DISINI test_09_sol
UNIVERSITY OF VERMONT
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-SECOND ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 11, 2009
7
8
1) Express
7
8
7
8
2) If
1
2
2
7
8
− 16
−
2
2
2
1
16
2
1
16
−
1
− 16
as a rational number in lowest terms.
7
8
7
1
− 16
8
=
1
− 16
2
7
1
+
8 16
2
x + 2 = 3, what is the value of x + 3
7
8
7
8
=
2
1
− 16
1
+ 16
·
16
16
=
14 – 1
14 + 1
=
13
15
?
Squaring x + 2 = 9 ï x = 7 ï x + 3 2 = 102 = 100
3) A rectangle consists of 12 squares, each of
which has an edge length of 1 inch.
Find the area of the shaded region.
Shaded area A = area of rectangle minus the areas of the unshaded triangles.
A = 12 –
1
2
1 6 –
1
2
2 6 = 12 – 3 – 6 = 3
4) The hour hand of a certain clock is 5 inches long. The minute hand of the clock is 7 inches long. What is the distance
between the tips of the hands of the clock at 4 p.m.?
5
120°
x
7
1
By the law of cosines x2 = 72 + 52 –2 7 5 cos 120 ° = 49 + 25–70 – 2 = 74 + 35 = 109
x=
109
5) Find the value of sin2 15 ° cos2 15 ° . Express your answer as a rational number in lowest terms.
sin2 15 ° cos2 15 ° = sin 15 ° cos 15 °
2
=
1
2
2
sin 30 °
=
1
2
1
2
2
=
1
16
6) Express 4.81 as a rational number in lowest terms.
s = 4.81
100s = 481.81
99s = 481.81 – 4.81 = 477 ï s =
477
99
=
53
11
7) Krisan can rake her yard in seven hours. Patti can rake the same yard in half that time while Nick would take four times as
long as Patti to rake the yard. If they all work together, how long will it take them to rake Krisan's yard?
Krisan’s rate =
1
7
, Patti’s rate =
1
7
=
2
7
4
2
If t is the time to complete the job t
1
7
+
1
, Nick’s rate =
2
7
+
1
14
7
=
1
14
2
= 1 ï t(2 + 4 + 1) = 14 ï 7t = 14 ï t = 2
8) Let x, y and z be positive integers satisfying x2 + y2 + z2 = 6 . Find x + y + z.
The maximum value for x, y or z is 2. At most one of x, y and z could be 2. Thus the only positive integer values
for x, y and z are 1, 1 and 2 ï x + y + z = 1 + 1 + 2 = 4
9) When the integer 72009 is written as a binary number, what are the three rightmost digits of the binary number?
The rightmost three digits of n in binary is n mod 8.
7 mod 8 = 7
72 mod 8 = 1
73 mod 8 = 7
···
72 k+1 mod 8 = 7 ï 72009 mod 8 = 7 = 1112
10) The Elk River Band is selling cheesecakes to raise money for a band trip to New Orleans. A large cheesecake sells
for $10.50 and a small cheesecake sells for $8.00. If the total sales were $945 and the ratio of large cheesecakes to small
cheesecakes was 2 to 3, how many large cheesecakes were sold?
Let L = number of large cheesecakes and S = number of small cheesecakes.
L
S
=
2
3
3
ï S = 2L
945 = 10.50 L + 8.00 S
3
945 = 10.50 L + 8.00 2 L
945 = (10.50 + 12)L ï 22.5 L = 945 ï L =
945
22.5
= 42
11) Consider the operation ∆, where (A ∆ B) = AB + B – 1. Find P so that (2 ∆ 7) ∆ 5 = (6 ∆ 3) ∆ P .
(2 ∆ 7) = 2(7) + 7 – 1 = 14 + 6 = 20
(6 ∆ 3) = 6(3) + 3 – 1 = 18 + 2 = 20
(2 ∆ 7) ∆ 5 = (6 ∆ 3) ∆ P ï 20 ∆ 5 = 20 ∆ P ï P = 5
2710 + 910
. Express your answer in simplest form.
911 + 274
12) Determine the value of
2710 + 910
=
911 + 274
1
2
1–
13) Evaluate
320 310 + 1
330 + 320
=
322 + 312
1
1
1
+ 24 – 25
23
1
1
1
1
1
+ 22 + 23 + 24 + 25 +
2
+
1+
1
22
–
1
1
22009
+ ⋅ ⋅ ⋅ + 22008 –
1
38 = 34 = 81
=
312 310 + 1
1
⋅ ⋅ ⋅+ 22008 + 22009
. Express your answer as a rational number in lowest terms.
1– 1 2 2010
1–
1
2
1
1
22
1
1
1
+
– 23 + 24 – 25
1
1
1
1
1
+ 2 + 22 + 23 + 24 + 25 +
1
+ ⋅ ⋅ ⋅ + 22008 –
1
1
⋅ ⋅ ⋅+ 22008 + 22009
1– – 1 2 2010
1
22009
1– – 1 2
=
1– 1 2 2010
3
=
1– 1 2
2
1– 1 2 2010
1
2
3
=
=
1
3
2
1
2
14) When three people are weighed two at a time, the weights are 280, 330 and 350 pounds. How much does the heaviest person weigh?
Assume w1 § w2 § w3
w1 + w2 = 280
w1 + w3 = 330
w2 + w3 = 350
(1)
(2)
(3)
(4)
(2) – (1) ï w3 – w2 = 50
(3) + (4) ï 2w3 = 400 ï w3 = 200
15) A circle with diameter AB has radius
5
π
. The curves
AXB and AYB are formed from semicircles
of radii
2
π
and
3
π
with centers
on AB. Find the area between the curves AXB and AYB.
