2020AMC8真题及答案

2020AMC8真题及答案

首发文字版，参考答案见文末（仅供参考）

2020 AMC8 Promblem

Problem 2

Four friends do yardwork for their neighbors over the weekend, earning  and  respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned  give to the others?

Problem 3

Carrie has a rectangular garden that measures  feet by  feet. She plants the entire garden with strawberry plants. Carrie is able to plant strawberry plants per square foot, and she harvests an average of  strawberries per plant. How many strawberries can she expect to harvest?

Problem 4

Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?

Problem 5

Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of  cups. What percent of the total capacity of the pitcher did each cup receive?

Problem 6

Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?

Problem 7

How many integers between  and  have four distinct digits arranged in increasing order? (For example,  is one integer.)

Problem 8

Ricardo has  coins, some of which are pennies (-cent coins) and the rest of which are nickels (-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?

Problem 9

Akash's birthday cake is in the form of a  inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into  smaller cubes, each measuring  inch, as shown below. How many of the small pieces will have icing on exactly two sides?

Problem 10

Zara has a collection of  marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?

 

Problem 11

After school, Maya and Naomi headed to the beach,  miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?

Problem 12

For a positive integer , the factorial notation  represents the product of the integers from  to . What value of  satisfies the following equation?

Problem 13

Jamal has a drawer containing  green socks,  purple socks, and  orange socks. After adding more purple socks, Jamal noticed that there is now a  chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

Problem 14

There are  cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all  cities?

Problem 15

Suppose  of  equals  of  What percentage of  is 

Problem 16

Each of the points  and  in the figure below represents a different digit from  to  Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is  What is the digit represented by 

 

Problem 17

How many positive integer factors of  have more than  factors? (As an example,  has  factors, namely  and )

Problem 18

Rectangle  is inscribed in a semicircle with diameter  as shown in the figure. Let  and let  What is the area of 

 

Problem 19

A number is called flippy if its digits alternate between two distinct digits. For example,  and  are flippy, but  and  are not. How many five-digit flippy numbers are divisible by 

 

Problem 20

A scientist walking through a forest recorded as integers the heights of  trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?

 

 

Problem 21

A game board consists of  squares that alternate in color between black and white. The figure below shows square  in the bottom row and square  in the top row. A marker is placed at  A step consists of moving the marker onto one of the adjoining white squares in the row above. How many -step paths are there from  to  (The figure shows a sample path.)

 

 

Problem 22

When a positive integer  is fed into a machine, the output is a number calculated according to the rule shown below.

For example, starting with an input of  the machine will output  Then if the output is repeatedly inserted into the machine five more times, the final output is When the same -step process is applied to a different starting value of  the final output is  What is the sum of all such integers 

Problem 23

Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?

Problem 24

A large square region is paved with  gray square tiles, each measuring  inches on a side. A border  inches wide surrounds each tile. The figure below shows the case for . When , the  gray tiles cover  of the area of the large square region. What is the ratio for this larger value of 

 

Problem 25

Rectangles  and  and squares  and  shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of  in units?

 

 

2020 AMC8 Answers

Problem 1

Solution 1

Let the side length of each square  be . Then, from the diagram, we can line up the top horizontal lengths of , , and  to cover the top side of the large rectangle, so . Similarly, the short side of  will be , and lining this up with the left side of  to cover the vertical side of the large rectangle gives . We subtract the second equation from the first to obtain , and thus .

Solution 2

Assuming that the problem is well-posed, it should be true in the particular case where  and . Let the sum of the side lengths of  and  be , and let the length of rectangle  be . We then have the systemwhich we solve to determine .

Solution 3 (faster version of solution 1)

Since, for each pair of rectangles, the side lengths have a sum of  or  and a difference of , the answer must be .

 

Problem 2

Solution

The friends earn  in total. Since they decided to split their earnings equally, it follows that each person will get . Since the friend who earned  will need to leave with , he will have to give  to the others.

Problem 3

Solution 1

The area of the garden is  square feet. Since Carrie plants  strawberry plants per square foot, there are a total of strawberry plants, each of which produces  strawberries on average. Accordingly, she can expect to harvest strawberries.

Solution 2

Looking at the units of each quantity, we observe that the answer will be the product of the number of square feet, the number of plants per square foot, and the number of strawberries per plant. This gives .

 

Problem 4

Solution 1

Looking at the rows of each hexagon, we see that the first hexagon has  dot, the second has  dots and the third has  dots, and given the way the hexagons are constructed, it is clear that this pattern continues. Hence the fourth hexagon has  dots.

