Notation. The notation ajb willmean that a and b are positive integers and a divides
b. Forexample, 31j2 10
1,sine 2 10
1=1023=(31)(33).
1. (b) Theprot perapple is$ 2=5 1=3 =$ 1/15. So, to makea protof $10 Tim
must buy(and sell)150 apples.
2. (d)The maximalvalue ofthe funtionsinxis1 anditsminimalvalue is-1. Sine
f(x)is ontinuous, itsrange is [ p
9; p
25℄=[3;5℄.
3. () The exponentof x is
3
2 2
3
3
5 =
1
2 .
4. () We get x=y6=0,thus 7x y
x+y =3.
5. (d)Wehave(1) CD+DA=14andfromthePythagoreantheoremCD 2
+BD 2
=
169andDA 2
+BD 2
=225. So,DA 2
CD 2
=56. Using(1)wegetDA CD=4,
and DA=9, CD =5. ThereforeBD=12.
6. (b)The last three digits of 2005 2005
are the same as the lastthree digits of 5 2005
.
Also, the lastthree digitsof the powersof 5 are 5;25;125;625;125;625;125:::,so
theystart repeating. Thus,the lastthreedigits of5 2005
are 125. (More rigorously,
1000 divides 5 k+2
5 k
=245 k
=310005 k 3
when k3.)
7. () From the seond equation we get that either x =0, or y =1. Substituting
x=0intherstequationwegetthesolutions(0;1)and(0; 1). Next,wheny =1
we get the solutions (0;1) and (2;1). Finally, when y = 1 we get the solutions
(0; 1),( 2; 1). So, the system has fourdistint solutions-(0;1);(0; 1);(2;1),
and ( 2; 1).
8. (e) Sine a+2b divides 10 and a+2b 3 we get that a+2b is either 5 or 10.
There are two ases. Case I: a+2b = 5 and a b = 2. Then a =3, b = 1, and
2a b = 5. Case II: a+2b = 10 and a b = 1. Then a = 4, b = 3, and again
2a b =5. So, inall ases 2a b=5.
9. () Note that 40000 = 410 4
= 2 6
5 4
. Sine mn = 2 6
5 4
and neither m nor n is
divisible by 10we get that one of the numbers m and n equals2 6
and the other
10. (d) Sine (log
x y)(log
y
x) = log
x
x = 1, we have (log
x
y) +(log
y
x) = (log
x y+ log y x) 2
2=47.
11. (a) Sine
3
2
2
>2and 3 2 3 >3, p 2and 3 p
3 are smallerthan 3
2
. Also,3 2 >2 3 , so log 2 3> 3 2
, and 5 2
<3 3
,implying log
3 5<
3
2 .
12. (a) Denoteby k thedistanewhihthe slowerpartiletravelsin50seonds. Then
50r=kand50R=k+300. Subtratingtheaboveequationsweget50(R r)=300
or R r=6 ft./se.
13. (e)Ifm=1thelinesareparallel. Ifm 6=1theyintersetatthepoint 5 1 m ; 2m+3 1 m .
Sinebothoordinatesofthepointofintersetionmustbepositive,weget 5
1 m >0,
so m<1. Also, 2m+3
1 m
>0,so m> 3=2.
14. (a)Substitutingx=
2
andthenx=
2
intheequationwegetf 2 +2f 2 =
1 and f 2 +2f 2
= 1. Solving the last two equations for f 2 we get f 2 = 1.
15. (b) First,weshowthat one an tthreenon-intersetingirles ofradius 1inside
the irle of radius 2.4(we allit the \big irle"). Denote by O the enter of the
\big" irle and let A, B, C be three points suh that 4ABC is an equilateral
triangle with O - the enter of its irumsribed irle, and with OA = OB =
OC = 1:3. Clearly,
6
AOB =
6
BOC =
6
COA = 120 Æ
. By the Law of Cosines,
AB 2 = OA 2 +OB 2
2OAOBos(120 Æ
) = 1:3 2
+1:3 2
+1:3 2
= 5:07 > 4. So,
AB = AC = BC > 2. Consider the three irles of radius 1 (we all them
\little irles") whose enters are the points A, B, and C. The little irles do
not intersetsine thedistanesbetweenany twoentersof distintlittleirlesis
> 2. Also, the three little irles are inside the big irle sine the distane from
O toany of the enters of the little irles is <1:4.