Y
A
Area = 2
1
2
p
3
p
2
–
1
2
p
2
p
2
X
=9–4=5
16) Find the integer value of the expression log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
1
4
= log 3 (7) ×
log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
1
4
= log 3 (7) × 2 log2 3 × –log 7 22
1
2
1
4
.
log2 32 + log2 3 × –log 7 22
B
log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
1
4
= log 3 (7) ×
log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
1
4
= 2 log 2 (7) – log
2
log3 2
× –log 7 22
2
2
7
=–4
17) In a group of 90 students, there are 48 who play football, 38 who play basketball and 54 who play baseball.
Twenty of the 90 students play all three sports. Each student plays at least one sport.
How many students play only one sport?
Football
Basketball
a
b
c
20
d
f
e
Baseball
Football a + b + d + 20 = 48 ï a + b + d = 28
Basketball b + c + f + 20 = 38 ï b + c + f = 18
Baseball d + e + f + 20 = 54 ï d + e + f = 34
(1)
(2)
(3)
Total a + b + c + d + e + f + 20 = 90 ï a + b + c + d + e + f = 70
(4)
(1) + (2) + (3) ï 2(d + e + f) + a + c + e = 80 ï b + d + f = 40 –
1
2
1
2
From (4) a + c + e + 40 –
10
a + c + e = 90 ï a + c + e = 60
= a + b i where a and b are real numbers and i2 = – 1, what is the value of a + b ?
18) If
1
1
i
2
= 1 – 2i – 1 = – 2i
1
i
4
=(– 2i 2 =– 4
1
i
8
= ( – 4 2 = 16
1
i
10
i
a +c+e
= 1
i
8
1
i
2
= 16(–2i) = – 32i ï a + b = – 32
19) A quadrilateral with sides of length 4, 6, 8 and x is
inscribed in a circle of radius 5 (see the figure).
What is x?
4
3
r
x
4
r2 = 32 + 42 ï r = 5
2 r 2 = 42 + x2 ï 102 = 16 + x2 ï x2 = 84 ï x =
84 = 2
20) ABC is an arc of a circle, AC is a chord of the circle,
B is the midpoint of arc ABC, D is the midpoint of
21
3
10
chord AC, AC = 20 and BD = 3. What is the
diameter of the circle containing ABC?
r–3
r
r2 = r–3 2 + 102
r2 = r2 – 6r + 9 + 100
6r = 109 ï r =
109
6
ï d=
109
3
1
. Define the sequence of functions fn x by
x
f1 x = f x and fn x = f fn 1 x for n ¥ 2.
21) Let f x = 1 –
What is f50 50 ?
f1 x = 1 –
f2 x = 1 –
f3 x = 1 –
1
x
1
1–
1
1
x–1
x
1
–
=–
1
=x
x–1
f3 k x = x
50 = 3(16) + 2 ï f50 x = –
1
x–1
ï f50 50 = –
1
50 – 1
=–
1
49
22) If S = { 4 , 1 , 12 , x , 7 }, find all real values of x such that the mean of S equals the median of S.
4 + 1 + 12 + x + 7 = 24 + x
S = { x , 1 , 4 , 7 , 12 }
24 + x
5
=4 ï x=–4
S = { 1 , x , 4 , 7 , 12 }
24 + x
5
=4 ï x=–4
S = { 1 , 4 , x , 7 , 12 }
24 + x
5
=x ï x=6
S = { 1 , 4 , 7 , x , 12 }
24 + x
5
= 7 ï x = 11
S = { 1 , 4 , 7 , 12 , x }
24 + x
5
= 7 ï x = 11
x = – 4 , 6 , 11
23) How many positive divisors does the integer 1440 have?
1440 = 25 ÿ32 ÿ5
number of divisors = (5+1)(2+1)(1+1) = 6(3)(2) = 36
24) Define the sequence an by a0 = 0, a1 = 1 and an+1 = an + –1
a0
a1
a2
a3
a4
a5
a6
a7
··
=0
= a1
= a1
= a2
= a3
= a4
= a5
= a6
·
n
an
1
for n ¥ 1. Find all values of n such that an = 144.
– a0 = a1
+ a1 = a1 + a1 = 2 a1
– a2 = 2 a1 – a1 = a1
+ a3 = a1 + 2 a1 = 3 a1
– a4 = 3 a1 – a1 = 2 a1
+ a5 = 2 a1 + 3 a1 = 5 a1
Let fn be the Fibonacci sequence.
fn = {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, · · ·}
a2 k = fk ÿ a1 = fk and a2 k+1 = fk+2 ÿa1 = fk+2
a2 k = 144 = f12 ï 2k = 21
a2 k+1 = 144 = f12 = f10+2 ï 2k+1 = 21
n = 21 and 24
25) Find all real numbers x such that | 2 x – 1 | + | x – 5 | = 8 .
x – 5
5 – x
2x – 1
1 – 2x
2x – 1 + x – 5
= 8
6 – 3x = 8
3x = –2
x = –
5 – x
x – 5
2x – 1
2x – 1
x + 4 = 8
3x – 6 = 8
x = 4
3x = 14
2
14
3
3
1
2
5
x=–
2
3
,4
26) The corners of a rectangle in the plane are located at the points (1,1), (1,10), (5,10) and (5,1). The line y = m x divides the
rectangle into two trapezoids of equal area. Find m.
