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2020AMC8真題及答案

Category: AMC美國數學競賽 Date: 2020年12月7日下午6:06

2020AMC8真題及答案

首發文字版，參考答案見文末（僅供參考）

2020 AMC8 Promblem

Problem 2

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

$textbf{(A) }$5 qquad textbf{(B) }$10 qquad textbf{(C) }$15 qquad textbf{(D) }$20 qquad textbf{(E) }$25$

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?

$textbf{(A) }560 qquad textbf{(B) }960 qquad textbf{(C) }1120 qquad textbf{(D) }1920 qquad textbf{(E) }3840$

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? $[asy] size(250); real side1 = 1.5; real side2 = 4.0; real side3 = 6.5; real pos = 2.5; pair s1 = (-10,-2.19); pair s2 = (15,2.19); pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); fill(circle(origin + s1, 1), grey1); for (int i = 0; i < 6; ++i) { draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25)); } fill(circle(origin, 1), grey1); for (int i = 0; i < 6; ++i) { fill(circle(pos*dir(60*i),1), grey2); draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25)); } fill(circle(origin+s2, 1), grey1); for (int i = 0; i < 6; ++i) { fill(circle(pos*dir(60*i)+s2,1), grey2); fill(circle(2*pos*dir(60*i)+s2,1), grey1); fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1); draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25)); } [/asy]$

$textbf{(A) }35 qquad textbf{(B) }37 qquad textbf{(C) }39 qquad textbf{(D) }43 qquad textbf{(E) }49$

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? $5$ ?cups. What percent of the total capacity of the pitcher did each cup receive?

$textbf{(A) }5 qquad textbf{(B) }10 qquad textbf{(C) }15 qquad textbf{(D) }20 qquad textbf{(E) }25$

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?

$textbf{(A) }text{Aaron} qquad textbf{(B) }text{Darren} qquad textbf{(C) }text{Karen} qquad textbf{(D) }text{Maren}qquad textbf{(E) }text{Sharon}$

Problem 7

How many integers between? $2020$ ?and? $2400$ ?have four distinct digits arranged in increasing order? (For example,? $2347$ ?is one integer.)

$textbf{(A) }text{9} qquad textbf{(B) }text{10} qquad textbf{(C) }text{15} qquad textbf{(D) }text{21}qquad textbf{(E) }text{28}$

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?

$textbf{(A) }text{8062} qquad textbf{(B) }text{8068} qquad textbf{(C) }text{8072} qquad textbf{(D) }text{8076}qquad textbf{(E) }text{8082}$

Problem 9

Akash's birthday cake is in the form of a? $4 times 4 times 4$ ?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? $1 times 1 times 1$ ?inch, as shown below. How many of the small pieces will have icing on exactly two sides?

$[asy] /* Created by SirCalcsALot and sonone Code modfied from https://artofproblemsolving.com/community/c3114h2152994_the_old__aops_logo_with_asymptote */ import three; currentprojection=orthographic(1.75,7,2); //++++ edit colors, names are self-explainatory ++++ //pen top=rgb(27/255, 135/255, 212/255); //pen right=rgb(254/255,245/255,182/255); //pen left=rgb(153/255,200/255,99/255); pen top = rgb(170/255, 170/255, 170/255); pen left = rgb(81/255, 81/255, 81/255); pen right = rgb(165/255, 165/255, 165/255); pen edges=black; int max_side = 4; //+++++++++++++++++++++++++++++++++++++++ path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle; path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle; path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle; for(int i=0; i<max_side; ++i){ for(int j=0; j<max_side; ++j){ draw(shift(i,j,-1)*surface(topface),top); draw(shift(i,j,-1)*topface,edges); draw(shift(i,-1,j)*surface(rightface),right); draw(shift(i,-1,j)*rightface,edges); draw(shift(-1,j,i)*surface(leftface),left); draw(shift(-1,j,i)*leftface,edges); } } picture CUBE; draw(CUBE,surface(leftface),left,nolight); draw(CUBE,surface(rightface),right,nolight); draw(CUBE,surface(topface),top,nolight); draw(CUBE,topface,edges); draw(CUBE,leftface,edges); draw(CUBE,rightface,edges); // begin made by SirCalcsALot int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}}; for (int i = 0; i < max_side; ++i) { for (int j = 0; j < max_side; ++j) { for (int k = 0; k < min(heights[i][j], max_side); ++k) { add(shift(i,j,k)*CUBE); } } } [/asy]$
$textbf{(A) }12 qquad textbf{(B) }16 qquad textbf{(C) }18 qquad textbf{(D) }20 qquad textbf{(E) }24$

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?

$textbf{(A) }6 qquad textbf{(B) }8 qquad textbf{(C) }12 qquad textbf{(D) }18 qquad textbf{(E) }24$

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?

$[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7}); for (int i = 0; i < 6; ++i) { for (int j = 0; j < 6; ++j) { draw((i,0)--(i,6), grey); draw((0,j)--(6,j), grey); } } for (int i = 1; i <= 6; ++i) { draw((-0.1,i)--(0.1,i),linewidth(1.25)); draw((i,-0.1)--(i,0.1),linewidth(1.25)); label(string(5*i), (i,0), 2*S); label(string(i), (0, i), 2*W); } draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25)); label(rotate(90) * "Distance (miles)", (-0.5,3), W); label("Time (minutes)", (3,-0.5), S); dot("Naomi", (2,6), 3*dir(305)); dot((6,6)); label("Maya", (4.45,3.5)); draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]$

$textbf{(A) }6 qquad textbf{(B) }12 qquad textbf{(C) }18 qquad textbf{(D) }20 qquad textbf{(E) }24$

Problem 12

For a positive integer? $n$ , the factorial notation? $n!$ ?represents the product of the integers from? $n$ ?to? $1$ . What value of? $N$ ?satisfies the following equation? $[5!cdot 9!=12cdot N!]$

$textbf{(A) }10qquadtextbf{(B) }11qquadtextbf{(C) }12qquadtextbf{(D) }13qquadtextbf{(E) }14qquad$

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?

$textbf{(A) }6 qquad textbf{(B) }9 qquad textbf{(C) }12 qquad textbf{(D) }18 qquad textbf{(E) }24$

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?

