(1)(a) The numbersbetween100 and1000that aredivisible by7 areof theform98+7k
where k is a positive integer satisfying 98+7k 1000. In other words, k is a positive
integer (1000 98)=7=902=7 (equivalently, 1k 128). Hene, the answer is 128.
(2) () If eah of the four friends aught 3 sh, then there would have to have been a
total of at least 43 = 12 sh aught. So () must be true. To see that the others are
not neessarily true, notethat (a), (b),and (e) would nothold if one person aught 8 sh
and the other three aught one sh eah; furthermore, (d) would not hold if three of the
friends aught three sh eah and the fourth friend aught two sh.
(3) (e) If T =( 2;3), then the slopes of !
PQ and !
R T are the same (namely, 2) and the
slopes of !
PRand !
QT are the same(namely, 1=2). So (e)is a orret answer. The other
hoiesfor T anbeseennottoprodueparallelogramsinasimilarmannerbyonsidering
slopes of the edges they wouldprodue.
(4) () If theequations hold for (x;y), then
y 2
y=(x+y 2
) (x+y)=4 2=2:
Thus, y 2
y 2=0 so that (y 2)(y+1)=0. We dedue that y=2 (and so x+y =2
impliesx=0)ory = 1(andsox+y =2impliesx=3). Thus, (0;2)and(3; 1)arethe
only possible pairs (x;y) satisfying the given system of equations. One heks that these
two possibilities aresolutions tothe equations, so thereare exatly2 pairs (x;y) as inthe
problem.
(5) (d) The number x 2
+x 3
=x 2
(x+1) is the square of an integer if and only if x+1
is a square (as an be seen by onsidering the prime fatorization of x 2
(x+1)). In other
words, we want to determine the number of x 2 f1;2;:::;100g suh that x+1 = n 2
for
some positiveintegern. This isequivalent tonding thenumber of positiveintegersnfor
whih 2n 2
101 whih is easily seento be 9.
(6) () If we expand (x 2
+ax+1)(x 2
+bx+2), we see that the oeÆient of x 3
is a+b
and the oeÆient of x is 2a+b. The equation in the problem implies a+b = 2 and
2a+b= 1. Solvingforaandb,wegeta= 3andb=5. Thevalueofk istheoeÆient
of x 2
in (x 2
3x+1)(x 2
+5x+2), whih is 12.
(7) (a) Suppose the prie of the ar whih made a 40% prot is x dollars and the prie
of the ar whih made a 20% loss is y dollars. Sine eah sold for $560, we dedue that
(1+0:40)x=560 and (1 0:20)y =560. Thedealer's net prot in dollars is
5602 (x+y)=1120
560
1:4
560
0:8
=1120 400 700=20:
(8) (a) The side opposite the angle has length p
3, and the hypotenuse of the triangle
is q
( p
3) 2
+2 2
= p
7. Hene, sin= p
3= p
7= p
(9) () Eah throw of the two oins, say A and B, an produe one of four equally likely
results: A and B an both show heads, A an show heads and B an show tails, A an
show tails andB anshowheads, or Aand B an both show tails. Theplayerloses when
one of these last three outomes ours on eah of the four throws. We dedue that the
player loses with probability
3333
4444 =
81
256 ;
this fration denoting the total number of possible outomes in whih a loss an our
divided by the total number of possible outomes for the four throws of the oins. The
player wins then with probability 1 (81=256)=175=25668%.
(10) () The distane from the enter of the equilateral triangle to eah vertex is 10. A
line fromthe enter of the equilateral triangle to a vertex bisets the angle at that vertex
formingtwoangleswiththesidesofthetriangle,eahmeasuring30 Æ
. Usingthatatriangle
withhypotenuse10andananglemeasuring30 Æ
hasitslegsoflength5and5 p
3,weobtain
that the height of the triangle is 15 and one-half the length of a side is 5 p
3. The area is
therefore 155 p
3=75 p
3.
(11) (d) The inequality 0 < sin12 Æ
< os12 Æ
< 1 implies that the hoies (a), (b), and
() are all < 1, that hoies (d) and (e) are >1, and that hoie (d) is larger than hoie
(e). (Thesamereasoningwouldapply with12 Æ
replaedbyanyangle measuringstritly
between 0 Æ
and 45 Æ
.)
(12) () Using properties of exponents, observe that y = 2 2
x
implies y 2
=4 2
x
. It follows
then thaty 2
+y 42 =0. On the otherhand, y 2
+y 42=(y+7)(y 6). Sine y=2 2
x
is learly positive, we dedue that y = 6. The problem is equivalent to determining the
value of p
2 y
= p
2 6
, whih is 8.
(13) (b) Let =sin (=7) and = os (=7), and note that 2
+ 2
= 1. The onditions
in the problem imply
x 2
bx+=(x )(x )=x 2
(+)x+:
We dedue that
b 2
=(+) 2
= 2
+ 2
+2 =1+2:
(14)() ThedistanefromB toC is2. WetranslatethepointD=( 5; 3)uptwounits
toD 0
=( 5; 1). Theshortestpathasintheproblemisequal inlengthto2(thedistane
fromB toC)plusthelengthof theshortestpathfromD 0
toA. TheshortestpathfromD 0
toAisalonga straight line,soits lengthisthedistanefromD 0
=( 5; 1) toA=(7;4).
This distane is p
(7+5) 2
+(4+1) 2
=13. Therefore, the answer is 2+13=15.
(15)(b) Letxrepresentthesizeof thebottleingallons. Initially,xgallonsofpurealohol
are emptied into B and stirred. At this point, B ontains a mixture that has a ratio of
aloholandwaterinthebottlethatwaspouredintoAisxx=(6+x)=x 2
=(6+x)gallons.
Therefore, A ends up with 6 x+(x 2
=(6+x)) gallons of alohol and x (x 2
=(6+x))
gallons of water. Theonditions in the problem imply
6 x+(x 2
=(6+x))
x (x 2
=(6+x)) =
4
1 :
Rewriting the above andsolving for x givesx =3=2=1:5.
(16) (a) Sine P(2) = P( 2) = P( 3) = 1, eah of the numbers 2, 2, and 3 is a
root of P(x)+1. Given that P(x) is a polynomialof degree 4, we dedue that there are
numbers a and bsuh that
P(x)=(x 2)(x+2)(x+3)(ax+b) 1:
UsingthatP(1)=P( 1)=1,weobtain 12a 12b 1=1and6a 6b 1=1. Solvingfor
aandb,wegetthata=1=12andb= 1=4(andP(x)=((x 2)(x+2)(x+3)(x 3)=12) 1).
We dedue thatP(0)= 12b 1=2.
(17) (e) Let t denotethe number of juniors (so t also is the number of seniors). Let y be
the number of respondents that answered \Yes", and let n be the number of respondents
that answered \No". Then 0:60y = 3y=5 is the number of seniors that said \Yes" and
0:80n = 4n=5 is the number of juniors that said \No". The number of juniors that said
\Yes" is 2y=5 so that (2y=5)+(4n=5) = t. The number of seniors that said \No" is n=5
so that (3y=5)+(n=5)=t. Thus,
2y+4n=5t and 3y+n=5t:
It follows that y = 3t=2 (and n = t=2). The proportion of juniors that said \Yes" is
(2y=5)=t =3=5=60%.
(18) (b) Let N =99999899999. ThenN =10 11
10 5
1. Thus,
N 2
=(10 11
10 5
1) 2
=10 22
210 16
210 11
+10 10
+210 5
+1
=9999979999810000200001:
The digit 9 appears nine times in the deimalexpansion of N 2
.
(19) () From the given information, =tan 1
(1=5) and =tan 1
(3=2). Observe also
that 0< <90 Æ
. We an obtain byomputing
tan( )=
tan tan
1+(tan)(tan) =
(3=2) (1=5)
1+(3=2)(1=5) =1:
(20)(b) Letndenotethe uniquepositiveintegerintheproblem. Wegivetwo approahes
to obtaining n. First, note that sine n ends in a 6, 4n ends in a 4. This means the last
two digits of n are 46, so the last two digits of 4n are 84. Hene, the last three digits of
n are 846, and the last three digits of 4n are 384. Continuing this way, we obtain quikly
that the last six digits of n are 153846 and that the last six digits of 4n are 615384. At
this point, werealize we have the solutionto the problem,namelyn=153846(sine then
4n is of the formindiated in the statement of theproblem).
Forthe seondapproah, we write ninthe formn=10m+6 wheremis some positive
integer. If r is the number of digits in m, then the onditions in the problem give that
610 r
+m=4n=40m+24. Thus, 610 r
=39m+24. Dividing by3and thenworking
modulo13,wededuethat210 r
8 (mod 13)sothat10 r
4 (mod 13). Oneomputes
the ongruenes
10 1
10 (mod 13)
10 2
1009 (mod 13)
10 3
9012 (mod 13)
10 4
1203 (mod 13)
10 5
304 (mod 13):
So r=5 is areasonable guess (andthis would give thesmallest possiblen). Taking r =5
in 610 r
= 39m+24 and solving for m, we obtain m = (600000 24)=39 = 15384.
Corresponding to this value of m, one heks that n = 153846 is in fat a number as in
the problem.
(21) (e) Let f(x) = x 200
2x 199
+x 50
2x 49
+x 2
+x+1. Observe that f(1) = 1 and
f(2)=7. Let q(x) andr(x)denote thequotientandremainder, respetively,when f(x) is
dividedby(x 1)(x 2). Sine(x 1)(x 2)isaquadrati, wededue thatr(x)=Ax+B
for some numbers Aand B. Thus,
f(x)=(x 1)(x 2)q(x)+Ax+B:
Wededue thatf(1)=A+B andf(2)=2A+B. Sine f(1)=1and f(2)=7,weobtain
A+B =1 and 2A+B=7. We dedue that A=6 and B= 5. Hene, r(x)=6x 5.
(22)(a) Weut theanalong alinepassing throughP,Q, andR , andwe streth outthe
an so that its side is in the shape of a retangle. Then P, Q, and Rappear as points on
two opposite edges of the retangle. The lengthof the retangleis 4, and its height is 6.
theedges asshown. The shortestpath fromP toQ toRis indiatedabove(the rsttime
around the an is indiated by the line segment fromP to Q the seond time around the
an is indiated by the line segment from Q to R ). Given that the vertial distane from
P to Qis 4andthe vertialdistanefromQ toRis2, we getfromtwoappliationsof the
Pythagoreantheorem that the length of the shortest path is
p
+21p 1 =71 whih is prime. Otherwise, p is not divisible by
3 (sine it is prime) and the remainder when p is divided by 3 is either 1 or 2. In other
words, either p = 3k+1 for some integer k or p = 3k+2 for some integer k. In either
(24) (e) Thereare 201 vertial lines and 201horizontal lines. Eah vertial line and eah
horizontallinetogetherintersetatexatlyonepoint,sothereare201 2
intersetionpoints
arisingfromtheselines. Foreveryintegerk 2[0;100℄,eahofthelinesx =kandy=k
intersetstheirlesenteredattheoriginwithradiik+(1=);k+1+(1=);:::;99+(1=)
and no others. For eah suh line and eah suh irle, there are preisely two points of
intersetion. Therefore, for eah integer k 2[1;100℄, there are preisely 8(100 k) points
of intersetion among the lines x = k and y = k and the given irles. For k = 0,
therearepreisely 4(100 k)=400points ofintersetionamongthelinesx=k =0and
y=k =0andthegivenirles. Theparallellinesdonotintersetamongthemselvesand
thesame istrue of theonentri irles. Hene,the totalnumber of intersetion points is
201
An alternativeapproah to obtaining this ubi isto onsider (x ab)(x a)(x b),
whih learly has ab as a root. Observe that
Thisimpliesa+b+= 3,ab+a+b =0,andab=1. Thus,a
With eitheroftheaboveapproahes,weseethat(a)isaorretanswer. Thepossibility
that one of the other answers is also orret should be eliminated. We illustrate how this
anbe done. Letf(x)=x
+ 2and,therefore,
ofx(x 2
+2) f(x)=5x+1. Butthenabisarootof25(x 2
+2) (5x 1)(5x+1)=51. But
learly nothinganbe aroot ofthepolynomial51, soourassumptionwaswrongand abis
not a root of x 3
+x 2
3x+1. Asimilar argument an be givento show thateah of the
remaining hoies is also not a orret answer. (The approah used here is an appliation
of a variation onthe Eulidean algorithm.)
(26)(d) Squaringbothsidesoftheequation p
222y is a rational
number. It follows that y = 222m 2
for some nonnegative integer m (see the omment
below). A similar argument gives that x = 222n 2
for some nonnegative integer n. Sine
p
1998=3 p
222, the equation p
We have used that if k is an integer for whih p
k is rational, then k is the square of
an integer. This an be seen by noting that if p
so that the prime fatorization of b 2
k and a 2
must be the same. The latter
implies that the highest power of anyprime p dividing k must be of the form p r
where r
is an even integer. Thus, k must be a square.
(27) (e) Sine xyz = 4000, the only prime divisors of x, y, and z are 2 and 5. Writing
(whereeah exponent isa nonnegative integer), we have
2
Thus, wewanttoknowthenumberofsolutionsinnonnegativeintegerstoa+b+=5and
d+e+f =3. Foreah a2f0;1;2;3;4;5g,thenumberof band suh that b+=5 a
We note that there is anotherapproah to ounting thenumber of nonnegative integer
solutions of a+b+=5. Consider seven suessive blanks numbered as follows:
1 2 3 4 5 6 7
Ifweputahekmarkonanytwoof them,weobtainahoiefora,b,andorresponding
tothe number of blanksbefore therst hek mark(for a),thenumber of blanks between
thetwohekmarks(forb),andthenumberof blanksafter theseondhekmark(for).
Forexample,if theblanks numbered3and4areheked, thenwewouldhavea=2,b=0,
=21 dierent possibilities for a, b, and , as
before. A similar argument shows that, more generally, if k and n are positive integers,
thenthenumberof solutionsinnonnegativeintegersa
1
(28) (d) In the drawing, put the y-axis along the line !
AB and put the x-axis along the
line !
BC. Then B is at the origin (0;0). The shaded region is bounded by four lines,
one through eah of the points A = (0;1), B = (0;0), C = (1;0), and D = (1;1). The
line through A also passes through (2=3;0) so its equation is y = ( 3=2)x+1. The line
through C is parallel andso has the sameslope;its equation is y=( 3=2)x+(3=2). The
line through B passes through (1;2=3) and has equation y = (2=3)x. The line through
D is parallel to the line through B. Sine the slope of the line through A is minus the
reiproal of the slope of theline through B, these linesare perpendiular. It follows that
the shaded region is a retangle. In fat, beause of the symmetry in the problem, eah
edge of the retangle will have the same length so that the retangle is a square. To nd
its area, we determine the length of a side of the square. The lines through A and B
(alongtwo sidesof the innersquare) intersetat thepoint (6=13;4=13)(whih isobtained
fromtheequations ofthese linesabove). Also,the linesthroughB andC (alongtwosides
of the inner square) interset at the point (9=13;6=13). The distane from (6=13;4=13)
to (9=13;6=13) is
13. Thus, the area of the shaded region is
(1= p
13) 2
=1=13.
(29) (d) Observe that
f(3)=log
We make use of the hange of base formula log
b
. Using this hange of base
formula and otherbasi properties of logarithms, we dedue
f(x)=log
Sine the denition of the logarithm implies log
2
x is an inreasing funtion for x > 0
to innity with x. Thus, f(3)<0, f(x) is inreasing for x 3, and f(x) tends to innity
with x. These imply(d) is true and eah of (a),(b), (), and (e) is not true.
(30) (d) Observe that
a
b =
1997
1998 +
1999
n =
1997n+19981999
1998n
:
Sine a is divisible by 1000, a is even. It follows that n must be even. We write n =2m
and simplify the aboveexpression for a=b to obtain
a
b =
1997m+9991999
1998m
:
Sine ais even, we see that m must beodd. Sine 5 divides a, we also see that m annot
be divisible by 5. So we onsider m now having no prime divisors in ommon with 10.
Observe that the denominator on the right-hand side above is divisible by 2 and not 4.
Also, this denominator is not divisible by 5. Thus, in order for a to be divisible by 1000,
it is neessary and suÆient for 1997m+9991999 to be divisible by2000. We solvefor
m by workingmodulo 2000. We seek mfor whih
01997m+9991999 3m+999( 1) (mod 2000)
whih is equivalent to
m 3331667 (mod 2000):
(Here, we have usedthat 3and 2000haveno ommonprimedivisors sothat division by3
inaongruene modulo2000ispermissible.) Notethatm1667 (mod 2000)impliesthat
m and 10 have no ommon prime divisors. Hene, the ondition m1667 (mod 2000) is
aneessaryandsuÆientonditionforn=2mtoresultinafration a=basintheproblem
with adivisible by1000. Therefore, thesmallest suh positive integer is 21667=3334.