1,10
5,10
10
5,5m
8
6
4
2
1,m
1,1
1
1
2
5,1
2
3
4
4 10 – m + 10 – 5 m =
5
1
2
6
4 m–1 + 5m–1
20 – 6m = 6m – 2 ï 12m = 22 ï m =
22
12
=
11
6
27) A jar contains 5 different pennies, 6 different nickels, 7 different dimes and 8 different quarters. How many ways can a set
of 3 coins be selected so that the total value of the selected coins is less than 15 cents?
pennies
nickels
dimes
ways
2
0
1
5
2
7
1
=
10 7
= 70
1
2
0
5
1
6
2
=
5 15
= 75
2
1
0
5
2
6
1
=
10 6
= 60
3
0
0
5
3
=
10
70 + 75 + 60 + 10 = 215
28) Semicircles are constructed on the legs and the base of
the right isosceles triangle as shown in the diagram.
If the length of the base is 4 units, find the sum of
the areas of the shaded regions.
1
2
2(a + b) = 2
1
2
2b + c =
c=
1
2
2
p
2
2
= 2p
p 2 2 = 2p
2
2
2 =4
2b = 2p – c = 2p – 4
2a + 2b = 2p ï 2a = 2p – 2b = 2p – (2p – 4) = 4
2
2+i
29) If i 2 = – 1, determine the value of i +
i+
2
2+i
1 + 2 i 2– i
· 2–i
2+i
i+
i 2+i +2
2+i
=
2
2+i
=
2–i+4i+2
4+1
2
=
. Express your answer in the form a + b i, where a and b are real numbers.
1+2i
2+i
4+3i
5
=
4+3i 2
5
=
2i–1+2
2+i
=
2
=
16 + 24 i – 9
25
=
7
25
+
24
i
25
30) Find the radius of the circle whose locus is the set of all points x, y) whose distance from (2 , 0) is
5 times the distance
from (– 1 , 0).
x–2 2 + y2 =
5
x + 1 2 + y2
x–2 2 + y2 = 5( x + 1 2 + y2 )
x2 – 4 x + 4 + y2 = 5 x2 + 10 x + 5 + y2
4 x2 + 14 x + 4 y2 = – 1
4 x2 +
x2 +
r=
7
2
7
2
x+
x+
45
16
=
49
16
49
16
3
+ 4 y2 = – 1 +
+ y2 = –
1
4
+
49
16
49
4
=
45
16
= r2
5
4
31) Suppose that x1 , x2 , x3 , · · ·, x9 are real numbers with the following property: for k = 1, 2, 3, · · ·, 9, the sum of all of the numbers
1
2
3
9
except xk is k. For example, x1 + x2 + x3 + x4 + x6 + x7 + x8 + x9 = 5. What is x3 ?
Each of the xi appears in 8 of the 9 sums ï
8( x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 ) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 =
x3 =
45
8
32) Let z =
45
8
– (x1 + x2 + x4 + x5 + x6 + x7 + x8 + x9 ) ï x3 =
1
a+
45
8
–3=
45
8
–
24
8
=
21
8
. If a, b, c, d and e are all either 1 or 2, what is the minimum value of z?
1
1
b+
1
c+
d+
1
e
To make the fraction as small as possible, pick a, b, c, d and e to make the denominator as large as possible.
Thus a = 2, b = 1, c = 2, d = 1 and e = 2
z=
1
2+
=
1
1
1+
2+
1
1+
1
2
1
2+
=
1
1+
1
2+
1
2+
1
3
2
=
1
1
1+
2+
1
2+
2
=
1
1
1+
1
1
1
=
1
1 =
8 =
2+ 3
2 + 11
2 + 11
1 +8
8
=
11
8
3
3
Y
33) ABC is a 3-4-5 right triangle. Side AB is
extended to point X so that B is the midpoint
1
30
p3
12
of AX, side CA is extended to point Z so
that A is the midpoint of CZ and side BC
is extended to point Y so that C is the
C
6
midpoint of BY. Find the area of triangle XYZ.
6
3
4
A
D
5
B
3
3
4
8
Z
p1
The indicated dimensions are computed using similar triangles.
Area = 12(9) –
1
2
9 4 + 8 3 + 6 12 = 108 –
1
2
X
36 + 24 + 72 = 108 –
1
2
132 = 108 – 66 = 42
p2
11
30
34) Let two circles of radii 1 and 3 be tangent at point
A and tangent to a common straight line at points B
and C. Find AB2 + BC2 + CA2 .
A
B
C
C2
π–α
3
2
A
1
C1
γ
β
E
α
β
γ
B
Let
—BC
1
C
—C
A = a. From trapezoid BC1 C2 C
From triangle ACC2 p – a + 2g = p ï g =
From triangle BC1 A a + 2b = p ï b =
1
C2 C = p – a.
a
2
p–a
2
=
p
2
–
a
2
=
p
2
–g ï b+g=
p
2
ï
—BAC =
p
2
From triangle C1 C2 E BC2 = 42 – 22 = 12
AB2 + AC2 = BC2 ï AB2 + BC2 + CA2 = 12 + 12 = 24
35) How many six-digit positive integers are there in which every digit occurs the same number of times as the value of the digit?
An example of such an integer is 133232.
This corresponds to the partitions of 6 into distinct positive integers.