$[asy] size(300); pen shortdashed=linetype(new real[] {6,6}); // axis draw((0,0)--(0,9300), linewidth(1.25)); draw((0,0)--(11550,0), linewidth(1.25)); for (int i = 2000; i < 9000; i = i + 2000) { draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey); label(string(i), (0,i), W); } for (int i = 500; i < 9300; i=i+500) { draw((0,i)--(150,i),linewidth(1.25)); if (i?% 2000 == 0) { draw((0,i)--(250,i),linewidth(1.25)); } } int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750}; int data_length = 20; int r = 550; for (int i = 0; i < data_length; ++i) { fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey); draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)); } draw((0,4750)--(11450,4750),shortdashed); label("Cities", (11450*0.5,0), S); label(rotate(90)*"Population", (0,9000*0.5), 10*W); [/asy]$

$textbf{(A) }65{,}000 qquad textbf{(B) }75{,}000 qquad textbf{(C) }85{,}000 qquad textbf{(D) }95{,}000 qquad textbf{(E) }105{,}000$

Problem 15

Suppose? $15%$ ?of? $x$ ?equals? $20%$ ?of? $y.$ ?What percentage of? $x$ ?is? $y?$

$textbf{(A) }5 qquad textbf{(B) }35 qquad textbf{(C) }75 qquad textbf{(D) }133 frac13 qquad textbf{(E) }300$

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? $B?$
$[asy] size(200); dotfactor = 10; pair p1 = (-28,0); pair p2 = (-111,213); draw(p1--p2,linewidth(1)); pair p3 = (-160,0); pair p4 = (-244,213); draw(p3--p4,linewidth(1)); pair p5 = (-316,0); pair p6 = (-67,213); draw(p5--p6,linewidth(1)); pair p7 = (0, 68); pair p8 = (-350,10); draw(p7--p8,linewidth(1)); pair p9 = (0, 150); pair p10 = (-350, 62); draw(p9--p10,linewidth(1)); pair A = intersectionpoint(p1--p2, p5--p6); dot("$A$", A, 2*W); pair B = intersectionpoint(p5--p6, p3--p4); dot("$B$", B, 2*WNW); pair C = intersectionpoint(p7--p8, p5--p6); dot("$C$", C, 1.5*NW); pair D = intersectionpoint(p3--p4, p7--p8); dot("$D$", D, 2*NNE); pair EE = intersectionpoint(p1--p2, p7--p8); dot("$E$", EE, 2*NNE); pair F = intersectionpoint(p1--p2, p9--p10); dot("$F$", F, 2*NNE); [/asy]$

$textbf{(A) }1 qquad textbf{(B) }2 qquad textbf{(C) }3 qquad textbf{(D) }4 qquad textbf{(E) }5$

Problem 17

How many positive integer factors of? $2020$ ?have more than? $3$ ?factors? (As an example,? $12$ ?has? $6$ ?factors, namely? $1,2,3,4,6,$ ?and? $12.$ )

$textbf{(A) }6 qquad textbf{(B) }7 qquad textbf{(C) }8 qquad textbf{(D) }9 qquad textbf{(E) }10$

Problem 18

Rectangle? $ABCD$ ?is inscribed in a semicircle with diameter? $overline{FE},$ ?as shown in the figure. Let? $DA=16,$ ?and let? $FD=AE=9.$ ?What is the area of? $ABCD?$

$[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); [/asy]$

$textbf{(A) }240 qquad textbf{(B) }248 qquad textbf{(C) }256 qquad textbf{(D) }264 qquad textbf{(E) }272$

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? $15?$

$textbf{(A) }3 qquad textbf{(B) }4 qquad textbf{(C) }5 qquad textbf{(D) }6 qquad textbf{(E) }8$

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?

$[begingroup setlength{tabcolsep}{10pt} renewcommand{arraystretch}{1.5} begin{tabular}{|c|c|} hline Tree 1 & rule{0.4cm}{0.15mm} meters \ Tree 2 & 11 meters \ Tree 3 & rule{0.5cm}{0.15mm} meters \ Tree 4 & rule{0.5cm}{0.15mm} meters \ Tree 5 & rule{0.5cm}{0.15mm} meters \ hline Average height & rule{0.5cm}{0.15mm}text{ .}2 meters \ hline end{tabular} endgroup]$

$textbf{(A) }22.2 qquad textbf{(B) }24.2 qquad textbf{(C) }33.2 qquad textbf{(D) }35.2 qquad textbf{(E) }37.2$

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.)

$[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j)?% 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]$

$textbf{(A) }28 qquad textbf{(B) }30 qquad textbf{(C) }32 qquad textbf{(D) }33 qquad textbf{(E) }35$

Problem 22

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

$[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy]$

For example, starting with an input of? $N=7,$ ?the machine will output? $3 cdot 7 +1 = 22.$ ?Then if the output is repeatedly inserted into the machine five more times, the final output is? $26.$ $[7 to 22 to 11 to 34 to 17 to 52 to 26]$ 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? $N?$ $[N to rule{0.5cm}{0.15mm} to rule{0.5cm}{0.15mm} to rule{0.5cm}{0.15mm} to rule{0.5cm}{0.15mm} to rule{0.5cm}{0.15mm} to 1]$ $textbf{(A) }73 qquad textbf{(B) }74 qquad textbf{(C) }75 qquad textbf{(D) }82 qquad textbf{(E) }83$

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?

$textbf{(A) }120 qquad textbf{(B) }150 qquad textbf{(C) }180 qquad textbf{(D) }210 qquad textbf{(E) }240$

Problem 24

A large square region is paved with? $n^2$ ?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? $fraccqle8lm{s}$ for this larger value of? $n?$

$[asy] draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); [/asy]$

$textbf{(A) }frac{6}{25} qquad textbf{(B) }frac{1}{4} qquad textbf{(C) }frac{9}{25} qquad textbf{(D) }frac{7}{16} qquad textbf{(E) }frac{9}{16}$

Problem 25

Rectangles? $R_1$ ?and? $R_2,$ ?and squares? $S_1,,S_2,,$ ?and? $S_3,$ ?shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of? $S_2$ ?in units?

$[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]$

$textbf{(A) }651 qquad textbf{(B) }655 qquad textbf{(C) }656 qquad textbf{(D) }662 qquad textbf{(E) }666$

2020 AMC8 Answers

Problem 1

Solution 1

Let the side length of each square? $S_k$ ?be? $s_k$ . Then, from the diagram, we can line up the top horizontal lengths of? $S_1$ ,? $S_2$ , and? $S_3$ ?to cover the top side of the large rectangle, so? $s_{1}+s_{2}+s_{3}=3322$ . Similarly, the short side of? $R_2$ ?will be? $s_1-s_2$ , and lining this up with the left side of? $S_3$ ?to cover the vertical side of the large rectangle gives? $s_{1}-s_{2}+s_{3}=2020$ . We subtract the second equation from the first to obtain? $2s_{2}=1302$ , and thus? $s_{2}=boxed{textbf{(A) }651}$ .

