2024 AMC 8

Solution 1

We can rewrite the expression as . We note that the units digit of is because all the units digits of the five numbers are and , which has a units digit of . Now, we have something with a units digit of subtracted from , and so the units digit of this expression is .

Solution 2

We only care about the units digits. Thus, ends in , after regrouping(10-2) ends in , ends in , ends in , and ends in .

We see that is ; simplifies to , which is ;

and simplifies to , which is ;

reveals

is .

Solution 1

We work inwards. The area of the outer shaded square is the area of the whole square minus the area of the second largest square. The area of the inner shaded region is the area of the third largest square minus the area of the smallest square. The sum of these areas is

Solution 2

In solution 1, we can use Difference of squares to get the answer, rather than calculating the squares of the numbers:

Adding numbers 1 through 9 gives us 45. This was her expected sum, but what she got was a perfect square. Since she got that perfect square sum by forgetting a number, that sum is less than 45. The square number right under 45 is 36. So 45 - 36 = 9. So the solution is E[9].

Solution 1

First, figure out all pairs of numbers whose product is 6. Then, using the process of elimination, we can find the following:

is possible:

is possible:

is possible:

The only integer that cannot be the sum is .

Solution 2

First, see that in order for the two numbers to have a product of 6, it must either have 6 as a factor or have factors that have a product of 6 (Notably ).

When we look at the answer choices, we see that answers all are more than 6, so they can all be added as a sum of 6. Answer works because , so 6 is the only answer choice that doesn't work. Thus, the answer is .

Solution 1

You can measure the lengths of the paths until you find a couple of guaranteed true inferred statements as such: is greater than , is greater than , and and are the smallest two, therefore the order is Thus we get the answer .

Solution 2

Obviously Path Q is the longest path, followed by Path S.

So, it is down to Paths P and R.

Notice that curved lines are always longer than the straight ones that meet their endpoints, therefore Path P is longer than Path R.

Thus, the order from shortest to longest is .

Solution 1

We can eliminate B, C, and D, because they are not subtracted by any multiple of . Finally, we see that there is no way to have A, so the solution is .

Solution 2

Let be the number of by tiles. There are squares and each by or by tile takes up 4 squares, so , so it is either or . Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are red squares and blue squares, but each by and by shape takes up an equal number of blue and red squares, so there must be more by tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is , which can easily be confirmed to work.

Solution 1

How many dollar values could be on the first day? Only dollars. The second day, you can either add dollars, or double, so you can have dollars, or . For each of these values, you have values for each. For dollars, you have dollars or , and for dollars, you have dollars or dollars. Now, you have values for each of these. For dollars, you have dollars or , for dollars, you have dollars or , for dollars, you have dollars or , and for dollars, you have dollars or .

On the final day, there are 11, 11, 16, and 16 repeating, leaving you with different values.

Solution 2

Continue as in Solution 1 to get , , or dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by or adding ) from here is if or which both aren't true. Hence our answer is .

Solution 1

Since she has half as many red marbles as green, we can call the number of red marbles , and the number of green marbles . Since she has half as many green marbles as blue, we can call the number of blue marbles . Adding them up, we have: marbles. The number of marbles therefore must be a multiple of , as represents an integer, so the only possible answer is

Solution 2

Suppose Maria has green marbles and total marbles. She then has red marbles and blue marbles. Altogether, Maria has

marbles, so so must be a multiple of . The only multiple of in the answer choices is

Solution 1

This is a time period of years, so we can expect the ppm to increase by ppm. Since we started with ppm, we have .

Solution 2

years. The ppm level in 2030 is .

Solution 1

Since the triangle has a base of , we can plug in that value as the base. Then, we can solve the equation for the height. Doing so gives us,

This means that , so that means that we have to add 4 to the -coordinate. So the answer is

Solution 2

By the Shoelace Theorem, has area

From the problem, this is equal to . We now solve for y.

OR

OR

OR

However, since, as stated in the problem, , our only valid solution is .

Solution 1

Let denote the number of guppies in the first tank.

Then, we have the following for the number of guppies in the rest of the tanks:

- The number of guppies in the second tank is

- The number of guppies in the third tank is

- The number of guppies in the fourth tank is

The number of guppies in all of the tanks combined is 90, so we can write the equation

Simplifying the equation gives

Solving the resulting equation gives , so the number of guppies in the fourth tank is .

Solution 2

Suppose there are no guppies in the first tank. Then, the number of guppies in the other tanks are and or guppies in total. We need to add guppies into tanks or guppies in each tank, so the number of guppies in the fourth tank is

Solution 1

Looking at the answer choices, you see that you can list them out. Doing this gets you:

Counting all the paths listed above gets you .

Solution 2

Any combination can be written as some re-arrangement of . Clearly we must end going down, and start going up, so we need the number of ways to insert 2 's and 2 's into . There are ways, but we have to remove the case , giving us .

Solution 1

We can simply see that path will give us the smallest value. Adding, . This is nice as it’s also the smallest value, solidifying our answer.

