2023 AMC 8

Solution 1

By the order of operations, we have

Solution 2

We can simplify the expression above in another way:

Solution 1

Notice that when we unfold the paper along the vertical fold line, we get the following shape:

It is clear that the answer is .

Solution 2

When you fold the paper in that specific way, it shows the top left square. That means that any cut you make on that folded part will look like that on that top left square. Since it is folded into four parts, the cut will reflect on all of the other 3 parts. Since there is a cut diagonally on the bottom right of that square, that will reflect on the other squares, making the shape of

Solution 1

By substitution, we have

Solution 2 (Estimation)

is very close to - therefore, we can substitute into the equation to get

As is slightly less than , the correct answer is slightly less than . Therefore, the answer is .

Solution 1

We fill out the grid, as shown below:

From the four numbers that appear in the shaded squares, of them are prime: and

Solution 2

Note that given time constraint, it's better to only count from perfect squares (in pink), as shown below:

From the four numbers that appear in the shaded squares, of them are prime: and

Note that

So, the total number of fish is times the number of trout. Since the lake contains trout, there are fish in the lake.

Solution 1

First, let us consider the case where is a base: This would result in the entire expression being Contrastingly, if is an exponent, we will get a value greater than is greater than and Therefore, the answer is

Solution 2

The maximum possible value of using the digits and : We can maximize our value by keeping the and together in one power (the biggest with the biggest and the smallest with the smallest). This shows (We don't want because that is .) It is going to be

Solution 1

If we extend the lines, we have the following diagram:

Therefore, we see that the answer is

Solution 2

Note that the -intercepts of line and line are and . If the analytic expression for line is , and the analytic expression for line is , we have equations: and . Solving these equations, we can find out that and . Therefore, we can determine that the expression for line is and the expression for line is . When , the coordinates that line and line pass through are and , and lies perfectly on one vertex of the rectangle while the coordinate of is out of the range (lower than the bottom left corner of the rectangle ). Considering that the value of the line will only decrease, and the value of the line will only increase, there will not be another point on the rectangle that lies on either of the two lines. Thus, we can conclude that the answer is

Solution 1

We can calculate the total number of wins ('s) by seeing how many matches were players, which is matches played. Then, we can calculate the # of wins already on the table, which is , so there are wins left in the mystery player. Now, we will make the key observation that there is only wins ('s) per column as there are winners and losers in each round. Strategically looking through the columns counting the 's and putting our own 's when the column isn't already full yields .

Solution 2

We can look one by one. We see that Lola and Lolo won the first game and Tiya lost. This shows that Tiyo must have lost as well because the results must be wins and loses. We use the same logic for games and , giving us 's again. We look at the choices, and we see A is the only that starts with 's.

This shows our answer is

Solution 1

We mark the time intervals in which Malaika's elevation is between and meters in red, as shown below:

The requested time intervals are:

- from the nd to the th seconds - from the th to the th seconds - from the th to the th seconds In total, Malaika spends seconds at such elevation.

Solution 2

Notice that the entire section between the second mark and the second mark is between the and feet elevation level except the seconds where she skis just under the feet mark and when she skis just above the feet mark, making the answer

Note that:

- Harold ate of the pie. After that, of the pie was left behind.

- The moose ate of the pie. After that, of the pie was left behind.

- The porcupine ate of the pie. After that, of the pie was left behind.

More simply, we can condense the solution above into the following equation:

Solution 1

Note that months is approximately hours. Therefore, the speed (in miles per hour) is

As the answer choices are far apart from each other, we can ensure that the approximation is correct.

Solution 2

Note that miles. We also know that months is approximately hours. Now, we can calculate the speed in miles per hour, which we find is about

Solution 1

First, the total area of the radius circle is simply just when using our area of a circle formula.

Now from here, we have to find our shaded area. This can be done by adding the areas of the -radius circles and add; then, take the area of the radius circles and subtract that from the area of the radius circle to get our resulting complex shape area. Adding these up, we will get

.

So, our answer is .

Solution 2

Pretend each circle is a square. The large shaded circle is a square with area , and the two white circles inside it each have areas of , which adds up to . The three small shaded circles become three squares with area , and add up to . Adding the areas of the shaded circles (19) and subtracting the areas of the white circles (8), we get . Since the largest white circle in which all these other circles are becomes a square that has area , our answer is .

Suppose that the race is miles long. The water stations are located at

miles from the start, and the repair stations are located at

miles from the start.

We are given that from which

Solution 1

Let's use the most stamps to make We have of each stamp, -cent (nickels), -cent (dimes), and -cent (quarters).