Partition
Positions
1+5
6 , 1
6
6
1+3+2
30 , 3
60
6
2+4
6 , 2
15
15
6
6 , 6
1
1
Total
Number of Ways
82
Number of integers = 82
36) An isosceles triangle whose sides are in the ratio 1 : 2 : 2 is inscribed in a circle of radius 1 and a second circle is inscribed in this triangle.
What is the radius of the second circle?
z
A
z
D
B
β
β
r
α
E
π–2α 4z
F
α
C
1
4
From triangle CDB, sin(a) =
By the law of cosines in triangle CFB, 4 z 2 = 12 + 12 – 2(1)(1)cos(p–2a)
16 z2 = 2 + 2cos(2a)
16 z2 = 2 + 2(1 – 2 sin2 a = 2 + 2(1 – 2
z2 =
From triangle CDA, cos(2b) =
From triangle ADE, tan( b =
tan(b) =
cos b
2
=
=
1 + cos 2 b
=
15
4
15
8
ï z=
1
4
r
z
1 – cos 2 b
sin b
15
64
1 2
)
4
1–
1+
1
4
1
3
5
=
4
2
3
5
tan(b) =
=
r
z
=
r
ï r=
15
3
5
·
15
8
6 j+
1
.
k
=
3
8
8
10
k
37) Determine the value of
k = 1 j= 1
10
k
6 j+
k = 1 j= 1
1
=
k
1
6 j+
j= 1
1
+
k
2
6 j+
j= 1
1
+
k
3
6 j+
j= 1
1
+· · ·
k
10
6 j+
j= 1
1
k
1
2
= 6(1) + 1 + 6(1 + 2) +
1
10
+
1
10
+
1
10
+
1
10
+
1
10
1
10
+
+
1
10
+
+
1
2
+ 6(1 + 2 + 3) +
1
3
+
1
3
+
1
3
+ · · · 6(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) +
1
10
= 6(1 + 1 + 2 + 1 + 2 + 3 + · · · 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) + 10(1)
= 6[10(1) + 9(2) + 8(3) + 7(4) + 6(5) + 5(6) + 4(7) + 3(8) + 2(9) + 10] + 10
= 6(10 + 18 + 24 + 28 + 30 + 30 + 42 + 24 + 18 + 10) + 10
= 6(220) + 10 = 1320 + 10 = 1330
38) Suppose that the lengths of the sides of a right triangle
ABC are 10, 24 and 26. A circle with center D and
radius 1 rolls inside the triangle, always tangent to
at least one side. How far has D traveled when it
first returns to its original position?
C
D
A
B
C
F
26
10
D
E
x
A
B
24
Triangles ABC and DEF are similar. ï the ratios of their corresponding sides is constant.
Let angle CAB = 2a ï angle DAB = a
tan(a) =
1
x
tan(2a) =
and tan(2a) =
2 tan a
=
1 – tan2 a
5
12
10
24
5
12
=
ï 24tan(a) = 5 – 5 tan2 a ï 5 tan2 a + 24tan(a) – 5 = 0
(5tan(a) – 1)(tan(a) + 5) = 0 ï tan(a) =
Then tan(a) =
1
x
=
1
5
1
5
, – 5. Since 0 < a <
p
2
tan(a) =
ï x=5
DE = 24 – 1 – 5 = 18 ï ratio of sides of triangles ABC and DEF is
EF =
3
4
10 =
15
2
and FD =
3
4
26 =
39
2
Distance traveled = DE + EF + FD = 18 +
15
2
+
39
2
= 45
18
24
=
3
4
1
5
1
10
+
1
10
+
10
A
39) The line segment AB is tangent at A to a circle with
B
center O. Point D is inside the circle and DB intersects
the circle at C. Given that BC = DC = 6, OD = 3
and AB = 10, find the radius of the circle.
6
r
C
O
6
3
D
x r–3
By the power of a point 6(12 + x) = 102 ï 6x = 28
By the power of a point 6x = (r + 3)(r – 3) ï 28 = r2 – 9 ï r2 = 37 ï r =
37
40) Suppose that p, q and r are the distinct roots of x3 – 2 x2 + 2 x – 3 = 0. Find the value of p3 + q3 + r3 .
x3 – 2 x2 + 2 x – 3 = (x – p)(x – q)(x – r) = x3 – (p + q + r) x2 + (pq + pr + qr) x – pqr ï p + q + r = 2, pq + pr + qr = 2 and pqr = 3
(p + q + r 3 = p3 + q3 + r3 + 3(p + q + r)(pq + qr + pr) – 3pqr = 23 ï 8 = p3 + q3 + r3 + 3(2)(2) – 3(3) ï p3 + q3 + r3 = 5
41) Express
1
– 2 sin 50 ° in simplest form.
2 cos 20 °
1
– 2 sin 50 ° =
2 cos 20 °
Using sin(A)cos(B) =
1
2
1 – 4 sin 50 ° cos 20 °
2 cos 20 °
sin A + B + sin A – B
1
– 2 sin 50 ° =
2 cos 20 °
1–4
1
2
sin 70 ° + sin 30 ° cos 20 °
2 cos 20 °
=
1 – 2 sin 70 ° – 1
2 cos 20 °
Since 70° and 20° are complimentary sin(70°) = cos(20°) ï
=
– 2 sin 70 °
2 cos 20 °
1
– 2 sin 50 ° =
2 cos 20 °
– 2 sin 70 °
2 cos 20 °
=
– 2 cos 70 °
2 cos 20 °
=–1
DEPARTMENT OF MATHEMATICS AND STATISTICS
FIFTY-SECOND ANNUAL HIGH SCHOOL PRIZE EXAMINATION
MARCH 11, 2009
7
8
1) Express
7
8
7
8
2) If
1
2
2
7
8
− 16
−
2
2
2
1
16
2
1
16
−
1
− 16
as a rational number in lowest terms.