Solution 2

The first hexagon has  dot, the second hexagon has  dots, the third hexagon  dots, and so on. The pattern continues since to go from hexagon  to hexagon , we add a new ring of hexagons around the outside of the existing ones, with each side of the ring having side length . Thus the number of hexagons added is  (we subtract  as each of the corner hexagons in the ring is counted as part of two sides), confirming the pattern. We therefore predict that that the fourth hexagon has  dots.

Solution 3 (variant of Solution 2)

Let the number of dots in the first hexagon be . By the same argument as in Solution 2, we have  for . Using this, we find that ,  and .

Problem 5

Solution 1

Each cup is filled with  of the amount of juice in the pitcher, so the percentage is .

Solution 2

The pitcher is  full, i.e.  full. Therefore each cup receives  percent of the total capacity.

Solution 3

Assume that the pitcher has a total capacity of  ounces. Since it is filled three fourths with pineapple juice, it contains  ounces of pineapple juice, which means that each cup will contain  ounces of pineapple juice. Since the total capacity of the pitcher was ounces, it follows that each cup received  of the total capacity of the pitcher, yielding  as the answer.

Problem 6

Solution

Write the order of the cars as , where the left end of the row represents the back of the train and the right end represents the front. Call the people , , , , and  respectively. The first condition gives , so we try , , and . In the first case, as  sat in front of , we must have  or , both of which do not comply with the last condition. In the second case, we obtain , which works, while the third case is obviously impossible, since it results in there being no way for  to sit in front of . It follows that, with the only possible arrangement being , the person sitting in the middle car is .

 

Problem 7

Solution 1

Firstly, observe that the second digit of such a number cannot be  or , because the digits must be distinct and increasing. The second digit also cannot be  as the number must be less than , so it must be . It remains to choose the latter two digits, which must be  distinct digits from . That can be done in  ways; there is then only  way to order the digits, namely in increasing order. This means the answer is .

Solution 2 (without using the "choose" function)

As in Solution 1, we find that the first two digits must be , and the third digit must be at least . If it is , then there are  choices for the last digit, namely , , , , or . Similarly, if the third digit is , there are  choices for the last digit, namely , , , and ; if , there are  choices; if , there are  choices; and if , there is  choice. It follows that the total number of such integers is .

Problem 8

Solution 1

Clearly, the amount of money Ricardo has will be maximized when he has the maximum number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is , giving a total of  cents. Analogously, the amount of money he has will be least when he has the greatest number of pennies; as he must have at least one nickel, the greatest number of pennies he can have is also , giving a total of  cents. Hence the required difference is

Solution 2

Suppose Ricardo has  pennies, so then he has  nickels. In order to have at least one penny and at least one nickel, we require  and , i.e. . The number of cents he has is , so the maximum is  and the minimum is , and the difference is therefore

Problem 9

Solution 1

Notice that, for a small cube which does not form part of the bottom face, it will have exactly  faces with icing on them only if it is one of the center cubes of an edge of the larger cube. There are  such edges (as we exclude the  edges of the bottom face), so this case yields  small cubes. As for the bottom face, we can see that only the  corner cubes have exactly  faces with icing, so the total is .

Solution 2

The following diagram shows  of the small cubes having exactly  faces with icing on them; that is all of them except for those on the hidden face directly opposite to the front face.

But the hidden face is an exact copy of the front face, so the answer is .

 

Problem 10

Solution 1

By the Georgeooga-Harryooga Theorem there are  way to arrange the marbles.

Solution by RedFireTruck

Solution 2

We can arrange our marbles like so .

To arrange the  and  we have  ways.

To place the  and  in the blanks we have  ways.

By fundamental counting principle our final answer is 

Solution by RedFireTruck

Solution 3

Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by  and , respectively. If we ignore the constraint that  and  cannot be next to each other, we get a total of  ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that  and  can be next to each other. If we place  and  next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. ). However, we could also have placed  and in the opposite order (i.e. ). Thus there are 6 ways of placing  and  directly next to each other. Next, notice that for each of these placements, we have two open slots for placing  and . Specifically, we can place  in the first open slot and  in the second open slot or switch their order and place  in the first open slot and  in the second open slot. This gives us a total of  ways to place  and  next to each other. Subtracting this from the total number of arrangements gives us  total arrangements .

