of squares folded on top of each other after the folding. The minimum number of gold pairs occurs when silver squares occupy the maximum number of pairs, and the maximum number of gold pairs occurs when silver squares occupy the minimum number of pairs. The latter case occurs when all silver squares are placed in different pairs, resulting in gold pairs. The former case occurs when the silver squares are paired up as much as possible, resulting in complete pairs and another square occupying another pair slot. Then there are gold pairs. Our answer is .

Solution 2

We can see that the least number of gold-on-gold pairs will be obtained when the 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 and on both sides, and then fold it, the number of gold-on-golds will be .

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 . Therefore, the answer is .

Solution 1

Note that for no two numbers to differ by , every number chosen must have a different units digit. To make computations easier, we can choose from the first group and from the second group. Then the sum evaluates to .

Solution 2

For , we can add the first term and the last term, which is . If we add the second term and the second-to-last term it is also . There are pairs that sum to , so the answer is which equals .

Solution 1

There are people who do not work in city that live in city , meaning that people who live in city work in city . There are people who live in city and work in , as well as people who live in city that work in city . Therefore, the answer is .

Solution 2

We could also make an equation. Let's denote the number of people that live in city as , as , and as . If denotes the number of people working in city , then . This is because of the people from City and of the people from City work in city as shown in the image. We also know that of the people living in work in other cities. We are already given the values of variables , , and as , , and respectively. Plug the values it into the main equation like this: . We solve for it and get our answer .

Solution 1

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 . The shaded area in the circle on the right is of the area of the circle minus the area of the square, or , which can be factored as . Since the shaded areas are equal to each other, we have , which simplifies to . Taking the square root, we have

Solution 2

We start with the first area. Since the square is inscribed, its diagonal is its side length is its area is , therefore the first area is . The second area is , found in a similar manner. Writing and solving the equation, we haveThe answer is .

Solution 1

The first car, moving from town at miles per hour, takes minutes. The second car, traveling another miles from town , takes minutes. The first car has traveled for 3 minutes or th of an hour at miles per hour when the second car has traveled 5 miles. The first car has traveled miles from the previous miles it traveled at miles per hour. They have miles left, and they travel at the same speed, so they meet miles through, so they are miles from town .

Solution 2

From the answer choices, the cars will meet somewhere along the mph stretch. Car travels mph for miles, so we can use dimensional analysis to see that it will be of an hour for this portion. Similarly, car spends of an hour on the mph portion.

Suppose that car travels miles along the mph portion-- then car travels miles along the mph portion. By identical methods, car travels for hours, and car travels for hours.

At their meeting point, cars and will have traveled for the same amount of time, so we have

so , and miles. This means that car will have traveled miles.

Solution 1

Let the total amount of cheese be . We will track the amount of cheese Sarika eats throughout the process.

First Round: Sarika eats half of the total cheese, so she eats:

Second Round: Dev eats half of what remains after Sarika's turn, which is:

Third Round: Rajiv eats half of the remaining cheese after Dev's turn, which is:

At the end of the first round, the total cheese eaten is:

We observe that Sarika's consumption follows a geometric sequence. In the first round, she eats , and in subsequent rounds, she eats half of what remains from the previous round. This gives the following series for Sarika;s total consumption:

This is a geometric series with first term and common ratio . The sum of this infinite geometric series is given by the formula:

where is the first term and is the common ratio. Substituting and :

Thus, Sarika eats of the original block of cheese. The correct answer is:

Solution 2

Sarika eats of the original cheese, and then because the others eat and , she eats next, and then , and then so on. Since the values later are going to be too small to make a huge difference, we can use these values. She ate . We can replace the with a for now, so , which simplifies to around . Since there is a little bit more of the cheese to be accounted for, the amount that she eats will be around .

Solution 1

The key observation for this solution is to observe that pods and both have degree 5; that is, they are connected to five other pods. This implies that and must contain grades 1 and 7 in either order (if or 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 and gets (see Solution 2 below), so we will assume that pod is assigned grade 1, and pod is assigned grade 7.

Next, pod is the only pod which is not adjacent to pod , so pod must be assigned grade 6. Similarly, pod must be assigned grade 2.

Lastly, we need to assign grades 3, 4, and 5 to pods , , and . Note that and are adjacent; therefore, pods and must contain grades 3 and 5. We conclude that must be assigned grade 4. The requested sum is .

Solution 2

Suppose that we replaced each label with . That is, we replaced the grade assigned to with the grade , the grade assigned to with , and so on. We claim that if the initial labeling is valid (i.e., each pair of connected pods has grades differing by 2 or more), then so will the new labeling. This follows as if , then . This shows that if there is an assignment of grades to the pods, where , then there is a second, valid assignment.

But because this is an AMC 8 problem and there is only one correct answer, we can conclude that . Simplifying and solving for gives .

Solution 1

Suppose there are coats on the rack. Notice that there are "gaps" formed by these coats, each of which must have the same number of empty spaces (say, ). Then the values and must satisfy . We now use Simon's Favorite Factoring Trick as follows:

Our only restrictions now are that and . Other than that, each factor pair of produces a valid solution , which in turn uniquely determines an arrangement. Since has factors, our answer is .

Solution 2

Say Paulina placed coats. That will divide the 35 hooks into spaces and empty hooks. Therefore,The values of that satisfy this areThe answer is .

The Condition (II) perfect square must end in "" because Condition (I). Four-digit perfect squares ending in "" are .