Solution 2

Assuming that the problem is well-posed, it should be true in the particular case where? $S_1 cong S_3$ ?and? $R_1 cong R_2$ . Let the sum of the side lengths of? $S_1$ ?and? $S_2$ ?be? $x$ , and let the length of rectangle? $R_2$ ?be? $y$ . We then have the system $[begin{dcases}x+y =3322 \x-y=2020end{dcases}]$ which we solve to determine? $y=boxed{textbf{(A) }651}$ .

Solution 3 (faster version of solution 1)

Since, for each pair of rectangles, the side lengths have a sum of? $3322$ ?or? $2020$ ?and a difference of? $S_2$ , the answer must be? $dfrac{3322 - 2020}{2} = dfrac{1302}{2} = boxed{textbf{(A) }651}$ .

Problem 2

Solution

The friends earn? $$left(15+20+25+40right)=$100$ ?in total. Since they decided to split their earnings equally, it follows that each person will get? $$left(frac{100}{4}right)=$25$ . Since the friend who earned? $$40$ ?will need to leave with? $$25$ , he will have to give? $$left(40-25right)=boxed{textbf{(C) }$15}$ ?to the others.

Problem 3

Solution 1

The area of the garden is? $6 cdot 8 = 48$ ?square feet. Since Carrie plants? $4$ ?strawberry plants per square foot, there are a total of? $48 cdot 4=192$ strawberry plants, each of which produces? $10$ ?strawberries on average. Accordingly, she can expect to harvest? $192 cdot 10 = boxed{textbf{(D) }1920}$ 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? $6 cdot 8 cdot 4 cdot 10 = boxed{textbf{(D) }1920}$ .

Problem 4

Solution 1

Looking at the rows of each hexagon, we see that the first hexagon has? $1$ ?dot, the second has? $2+3+2$ ?dots and the third has? $3+4+5+4+3$ ?dots, and given the way the hexagons are constructed, it is clear that this pattern continues. Hence the fourth hexagon has? $4+5+6+7+6+5+4=boxed{textbf{(B) }37}$ ?dots.

Solution 2

The first hexagon has? $1$ ?dot, the second hexagon has? $1+6$ ?dots, the third hexagon? $1+6+12$ ?dots, and so on. The pattern continues since to go from hexagon? $n$ ?to hexagon? $(n+1)$ , we add a new ring of hexagons around the outside of the existing ones, with each side of the ring having side length? $(n+1)$ . Thus the number of hexagons added is? $6(n+1)-6 = 6n$ ?(we subtract? $6$ ?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? $1+6+12+18=boxed{textbf{(B) }37}$ ?dots.

Solution 3 (variant of Solution 2)

Let the number of dots in the first hexagon be? $h_0 = 1$ . By the same argument as in Solution 2, we have? $h_n=h_{n-1}+6n$ ?for? $n > 0$ . Using this, we find that? $h_1=7$ ,? $h_2=19,$ ?and? $h_3=boxed{textbf{(B) }37}$ .

Problem 5

Solution 1

Each cup is filled with? $frac{3}{4} cdot frac{1}{5} = frac{3}{20}$ ?of the amount of juice in the pitcher, so the percentage is? $frac{3}{20} cdot 100 = boxed{textbf{(C) }15}$ .

Solution 2

The pitcher is? $frac{3}{4}$ ?full, i.e.? $75%$ ?full. Therefore each cup receives? $frac{75}{5}=boxed{textbf{(C) }15}$ ?percent of the total capacity.

Solution 3

Assume that the pitcher has a total capacity of? $100$ ?ounces. Since it is filled three fourths with pineapple juice, it contains? $75$ ?ounces of pineapple juice, which means that each cup will contain? $frac{75}{5}=15$ ?ounces of pineapple juice. Since the total capacity of the pitcher was? $100$ ounces, it follows that each cup received? $15%$ ?of the total capacity of the pitcher, yielding? $boxed{textbf{(C) }15}$ ?as the answer.

Problem 6

Solution

Write the order of the cars as? $squaresquaresquaresquaresquare$ , where the left end of the row represents the back of the train and the right end represents the front. Call the people? $A$ ,? $D$ ,? $K$ ,? $M$ , and? $S$ ?respectively. The first condition gives? $Msquaresquaresquaresquare$ , so we try? $MASsquaresquare$ ,? $Msquare ASsquare$ , and? $Msquaresquare AS$ . In the first case, as? $D$ ?sat in front of? $A$ , we must have? $MASDK$ ?or? $MASKD$ , both of which do not comply with the last condition. In the second case, we obtain? $MKASD$ , which works, while the third case is obviously impossible, since it results in there being no way for? $D$ ?to sit in front of? $A$ . It follows that, with the only possible arrangement being? $MKASD$ , the person sitting in the middle car is? $boxed{textbf{(A) }text{Aaron}}$ .

Problem 7

Solution 1

Firstly, observe that the second digit of such a number cannot be? $1$ ?or? $2$ , because the digits must be distinct and increasing. The second digit also cannot be? $4$ ?as the number must be less than? $2400$ , so it must be? $3$ . It remains to choose the latter two digits, which must be? $2$ ?distinct digits from? $left{4,5,6,7,8,9right}$ . That can be done in? $binom{6}{2} = frac{6 cdot 5}{2 cdot 1} = 15$ ?ways; there is then only? $1$ ?way to order the digits, namely in increasing order. This means the answer is? $boxed{textbf{(C) }15}$ .

Solution 2 (without using the "choose" function)

As in Solution 1, we find that the first two digits must be? $23$ , and the third digit must be at least? $4$ . If it is? $4$ , then there are? $5$ ?choices for the last digit, namely? $5$ ,? $6$ ,? $7$ ,? $8$ , or? $9$ . Similarly, if the third digit is? $5$ , there are? $4$ ?choices for the last digit, namely? $6$ ,? $7$ ,? $8$ , and? $9$ ; if? $6$ , there are? $3$ ?choices; if? $7$ , there are? $2$ ?choices; and if? $8$ , there is? $1$ ?choice. It follows that the total number of such integers is? $5+4+3+2+1=boxed{textbf{(C) }15}$ .

