華南師范大學(xué)數(shù)學(xué)科學(xué)學(xué)院(510631) 李湖南

1.What is the value of 1-(-2)-3-(-4)-5-(-6)?

(A)-20 (B)-3 (C) 3 (D) 5 (E) 21

譯文式子1-(-2)-3-(-4)-5-(-6)的值是多少?

解直接計算,原式=1+2-3+4-5+6=5,故(D)正確.

2.Carl has 5 cubes having side length 1,and Kate has 5 cubes each having side length 2.What is the total volume of the 10 cubes?

(A) 24 (B) 25 (C) 28 (D) 40 (E) 45

譯文卡爾有5 個棱長為1 的立方體,凱特有5 個棱長為2 的立方體.則這10 個立方體的總體積是多少?

解依題意,總體積=5×13+5×23=45,故(E)正確.

3.The ratio ofwtoxis 4:3,the ratio ofytozis 3:2,and the ratio ofztoxis 1:6.What is the ratio ofwtoy?

(A) 4:3 (B) 3:2 (C) 8:3 (D) 4:1 (E) 16:3

譯文已知w與x之比為4:3,y與z之比為3:2,z與x之比為1:6.則w與y之比是多少?

解化簡即得故(E)正確.

4.The acute angles of a right triangle area°andb°,wherea＞band bothaandbare prime numbers.What is the least possible value ofb?

(A) 2 (B) 3 (C) 5 (D) 7 (E) 11

譯文一個直角三角形的兩個銳角分別是a°和b°,其中a＞b且a和b均為素數(shù).則b最小可能的值是多少?

解問題轉(zhuǎn)化為求a+b=90 的最小素數(shù)解,可解得a=83,b=7,故(D)正確.

5.How many distinguishable arrangements are there of 1 brown tile,1 purple tile,2 green tiles,and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable)

(A) 210 (B) 420 (C) 630 (D) 840 (E) 1050

譯文把1 塊棕色瓷磚,1 塊紫色瓷磚,2 塊綠色瓷磚和3 塊黃色瓷磚從左到右排成一行有多少種不同的方法? (同種顏色的瓷磚不可分辨)

解這是一個組合問題,相當于從7 塊瓷磚中選出1 塊棕色的,再選出1 塊紫色的,然后選出2 塊綠色的,剩下的全是黃色,即有=7×6×10×1=420 種方法,故(B)正確.

6.Driving along a highway,Megan noticed that her odometer showed 15951(miles).This number is a palindrome—it reads the same forward and backward.Then 2 hours later,the odometer displayed the next higher palindrome.What was her average speed,in miles per hour,during this 2-hour period?

(A) 50 (B) 55 (C) 60 (D) 65 (E) 70

譯文梅根正開車在一條高速公路上,她注意到里程表正好顯示著15951 英里.這是一個回文數(shù)—正反讀起來是一樣的.2 個小時后,里程表顯示著下一個更大的回文數(shù).則在這2 小時期間她的平均速度是多少英里/小時?

解下一個回文數(shù)是16061,在2 小時內(nèi),梅根行駛了16061-15951=110 英里,從而她的平均速度是55 英里/小時,故(B)正確.

7.How many positive even multiples of 3 less than 2020 are perfect squares?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 12

譯文有多少個小于2020 的平方數(shù)是正的3 的偶數(shù)倍?

解依題意,正的3 的偶數(shù)倍的平方數(shù)必須是62的倍數(shù),因此滿足條件的數(shù)有以下這些:62,122,182,242,302,362,422,共7 個,故(A)正確.

