The 2020 AMC 8 is officially over! Now it’s time to discuss the 2020 AMC 8 answers and problems.

Before we dive into the answers, if you participated in 2020 AMC 8, congrats! You did it.

Taking the AMC 8 is no easy feat. No matter what score you received, be proud of yourself! All those long hours you spent studying and preparing for the exam deserve applause.

Once you’re done celebrating, learn where you can improve on next year’s exam! Scroll below to view the 2020 AMC answers and problems. This answer key is a great tool you can use to prepare for next year’s AMC 8 and other math competitions like AMC 10.

With proven successful test results, Think Academy offers top-notch and comprehensive AMC 8 training resources for the 2021/22 exam prep season. 

2020 AMC 8 Problems & Answers Key

AMC 8 Problem 1. Luka is making lemonade to sell at a school fundraiser. His recipe requires 4 times as much water as sugar and twice as much sugar as lemon juice. He uses 3 cups of lemon juice. How many cups of water does he need? ( ).

A. 6 B. 8 C. 12 D. 18 E. 24

Answer: E

The ratio of water to sugar to lemon juice, in general, is 8:2:1. So if 3 cups of lemon juice is used, 8×3=24 cups of water would be needed.

AMC 8 Problem 2. Four friends do yardwork for their neighbors over the weekend, earning 15 dollars, 20 dollars, 25 dollars, and 40 dollars, respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned 40 dollars give to the others? ( ).

A. 5 dollars B. 10 dollars C. 15 dollars D. 20 dollars E. 25 dollars

Answer: C

The average is 15 + 20 + 25 + 40 / 4 = 25 dollars. Thus the person with 40 dollars would redistribute 40 − 25 = 15 dollars.

AMC 8 Problem 3. Carrie has a rectangular garden that measures 6 feet by 8 feet. She plants the entire garden with strawberry plants. Carrie is able to plant 4 strawberry plants per square foot, and she harvests an average of 10 strawberries per plant. How many strawberries can she expect to harvest? ( ).

A. 560 B. 960 C. 1120 D. 1920 E. 3840

Answer: D

There are 6 × 8 = 48 square foot, which results in 4×6×8 = 192 plants. Consequently, 192 × 10 = 1920 strawberries that Carrie harvests.

AMC 8 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? ( ).

A. 35 B. 37 C. 39 D. 43 E.49

Answer: B

Except the very center, moving to a next hexagon includes one more band of dots, i.e. the fourth graph would include a fourth layer, yielding (4 − 1) × 6 = 18 new dots. In total:

1 + (2 − 1) × 6 + (3 − 1) × 6 + (4 − 1) × 6 = 1 + 6 + 12 + 18 = 37

AMC 8 Problem 5. Three-fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into 5 cups. What percent of the total capacity of the pitcher did each cup receive? ( ).

A. 5 B. 10 C. 15. D. 20. E. 25

Answer: C
Equally distribute three quarters into five cups:

43 ÷ 5 × 100% = 15%

AMC 8 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? ( ).

A. Aaron B. Darren. C. Karen D. Maren E. Sharon

Answer: A

Since Maren sat in the last car, she is not the person who sat between Karen and Darren:

___ , ____ , ____ , _____ , M

Now Aaron sat directly behind Sharon, and they can be considered as a small team. Since Darren sat somewhere in front of Aaron, the order of cars now looks like:

___ , D , ___ , S, A, ___ , M

The dots indicate the possibility of other people. Now at least one person sat between Karen and Daren, thus Karen could only be the second to the last, indicating:

D, S, A, K, M

AMC 8 Problem 7. How many integers between 2020 and 2400 have four distinct digits arranged in increasing order? (For example, 2357 is one such integer.) ( )

A. 9 B. 10 C. 15 D. 21 E. 28

Answer: C

In strictly increasing order digits, such number must start from 2300 to 2399, i.e. only 3 could be the hundred digit. Thus, it remains to pick two distinct numbers from 4 to 9. With natural order, we don’t need a permutation. Note that there are 9 − 4 + 1 = 6 digits.

( 6 /2) = 15

AMC 8 Problem 8. Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-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? ( ).

A. 8062 B. 8068 C. 8072 D. 8076 E. 8082

Answer: C

Let’s figure out the greatest and least possible amount of money with at least one penny and at least one nickel. The greatest amount would be the case where one coin is a penny and the rest are nickels. The least amount would be the case where one nickel and all the rest pennies. Hence,

(5×2019+1)−(1×2019+5) = (5−1)×2019+(1−5)

= 4×2019−4 =

4×2018 = 8072

AMC 8 Problem 9. Akash’s birthday cake is in the form of a 4 × 4 × 4 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 64 smaller cubes, each measuring 1 × 1 × 1 inch, as shown below. How many of the small pieces will have icing on exactly two sides? ( ).