Next, weshow thatone annott fournon-interseting little irlesinside the big
irle. Indeed, let A, B, C, and D be the enters of four little irles inside the
big irle. Then OA, OB, OC, and OD are all < 1:4. If the points are labeled
in a suitable way, we have
6 AOB + 6 BOC+ 6 COD+ 6
DOA =360 Æ
(if not, we
relabelthepoints). Then, atleast oneof theangles
6 AOB; 6 BOC; 6 COD; 6 DOA
is 90 Æ
. Suppose
6
AOB 90 Æ
. Again, by the Law of Cosines we get AB 2 = OA 2 +OB 2
2OAOBos(
6
AOB)OA 2 +OB 2 <1:4 2 +1:4 2
=3:92 <4. Thus,
AB <2and theirleswhoseenters are AandB interset. Weproeedsimilarly
if 6 BOC; 6 COD,or 6
DOA is90 Æ
AB. ThenCDisthe heightof4ABC. Sinethe baseof4ABC is p
3 1andits
areais p
3 1
2
,wegetCD=1. Now,onsideringtherighttriangle4ADCweobtain
AD = p
3 and
6
CAD = 30 Æ
. Also, sine AD = p
3 and AB = p
3 1, we get
BD = 1. Finally, onsidering the right triangle 4BDC we obtain
6
BCD =45 Æ
.
So,
6
ACB =15 Æ
.
17. (b) LetN =1234424344. Then N is divisibleby9 sine1+2+3++41+
42+43+44=990 is divisible by 9. Also, the remainder when N isdivided by 5
is 4. The onlyinteger between 0and 44whihis divisibleby9 and has remainder
4 when divided by 5 is9.
18. ()Sinesin(90 Æ
x)=osxandos(90 Æ
x)=sinxwegettan(90 Æ
x)tanx=1.
Applying the above formula for x=5 Æ
;15 Æ
;25 Æ
; and 35 Æ
, and taking into aount
tan45 Æ
=1,weget that the value of the produt is 1.
19. (b) IfP is apointon the x-axis,
PA+PB =PA+PB 0
AB
0
=13;
where B 0
=(6:5; 3)(the reetion of B with respet tothe x-axis),and equality
holds when P isthe pointof intersetion of the straight line AB 0
and the x-axis.
Similarly,if P is apoint onthe y-axis,
PA+PB =PA+PB 00
AB
00
=10;
whereB 00
=( 6:5;3)(the reetion ofB with respettothe y-axis),and equality
holds when P isthe pointof intersetion of the straight line AB 00
and the y-axis.
20. ()Oneanpaintvearsusing3olorsin3 5
=243ways. Also,oneanpaintve
ars using only2out of3 olorsin32 5
=96ways (thereare 3 ways topik2 out
of3olors). Finally,oneanpaintvearsusingonly1outof3olorsin31 5
=3
ways. Bythe inlusion-exlusion priniple, the answer is243 96+3=150.
21. (b) Let d be the greatest ommondivisor of a and b. Then 6d 2
j36ma 2
6nb 2
=
1008, sod 2
j168. Sine168=2 3
37, theonlysquares ofintegerswhihdivide 168
are 1 and 4. Thus,d2, and when a=b =2, m=7, n=0we have d=2.
22. (d) The median an not be 1 (there are at least three numbers in the set larger
than 1). For similar reasons, the median is not 9. The medianould be 6. Then
6=(x+6+4+1+9)=5,sox=10. Sine6isthemedianofthesetf1;4;6;9;10gand
the mean ofthe set isalso6,the number6has the desribedproperty. Further, if
themedianis4,then4=(x+6+4+1+9)=5,sox=0. Again,oneheksthatx=0
23. (d) Sine x 1jx 1 when x and n are positive integers, then 2 1j2 1
whenever a and b are positive integers and ajb (set x = 2 a
and n = b=a). Thus,
3= 2 2
1 divides 2 1650
1 sine 2j1650. Similarly, 7= 2 3
1 divides 2 1650
1 (
3j1650). In the sameway, 31=2 5
1divides2 1650
1,and 2047=2 11
1divides
2 1650
1. On the other hand, 127 = 2 7
1 does not divide 2 1650
1. (Indeed,
7j1652so127j2 1652
1. Thus127j2 1650
1wouldimply127j(2 1652
1) 4(2 1650
1)
or 127j3.)