經(jīng)過上述的分析，不難發(fā)現(xiàn)：傳統(tǒng)的IoT技術(shù)和中心化的系統(tǒng)框架已經(jīng)很難滿足未來的發(fā)展需求。針對IoT平臺的特點，區(qū)塊鏈技術(shù)能解決大量的智能設(shè)備數(shù)據(jù)在中心化的系統(tǒng)框架中會出現(xiàn)的安全和管理問題。區(qū)塊鏈是一種集成分布式數(shù)據(jù)庫、共識機制、點對點（P2P）傳輸和非對稱加密算法等新型應(yīng)用模式，具有去中心化、開放性、自治性、匿名性和信息不可篡改的特點。本文主要對IoT平臺之一的體域網(wǎng)展開了詳細研究，結(jié)合新興的區(qū)塊鏈技術(shù)，針對傳輸?shù)陌踩阅芎蛿?shù)據(jù)傳輸性能差的特點，研究分析了適用于體域網(wǎng)的身份認證技術(shù)，利用區(qū)塊鏈技術(shù)在體域網(wǎng)平臺下設(shè)計一個新興的系統(tǒng)框架，主要解決以下問題:

8.PointsPandQlie on the number line withPQ=8.How many locations for pointRin this plane are there such triangle with verticesP,Q,andRis a right triangle with area 12 square units?

(A) 2 (B) 4 (C) 6 (D) 8 (E) 10

譯文點P,Q在數(shù)軸上,PQ=8.在平面上有多少個點R使得以P,Q,R為頂點的直角三角形面積為12 個平方單位?

解以PQ為底,則ΔPQR的高為h=12×2÷8=3,從而點R在與PQ距離為3 的平行線上.如圖示,作距離為3 的上下兩條平行線,分三種情況:(1)若∠RPQ為直角,則R在過點P且垂直于PQ的直線與平行線的交點上,共2 個點;(2)若∠RQP為直角,則R在過點Q且垂直于PQ的直線與平行線的交點上,共2 個點;(3)若∠PRQ為直角,則R在以PQ為直徑的圓與平行線的交點上,共4 個點.故滿足條件的點R有8 個,(D)正確.

9.How many ordered pairs of integers(x,y)satisfy the equationx2020+y2=2y?

(A) 1 (B) 2 (C) 3 (D) 4 (E) infinitely many

譯文有多少個整數(shù)對(x,y)滿足方程x2020+y2=2y?

解原方程可化為x2020+(y-1)2=1,容易解得故滿足方程的解共有4 對,(D)正確.

10.A three-quarter sector of a circle of radius 4 inches together with its interior can be rolled up to form the lateral surface of a right circular cone by taping together along the two radii shown.What is the volume of the cone in cubic inches?

譯文如圖示,有一個半徑為4 英寸的圓的四分之三部分連同它的內(nèi)部,將其沿著兩條半徑粘合,正好可以卷成一個圓錐的側(cè)面.則這個圓錐的體積是多少立方英寸?

解設(shè)這個圓錐的底面半徑和高分別為r和h,則有2πr=·2π·4,r2+h2=42,解得r=3,h=于是該圓錐的體積為V==故(C)正確.

11.Ms.Carr asks her students to read any 5 of the 10 books on a reading list.Harold randomly selects 5 books from this list,and Betty does the same.What is the probability that there are exactly 2 books that they both select?

譯文卡爾夫人要求她的學(xué)生們從一個含有10 本書的書單中閱讀其中的5 本.哈羅德從書單中隨機選擇了5 本書,貝蒂也隨機選擇了5 本書.那么他們同時選擇了其中2 本書的概率是多少?

解假定哈羅德選擇的5 本書為A,B,C,D,E,則符合條件的結(jié)果是貝蒂從中選擇了任意2 本,且在另外5 本書中選擇了任意3 本,即有=100 種選擇,概率為故(D)正確.

12.The decimal representation ofconsists of a string of zeros after the decimal point,followed by a 9 and then several more digits.How many zeros are in that initial string of zeros after the decimal point?

(A) 23 (B) 24 (C) 25 (D) 26 (E) 27

譯文的小數(shù)表示為小數(shù)點后面一串的0,接下來是9 和其它幾位數(shù)字.則在小數(shù)點后面那一串有多少個0?