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? $2019$ , giving a total of? $(2019cdot 5 + 1)$ ?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? $2019$ , giving a total of? $(2019cdot 1 + 5)$ ?cents. Hence the required difference is $[(2019cdot 5 + 1)-(2019cdot 1 + 5)=2019cdot 4-4=4cdot 2018=boxed{textbf{(C) }8072}]$

Solution 2

Suppose Ricardo has? $p$ ?pennies, so then he has? $(2020-p)$ ?nickels. In order to have at least one penny and at least one nickel, we require? $p geq 1$ ?and? $2020 - p geq 1$ , i.e.? $1 leq p leq 2019$ . The number of cents he has is? $p+5(2020-p) = 10100-4p$ , so the maximum is? $10100-4 cdot 1$ ?and the minimum is? $10100 - 4 cdot 2019$ , and the difference is therefore $[(10100 - 4cdot 1) - (10100 - 4cdot 2019) = 4cdot 2019 - 4 = 4cdot 2018 = boxed{textbf{(C) }8072}]$

Problem 9

Solution 1

Notice that, for a small cube which does not form part of the bottom face, it will have exactly? $2$ ?faces with icing on them only if it is one of the? $2$ center cubes of an edge of the larger cube. There are? $12-4 = 8$ ?such edges (as we exclude the? $4$ ?edges of the bottom face), so this case yields? $2 cdot 8 = 16$ ?small cubes. As for the bottom face, we can see that only the? $4$ ?corner cubes have exactly? $2$ ?faces with icing, so the total is? $16+4 = boxed{textbf{(D) }20}$ .

Solution 2

The following diagram shows? $12$ ?of the small cubes having exactly? $2$ ?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? $12+8=boxed{textbf{(D) }20}$ .

Problem 10

Solution 1

By the?Georgeooga-Harryooga Theorem?there are? $frac{(4-2)!(4-2+1)!}{(4-2cdot2+1)!}=boxed{textbf{(C) }12}$ ?way to arrange the marbles.

Solution by?RedFireTruck

Solution 2

We can arrange our marbles like so? $square Asquare Bsquare$ .

To arrange the? $A$ ?and? $B$ ?we have? $2!=2$ ?ways.

To place the? $S$ ?and? $T$ ?in the blanks we have? $_3P_2=6$ ?ways.

By fundamental counting principle our final answer is? $2cdot6=boxed{textbf{(C) }12}$

Solution by?RedFireTruck

Solution 3

Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by? $A,B,S,$ ?and? $T$ , respectively. If we ignore the constraint that? $S$ ?and? $T$ ?cannot be next to each other, we get a total of? $4!=24$ ?ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that? $S$ ?and? $T$ ?can be next to each other. If we place? $S$ ?and? $T$ ?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.? $STsquaresquare, square STsquare, squaresquare ST$ ). However, we could also have placed? $S$ ?and? $T$ in the opposite order (i.e.? $TSsquaresquare, square TSsquare, squaresquare TS$ ). Thus there are 6 ways of placing? $S$ ?and? $T$ ?directly next to each other. Next, notice that for each of these placements, we have two open slots for placing? $A$ ?and? $B$ . Specifically, we can place? $A$ ?in the first open slot and? $B$ ?in the second open slot or switch their order and place? $B$ ?in the first open slot and? $A$ ?in the second open slot. This gives us a total of? $6times 2=12$ ?ways to place? $S$ ?and? $T$ ?next to each other. Subtracting this from the total number of arrangements gives us? $24-12=12$ ?total arrangements? $impliesboxed{textbf{(C) }12}$ .

We can also solve this problem directly by looking at the number of ways that we can place? $S$ ?and? $T$ ?such that they are not directly next to each other. Observe that there are three ways to place? $S$ ?and? $T$ ?(in that order) into the four slots so they are not next to each other (i.e.? $Ssquare Tsquare, square Ssquare T, Ssquaresquare T$ ). However, we could also have placed? $S$ ?and? $T$ ?in the opposite order (i.e.? $Tsquare Ssquare, square Tsquare S, Tsquaresquare S$ ). Thus there are 6 ways of placing? $S$ ?and? $T$ ?so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing? $A$ ?and? $B$ . Specifically, we can place? $A$ ?in the first open slot and? $B$ ?in the second open slot or switch their order and place? $B$ ?in the first open slot and? $A$ ?in the second open slot. This gives us a total of? $6times 2=12$ ?ways to place? $S$ ?and? $T$ ?such that they are not next to each other? $impliesboxed{textbf{(C) }12}$ .
~junaidmansuri

Solution 4

Let's try complementary counting. There? $4!$ ?ways to arrange the 4 marbles. However, there are? $2cdot3!$ ?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, $[4!-2cdot3!=boxed{12 textbf{(C)}}]$

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? $2!$ ?ways to arrange Steelie and Tiger within this "super marble." Then there are? $3!$ ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are? $4!$ ?ways to arrange the marbles without any restrictions, the answer is given by? $4!-2!cdot 3!=textbf{(C) }12$

-franzliszt

Solution 6

We will use the following

$textbf{Georgeooga-Harryooga Theorem:}$ ?The?Georgeooga-Harryooga Theorem?states that if you have? $a$ ?distinguishable objects and? $b$ ?of them cannot be together, then there are? $frac{(a-b)!(a-b+1)!}{b!}$ ?ways to arrange the objects.

$textit{Proof. (Created by AoPS user RedFireTruck)}$

Let our group of? $a$ ?objects be represented like so? $1$ ,? $2$ ,? $3$ , ...,? $a-1$ ,? $a$ . Let the last? $b$ ?objects be the ones we can't have together.

Then we can organize our objects like so? $square1square2square3square...square a-b-1square a-bsquare$ .

We have? $(a-b)!$ ?ways to arrange the objects in that list.

Now we have? $a-b+1$ ?blanks and? $b$ ?other objects so we have? $_{a-b+1}P_{b}=frac{(a-b+1)!}{(a-2b+1)!}$ ?ways to arrange the objects we can't put together.

By fundamental counting principal our answer is? $frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$ .

Proof by?RedFireTruck

Back to the problem. By the?Georgeooga-Harryooga Theorem, our answer is? $frac{(4-2)!(4-2+1)!}{(4-2cdot2+1)!}=textbf{(C) }12$ .

Problem 11

Solution 1

Naomi travels? $6$ ?miles in a time of? $10$ ?minutes, which is equivalent to? $dfrac{1}{6}$ ?of an hour. Since? $text{speed} = frac{text{distance}}{text{time}}$ , her speed is? $frac{6}{left(frac{1}{6}right)} = 36$ ?mph. By a similar calculation, Maya's speed is? $12$ ?mph, so the answer is? $36-12 = boxed{textbf{(E) }24}$ .