7
8
7
1
− 16
8
=
1
− 16
2
7
1
+
8 16
2
x + 2 = 3, what is the value of x + 3
7
8
7
8
=
2
1
− 16
1
+ 16
·
16
16
=
14 – 1
14 + 1
=
13
15
?
Squaring x + 2 = 9 ï x = 7 ï x + 3 2 = 102 = 100
3) A rectangle consists of 12 squares, each of
which has an edge length of 1 inch.
Find the area of the shaded region.
Shaded area A = area of rectangle minus the areas of the unshaded triangles.
A = 12 –
1
2
1 6 –
1
2
2 6 = 12 – 3 – 6 = 3
4) The hour hand of a certain clock is 5 inches long. The minute hand of the clock is 7 inches long. What is the distance
between the tips of the hands of the clock at 4 p.m.?
5
120°
x
7
1
By the law of cosines x2 = 72 + 52 –2 7 5 cos 120 ° = 49 + 25–70 – 2 = 74 + 35 = 109
x=
109
5) Find the value of sin2 15 ° cos2 15 ° . Express your answer as a rational number in lowest terms.
sin2 15 ° cos2 15 ° = sin 15 ° cos 15 °
2
=
1
2
2
sin 30 °
=
1
2
1
2
2
=
1
16
6) Express 4.81 as a rational number in lowest terms.
s = 4.81
100s = 481.81
99s = 481.81 – 4.81 = 477 ï s =
477
99
=
53
11
7) Krisan can rake her yard in seven hours. Patti can rake the same yard in half that time while Nick would take four times as
long as Patti to rake the yard. If they all work together, how long will it take them to rake Krisan's yard?
Krisan’s rate =
1
7
, Patti’s rate =
1
7
=
2
7
4
2
If t is the time to complete the job t
1
7
+
1
, Nick’s rate =
2
7
+
1
14
7
=
1
14
2
= 1 ï t(2 + 4 + 1) = 14 ï 7t = 14 ï t = 2
8) Let x, y and z be positive integers satisfying x2 + y2 + z2 = 6 . Find x + y + z.
The maximum value for x, y or z is 2. At most one of x, y and z could be 2. Thus the only positive integer values
for x, y and z are 1, 1 and 2 ï x + y + z = 1 + 1 + 2 = 4
9) When the integer 72009 is written as a binary number, what are the three rightmost digits of the binary number?
The rightmost three digits of n in binary is n mod 8.
7 mod 8 = 7
72 mod 8 = 1
73 mod 8 = 7
···
72 k+1 mod 8 = 7 ï 72009 mod 8 = 7 = 1112
10) The Elk River Band is selling cheesecakes to raise money for a band trip to New Orleans. A large cheesecake sells
for $10.50 and a small cheesecake sells for $8.00. If the total sales were $945 and the ratio of large cheesecakes to small
cheesecakes was 2 to 3, how many large cheesecakes were sold?
Let L = number of large cheesecakes and S = number of small cheesecakes.
L
S
=
2
3
3
ï S = 2L
945 = 10.50 L + 8.00 S
3
945 = 10.50 L + 8.00 2 L
945 = (10.50 + 12)L ï 22.5 L = 945 ï L =
945
22.5
= 42
11) Consider the operation ∆, where (A ∆ B) = AB + B – 1. Find P so that (2 ∆ 7) ∆ 5 = (6 ∆ 3) ∆ P .
(2 ∆ 7) = 2(7) + 7 – 1 = 14 + 6 = 20
(6 ∆ 3) = 6(3) + 3 – 1 = 18 + 2 = 20
(2 ∆ 7) ∆ 5 = (6 ∆ 3) ∆ P ï 20 ∆ 5 = 20 ∆ P ï P = 5
2710 + 910
. Express your answer in simplest form.
911 + 274
12) Determine the value of
2710 + 910
=
911 + 274
1
2
1–
13) Evaluate
320 310 + 1
330 + 320
=
322 + 312
1
1
1
+ 24 – 25
23
1
1
1
1
1
+ 22 + 23 + 24 + 25 +
2
+
1+
1
22
–
1
1
22009
+ ⋅ ⋅ ⋅ + 22008 –
1
38 = 34 = 81
=
312 310 + 1
1
⋅ ⋅ ⋅+ 22008 + 22009
. Express your answer as a rational number in lowest terms.
1– 1 2 2010
1–
1
2
1
1
22
1
1
1
+
– 23 + 24 – 25
1
1
1
1
1
+ 2 + 22 + 23 + 24 + 25 +
1
+ ⋅ ⋅ ⋅ + 22008 –
1
1
⋅ ⋅ ⋅+ 22008 + 22009
1– – 1 2 2010
1
22009
1– – 1 2
=
1– 1 2 2010
3
=
1– 1 2
2
1– 1 2 2010
1
2
3
=
=
1
3
2
1
2
14) When three people are weighed two at a time, the weights are 280, 330 and 350 pounds. How much does the heaviest person weigh?