解又因此,的小數(shù)表示中,小數(shù)點后面有26 個0,故(D)正確.

13.Andy the Ant lives on a coordinate plane and is currently at (-20,20) facing east (that is,in the positivex-direction).Andy moves 1 unit and then turns 90°degrees left.From there,Andy moves 2 units (north) and the turns 90°degrees left.He then moves 3 units(west)and again turns 90°degrees left.Andy continues his progress,increasing his distance each time by 1 unit and always turning left.What is the location of the point at which Andy makes the 2020th left turn?

(A)(-1030,-994) (B)(-1030,-990) (C)(-1026,-994)(D)(-1026,-990) (E)(-1022,-994)

譯文螞蟻安迪住在坐標平面上,且在點(-20,20)處面朝東(即x軸正方向).安迪爬了1 個單位然后向左轉(zhuǎn)90°,跟著安迪爬了2 個單位(向北)然后向左轉(zhuǎn)90°,接著他又爬了3 個單位(向西)然后向左轉(zhuǎn)90°.安迪重復(fù)他這種動作,每次都比上次多爬1 個單位然后向左轉(zhuǎn).則安迪向左轉(zhuǎn)了2020次之后處在什么位置?

解從方向上來講,安迪爬4 次就是一個周期.他的爬行軌跡為:由此可見,每爬一個周期,安迪會向西移動2 個單位,向南移動2 個單位.于是,經(jīng)過2020 次爬行即505 個周期之后,安迪所處的位置為(-20-505×2,20-505×2) 即(-1030,-990),故(B)正確.

14.As shown in the figure below,six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon.What is the area of the shaded region—inside the hexagon but outside all of the semicircles?

譯文如圖所示,一個邊長為2 的正六邊形的內(nèi)部有6個半圓,這些半圓的直徑恰好是該六邊形的6 條邊.則圖中陰影部分——六邊形內(nèi)部半圓以外的面積是多少?

解S六邊形=6·S半圓+S陰影-6·S樹葉,其中樹葉為兩個半圓的交集部分.而S半圓=樹葉恰好是半圓內(nèi)一個60°的扇形去掉一個正三角形的部分的兩倍,于是有

故(D)正確.

15.Steve wrote the digits 1,2,3,4,and 5 in order repeatedly from left to right,forming a list of 10,000 digits,beginning 123451234512….He then erased every third digit from his list(that is,the 3rd,6th,9th,…digits from the left),then erased every fourth digit from the resulting list(that is,the 4th,8th,12th,…digits from the left in what remained),and then erased every fifth digit from what remained at that point.What is the sum of the three digits that were then in the positions 2019,2020,2021?

(A) 7 (B) 9 (C) 10 (D) 11 (E) 12

譯文史蒂夫從左至右依次按順序地重復(fù)寫下數(shù)字1,2,3,4,5,形成了一個以123451234512…開頭的10,000 位數(shù)字.然后他每三位擦掉最后一位數(shù)(也就是從左邊開始的第3 位,第6 位,第9 位,…),接著他在剩下的數(shù)字中每四位擦掉最后一位數(shù)(也就是剩下從左邊開始的第4 位,第8 位,第12 位,…),最后他在剩下的數(shù)字中每五位擦掉最后一位數(shù).則在剩下的數(shù)字中第2019,2020 和2021 位數(shù)字之和是多少?

解我們可選擇30 位數(shù)字為一個周期進行分析:123451234512345123451234512345,第一次操作后得12452351341245235134,第二次操作后得124235341452513,第三次操作后得124253415251,剩下12 位數(shù)字.而2019≡3(mod 12),即剩下數(shù)字中的第2019,2020 和2021 位數(shù)字就是最后12 位數(shù)字中的第3,4,5 位,和為11,故(D)正確.

16.Bela and Jenn play the following game on the closed interval[0,n]of the real number line,wherenis a fixed integer greater than 4.They take turns playing,with Bela going first.At his first turn,Bela chooses any real number in the[0,n].Thereafter,the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player.A player unable to choose such a number loses.Using optimal strategy,which player will win the game?