You can also simply brute-force it or sort of think ahead - for example, getting from A to M can be done ways; () or , so you should take the shorter route (). Another example is M to C, two ways - one is and the other is . Take the shorter route. After this, you need to consider a few more times - consider if () is greater than ), which it is not, and consider if ) is greater than () or () which it is not. TLDR: .

Solution 2

We can execute Dijkstra's algorithm by hand to find the shortest path from to every other town, including . This effectively proves that, assuming we execute the algorithm correctly, that we will have found the shortest distance. The distance estimates for each step of the algorithm (from to each node) are shown below:

The steps are as follows: starting with the initial node , set and for all where indicates the distance from to . Consider the outgoing edges and and update the distance estimates and , completing the first row of the table.

The node is the unvisited node with the lowest distance estimate, so we will consider and its outgoing edges and . The distance estimate equals , and the distance estimate updates to , because . This completes the second row of the table. Repeating this process for each unvisited node (in order of its distance estimate) yields the correct distance once the algorithm is complete.

Solution 1

The highest that can be would have to be , and it cannot be higher than that, because then it would be , and multiplied by 8 is , and then it would exceed the - digit limit set on .

So, if we start at , we get , which would be wrong because both would be , and the numbers cannot be repeated between different letters.

If we move on to the next highest, , and multiply by , we get . All the digits are different, so would be , which is . So, the answer is .

Solution 2

Notice that .

Likewise, .

Therefore, we have the following equation:

.

Simplifying the equation gives

.

We can now use our equation to test each answer choice.

We have that , so we can find the sum:

.

So, the correct answer is .

Solution 1

Note you can swap/rotate any configuration of rows, such that all the rows and columns that have a product of 3 are in the top left. Hence the points are bounded by a rectangle. This has area and rows and columns divisible by . We want and minimized.

If , we achieve minimum with .

If ,our best is . Note if , . Because , there is no smaller answer, and we get .

Solution 2

For a row or column to have a product divisible by , there must be a multiple of in the row or column. To create the least amount of rows and columns with multiples of , we must find a way to keep them all together, to minimize the total number of rows and columns with multiples of threes in it. From to , there are multiples of . So we have to fill cells with numbers that are multiples of . If we put of these numbers in a grid, there would be rows and columns ( in total), with products divisible by . However, we have numbers, so numbers still need to be put in the grid. If we put both numbers in the th column, but one in the first row, and one in the second row, (next to the by grid already filled), we would have a total of columns now, and still rows with products that are multiples of . Since , the answer is

Solution 1

If you place a king in any of the corners, the other king will have spots to go and there are corners, so . If you place a king in any of the edges, the other king will have spots to go and there are edges so . That gives us spots for the other king to go into in total. So is the answer.

Solution 2

We see that the center is not a viable spot for either of the kings to be in, as it would attack all nearby squares.

This gives three combinations:

Corner-corner: There are 4 corners, and none of them are touching orthogonally or diagonally, so it's

Corner-edge: For each corner, there are two edges that don't border it,

Edge-edge: The only possible combinations of this that work are top-bottom and left-right edges, so for this type

Multiply by two to account for arrangements of colors to get .

Solution 1

Let .

We see that the shaded region is the inner ring plus a sector of the outer ring. Using the formula for the area of a circle (), we find that the area of is . This simplifies to .

The unshaded portion is comprised of the smallest circle plus the sector of the outer ring, which evaluates to .

We are told these are equal. Therefore, . Solving for reveals .

Solution 2

Notice that for the 3rd most outer ring of the circle, the ratio of the shaded region to non-shaded region is the ratio of to . With that, all we need to do is solve for the shaded region.

The inner most circle has radius , and the second circle has radius 2. Therefore, the first shaded area has area. The circle has total area , so the other shaded region must have area, as the non-shaded and shaded area is equivalent. So for the 3rd outer ring, the total area is , so the non-shaded part of the outer ring is .

Now as said before, the ratio of these two areas is the ratio of and . So, . We have where , , so our answer is .

Solution 1

Jordan has high top sneakers, and white sneakers. We would want as many white high-top sneakers as possible, so we set high-top sneakers to be white. Then, we have red high-top sneakers, so the answer is

Solution 2

We first start by finding the amount of red and white sneakers. 3/5 * 15=9 red sneakers, so 6 are white sneakers. Then 2/3 * 15=10 are high top sneakers, so 5 are low top sneakers. Now think about 15 slots and the first 10 are labeled high top sneakers. if we insert the last 5 sneakers as red sneakers there are 4 leftover over red sneakers. Putting those four sneakers as high top sneakers we have our answer as C, or

Solution 1

The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are possible triangles. So the answer is .

Solution 2

Each other compatible point must be an even number of edges away from P, so the compatible points are R, V, and T. Therefore, we must choose two of the three points, because P must be a point in the triangle. So, the answer is .

Solution 1

Let the initial number of green frogs be and the initial number of yellow frogs be . Since the ratio of the number of green frogs to yellow frogs is initially , . Now, green frogs move to the sunny side and yellow frogs move to the shade side, thus the new number of green frogs is and the new number of yellow frogs is . We are given that , so , since , we have , so and . Thus the answer is