We can also solve this problem directly by looking at the number of ways that we can place  and  such that they are not directly next to each other. Observe that there are three ways to place  and  (in that order) into the four slots so they are not next to each other (i.e. ). However, we could also have placed  and  in the opposite order (i.e. ). Thus there are 6 ways of placing  and  so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing  and . Specifically, we can place  in the first open slot and  in the second open slot or switch their order and place  in the first open slot and  in the second open slot. This gives us a total of  ways to place  and  such that they are not next to each other .
~junaidmansuri

Solution 4

Let's try complementary counting. There  ways to arrange the 4 marbles. However, there are  arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus,

Solution 5

We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a "super marble." There are  ways to arrange Steelie and Tiger within this "super marble." Then there are ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are  ways to arrange the marbles without any restrictions, the answer is given by 

-franzliszt

Solution 6

We will use the following

 The Georgeooga-Harryooga Theorem states that if you have  distinguishable objects and  of them cannot be together, then there are  ways to arrange the objects.

Let our group of  objects be represented like so , , , ..., , . Let the last  objects be the ones we can't have together.

Then we can organize our objects like so .

We have  ways to arrange the objects in that list.

Now we have  blanks and  other objects so we have  ways to arrange the objects we can't put together.

By fundamental counting principal our answer is .

Proof by RedFireTruck

Back to the problem. By the Georgeooga-Harryooga Theorem, our answer is .

 

Problem 11

Solution 1

Naomi travels  miles in a time of  minutes, which is equivalent to  of an hour. Since , her speed is  mph. By a similar calculation, Maya's speed is  mph, so the answer is .

Solution 2 (variant of Solution 1)

Naomi's speed of  miles in  minutes is equivalent to  miles per hour, while Maya's speed of  miles in  minutes (i.e. half an hour) is equivalent to  miles per hour. The difference is consequently .

Problem 12

Solution 1

We have , and . Therefore the equation becomes , and so . Cancelling the s, it is clear that .

Solution 2 (variant of Solution 1)

Since , we obtain , which becomes  and thus . We therefore deduce .

Solution 3 (using answer choices)

We can see that the answers  to  contain a factor of , but there is no such factor of  in . Therefore, the answer must be

Problem 13

Solution 1

After Jamal adds  purple socks, he has  purple socks and  total socks. This means the probability of drawing a purple sock is , so we obtainSince , the answer is .

Solution 2 (variant of Solution 1)

As in Solution 1, we have the equation . Cross-multiplying yields . Thus, Jamal added  purple socks.

Solution 3 (non-algebraic)

 green socks and  orange socks together should be  of the new total number of socks, so that new total must be . It follows that  purple socks were added.

Problem 14

Solution 1

We can see that the dotted line is exactly halfway between  and , so it is at . As this is the average population of all  cities, the total population is simply .

Solution 2 (estimation)

The dashed line, which represents the average population of each city, is slightly below . Since there are  cities, the answer is slightly less than , which is closest to .

Problem 15

Solution 1

Since , multiplying the given condition by  shows that  is  percent of .

Solution 2

Letting  (without loss of generality), the condition becomes . Clearly, it follows that  is  of , so the answer is .

Solution 3

We have  and , so . Solving for , we multiply by  to give , so the answer is .

Solution 4

We are given , so we may assume without loss of generality that  and . This means , and thus the answer is .

Problem 16

Solution 1

We can form the following expressions for the sum along each line:Adding these together, we must have , i.e. . Since  are unique integers between  and , we obtain  (where the order doesn't matter as addition is commutative), so our equation simplifies to . This means . ~RJ5303707

Solution 2

Following the first few steps of Solution 1, we have . Because an even number () subtracted from an odd number (47) is always odd, we know that  is odd, showing that  is odd. Now we know that  is either 1, 3, or 5. If we try , we get  which is not true. Testing , we get , which is also not true. Therefore, we have .

Problem 17

Solution 1

Since , we can simply list its factors:There are  of these; only  (i.e.  of them) have at most  factors, so the remaining  factors have more than factors.

Solution 2

As in Solution 1, we prime factorize  as , and we recall the standard formula that the number of positive factors of an integer is found by adding  to each exponent in its prime factorization, and then multiplying these. Thus  has  factors. The only number which has one factor is . For a number to have exactly two factors, it must be prime, and the only prime factors of  are , , and . For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of  is . Thus, there are  factors of  which themselves have , , or  factors (namely , , , , and ), so the number of factors of  that have more than  factors is .

Problem 18

Solution 1

Let  be the center of the semicircle. The diameter of the semicircle is , so . By symmetry,  is in fact the midpoint of , so . By the Pythagorean theorem in right-angled triangle  (or ), we have that  (or ) is . Accordingly, the area of  is .