(A) Bela will always win.

(B) Jenn will always win.

(C) Bela will win if and only ifnis odd.

(D) Jenn will win if and only ifnis odd.

(E) Jenn will win if and only ifn＞8.

譯文貝拉和珍在實數(shù)軸的閉區(qū)間[0,n]上玩如下游戲,其中給定整數(shù)n＞4.他們輪流玩,貝拉先開始.最開始,貝拉在[0,n]上選擇一個任意實數(shù).然后,接下來輪到的選手要選擇一個與他們之前選擇的所有實數(shù)距離都要大1 個單位的實數(shù).無法選擇的選手就算輸.使用最優(yōu)策略,哪位選手會贏得這個游戲?

解貝拉將贏得這個游戲.因為最開始貝拉選擇將區(qū)間分成兩個對稱的小區(qū)間和接下來無論珍在哪個區(qū)間中選擇什么數(shù),貝拉都將選擇另一個區(qū)間中關(guān)于中心對稱的那個數(shù).只要珍能夠選擇數(shù),貝拉就可以繼續(xù)游戲.到最后,無法選擇的一定是珍,即珍必敗,故(A)正確.

17.There are 10 people standing equally spaced around a circle.Each person knows exactly 3 of the other 9 people:the 2 people standing next to her or him,as well as the person directly across the circle.How many ways are there for the 10 people to split up into 5 pairs so that the members of each pair know each other?

(A) 11 (B) 12 (C) 13 (D) 14 (E) 15

譯文10 個人等距地圍坐在一個圓周上,每個人都只認識其余9 人中的3 人:與他或她相鄰的2 人,以及圓周上相對的那個人.問有多少種方式把這10 個人分成5 對,使得每對的兩個人都互相認識?

解如圖示,分別給這10 個人編號為1,2,3,…,10,互相認識的人用線段或弧線連接,先考慮1 號,分如下三種情況:

(1)1 與2 配對:若3 與4 配對,則有{5,6},{7,8},{9,10},或{5,10},{6,7},{8,9},共2 種方式;若3 與8 配對,則7 必與6 配對,有{4,5},{9,10},或{4,9},{5,10},也是2 種方式.

(2)1 與10 配對:同理可得4 種方式;

(3)1 與6 配對:若2 與3 配對,則7 必與8 配對,有{4,5},{9,10},或{4,9},{5,10},共2 種方式;若2 與7 配對,則有{3,4},{5,10},{8,9},或{3,8},{4,5},{9,10},或{3,8},{4,9},{5,10},共3 種方式.

綜上可得,符合條件的分法共有13 種.故(C)正確.

18.An urn contains one red ball and one blue ball.A box of extra red and blue balls lies nearby.George performs the following operation four times:he draws a ball from the urn at random and that takes a ball of the same color from the box and returns those two matching balls to the urn.After the four iterations the urn contains six balls.What is the probability that the urn contains three balls of each color?

譯文缸里有1 個紅球和1 個藍球,旁邊另有一盒子紅球和藍球.喬治進行如下操作四次:首先他從缸里隨機拿起一個球,然后他從盒子里拿起一個相同顏色的球,接著把兩個球放回缸里.四次操作之后,缸里有6 個球.則缸里有兩種顏色各3 個球的概率是多少?

解四次操作之后,缸里可能有1,2,3,4,5 個紅球.根據(jù)對稱性,剩1 個紅球和1 個藍球即5 個紅球的概率是一樣的,2 個紅球和4 個紅球的概率是一樣的.

剩1 個紅球,也即喬治每次拿起的都是藍球,從而P(1個紅球)=

剩2 個紅球,也即喬治四次操作中只拿起一次紅球,其中第1,2,3,4 次均有可能,從而

因 此,P(3個紅球)=1-2× P(1個紅球)-2×P(2個紅球)=故(B)正確.