If we want the highest number of stamps, we must have the highest number of the smaller value stamps (like the coins above). We can use nickels and dimes to bring our total cost to . However, when we try to use quarters, the cents don’t fit evenly, so we have to give back cents to make the quarter amount . The most efficient way to do this is to give back a -cent (dime) stamp and a -cent (nickel) stamp to have stamps used so far. Now, we just use quarters to get a grand total of .

Solution 2

The value of his entire stamp collection is dollars. To make with stamps, he should remove cents worth of stamps with as few stamps as possible. To do this, he should start by removing as many cent stamps as possible as they have the greatest denomination. He can remove at most of these stamps. He still has to remove cents worth of stamps. This can be done with one and cent stamp. In total, he has stamps in his entire collection. As a result, the maximum number of stamps he can use is .

Solution 1

Note that Viswam walks at a constant speed of blocks per hour as he takes minute to walk each block. After walking blocks, he has taken minutes and has minutes remaining to walk blocks. Therefore, he must walk at a speed of blocks per hour to get to school on time, from the time he starts his detour. Since he normally walks mile, which is equal to blocks, mile is equal to blocks. Therefore, he must walk at mph from the time he starts his detour to get to school on time, so the answer is .

Solution 2

Viswam walks blocks, or half a mile, in minutes. Therefore, he walks at a rate of mph. When he takes his detour, he must travel blocks instead of . Our final equation is .

Solution 1

In our grid, there are and of the letters and , respectively, and in a grid, there are and of the letters and , respectively. We see that in both grids, there are and of the and , respectively. This is because in any grid with , there are and of the and , respectively. We can see that the only answer choice which satisfies this condition is

Solution 2

Since and is in the 2nd diagonal, it is also in the 20th diagonal, and so we find that there are . Since all the 's and 's are symmetric, the answer is

Solution 1

We color face red and face yellow. Note that from the octahedron, face and face do not share anything in common. From the net, face shares at least one vertex with all other faces except face which is shown in green:

Therefore, the answer is

Solution 2

We label the octohedron going triangle by triangle until we reach the triangle. The triangle to the left of the should be labeled with a . Underneath triangle is triangle . The triangle to the right of triangle is triangle and further to the right is triangle . Finally, the side of triangle under triangle is , so the triangle to the right of is .

Solution 1

We have directions going right or left. We can assign a variable to each of these directions. We can call going right direction and we can call going left . We can build an equation of , where we have to limit the number of moves we do. We can do this by making more of our moves the move turn than the move turn. The first obvious step is to go some amount of moves in the right direction then subtract off in the left direction to land on . The least amount of ’s added to to make a multiple of is as . So now, we have solved the problem as we just go hops right, and just do 4 more hops left. Yielding as our answer.

Solution 2

Notice that , and jumping to the left increases the value of Greta's position by . Therefore, the number of jumps to the left must be . As the number of jumps to the left increases, so does the number of jumps to the right, we must minimize both, which occurs when we jump to the left and to the right. The answer is .

Solution 1

All equilateral triangles are similar. For the outer equilateral triangle to the inner equilateral triangle, since their side-length ratio is their area ratio is It follows that the area ratio of three trapezoids to the inner equilateral triangle is so the area ratio of one trapezoid to the inner equilateral triangle is

Solution 2

Subtracting the larger equilateral triangle from the smaller one yields the sum of the three trapezoids. Since the ratio of the side lengths of the larger to the smaller one is , we can set the side lengths as and , respectively. So, the sum of the trapezoids is . We are also told that the three trapezoids are congruent, thus the area of each of them is . Hence, the ratio is .

To double the range, we must find the current range, which is , to then double to: . Since we do not want to change the median, we need to get a value less than (as would change the mode) for the smaller, making fixed for the larger. Anything less than is not beneficial to the optimization because you want to get the largest range without changing the mode. So, taking our optimal values of and , we have an answer of .

Solution 1

First, we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. . Then, dividing by , we have , so each group of must have a sum of 15. To make the counting easier, we will just see the possible groups 9 can be with. The possible groups 9 can be with 2 distinct numbers are and . Going down both of these avenues, we will repeat the same process for using the remaining elements in the list. Where there is only 1 set of elements getting the sum of , needs in both cases. After is decided, the remaining 3 elements are forced in a group, yielding us an answer of as our sets are and .

Solution 2

The group with must have the two other numbers adding up to , since the sum of all the numbers is . The sum of the numbers in each group must therefore be . We can have , , , or . With the first group, we have left over. The only way to form a group of numbers that add up to is with or . One of the possible arrangements is therefore . Then, with the second group, we have left over. With these numbers, there is no way to form a group of numbers adding to . Similarly, with the third group there is left over and we can make a group of numbers adding to with or . Another arrangement is . Finally, the last group has left over. There is no way to make a group of numbers adding to with this, so the arrangements are and . So, there are sets that can be formed.