Solution 2 (coordinate geometry)

Let the midpoint of segment  be the origin. Evidently, point  and . Since points  and  share -coordinates with  and  respectively, it suffices to find the -coordinate of  (which will be the height of the rectangle) and multiply this by  (which we know is ). The radius of the semicircle is , so the whole circle has equation ; as already stated,  has the same -coordinate as , i.e. , so substituting this into the equation shows that . Since  at , the y-coordinate of  is . Therefore, the answer is .

(Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.)

Solution 3

We can use a result from the Art of Problem Solving Introduction to Algebra book Sidenote: for a semicircle with diameter , such that the  part is on one side and the  part is on the other side, the height from the end of the  side (or the start of the  side) is . To use this formula, we scale the figure down by ; this will give the height a length of . Now, scaling back up by , the height is . The answer is then .

Problem 19

Solution 1

A number is divisible by  precisely if it is divisible by  and . The latter means the last digit must be either  or , and the former means the sum of the digits must be divisible by . If the last digit is , the first digit would be  (because the digits alternate), which is not possible. Hence the last digit must be , and the number is of the form . If the unknown digit is , we deduce . We know  exists modulo  because 2 is relatively prime to 3, so we conclude that  (i.e. the second and fourth digit of the number) must be a multiple of . It can be , , , or , so there are  options: , , , and .

Solution 2 (variant of Solution 1)

As in Solution 1, we find that such numbers must start with  and alternate with  (i.e. must be of the form ), where the two digits between the s need to be the same. Call that digit . For the number to be divisible by , the sum of the digits must be divisible by ; since the sum of the three s is , which is already a multiple of , it must also be the case that  is a multiple of . Thus, the problem reduces to finding the number of digits from  to  for which  is a multiple of . This leads to , , , or , so there are  possible numbers (namely , , , and ).

Problem 20

Solution 1

We will show that , , , , and  meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height  meters, we can deduce that Trees 1 and 3 both have a height of  meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of  and ,  and , or  and . Checking each of these, in the first case, the average is  meters, which doesn't end in  as the problem requires. Therefore, we consider the other cases. With  and , the average is  meters, which again does not end in , but with  and , the average is  meters, which does. Consequently, the answer is .

Solution 2

Notice the average height of the trees ends with  therefore the sum of all five heights of the trees must end with . ( *  = ) We already know Tree  is  meters tall. Both Tree  and Tree  must  meters tall - since neither can be . Once again, apply our observation for solving for the Tree 's height. Tree  can't be  meters for the sum of the five tree heights to still end with . Therefore, the Tree  is  meters tall. Now the Tree  can either be  or . Find the average height for both cases of Tree . Doing this, we realize the Tree  must be  for the average height to end with  and that the average height is .

Solution 3

As in Solution 1, we shall show that the heights of the trees are , , , , and  meters. Let  be the sum of the heights, so that the average height will be  meters. We note that , so in order for  to end in ,  must be one more than a multiple of . Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both  meters. At this point, our table looks as follows:

If Tree 4 now has a height of , then Tree 5 would need to have height , but in that case  would equal , which is not  more than a multiple of . So we instead take Tree 4 to have height . Then the sum of the heights of the first 4 trees is , so using a height of  for Tree 5 gives , which is  more than a multiple of  (whereas  gives , which is not). Thus the average height of the trees is  meters.

Problem 21

Solution 1

Notice that, in order to step onto any particular white square, the marker must have come from one of the  or  white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from  to that square is the sum of the numbers of ways to move from  to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from  to that square, as already stated).
The answer is therefore .

Solution 2

Suppose we "extend" the chessboard infinitely with  additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.

The total number of paths from  to , including invalid paths which cross over the red line, is then the number of paths which make  steps up-and-right and  steps up-and-left, which is . We need to subtract the number of invalid paths, i.e. the number of paths that pass through  or . To get to , the marker has to make  up-and-right steps, after which it can proceed to  with  steps up-and-left and  step up-and-right. Thus, the number of paths from  to  that pass through  is . Similarly, the number of paths that pass through  is . However, we have now double-counted the invalid paths which pass through both  and ; from the diagram, it is clear that there are only  of these (as the marker can get from  to  by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is , and the number of valid paths from  to  is .

Problem 22

Solution 1

We start with the final output of  and work backwards, taking care to consider all possible inputs that could have resulted in any particular output. This produces the following set of possibilities at each stage:where, for example,  must come from  (as there is no integer  satisfying ), but  could come from  or  (as , and  is even while  is odd). By construction, the last set in this sequence contains all the numbers which will lead to the number  at the end of the -step process, and their sum is .

Solution 2 (variant of Solution 1)

As in Solution 1, we work backwards from , this time showing the possible cases in a tree diagram:

The possible numbers are those at the "leaves" of the tree (the ends of the various branches), which are , , , and . Thus the answer is .