Solution 2 (variant of Solution 1)

Naomi's speed of? $6$ ?miles in? $10$ ?minutes is equivalent to? $6 cdot 6 = 36$ ?miles per hour, while Maya's speed of? $6$ ?miles in? $30$ ?minutes (i.e. half an hour) is equivalent to? $6 cdot 2 = 12$ ?miles per hour. The difference is consequently? $36-12=boxed{textbf{(E) }24}$ .

Problem 12

Solution 1

We have? $5! = 2 cdot 3 cdot 4 cdot 5$ , and? $2 cdot 5 cdot 9! = 10 cdot 9! = 10!$ . Therefore the equation becomes? $3 cdot 4 cdot 10! = 12 cdot N!$ , and so? $12 cdot 10! = 12 cdot N!$ . Cancelling the? $12$ s, it is clear that? $N=boxed{textbf{(A) }10}$ .

Solution 2 (variant of Solution 1)

Since? $5! = 120$ , we obtain? $120cdot 9!=12cdot N!$ , which becomes? $12cdot 10cdot 9!=12cdot N!$ ?and thus? $12 cdot 10!=12cdot N!$ . We therefore deduce? $N=boxed{textbf{(A) }10}$ .

Solution 3 (using answer choices)

We can see that the answers? $textbf{(B)}$ ?to? $textbf{(E)}$ ?contain a factor of? $11$ , but there is no such factor of? $11$ ?in? $5! cdot 9!$ . Therefore, the answer must be

Problem 13

Solution 1

After Jamal adds? $x$ ?purple socks, he has? $(18+x)$ ?purple socks and? $6+18+12+x=(36+x)$ ?total socks. This means the probability of drawing a purple sock is? $frac{18+x}{36+x}$ , so we obtain $[frac{18+x}{36+x}=frac{3}{5}]$ Since? $frac{18+9}{36+9}=frac{27}{45}=frac{3}{5}$ , the answer is? $boxed{textbf{(B) }9}$ .

Solution 2 (variant of Solution 1)

As in Solution 1, we have the equation? $frac{18+x}{36+x}=frac{3}{5}$ . Cross-multiplying yields? $90+5x=108+3x Rightarrow 2x=18 Rightarrow x=9$ . Thus, Jamal added? $boxed{textbf{(B) }9}$ ?purple socks.

Solution 3 (non-algebraic)

$6$ ?green socks and? $12$ ?orange socks together should be? $100%-60% = 40%$ ?of the new total number of socks, so that new total must be? $frac{6+12}{0.4}= 45$ . It follows that? $45-6-18-12=boxed{textbf{(B) }9}$ ?purple socks were added.

Problem 14

Solution 1

We can see that the dotted line is exactly halfway between? $4{,}500$ ?and? $5{,}000$ , so it is at? $4{,}750$ . As this is the average population of all? $20$ ?cities, the total population is simply? $4{,}750 cdot 20 = boxed{textbf{(D) }95{,}000}$ .

Solution 2 (estimation)

The dashed line, which represents the average population of each city, is slightly below? $5{,}000$ . Since there are? $20$ ?cities, the answer is slightly less than? $20cdot 5{,}000 approx 100{,}000$ , which is closest to? $boxed{textbf{(D) }95{,}000}$ .

Problem 15

Solution 1

Since? $20% = frac{1}{5}$ , multiplying the given condition by? $5$ ?shows that? $y$ ?is? $15 cdot 5 = boxed{textbf{(C) }75}$ ?percent of? $x$ .

Solution 2

Letting? $x=100$ ?(without loss of generality), the condition becomes? $0.15cdot 100 = 0.2cdot y Rightarrow 15 = frac{y}{5} Rightarrow y=75$ . Clearly, it follows that? $y$ ?is? $75%$ ?of? $x$ , so the answer is? $boxed{textbf{(C) }75}$ .

Solution 3

We have? $15%=frac{3}{20}$ ?and? $20%=frac{1}{5}$ , so? $frac{3}{20}x=frac{1}{5}y$ . Solving for? $y$ , we multiply by? $5$ ?to give? $y = frac{15}{20}x = frac{3}{4}x$ , so the answer is? $boxed{textbf{(C) }75}$ .

Solution 4

We are given? $0.15x = 0.20y$ , so we may assume without loss of generality that? $x=20$ ?and? $y=15$ . This means? $frac{y}{x}=frac{15}{20}=frac{75}{100}$ , and thus the answer is? $boxed{textbf{(C) }75}$ .

Problem 16

Solution 1

We can form the following expressions for the sum along each line: $[begin{dcases}A+B+C\A+E+F\C+D+E\B+D\B+Fend{dcases}]$ Adding these together, we must have? $2A+3B+2C+2D+2E+2F=47$ , i.e.? $2(A+B+C+D+E+F)+B=47$ . Since? $A,B,C,D,E,F$ ?are unique integers between? $1$ ?and? $6$ , we obtain? $A+B+C+D+E+F=1+2+3+4+5+6=21$ ?(where the order doesn't matter as addition is commutative), so our equation simplifies to? $42 + B = 47$ . This means? $B = boxed{textbf{(E) }5}$ . ~RJ5303707

Solution 2

Following the first few steps of Solution 1, we have? $2(A+C+D+E+F)+3B=47$ . Because an even number ( $2(A+C+D+E+F)$ ) subtracted from an odd number (47) is always odd, we know that? $3B$ ?is odd, showing that? $B$ ?is odd. Now we know that? $B$ ?is either 1, 3, or 5. If we try? $B=1$ , we get? $43=47$ ?which is not true. Testing? $B=3$ , we get? $45=47$ , which is also not true. Therefore, we have? $B = boxed{textbf{(E) }5}$ .

Problem 17

Solution 1

Since? $2020 = 2^2 cdot 5 cdot 101$ , we can simply list its factors: $[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.]$ There are? $12$ ?of these; only? $1, 2, 4, 5, 101$ ?(i.e.? $5$ ?of them) have at most? $3$ ?factors, so the remaining? $12-5 = boxed{textbf{(B) }7}$ ?factors have more than? $3$ factors.

Solution 2

As in Solution 1, we prime factorize? $2020$ ?as? $2^2cdot 5cdot 101$ , and we recall the standard formula that the number of positive factors of an integer is found by adding? $1$ ?to each exponent in its prime factorization, and then multiplying these. Thus? $2020$ ?has? $(2+1)(1+1)(1+1) = 12$ ?factors. The only number which has one factor is? $1$ . For a number to have exactly two factors, it must be prime, and the only prime factors of? $2020$ ?are? $2$ ,? $5$ , and? $101$ . 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? $2020$ ?is? $4$ . Thus, there are? $5$ ?factors of? $2020$ ?which themselves have? $1$ ,? $2$ , or? $3$ ?factors (namely? $1$ ,? $2$ ,? $4$ ,? $5$ , and? $101$ ), so the number of factors of? $2020$ ?that have more than? $3$ ?factors is? $12-5=boxed{textbf{(B) }7}$ .