A. 12 B. 16 C. 18 D. 20 E. 24

Answer: D

Exactly two sides have icing which means that they are the units on the edges, but not the top corners (only top corners have three sides). BE CAREFUL the bottom corners have two sides as well. Therefore, two on each of the top edges, and three on each of the side edges:

4 × 2 + 4 × 3 = 20.

AMC 8 Problem 10. Zara has a collection of 4 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? ( ).

A. 6 B.8 C.12 D.18 E.24

Answer: E

By complement, first there are 4! = 24 ways to arrange all without restrictions. Next the cases where Steelie is next to Tiger are counted as:

2! x 3! = 12

by grouping them together. Their bundle with the other two forms 3! arrangements, and 2! as a factor of themselves arranged in the bundle. Therefore the complement is:

24−12 = 12

AMC 8 Problem 11. After school, Maya and Naomi headed to the beach, 6 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? ( )

A. 6 B.12 C. 18 D. 2. E. 24

Answer: E

The question asks for the difference in miles per hour, instead of miles per minute:

AMC 8 Problem 12. For positive integer n, the factorial notation n! represents the product of the integers from n to 1. (For example, 6! = 6·5·4·3·2·1.) What value of N satisfies the following equation? ( ).

5! · 9! = 12 · N!

A. 10 B. 11 C. 12 D. 13

Answer: A

AMC 8 Problem 13. Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add? ( ).

A. 6 B. 9 C. 12 D.18 E. 24

Answer: B

Denote x the number of added purple socks. Then:

AMC 8 Problem 14. There are 20 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 20 cities? ( ).

A. 65,000 B. 75,000 C. 85,000 D. 95,000 E. 105,000

Answer: D

The dashed line is very close to 4750. Thus the total is close to:

4750 × 20 = 95000

AMC 8 Problem 15: Suppose 15% of x equals 20% of y. What percentage of x is y? ( ).

A. 5 B. 35 C. 75 D. 133 1/3 E. 300

Answer: C

AMC 8 Problem 16. Each of the points A, B, C, D, E, and F in the figure below represents a different digit from 1 to 6. 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 47. What is the digit represented by B?

A. 1 B. 2 C. 3 D. 4 E. 5

Answer: E

If we count carefully for each digit how many times it is added, then we would obtain:

2A + 3B + 2C + 2D + 2E + 2F =47 = B + 2 (A + B + C + D + E + ).

Note that without ordering, A + B + C + D + E + F =1 + 2 + 3 + 4 + 5 + 6 = 21. Thus:

B = 47 − 2 × 21 = 5

AMC 8 Problem 17: How many factors of 2020 have more than 3 factors? (As an example, 12 has 6 factors, namely 1, 2, 3, 4, 6, and 12. ( ).

A. 6 B. 7 C. 8 D. 9 E. 10

Answer: B

Note that 2020 = 22 × 5 × 101. There are (2 + 1) × ( 1 + 1) × ( 1 + 1 ) = 12 factors of itself.

Way I: Observe the divisor count function. If any number contains at least 2 different prime factors, then it must have at least ( 1 + 1)( 1 + 1) different factors. However, with unique prime factor it could be at most 2+1 = 3.

2 × 5 , 2 × 101 , 2 × 5 × 101 , 22 × 5 , 22 × 101 , 22 × 5 × 101 , 5 × 101

There are 7 of such.

Way II: By complement, exhaust all those numbers with at most 3 factors. Similar to Way I argument, it could only be a power of at most one prime, namely

1, 2, 4, 5, 101

Therefore 12 − 5 = 7 gives us the answer as well.

AMC 8 Problem 18. Rectangle ABCD is inscribed in a semicircle with diameter FE, as shown in the figure. Let DA = 16, and let FD = AE = 9. What is the area of ABCD? ( ).

A. 240 B. 248 C. 256 D. 264 E. 272

Answer: A

The center of the semicircle is the midpoint of EF and given the measures, the midpoint of AD as well, denoted as O. Construct OC.

So the radius is 16 + 9 + 9 / 2 = 17 = OE = OF = OC. Now obviously OD = OA = 8. Then given ABCD is a ∠CDO = 90◦ Pythagorean Theorem indicates:

The area of the rectangle then becomes:

15 × 16 = 240

AMC 8 Problem 19. A number is called flippy if its digits alternate between two distinct digits. For example, 2020 and 37373 are flippy, but 3883 and 123123 are not. How many five-digit flippy numbers are divisible by 15? ( ).

A. 3 B. 4 C. 5 D. 6 E. 8

Answer: B

A flippy five-digit number must be of form:

If n is divisible by 15, then it is divisible by 3 and 5 simultaneously.