Solution 3 (algebraic)

We begin by finding the inverse of the function that the machine uses. Call the input  and the output . If  is even, , and if  is odd, . We can therefore see that  when  is even and  when  is odd. Therefore, starting with , if  is even, , and if  is odd, , but the latter is not valid since  is not actually odd. This means that the 2nd-to-last number in the sequence has to be . Now, substituting  into the inverse formulae, if  is even,  (which is indeed even), and if  is odd, , which is not an integer. This means the 3rd-to-last number in the sequence has to be . Substituting in , if  is even, , but if  is odd, . Both of these are valid solutions, so the 4th-to-last number can be either  or . If it is , then by the argument we have just made, the 5th-to-last number has to be , the 6th-to-last number has to be , and the 7th-to-last number, which is the first number, must be either  or . In this way, we have ultimately found two solutions:  and .

On the other hand, if the 4th-to-last number is , substituting  into the inverse formulae shows that the 5th-to-last number is either  or , but the latter is not an integer. Substituting  shows that if  is even, , but if I is odd, , and both of these are valid. If the 6th-to-last number is , then the first number must be , since  is not an integer; if the 6th-to-last number is  then the first number has to be , as is not an integer. This means that, in total, there are  solutions for , specifically, , , , and , which sum to .

Problem 23

Solution 1 (complementary counting)

Without the restriction that each student receives at least one award, we could simply take each of the  awards and choose one of the students to give it to, so that there would be  ways to distribute the awards. We now need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are  choices for which student that is, then  ways of choosing a student to receive each of the awards, for a total of . However, if  students both don't receive an award, then such a case would be counted twice among our , so we need to add back in these cases. Of course,  students both not receiving an award is equivalent to only  student receiving all  awards, so there are simply  choices for which student that would be. It follows that the total number of ways of distributing the awards is .

Solution 2

Firstly, observe that it is not possible for a single student to receive  or  awards because this would mean that one of the other students receives no awards. Thus, each student must receive either , , or  awards. If a student receives  awards, then the other two students must each receive  award; if a student receives  awards, then another student must also receive  awards and the remaining student must receive award. We consider each of these two cases in turn. If a student receives three awards, there are  ways to choose which student this is, and  ways to give that student  out of the  awards. Next, there are  students left and  awards to give out, with each student getting one award. There are clearly just  ways to distribute these two awards out, giving  ways to distribute the awards in this case.

In the other case, a student receives  awards. We first have to choose which of the two students we will select to give two awards each to. There are  ways to do this, after which there are  ways to give the first student his two awards, leaving  awards yet to distribute. There are then  ways to give the second student his  awards. Finally, there is only  student and  award left, so there is only  way to distribute this award. This results in  ways to distribute the awards in this case. Adding the results of these two cases, we get .

Solution 3 (variation of Solution 2)

If each student must receive at least one award, then, as in Solution 2, we deduce that the only possible ways to split up the  awards are and  (i.e. one student gets three awards and the others get one each, or two students each get two awards and the other student is left with the last one). In the first case, there are  choices for which student gets  awards, and  choices for which  awards they get. We are then left with  awards, and there are exactly  choices depending on which remaining student gets which. This yields a total for this case of . For the second case, there are similarly  choices for which student gets only  award, and  choices for which award he gets. There are then  remaining awards, from which we choose  to give to one student and  to give to the other, which can be done in  ways (and we can say that e.g. the  chosen this way go to the first remaining student and the other  go to the second remaining student, which counts all possibilities). This means the total for the second case is , and the answer is .

Problem 24

Solution 1

The area of the shaded region is . To find the area of the large square, we note that there is a -inch border between each of the  pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of  times the length of the border, i.e. . Adding this to the total length of the consecutive squares, which is , the side length of the large square is , yielding the equation . Taking the square root of both sides (and using the fact that lengths are non-negative) gives , and cross-multiplying now gives .

Solution 2

Without loss of generality, we may let  (since  will be determined by the scale of , and we are only interested in the ratio ). Then, as the total area of the  gray tiles is simply , the large square has area , making the side of the large square . As in Solution 1, the the side length of the large square consists of the total length of the gray tiles and  lots of the border, so the length of the border is . Since  if , the answer is .

Solution 3 (using answer choices)

As in Solution 2, we let  without loss of generality. For sufficiently large , we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown:
Each red square has side length , so by solving , we obtain . The actual fraction of the total area covered by the gray tiles will be slightly less than , which implies . Hence (and thus , since we are assuming ) is less than , and the only choice that satisfies this is .

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