Assume w1 § w2 § w3
w1 + w2 = 280
w1 + w3 = 330
w2 + w3 = 350
(1)
(2)
(3)
(4)
(2) – (1) ï w3 – w2 = 50
(3) + (4) ï 2w3 = 400 ï w3 = 200
15) A circle with diameter AB has radius
5
π
. The curves
AXB and AYB are formed from semicircles
of radii
2
π
and
3
π
with centers
on AB. Find the area between the curves AXB and AYB.
Y
A
Area = 2
1
2
p
3
p
2
–
1
2
p
2
p
2
X
=9–4=5
16) Find the integer value of the expression log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
1
4
= log 3 (7) ×
log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
1
4
= log 3 (7) × 2 log2 3 × –log 7 22
1
2
1
4
.
log2 32 + log2 3 × –log 7 22
B
log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
1
4
= log 3 (7) ×
log 3 (7) × log 4 (9) + log 2 (3)
] × log 7
1
4
= 2 log 2 (7) – log
2
log3 2
× –log 7 22
2
2
7
=–4
17) In a group of 90 students, there are 48 who play football, 38 who play basketball and 54 who play baseball.
Twenty of the 90 students play all three sports. Each student plays at least one sport.
How many students play only one sport?
Football
Basketball
a
b
c
20
d
f
e
Baseball
Football a + b + d + 20 = 48 ï a + b + d = 28
Basketball b + c + f + 20 = 38 ï b + c + f = 18
Baseball d + e + f + 20 = 54 ï d + e + f = 34
(1)
(2)
(3)
Total a + b + c + d + e + f + 20 = 90 ï a + b + c + d + e + f = 70
(4)
(1) + (2) + (3) ï 2(d + e + f) + a + c + e = 80 ï b + d + f = 40 –
1
2
1
2
From (4) a + c + e + 40 –
10
a + c + e = 90 ï a + c + e = 60
= a + b i where a and b are real numbers and i2 = – 1, what is the value of a + b ?
18) If
1
1
i
2
= 1 – 2i – 1 = – 2i
1
i
4
=(– 2i 2 =– 4
1
i
8
= ( – 4 2 = 16
1
i
10
i
a +c+e
= 1
i
8
1
i
2
= 16(–2i) = – 32i ï a + b = – 32
19) A quadrilateral with sides of length 4, 6, 8 and x is
inscribed in a circle of radius 5 (see the figure).
What is x?
4
3
r
x
4
r2 = 32 + 42 ï r = 5
2 r 2 = 42 + x2 ï 102 = 16 + x2 ï x2 = 84 ï x =
84 = 2
20) ABC is an arc of a circle, AC is a chord of the circle,
B is the midpoint of arc ABC, D is the midpoint of
21
3
10
chord AC, AC = 20 and BD = 3. What is the
diameter of the circle containing ABC?
r–3
r
r2 = r–3 2 + 102
r2 = r2 – 6r + 9 + 100
6r = 109 ï r =
109
6
ï d=
109
3
1
. Define the sequence of functions fn x by
x
f1 x = f x and fn x = f fn 1 x for n ¥ 2.
21) Let f x = 1 –
What is f50 50 ?
f1 x = 1 –
f2 x = 1 –
f3 x = 1 –
1
x
1
1–
1
1
x–1
x
1
–
=–
1
=x
x–1
f3 k x = x
50 = 3(16) + 2 ï f50 x = –
1
x–1
ï f50 50 = –
1
50 – 1
=–
1
49
22) If S = { 4 , 1 , 12 , x , 7 }, find all real values of x such that the mean of S equals the median of S.
4 + 1 + 12 + x + 7 = 24 + x
S = { x , 1 , 4 , 7 , 12 }
24 + x
5
=4 ï x=–4
S = { 1 , x , 4 , 7 , 12 }
24 + x
5
=4 ï x=–4
S = { 1 , 4 , x , 7 , 12 }
24 + x
5
=x ï x=6
S = { 1 , 4 , 7 , x , 12 }
24 + x
5
= 7 ï x = 11
S = { 1 , 4 , 7 , 12 , x }
24 + x
5
= 7 ï x = 11
x = – 4 , 6 , 11
23) How many positive divisors does the integer 1440 have?
1440 = 25 ÿ32 ÿ5
number of divisors = (5+1)(2+1)(1+1) = 6(3)(2) = 36
24) Define the sequence an by a0 = 0, a1 = 1 and an+1 = an + –1
a0
a1
a2
a3
a4
a5
a6
a7
··
=0
= a1
= a1
= a2
= a3
= a4
= a5
= a6
·
n
an
1
for n ¥ 1. Find all values of n such that an = 144.
– a0 = a1
+ a1 = a1 + a1 = 2 a1
– a2 = 2 a1 – a1 = a1
+ a3 = a1 + 2 a1 = 3 a1
– a4 = 3 a1 – a1 = 2 a1
+ a5 = 2 a1 + 3 a1 = 5 a1
Let fn be the Fibonacci sequence.
fn = {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, · · ·}
a2 k = fk ÿ a1 = fk and a2 k+1 = fk+2 ÿa1 = fk+2
a2 k = 144 = f12 ï 2k = 21
a2 k+1 = 144 = f12 = f10+2 ï 2k+1 = 21
n = 21 and 24
25) Find all real numbers x such that | 2 x – 1 | + | x – 5 | = 8 .
x – 5
5 – x
2x – 1
1 – 2x
2x – 1 + x – 5
= 8
6 – 3x = 8
3x = –2
x = –
5 – x
x – 5
2x – 1
2x – 1
x + 4 = 8
3x – 6 = 8
x = 4
3x = 14
2
14
3
3
1
2
5
x=–
2
3
,4
26) The corners of a rectangle in the plane are located at the points (1,1), (1,10), (5,10) and (5,1). The line y = m x divides the
rectangle into two trapezoids of equal area. Find m.
