Download 1. Triangle ACD is divided into triangles ABC and BCD, where B is

1. Triangle ACD is divided into triangles ABC and BCD, where B is on AD. If AB, BC, and CD
all have the same length, and angle BAC = 18°, what is the measure of angle DCB?
If BAC = 18°, then:
BCA = 18° (opposite sides of a triangle with the same length)
ABC = 144° (sum of angles of a triangle = 180°)
DBC = 36° (B is on AD, so ABC + DBC = 180°)
BDC = 36° (opposite sides of a triangle with the same length)
Therefore, DCB = 180 - (36 + 36) = 108°.
2. If x and z are positive integers such that x2 + 40 = z2, what is the largest value of x?
Let z = x + n, where n is a positive integer; the problem is now x2 + 40 = (x + n)2.
40 = 2xn + n2
2xn = 40 - n2
x = (40 - n2) / 2n = 20/n - n/2
If n = 1, then x = 19/2, which is not a positive integer
If n = 2, then x = 9
Since 20/n and -n/2 both decrease as n increases, subsequent values of n will result in lower
values of x, so x = 9 is the largest value.
3. Square ABCD has sides of length 10. Points P, Q, R and S are on sides AB, BC, CD, and DA,
respectively. If AP, BQ, CR, and DS all have length 3, what is the area of PQRS?
The area of PQRS = the area of ABCD - the sum of the areas of triangles SAP, PBQ, QCR, and
RDS. Each triangle is a right triangle with one leg of length 3 and the other of length 7, so the
area of PQRS = 10 • 10 - 4 • (3 • 7 / 2) = 100 - 42 = 58.
4. For what values of X is | X + 3 | + | X - 4 | = 18 true?
If X ≥ 4, then both (X + 3) and (X - 4) are non-negative, so it becomes X + 3 + X - 4 = 18, or
2X = 19. X = 19/2 ≥ 4, so this is a solution.
If -3 < X < 4, then (X + 3) is positive but (X - 4) is negative, so it becomes X + 3 + 4 - X = 18,
or 7 = 18, which has no solutions.
If X ≤ -3, then both (X + 3) and (X - 4) are negative, so it becomes -X - 3 + 4 - X = 18, or
-2X = 17; X = -17/2 ≤ -3, so this is a solution.
The only two values of X are 19/2 and -17/2.
5. ABC is a triangle. Points D, E, and F bisect sides BC, AC, and AB respectively. AD, BE, and
CF intersect at point O. DE intersects CF at point X. If triangle DOX has area 1, what is the
area of triangle ABC?
Since D, E, anf F are bisectors, then triangles AEF, BDF, and CDE each have an area 1/4 of the
1
area of ABC, which means triangle DEF has an area 1−(3 • ) = 1/4 of the area of ABC.
4
AD, BE, and CF divide triangle DEF into six smaller triangles; label DOX "a", and the others
"b", "c", "d", "e", and "f" in order clockwise from "a".
Each bisector divides DEF into two triangles, each consisting of three of the smaller triangles.
Since each of the bisectors also bisects a line of DEF, each group of three has the same area as
the other group of three (as the two triangles made by the bisector has the same base length and
the same height), and since each "half" of DEF has area equal to 1/2 of the area of DEF, the area
of each group of three consecutive areas is the same - in other words:
a+b+c=b+c+d=c+d+e=d+e+f=e+f+a=f+a+b
a = d, b = e, and c = f
Also, the two smaller triangles that share a side of DEF have the same area, as they have the
same base length and height (to point O):
a = b, c = d, and e = f
Therefore, a = b = e = f = c = d, so a's area, which is the area of DOX, is 1/6 of the area of DEF,
or 1/6 • 1/4 = 1/24 of the area of ABC.
If DOX has area 1, then ABC has area 24.
6. What is the smallest positive integer N such that N/2 is a perfect cube and N/3 is a perfect
square?
There must exist positive integers A and B such that N = 2A3 = 3B2.
The prime factor 2 must appear (a multiple of 3) + 1 times as well as a multiple of 2 times; the
smallest such positive number is 4.
The prime factor 3 must appear a multiple of 3 times as well as (a multiple of 2) + 1 times; the
smallest such positive number is 3.
The smallest N that contains the prime factor 2 4 times and the prime factor 3 3 times is
24 • 33 = 432.
7. In a group of three people chosen at random, what is the probability that at least two were born
on the same day of the week? (Assume that a "person chosen at random" has a 1/7 chance of
being born on any particular day of the week.)
The probability = 1 - the probability that all three were born on different days of the week,
6 5 19
which is 1−( • )=
7 7 49
8. A person travelling at 55 km/hr discovers with 30 km remaining on his trip that he will be three
minutes late for a meeting. At what speed (in km/hr, expressed as a fraction) does he need to
travel in order to get to the meeting on time?
30
360
• 60=
minutes.
55
11
327
360
327
109
−33 , or
, minutes, which is
11
The meeting starts in
=
hours.
11
11
220
60
30
109
6600
Therefore, the person must travel 30 km in
hours, which is 119 =
km/hr.
220
109
220
A person travelling at 55 km/hr will move 30 km in
( )
( )
9. The vertices of regular pentagon ABCDE are the points of a five-pointed star. AC intersects BE
and BD at points X and Y, respectively. If XY has length 1 and XA has length R, then express
the length of AB as the ratio of two integral polynomials in R.