24. (b) Thepolynomial Q(x)must bea quadratipolynomialwith leadingoeÆient
1 or 1. We an assume that the leading oeÆient of Q(x) is 1 (if not, we an
work with Q(x)sine (Q(x)) 2
=( Q(x)) 2
). So, Q(x)=x 2
+x+d and
(Q(x)) 2
=x 4
+(2)x 3
+( 2
+2d)x 2
+(2d)x+d 2
=x 4
+2x 3
x 2
+ax+b:
So, 2 = 2, 2
+2d = 1, 2d = a, and d 2
= b. We get =1, d = 1, a = 2,
b =1, and a+b= 1.
25. () Suppose B has oordinates (p;q) and C -(r;s). Then
(1) p>0;r >0; p 2
q 2
=1; r 2
s 2
=1;
and
(2) AB 2
=(p+1) 2
+q 2
=AC 2
=(r+1) 2
+s 2
=BC 2
=(p r) 2
+(q s) 2
:
From (1) q 2
= p 2
1 and s 2
= r 2
1. Substituting in (2) we get (p+1) 2
+
p 2
1 = (r+ 1) 2
+r 2
1, or 2(p r)(p +r + 1) = 0 whih implies p = r.
Now, substituting in (1) we obtain p 2
q 2
= p 2
s 2
= 1, so q = s. Therefore
C is a reetion of B with respet to the x-axis. Referring to (2) again we get
(p+1) 2
+p 2
1=(p p) 2
+(q ( q)) 2
=4q 2
=4(p 2
1). Simplifyingweobtain
p(p+1)=2(p+1)(p 1),sop=2,q= p
3. LetD=( p
3;0)(themidpointofBC).
Then AD isperpendiular toBC and the area of 4ABC is (BC)(AD)=2=3 p
3.
26. (d) Clearly,x6=0and y6=0. Solving for y we get
y = 4x
x 4
=4+ 16
x 4 :
So, x 4 divides 16and sine x6=0, y6=0,x 4 must be in the set
f 16; 8; 2; 1;1;2;4;8;16g. So,therearenineintegersolutions( 12;3),( 4;2),
(2; 4),(3; 12), (5;20), (6;12), (8;8),(12;6),and (20;5).
27. (a) There (5)(4) = 20 ways to pik distint kinds of owers for the north and
southplots. Also,if distintkindsof owersare planted onplotsnorth andsouth,
whihshare a ommonedge are dierent.
Next, there are 5 ways to pik the same kind of ower for the north and south
plots. In this ase,there are 4kinds ofowerslefttoplant inplots westand east.
So, intheasewhenthe samekindofowersareplantedonplotsnorthandsouth,
thereareatotalof(5)(4)(4)=80waystoplantowersinthe plots,sothatowers
inplots whih sharea ommonedge are dierent. So, the total number of hoies
is 260.
28. (e) Denote Amy's answers by (A), Bob's by (B), Cathy's by (C), Dave's by (D),
andEva'sby(E).Assumev =2. Usingthateahontestanthadfoundtheorret
value of exatlyoneunknown and (B),(D) weget y6=3,x6=4. Next,(E)implies
y=1, soz 6=1whih ontradits (A).
Therefore, v 6=2.
Again, using (B) and (D) we get y = 3, x = 4. Next (C) implies z = 5. Sine
v 6=2,weobtain v =1,u=2.
29. (a) Sinethetwotrianglesaresimilarweget, DC
DA =
AC
AB =
AD
DB
. SineAC =15,
and AB =20,weget DC = 3
4
DA and DB = 4
3
DA. The lasttwoequations imply
DB DC =
7
12
DA. Now,DB DC =CB =7, so DA=12 and DC =9.
30. (e)Wehavethatf(x)=2x 1forx=1;2;3. Thus,f(x) (2x 1)isapolynomial
of degree 4, with leading oeÆient 1, and three of its rootsare the numbers 1,2,
and 3. Therefore
() f(x) (2x 1)=(x 1)(x 2)(x 3)(x k);
where k is some real number. Substituting x = 0, and x = 4 in (*) we get
f(0) ( 1)=6k and f(4) 7=6(4 k). Adding the lasttwo equations we get