19.In a certain card game,a player is dealt a hand of 10 cards from a deck of 52 distinct cards.The number of distinct(unordered) hands that can be dealt to the player can be written as 158A00A4AA0.What is the digitA?

(A) 2 (B) 3 (C) 4 (D) 6 (E) 7

譯文在某撲克游戲中,一位選手要從裝有52 張不同撲克的盒子中抽取一手10 張牌.抽取這一手牌(無序)的方式可以寫成158A00A4AA0 種,則數(shù)字A是多少?

解從52 張牌中抽取10 張牌的方式為種,而==15820024220,故A=2,(A)正確.

20.LetBbe a right rectangular prism (box) with edges lengths 1,3,and 4,together with its interior.For realr＞0,letS(r)be the set of points in 3-dimensional space that lie within a distancerof some pointB.The volume ofS(r) can be expressed asar3+br2+cr+d,wherea,b,c,dare positive real numbers.What is

(A) 6 (B) 19 (C) 24 (D) 26 (E) 38

譯文設(shè)B是一個棱長分別為1,3,4 的長方體.對于實數(shù)r＞0,令S(r)為三維空間中與B中的點距離為r的所有點的集合,S(r)的體積可以表示成ar3+br2+cr+d,其中a,b,c,d均為正實數(shù).則是多少?

解實際上,S(r)就是長方體往外擴張距離為r的所有內(nèi)部點集,平面往外擴張形成一個高為r的長方體,棱往外擴張形成個底面半徑為r的圓柱體,點往外擴張形成個半徑為r的球體,從而

于是有a=b=8π,c=38,d=12.故==19,(B)正確.

21.In squareABCD,pointsEandHlie onand,respectively,so thatAE=AH.PointsFandGlie onand,respectively,and pointsIandJlie onso thatandSee the figure below.TriangleAEH,quadrilateralBFIE,quadrilateralDHJG,and pentagonFCGJIeach has area 1.What isFI2?

譯文在正方形ABCD中,點E,H分別在邊和上,AE=AH,點F,G分別在邊和上,點I,J在邊上,使得如圖所示,三角形AEH,四邊形BFIE,DHJG和五邊形FCGJI的面積均為1.則FI2是多少?

解由題意知,SABCD=4,AE=從而AB=2,BE=2-作四邊形BFIE關(guān)于BF的對稱圖形BFKL,分別延長IE和KL且相交于點M,連接BM,則可得四邊形FKMI是一個正方形,如圖示.于是

故(B)正確.

22.What is the remainder when 2202+202 is divided by 2101+251+1?

(A) 100 (B) 101 (C) 200 (D) 201 (E) 202

譯文2202+202 被2101+251+1 除后的余數(shù)是多少?

解令250=x,則2202+202=4x4+202,2101+251+1=2x2+2x+1,由多項式的帶余除法,可得4x4+202=(2x2+2x+1)·(2x2-2x+1)+201.故余數(shù)為201,(D)正確.

23.SquareABCDin the coordinate plane has vertices at the pointsA(1,1),B(-1,1),C(-1,-1),D(1,-1).Consider the following four transformations:

L,a rotation of 90°counterclockwise around tic origin;

R,a rotation of 90°clockwise around tic origin;

H,a reflection across thex-axis;

V,a reflection across they-axis,

Each of these transformations maps the square onto itself,but the positions of the labeled will change.For example,applyingRand thenVwould send the vertexAat(1,1)to(-1,-1)and would send the vertexBat (-1,1)to itself.How many sequences of 20 transformations chosen from{L,R,H,V}will send all of the labeled vertices back to their positions? (For example,R,R,V,His one sequence of 4 transformations that will send the vertices back to their original positions.)