Problem 18

Solution 1

Let? $O$ ?be the center of the semicircle. The diameter of the semicircle is? $9+16+9=34$ , so? $OC = 17$ . By symmetry,? $O$ ?is in fact the midpoint of? $DA$ , so? $OD=OA=frac{16}{2}= 8$ . By the Pythagorean theorem in right-angled triangle? $ODC$ ?(or? $OBA$ ), we have that? $CD$ ?(or? $AB$ ) is? $sqrt{17^2-8^2}=15$ . Accordingly, the area of? $ABCD$ ?is? $16cdot 15=boxed{textbf{(A) }240}$ .

Solution 2 (coordinate geometry)

Let the midpoint of segment? $FE$ ?be the origin. Evidently, point? $D=(-8,0)$ ?and? $A=(8,0)$ . Since points? $C$ ?and? $B$ ?share? $x$ -coordinates with? $D$ ?and? $A$ ?respectively, it suffices to find the? $y$ -coordinate of? $B$ ?(which will be the height of the rectangle) and multiply this by? $DA$ ?(which we know is? $16$ ). The radius of the semicircle is? $frac{9+16+9}{2} = 17$ , so the whole circle has equation? $x^2+y^2=289$ ; as already stated,? $B$ ?has the same? $x$ -coordinate as? $A$ , i.e.? $8$ , so substituting this into the equation shows that? $y=pm15$ . Since? $y>0$ ?at? $B$ , the y-coordinate of? $B$ ?is? $15$ . Therefore, the answer is? $16cdot 15 = boxed{textbf{(A) }240}$ .

(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? $(1+n)$ , such that the? $1$ ?part is on one side and the? $n$ ?part is on the other side, the height from the end of the? $1$ ?side (or the start of the? $n$ ?side) is? $sqrt{n}$ . To use this formula, we scale the figure down by? $9$ ; this will give the height a length of? $sqrt{frac{16+9}{9}} = sqrt{frac{25}{9}} = frac{5}{3}$ . Now, scaling back up by? $9$ , the height? $DC$ is? $9 cdot frac{5}{3} = 15$ . The answer is then? $15 cdot 16 = boxed{textbf{(A) }240}$ .

Problem 19

Solution 1

A number is divisible by? $15$ ?precisely if it is divisible by? $3$ ?and? $5$ . The latter means the last digit must be either? $5$ ?or? $0$ , and the former means the sum of the digits must be divisible by? $3$ . If the last digit is? $0$ , the first digit would be? $0$ ?(because the digits alternate), which is not possible. Hence the last digit must be? $5$ , and the number is of the form? $5square 5square 5$ . If the unknown digit is? $x$ , we deduce? $5+x+5+x+5 equiv 0 pmod{3} Rightarrow 2x equiv 0 pmod{3}$ . We know? $2^{-1}$ ?exists modulo? $3$ ?because 2 is relatively prime to 3, so we conclude that? $x$ ?(i.e. the second and fourth digit of the number) must be a multiple of? $3$ . It can be? $0$ ,? $3$ ,? $6$ , or? $9$ , so there are? $boxed{textbf{(B) }4}$ ?options:? $50505$ ,? $53535$ ,? $56565$ , and? $59595$ .

Solution 2 (variant of Solution 1)

As in Solution 1, we find that such numbers must start with? $5$ ?and alternate with? $5$ ?(i.e. must be of the form? $5square 5square 5$ ), where the two digits between the? $5$ s need to be the same. Call that digit? $x$ . For the number to be divisible by? $3$ , the sum of the digits must be divisible by? $3$ ; since the sum of the three? $5$ s is? $15$ , which is already a multiple of? $3$ , it must also be the case that? $x+x=2x$ ?is a multiple of? $3$ . Thus, the problem reduces to finding the number of digits from? $0$ ?to? $9$ ?for which? $2x$ ?is a multiple of? $3$ . This leads to? $x=0$ ,? $3$ ,? $6$ , or? $9$ , so there are? $boxed{textbf{(B) }4}$ ?possible numbers (namely? $50505$ ,? $53535$ ,? $56565$ , and? $59595$ ).

Problem 20

Solution 1

We will show that? $22$ ,? $11$ ,? $22$ ,? $44$ , and? $22$ ?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? $11$ ?meters, we can deduce that Trees 1 and 3 both have a height of? $22$ ?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? $11$ ?and? $22$ ,? $44$ ?and? $88$ , or? $44$ ?and? $22$ . Checking each of these, in the first case, the average is? $17.6$ ?meters, which doesn't end in? $.2$ ?as the problem requires. Therefore, we consider the other cases. With? $44$ ?and? $88$ , the average is? $37.4$ ?meters, which again does not end in? $.2$ , but with? $44$ ?and? $22$ , the average is? $24.2$ ?meters, which does. Consequently, the answer is? $boxed{textbf{(B) }24.2}$ .

Solution 2

Notice the average height of the trees ends with? $0.2$ ?therefore the sum of all five heights of the trees must end with? $1$ . ( $0.2$ ?*? $5$ ?=? $1$ ) We already know Tree? $2$ ?is? $11$ ?meters tall. Both Tree? $1$ ?and Tree? $3$ ?must? $22$ ?meters tall - since neither can be? $5.5$ . Once again, apply our observation for solving for the Tree? $4$ 's height. Tree? $4$ ?can't be? $11$ ?meters for the sum of the five tree heights to still end with? $1$ . Therefore, the Tree? $4$ ?is? $44$ ?meters tall. Now the Tree? $5$ ?can either be? $22$ ?or? $88$ . Find the average height for both cases of Tree? $5$ . Doing this, we realize the Tree? $5$ ?must be? $22$ ?for the average height to end with? $0.2$ ?and that the average height is? $boxed{textbf{(B) }24.2}$ .

Solution 3

As in Solution 1, we shall show that the heights of the trees are? $22$ ,? $11$ ,? $22$ ,? $44$ , and? $22$ ?meters. Let? $S$ ?be the sum of the heights, so that the average height will be? $frac{S}{5}$ ?meters. We note that? $0.2 = frac{1}{5}$ , so in order for? $frac{S}{5}$ ?to end in? $.2$ ,? $S$ ?must be one more than a multiple of? $5$ . Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both? $22$ ?meters. At this point, our table looks as follows: $[begingroup setlength{tabcolsep}{10pt} renewcommand{arraystretch}{1.5} begin{tabular}{|c|c|} hline Tree 1 & 22 meters \ Tree 2 & 11 meters \ Tree 3 & 22 meters \ Tree 4 & rule{0.5cm}{0.15mm} meters \ Tree 5 & rule{0.5cm}{0.15mm} meters \ hline Average height & rule{0.5cm}{0.15mm}text{ .}2 meters \ hline end{tabular} endgroup]$

If Tree 4 now has a height of? $11$ , then Tree 5 would need to have height? $22$ , but in that case? $S$ ?would equal? $88$ , which is not? $1$ ?more than a multiple of? $5$ . So we instead take Tree 4 to have height? $44$ . Then the sum of the heights of the first 4 trees is? $22+11+22+44 = 99$ , so using a height of? $22$ ?for Tree 5 gives? $S=121$ , which is? $1$ ?more than a multiple of? $5$ ?(whereas? $88$ ?gives? $S = 187$ , which is not). Thus the average height of the trees is? $frac{121}{5} = boxed{textbf{(B) }24.2}$ ?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? $1$ ?or? $2$ ?white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from? $P$ ?to that square is the sum of the numbers of ways to move from? $P$ ?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? $P$ ?to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j)?% 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label("$1$", (5.5, .5)); label("$1$", (4.5, 1.5)); label("$1$", (6.5, 1.5)); label("$1$", (3.5, 2.5)); label("$1$", (7.5, 2.5)); label("$2$", (5.5, 2.5)); label("$1$", (2.5, 3.5)); label("$3$", (6.5, 3.5)); label("$3$", (4.5, 3.5)); label("$4$", (3.5, 4.5)); label("$3$", (7.5, 4.5)); label("$6$", (5.5, 4.5)); label("$10$", (4.5, 5.5)); label("$9$", (6.5, 5.5)); label("$19$", (5.5, 6.5)); label("$9$", (7.5, 6.5)); label("$28$", (6.5, 7.5)); [/asy]$
The answer is therefore? $boxed{textbf{(A) }28}$ .

Solution 2

Suppose we "extend" the chessboard infinitely with? $2$ ?additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.

$[asy] int N = 7; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j)?% 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } draw((8,0) -- (8,8),red); label("$P$", (5.5,.5)); label("$Q$", (6.5,7.5)); label("$X$", (8.5,3.5)); label("$Y$", (8.5,5.5)); [/asy]$
The total number of paths from? $P$ ?to? $Q$ , including invalid paths which cross over the red line, is then the number of paths which make? $4$ ?steps up-and-right and? $3$ ?steps up-and-left, which is? $binom{4+3}{3} = binom{7}{3} = 35$ . We need to subtract the number of invalid paths, i.e. the number of paths that pass through? $X$ ?or? $Y$ . To get to? $X$ , the marker has to make? $3$ ?up-and-right steps, after which it can proceed to? $Q$ ?with? $3$ ?steps up-and-left and? $1$ ?step up-and-right. Thus, the number of paths from? $P$ ?to? $Q$ ?that pass through? $X$ ?is? $1 cdot binom{3+1}{3} = 4$ . Similarly, the number of paths that pass through? $Y$ ?is? $binom{4+1}{1}cdot 1 = 5$ . However, we have now double-counted the invalid paths which pass through both? $X$ ?and? $Y$ ; from the diagram, it is clear that there are only? $2$ ?of these (as the marker can get from? $X$ ?to? $Y$ ?by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is? $4+5-2=7$ , and the number of valid paths from? $P$ ?to? $Q$ ?is? $35-7 = boxed{textbf{(A) }28}$ .

Problem 22

Solution 1

We start with the final output of? $1$ ?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: $[{1}leftarrow{2}leftarrow{4}leftarrow{1,8}leftarrow{2,16}leftarrow{4,5,32}leftarrow{1,8,10,64}]$ where, for example,? $2$ ?must come from? $4$ ?(as there is no integer? $n$ ?satisfying? $3n+1=2$ ), but? $16$ ?could come from? $32$ ?or? $5$ ?(as? $frac{32}{2} = 3 cdot 5 + 1 = 16$ , and? $32$ ?is even while? $5$ ?is odd). By construction, the last set in this sequence contains all the numbers which will lead to the number? $1$ ?at the end of the? $6$ -step process, and their sum is? $1+8+10+64=boxed{textbf{(E) }83}$ .

Solution 2 (variant of Solution 1)

As in Solution 1, we work backwards from? $1$ , 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? $1$ ,? $8$ ,? $64$ , and? $10$ . Thus the answer is? $1+8+64+10=boxed{textbf{(E) }83}$ .

Solution 3 (algebraic)

We begin by finding the inverse of the function that the machine uses. Call the input? $I$ ?and the output? $O$ . If? $I$ ?is even,? $O=frac{I}{2}$ , and if? $I$ ?is odd,? $O=3I+1$ . We can therefore see that? $I=2O$ ?when? $I$ ?is even and? $I=frac{O-1}{3}$ ?when? $I$ ?is odd. Therefore, starting with? $1$ , if? $I$ ?is even,? $I=2$ , and if? $I$ ?is odd,? $I=0$ , but the latter is not valid since? $0$ ?is not actually odd. This means that the 2nd-to-last number in the sequence has to be? $2$ . Now, substituting? $2$ ?into the inverse formulae, if? $I$ ?is even,? $I=4$ ?(which is indeed even), and if? $I$ ?is odd,? $I=frac{1}{3}$ , which is not an integer. This means the 3rd-to-last number in the sequence has to be? $4$ . Substituting in? $4$ , if? $I$ ?is even,? $I=8$ , but if? $I$ ?is odd,? $I=1$ . Both of these are valid solutions, so the 4th-to-last number can be either? $1$ ?or? $8$ . If it is? $1$ , then by the argument we have just made, the 5th-to-last number has to be? $2$ , the 6th-to-last number has to be? $4$ , and the 7th-to-last number, which is the first number, must be either? $1$ ?or? $8$ . In this way, we have ultimately found two solutions:? $N=1$ ?and? $N=8$ .