1,10
5,10
10
5,5m
8
6
4
2
1,m
1,1
1
1
2
5,1
2
3
4
4 10 – m + 10 – 5 m =
5
1
2
6
4 m–1 + 5m–1
20 – 6m = 6m – 2 ï 12m = 22 ï m =
22
12
=
11
6
27) A jar contains 5 different pennies, 6 different nickels, 7 different dimes and 8 different quarters. How many ways can a set
of 3 coins be selected so that the total value of the selected coins is less than 15 cents?
pennies
nickels
dimes
ways
2
0
1
5
2
7
1
=
10 7
= 70
1
2
0
5
1
6
2
=
5 15
= 75
2
1
0
5
2
6
1
=
10 6
= 60
3
0
0
5
3
=
10
70 + 75 + 60 + 10 = 215
28) Semicircles are constructed on the legs and the base of
the right isosceles triangle as shown in the diagram.
If the length of the base is 4 units, find the sum of
the areas of the shaded regions.
1
2
2(a + b) = 2
1
2
2b + c =
c=
1
2
2
p
2
2
= 2p
p 2 2 = 2p
2
2
2 =4
2b = 2p – c = 2p – 4
2a + 2b = 2p ï 2a = 2p – 2b = 2p – (2p – 4) = 4
2
2+i
29) If i 2 = – 1, determine the value of i +
i+
2
2+i
1 + 2 i 2– i
· 2–i
2+i
i+
i 2+i +2
2+i
=
2
2+i
=
2–i+4i+2
4+1
2
=
. Express your answer in the form a + b i, where a and b are real numbers.
1+2i
2+i
4+3i
5
=
4+3i 2
5
=
2i–1+2
2+i
=
2
=
16 + 24 i – 9
25
=
7
25
+
24
i
25
30) Find the radius of the circle whose locus is the set of all points x, y) whose distance from (2 , 0) is
5 times the distance
from (– 1 , 0).
x–2 2 + y2 =
5
x + 1 2 + y2
x–2 2 + y2 = 5( x + 1 2 + y2 )
x2 – 4 x + 4 + y2 = 5 x2 + 10 x + 5 + y2
4 x2 + 14 x + 4 y2 = – 1
4 x2 +
x2 +
r=
7
2
7
2
x+
x+
45
16
=
49
16
49
16
3
+ 4 y2 = – 1 +
+ y2 = –
1
4
+
49
16
49
4
=
45
16
= r2
5
4
31) Suppose that x1 , x2 , x3 , · · ·, x9 are real numbers with the following property: for k = 1, 2, 3, · · ·, 9, the sum of all of the numbers
1
2
3
9
except xk is k. For example, x1 + x2 + x3 + x4 + x6 + x7 + x8 + x9 = 5. What is x3 ?
Each of the xi appears in 8 of the 9 sums ï
8( x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 ) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 =
x3 =
45
8
32) Let z =
45
8
– (x1 + x2 + x4 + x5 + x6 + x7 + x8 + x9 ) ï x3 =
1
a+
45
8
–3=
45
8
–
24
8
=
21
8
. If a, b, c, d and e are all either 1 or 2, what is the minimum value of z?
1
1
b+
1
c+
d+
1
e
To make the fraction as small as possible, pick a, b, c, d and e to make the denominator as large as possible.
Thus a = 2, b = 1, c = 2, d = 1 and e = 2
z=
1
2+
=
1
1
1+
2+
1
1+
1
2
1
2+
=
1
1+
1
2+
1
2+
1
3
2
=
1
1
1+
2+
1
2+
2
=
1
1
1+
1
1
1
=
1
1 =
8 =
2+ 3
2 + 11
2 + 11
1 +8
8
=
11
8
3
3
Y
33) ABC is a 3-4-5 right triangle. Side AB is
extended to point X so that B is the midpoint
1
30
p3
12
of AX, side CA is extended to point Z so
that A is the midpoint of CZ and side BC
is extended to point Y so that C is the
C
6
midpoint of BY. Find the area of triangle XYZ.
6
3
4
A
D
5
B
3
3
4
8
Z
p1
The indicated dimensions are computed using similar triangles.
Area = 12(9) –
1
2
9 4 + 8 3 + 6 12 = 108 –
1
2
X
36 + 24 + 72 = 108 –
1
2
132 = 108 – 66 = 42
p2
11
30
34) Let two circles of radii 1 and 3 be tangent at point
A and tangent to a common straight line at points B
and C. Find AB2 + BC2 + CA2 .
A
B
C
C2
π–α
3
2
A
1
C1
γ
β
E
α
β
γ
B
Let
—BC
1
C
—C
A = a. From trapezoid BC1 C2 C
From triangle ACC2 p – a + 2g = p ï g =
From triangle BC1 A a + 2b = p ï b =
1
C2 C = p – a.
a
2
p–a
2
=
p
2
–
a
2
=
p
2
–g ï b+g=
p
2
ï
—BAC =
p
2
From triangle C1 C2 E BC2 = 42 – 22 = 12
AB2 + AC2 = BC2 ï AB2 + BC2 + CA2 = 12 + 12 = 24
35) How many six-digit positive integers are there in which every digit occurs the same number of times as the value of the digit?
An example of such an integer is 133232.
This corresponds to the partitions of 6 into distinct positive integers.