Divisible by 5: a = 0 or 5. However, as the leading digit, it couldn’t be 0. Thus a = 5.
Divisible by 3: Divisibility of 3 only relates to the sum of digits.

Note that it requires two distinct digits. Luckily 5 is not one of the options. Thus 1 × 4 = 4 in total. Namely:

50505, 53535, 56565, 59595

AMC 8 Problem 20. A scientist walking through a forest recorded as integers the heights of 5 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? ( ).

A. 22.2 B. 24.2 C. 33.2 D. 35.2 E. 37.2

Answer: B

Tree 1 must be of height 22 meters. It is either half or twice of 11, however it is also an integer. Similar idea implies that Tree 3 must be of height 22 meters. Tree 4 could be 11 or 44 meters.

11 meters high: Then Tree 5 must be of 22 meters, as discussed. Now the average is:

The tenth digit doesn’t match.

44 meters high: Tree 5 could be of height 22 or 88 meters. Let’s check how the tenth digit is generated: it must be the unit digit of the sum dividing 5.

If it is 2: 1/5 (2 + 1 + 2 + 4 + 2) = 2.2, satisfied.

If it is 8: 1/5(2 + 1 + 2 + 4 + 8) = 3.4, unsatisfied.

These five trees are of heights 22, 11, 22, 44, 22 meters, respectively, which gives the average height:

AMC 8 Problem 21. A game board consists of 64 squares that alternate in color between black and white. The figure below shows square P in the bottom row and square Q in the top row. A marker is placed at P. A step consists of moving the marker onto one of the adjoining white squares in the row above. How many 7-step paths are there from P to Q ? (The figure shows a sample path.) ( ).

A. 28 B. 30 C. 32 D. 33 E. 35

Answer: A

7-step paths imply that each step must be moving upward, either left or right, to an adjoint white square. Thus labeling a possible number of ways starting from 1 at P would be enough:

In each upper level white square, the number of paths is the sum of number of ways in its previous black squares. It is 28 to get to Q.

AMC 8 Problem 22. When a positive integer N is fed into a machine, the output is a number calculated according to the rule shown below.

For example, starting with an input of N = 7, the machine will output 3 · 7 + 1 = 22. Then if the output is repeatedly inserted into the machine five more times, the final output is 26.

7 → 22 → 11 → 34 → 17 → 52 → 26

When the same 6-step process is applied to a different starting value of N, the final output is 1. What is the sum of all such integers N? ( ).

N→___→___→___→___→___→1

A. 73 B. 74 C. 75 D. 82 E. 83

Answer: E

a1 and a2 are easy. To obtain 1, the previous number of a2 must be 2. Although 1 is in 3N + 1 format, that N is 0, which is not a positive number, not to mention an odd positive number. Similarly the one in front of 2, a2 is 4, since 2 is not in 3N + 1 format.

N→ ____→ ____→ ____→4→2→1

However, 4 is in 3N + 1 format. So now a3 could be 1 o r8.

If a3 =1, then follow the analysis above. a4 =2, a5 =4. We obtain N1 =1, N2 =8.

If a3 =8, then a4 = 16 since 8 is not in 3N + 1format. a5 could be 5 or 32.

If a5 = 5, then N3 =10, as5 is not in 3N + 1format.

If a5 = 32, then N4 = 64, as 32 is also not in 3N +1format.

N1 + N2 + N3 + N4 = 1 + 8 + 10 + 64 = 83.

The setting refers to the famous 3N + 1 problem.

AMC 8 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? ( ).

A. 120 B. 150 C. 180 D. 210 E. 240

Answer: B

Note that awards and students differ. Each student receive at least one award, then five awards could be separated into 3+1+1 or 2+2+1 awards.

3+1+1 assignments:

2 + 2 + 1 assignments:

Together 60 + 90 = 150 different ways.

AMC 8 Problem 24. A large square region is paved with n2 gray square tiles, each measuring s inches on a side. A border d inches wide surrounds each tile. The figure below shows the case for n = 3. When n = 24, the 576 gray tiles cover 64% of the area of the large square region. What is the ratio ds for this larger value of n? ( ).

A. 6/25 B. 1/4 C. 9/25 D. 7/16 E. 9/16

Answer: A

The large square is of side length √576s + (√576 + 1)d. Here, √576 = 24.

AMC 8 Problem 25. Rectangles R1 and R2, and squares S1, S2, and S3, shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of S2 in units? ( ).

A. 651 B. 655 C. 656 D. 662 E. 666

Answer: A

Let the side lengths of S1, S2, and S3 be a, b, and c respectively. Then the length and width of the large rectangle could be expressed in terms of a, b and c: