S
OLUTIONS TO
USC’
S
2000 H
IGH
S
CHOOL
M
ATH
C
ONTEST
1. a Since
2 3
3 2
x 2
= 72x
2 =
9800 =
2 3
5 2
7 2
, we obtain
x 2
= 5
2 7
2 =3
2
. Since
x
, we deduce that
x =
5 7=3
= 35=3
.
2. b Since
log 2
log 2
log 2
16 =
log 2
log 2
4 =
log 2
2 =
1
, the answer is
1
.
3. e Using that
2 600
= 64
100
,
3 500
= 243
100
,
4 400
= 256
100
,
5 300
= 125
100
, and
6 200
= 36
100
, we see that the answer is
6 200
.
4. b Let
x =
35811231
and observe that
x 10
8
. Hence,
10 14
+ x
10 14
2 =
10 28
+ 2x10
14 +
x 2
10 28
= 100000071622462
10 14
+ x
2 10
28 :
Since
x 2
10 16
, we see that the expression
x 2
does not effect the leading 12 digits of the integer in the numerator. Thus,
1:00000035811231 2
= 1:000000716224
: :
:
. Hence,
x =
7
,
y =
1
, and
z =
6
so that
x +
y +
z =
14
.
5. e Given
x 3
x 2
5x 2x
4 +
x 3
+ k
x 2
5x +
2 =
x 7
4x 5
14x 4
5x 3
+ 19x
2 4;
the coefficient of
x 2
on the left is
2k +
25 2
= 2k
+ 23
and the coefficient of
x 2
on the right is
19
. These must be equal so that
2k +
23 =
19
. Hence,
k =
2
.
6. d The 100 horizontal lines divide the plane into 101 regions. If each vertical line is considered in turn, each divides
the plane into 101 additional regions. The total number of regions obtained is
101 2
= 10201
.
7. c The answer is
2 15
= 32768
since it is
2 3
5 =
8 5
as well as
2 5
3 =
32 3
. To see that
N
cannot be smaller, observe that the conditions in the problem imply
N =
a 5
for some positive integer
a
and that for
1 a
7
the number
a 5
is not a cube of a number different from
a
.
8. d Recall that
2 sin
os =
sin 2
. Hence,
sin 15
Æ 2
os 15
Æ 2
= sin
15 Æ
os 15
Æ 2
= sin
30 Æ
2 2
= 1
4 2
= 1
16 :
9. a Let
x =
B C
. Since
D B
C
is a
45
-
90
-
45
degree triangle,
D B
= x
. Since
AB C
is a
30
-
90
-
60
degree triangle and
tan 30
Æ =
1= p
3
, we deduce that
1 p
3 =
tan 30
Æ =
B C
AB =
B C
AD +
D B
= x
2 +
x :
Therefore,
2 +
x =
x p
3
, from which we deduce that
B C
= x
= 2=
p 3
1 =
p 3
+ 1
.
10. b Since
9 3
= 729
and
10 3
= 1000
, we see that
3 p
829 =
9 +
r
where
r 1
. Since
10 2
= 100
and
10 3
= 1000
, we see that
log 10
829 =
2 +
s
where
s 1
. Thus,
3 p
829 +
log 10
829 2
= 9
+ r
+ 2
+ s
2 =
11 +
r +
s 2
= 5:5
+ t;
where
t =
r +
s=2 1
. In other words,
3 p
829 +
log 10
829=2
is strictly between
5:5
and
6:5
. The nearest integer is therefore
6
.
11. d The graph of
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