Partition
Positions
1+5
6 , 1
6
6
1+3+2
30 , 3
60
6
2+4
6 , 2
15
15
6
6 , 6
1
1
Total
Number of Ways
82
Number of integers = 82
36) An isosceles triangle whose sides are in the ratio 1 : 2 : 2 is inscribed in a circle of radius 1 and a second circle is inscribed in this triangle.
What is the radius of the second circle?
z
A
z
D
B
β
β
r
α
E
π–2α 4z
F
α
C
1
4
From triangle CDB, sin(a) =
By the law of cosines in triangle CFB, 4 z 2 = 12 + 12 – 2(1)(1)cos(p–2a)
16 z2 = 2 + 2cos(2a)
16 z2 = 2 + 2(1 – 2 sin2 a = 2 + 2(1 – 2
z2 =
From triangle CDA, cos(2b) =
From triangle ADE, tan( b =
tan(b) =
cos b
2
=
=
1 + cos 2 b
=
15
4
15
8
ï z=
1
4
r
z
1 – cos 2 b
sin b
15
64
1 2
)
4
1–
1+
1
4
1
3
5
=
4
2
3
5
tan(b) =
=
r
z
=
r
ï r=
15
3
5
·
15
8
6 j+
1
.
k
=
3
8
8
10
k
37) Determine the value of
k = 1 j= 1
10
k
6 j+
k = 1 j= 1
1
=
k
1
6 j+
j= 1
1
+
k
2
6 j+
j= 1
1
+
k
3
6 j+
j= 1
1
+· · ·
k
10
6 j+
j= 1
1
k
1
2
= 6(1) + 1 + 6(1 + 2) +
1
10
+
1
10
+
1
10
+
1
10
+
1
10
1
10
+
+
1
10
+
+
1
2
+ 6(1 + 2 + 3) +
1
3
+
1
3
+
1
3
+ · · · 6(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) +
1
10
= 6(1 + 1 + 2 + 1 + 2 + 3 + · · · 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) + 10(1)
= 6[10(1) + 9(2) + 8(3) + 7(4) + 6(5) + 5(6) + 4(7) + 3(8) + 2(9) + 10] + 10
= 6(10 + 18 + 24 + 28 + 30 + 30 + 42 + 24 + 18 + 10) + 10
= 6(220) + 10 = 1320 + 10 = 1330
38) Suppose that the lengths of the sides of a right triangle
ABC are 10, 24 and 26. A circle with center D and
radius 1 rolls inside the triangle, always tangent to
at least one side. How far has D traveled when it
first returns to its original position?
C
D
A
B
C
F
26
10
D
E
x
A
B
24
Triangles ABC and DEF are similar. ï the ratios of their corresponding sides is constant.
Let angle CAB = 2a ï angle DAB = a
tan(a) =
1
x
tan(2a) =
and tan(2a) =
2 tan a
=
1 – tan2 a
5
12
10
24
5
12
=
ï 24tan(a) = 5 – 5 tan2 a ï 5 tan2 a + 24tan(a) – 5 = 0
(5tan(a) – 1)(tan(a) + 5) = 0 ï tan(a) =
Then tan(a) =
1
x
=
1
5
1
5
, – 5. Since 0 < a <
p
2
tan(a) =
ï x=5
DE = 24 – 1 – 5 = 18 ï ratio of sides of triangles ABC and DEF is
EF =
3
4
10 =
15
2
and FD =
3
4
26 =
39
2
Distance traveled = DE + EF + FD = 18 +
15
2
+
39
2
= 45
18
24
=
3
4
1
5
1
10
+
1
10
+
10
A
39) The line segment AB is tangent at A to a circle with
B
center O. Point D is inside the circle and DB intersects
the circle at C. Given that BC = DC = 6, OD = 3
and AB = 10, find the radius of the circle.
6
r
C
O
6
3
D
x r–3
By the power of a point 6(12 + x) = 102 ï 6x = 28
By the power of a point 6x = (r + 3)(r – 3) ï 28 = r2 – 9 ï r2 = 37 ï r =
37
40) Suppose that p, q and r are the distinct roots of x3 – 2 x2 + 2 x – 3 = 0. Find the value of p3 + q3 + r3 .
x3 – 2 x2 + 2 x – 3 = (x – p)(x – q)(x – r) = x3 – (p + q + r) x2 + (pq + pr + qr) x – pqr ï p + q + r = 2, pq + pr + qr = 2 and pqr = 3
(p + q + r 3 = p3 + q3 + r3 + 3(p + q + r)(pq + qr + pr) – 3pqr = 23 ï 8 = p3 + q3 + r3 + 3(2)(2) – 3(3) ï p3 + q3 + r3 = 5
41) Express
1
– 2 sin 50 ° in simplest form.
2 cos 20 °
1
– 2 sin 50 ° =
2 cos 20 °
Using sin(A)cos(B) =
1
2
1 – 4 sin 50 ° cos 20 °
2 cos 20 °
sin A + B + sin A – B
1
– 2 sin 50 ° =
2 cos 20 °
1–4
1
2
sin 70 ° + sin 30 ° cos 20 °
2 cos 20 °
=
1 – 2 sin 70 ° – 1
2 cos 20 °
Since 70° and 20° are complimentary sin(70°) = cos(20°) ï
=
– 2 sin 70 °
2 cos 20 °
1
– 2 sin 50 ° =
2 cos 20 °
– 2 sin 70 °
2 cos 20 °
=
– 2 cos 70 °
2 cos 20 °
=–1