On the other hand, if the 4th-to-last number is? $8$ , substituting? $8$ ?into the inverse formulae shows that the 5th-to-last number is either? $16$ ?or? $frac{7}{3}$ , but the latter is not an integer. Substituting? $16$ ?shows that if? $I$ ?is even,? $I=32$ , but if I is odd,? $I=5$ , and both of these are valid. If the 6th-to-last number is? $32$ , then the first number must be? $64$ , since? $frac{31}{3}$ ?is not an integer; if the 6th-to-last number is? $5,$ ?then the first number has to be? $10$ , as? $frac{4}{3}$ is not an integer. This means that, in total, there are? $4$ ?solutions for? $N$ , specifically,? $1$ ,? $8$ ,? $10$ , and? $64$ , which sum to? $boxed{textbf{(E) }83}$ .

Problem 23

Solution 1 (complementary counting)

Without the restriction that each student receives at least one award, we could simply take each of the? $5$ ?awards and choose one of the? $3$ students to give it to, so that there would be? $3^5=243$ ?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? $3$ ?choices for which student that is, then? $2^5 = 32$ ?ways of choosing a student to receive each of the awards, for a total of? $3 cdot 32 = 96$ . However, if? $2$ ?students both don't receive an award, then such a case would be counted twice among our? $96$ , so we need to add back in these cases. Of course,? $2$ ?students both not receiving an award is equivalent to only? $1$ ?student receiving all? $5$ ?awards, so there are simply? $3$ ?choices for which student that would be. It follows that the total number of ways of distributing the awards is? $243-96+3=boxed{textbf{(B) }150}$ .

Solution 2

Firstly, observe that it is not possible for a single student to receive? $4$ ?or? $5$ ?awards because this would mean that one of the other students receives no awards. Thus, each student must receive either? $1$ ,? $2$ , or? $3$ ?awards. If a student receives? $3$ ?awards, then the other two students must each receive? $1$ ?award; if a student receives? $2$ ?awards, then another student must also receive? $2$ ?awards and the remaining student must receive? $1$ award. We consider each of these two cases in turn. If a student receives three awards, there are? $3$ ?ways to choose which student this is, and? $binom{5}{3}$ ?ways to give that student? $3$ ?out of the? $5$ ?awards. Next, there are? $2$ ?students left and? $2$ ?awards to give out, with each student getting one award. There are clearly just? $2$ ?ways to distribute these two awards out, giving? $3cdotbinom{5}{3}cdot 2=60$ ?ways to distribute the awards in this case.

In the other case, a student receives? $2$ ?awards. We first have to choose which of the two students we will select to give two awards each to. There are? $binom{3}{2}$ ?ways to do this, after which there are? $binom{5}{2}$ ?ways to give the first student his two awards, leaving? $3$ ?awards yet to distribute. There are then? $binom{3}{2}$ ?ways to give the second student his? $2$ ?awards. Finally, there is only? $1$ ?student and? $1$ ?award left, so there is only? $1$ ?way to distribute this award. This results in? $binom{3}{2}cdotbinom{5}{2}cdotbinom{3}{2}cdot 1=90$ ?ways to distribute the awards in this case. Adding the results of these two cases, we get? $60+90=boxed{textbf{(B) }150}$ .

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? $5$ ?awards are? $3,1,1$ and? $2,2,1$ ?(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? $3$ ?choices for which student gets? $3$ ?awards, and? $binom{5}{3} = 10$ ?choices for which? $3$ ?awards they get. We are then left with? $2$ ?awards, and there are exactly? $2$ ?choices depending on which remaining student gets which. This yields a total for this case of? $3 cdot 10 cdot 2 = 60$ . For the second case, there are similarly? $3$ ?choices for which student gets only? $1$ ?award, and? $5$ ?choices for which award he gets. There are then? $4$ ?remaining awards, from which we choose? $2$ ?to give to one student and? $2$ ?to give to the other, which can be done in? $binom{4}{2} = 6$ ?ways (and we can say that e.g. the? $2$ ?chosen this way go to the first remaining student and the other? $2$ ?go to the second remaining student, which counts all possibilities). This means the total for the second case is? $3 cdot 5 cdot 6 = 90$ , and the answer is? $60 + 90 = boxed{textbf{(B) }150}$ .

Problem 24

Solution 1

The area of the shaded region is? $(24s)^2$ . To find the area of the large square, we note that there is a? $d$ -inch border between each of the? $23$ ?pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of? $23+2 = 25$ ?times the length of the border, i.e.? $25d$ . Adding this to the total length of the consecutive squares, which is? $24s$ , the side length of the large square is? $(24s+25d)$ , yielding the equation? $frac{(24s)^2}{(24s+25d)^2}=frac{64}{100}$ . Taking the square root of both sides (and using the fact that lengths are non-negative) gives? $frac{24s}{24s+25d}=frac{8}{10} = frac{4}{5}$ , and cross-multiplying now gives? $120s = 96s + 100d Rightarrow 24s = 100d Rightarrow fraccqle8lm{s} = frac{24}{100} = boxed{textbf{(A) }frac{6}{25}}$ .

Solution 2

Without loss of generality, we may let? $s=1$ ?(since? $d$ ?will be determined by the scale of? $s$ , and we are only interested in the ratio? $fraccqle8lm{s}$ ). Then, as the total area of the? $576$ ?gray tiles is simply? $576$ , the large square has area? $frac{576}{0.64} = 900$ , making the side of the large square? $sqrt{900}=30$ . As in Solution 1, the the side length of the large square consists of the total length of the gray tiles and? $25$ ?lots of the border, so the length of the border is? $d = frac{30-24}{25} = frac{6}{25}$ . Since? $fraccqle8lm{s}=d$ ?if? $s=1$ , the answer is? $boxed{textbf{(A) }frac{6}{25}}$ .

Solution 3 (using answer choices)

As in Solution 2, we let? $s = 1$ ?without loss of generality. For sufficiently large? $n$ , we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown: $[asy] draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); for(int i = 1; i <= 13; i += 4){ draw((1,i)--(13,i), red); draw((i,1)--(i,13), red); } [/asy]$
Each red square has side length? $(1+d)$ , so by solving? $frac{1^2}{(1+d)^2} = frac{64}{100} iff frac{1}{1+d} = frac{4}{5}$ , we obtain? $d = frac{1}{4}$ . The actual fraction of the total area covered by the gray tiles will be slightly less than? $frac{1}{(1+d)^2}$ , which implies? $frac{1}{(1+d)^2} > frac{64}{100} iff frac{1}{1+d} > frac{4}{5} iff d < frac{1}{4}$ . Hence? $d$ (and thus? $fraccqle8lm{s}$ , since we are assuming? $s=1$ ) is less than? $frac{1}{4}$ , and the only choice that satisfies this is? $boxed{textbf{(A) }frac{6}{25}}$ .

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