(A) 237(B) 3·236(C) 238(D) 3·237(E) 239

譯文坐標平面中的正方形ABCD頂點分別為A(1,1),B(-1,1),C(-1,-1),D(1,-1).考慮以下四種變換:

L,繞原點做90°的逆時針旋轉(zhuǎn);

R,繞原點做90°的順時針旋轉(zhuǎn);

H,關(guān)于x軸的反射;

V,關(guān)于y軸的反射.

每一種變換都將這個正方形映回自身,但頂點的位置會發(fā)生變化.比如,施行R和V會將點A由(1,1) 映到(-1,-1),將點B由(-1,1)映回自身.從集合{L,R,H,V}中選擇20 個變換,有多少種順序方式可以將四個頂點均映回自身? (比如,R,R,V,H就是一種4 個變換的順序方式,它將四個頂點均映回自身.)

解我們記正方形ABCD經(jīng)過變換后得到的正方形分別在第一、二、三、四象限的頂點順序.例如:變換L將ABCD映成DABC,LL將ABCD映成CDAB.經(jīng)過一對變換之后,結(jié)果只可能為如下四種:

(1)ABCD映成ABCD:恒等變換,通過4 種方式LR,RL,HH,V V可得;

(2)ABCD映成CDAB:相當于中心對稱變換,通過4種方式LL,RR,HV,V H可得;

(3)ABCD映成ADCB:相當于交換B,D兩點,通過4種方式LV,RH,HL,V R可得;

(4)ABCD映成CBAD:相當于交換A,C兩點,通過4種方式LH,RV,HR,V L可得.

因此,首先經(jīng)過18 個任意變換之后,其結(jié)果也只可能是ABCD,CDAB,ADCB,CBAD中的一種,且是等概率的.此時,只需要再經(jīng)過一對變換即可映回ABCD,且均有4 種方式.

故將ABCD映回自身的方式共有418×4=238種,(C)正確.

24.How many positive integers satisfy=(recall thatis the greatest integer not exceedingx.)

(A) 2 (B) 4 (C) 6 (D) 30 (E) 32

譯文有多少個正整數(shù)滿足(注意表示不超過x的最大整數(shù).)

解設(shè)則k≥1,0≤r＜1.于是且n=70k-1000,從而原方程可化為k2-70k+1000+2kr+r2=0,即(k-20)(k-50)+2kr+r2=0.當r=0時,解得當r＞0 時,解得故共有6 組解,(C)正確.

25.LetD(n)denote the number of ways of writing the positive integernas a productn=f1·f2·····fk,wherek≥1,thefiare integers strictly greater than 1,and the order in which the factors are listed matters (that is,two representations that differ only in the order of the factors are counted as distinct).For example,the number 6 can be written as 6,2·3,3·2,soD(6)=3.What isD(96)?

(A) 112 (B) 128 (C) 144 (D) 172 (E) 184

譯文設(shè)D(n)表示正整數(shù)n寫成乘積n=f1·f2·····fk的方式數(shù)目,其中k≥1,且fi是嚴格大于1 的整數(shù),因子的排列順序也考慮在內(nèi)(也就是說,兩種僅因子順序不同的表示被算作是不一樣的).比如,6 可以表示成6,2·3,3·2,所以D(6)=3.則D(96)是多少?

解96=3×25,分情況討論:

(1)1 個因子:就是表示成96,1 種方式;

(2)2 個因子:可以表示成3×32,6×16,12×8,24×4,48×2,每一種表示可以交換順序,共10 種方式;

(3)3 個因子:可以表示成3×2×16,3×4×8,6×2×8,6×4×4,12×2×4,24×2×2,因子的順序可以重新排列,共有4+2=30 種方式;

(4)4 個因子:可以表示成3×2×2×8,3×2×4×4,6×2×2×4,12×2×2×2,重新排列后共有2×3+=40種方式;

(5)5 個因子:可以表示成3×2×2×2×4,6×2×2×2×2,重新排列后共有=25 種方式;

(6)6 個因子:可以表示成3×2×2×2×2×2,重新排列后共有=6 種方式.

由加法原理,可得112 種方式,故(A)正確.