BE intersects AD at E. F is the midpoint of CD. Since ABCDE is regular, BE is parallel to CD,
AF bisects HX (let G be the intersection point), and AGX is simular to AFC.
1
AC AX
XG
1/2 2R+1
, and
=
=( 2R+1) •
=
HX = XY = 1, so GX =
→ CF= AC •
2
CF XG
AX
R
2R
2R+1
Length AB = CD = 2 CF =
R
10. The three solutions of X3 + 18 X2 + 92 X + 120 = 0 form an arithmetic sequence whose
consecutive terms increase by 4. What are the three solutions?
Let A, A+4, and A+8 be the solutions.
(X - A) (X - (A+4)) (X - (A+8)) has an X2 component of 3A + 12, so 3A + 12 = 18, or A = 2.
The solutions are 2, 6, and 10.
11. An equilateral triangle with area 1 has each of its sides trisected. Each of the middle segments
is also one side of an equilateral triangle that extends outward from the original triangle. Each
of the 12 sides of the resulting figure is trisected, and each of the middle segments of these 12
sides is also one side of an equilateral triangle that extends outward from the large figure. What
is the area of the resulting figure?
Since each of the three "extensions" from the main triangle has side length that is 1/3 of the side
length of the main triangle, each has an area 1/9 of the area of the main triangle. Each of the 12
extensions from the resulting figure has a side length that is 1/9 of the side length of the main
triangle, so each has area 1/81 of the area of the main triangle.
The main triangle has area 1, so the figure has area 1 + (3 • 1/9) + (12 • 1/81) = 40/27.
12. What is the sum of the digits of the smallest positive integer N such that
B=
A−1
,
3
C=
B−1
,
3
D=
C −1
,
3
E=
D−1
, and
3
F=
A=
N −1
,
3
E−1
, where A, B, C, D,
3
E, and F are positive integers?
N > A > B > C > D > E > F, so the smallest N comes from the smallest F.
F = 1 → E = 4 → D = 13 → C = 40 → B = 121 → A = 364 → N = 1093; the sum of the digits
is 1 + 0 + 9 + 3 = 13
13. 5, 12, 19, 26,... and 103, 106, 109, 112,... are arithmetic sequences. What is the smallest
number N such that the sum of the first N terms of the first sequence is greater than the sum of
the first N terms of the second sequence?
The first N terms of the first sequence are 5, 5 + 7, 5 + 7 • 2,..., 5 + 7 • (N-1); the sum is
( N −1)• N
7
7 2 17
•( N 2+N ) =
N + N.
5N + 7 •
= 5N +
2
2
2
2
( N −1)• N
3 2 209
N +
N. .
The first N terms of the second sequence are 103 N + 3 •
=
2
2
2
The two are equal when 7 N 2+ 17 N = 3 N 2+ 209 N , or 2N = 51.
2
2
2
2
Since N needs to be a positive integer, the first sum < the second sum for N = 25, but > the
second sum for N = 26.
14. 0.122 in base ten equals what in base 7 (rounded to three digits to the right of the "decimal"
point)?
0.122 base 10 = 122/1000 = 41.846 / 343 = (0 • 49 + 5 • 7 + 6.846) / 343, which rounds to the
base 7 number 0.060.
15. A "Brother Alfred" set is a subset of three elements of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
that contains no more than one odd number. How many different Brother Alfred sets are there?
(53)=10
()
5 =60
sets containing no odd numbers, and 6 •
sets containing one
2
odd number; the total number of Brother Alfred sets = 10 + 60 = 70
There are
16. y and z are the solutions of x2 + Ax + B = 0, and (y-C) and (z-C) are the solutions of x2 + Dx +
E = 0. Express E in terms of A, B, and C.
(x-y)(x-z) = x2 + (-y - z)x + yz = x2 + Ax + B
A = -y - z
B = yz
(x-(y-C))(x-(z-C)) = x2 + (2C - y - z) x + (C2 + (-y - z)C + yz) = x2 + Dx + E
E = C2 + (-y - z)C + yz
E = C2 + AC + B
17. A table similar to Pascal's Triangle is formed as follows: the first row has two numbers, 1 and 2
(in order from left to right), and each row below the first consists of 1 followed by the sums of
each pair of consecutive numbers in the row above it followed by 2. (For example, row 2 is 1,
(1+2) = 3, and 2, and row 3 = 1, (1+3) = 4, (3+2)= 5, and 2.) What is the sum of the numbers in
the 11th row?
To determine the sum of a row, put a zero at the beginning and end of each row (this will not
change the sum of any row). Each number in a row appears in two sums below it except for the
two zeroes, so each row's sum is twice the sum of the row above it.
The top row has sum 3, so the 11th row, which is 10 rows below it, has sum 3 • 210 = 3072.
18. x4 + x2y2 + y4 is the product of what positive degree polynomials in x and y with integral
coefficients?
Let x4 + x2y2 + y4 = (x2 + a xy + y2) (x2 + b xy + y2)
= x4 + (a + b) x3y + (2 + ab) x2y2 + (a + b) xy3 + y4
Therefore, a + b = 0, or a = -b, and 2 + ab = 2 - a2 = 1, so a = 1 and b = -1 (or a = -1 and b = 1).
The solution is (x2 + xy + y2) (x2 - xy + y2)
19. 13 + 23 + ... + 203 = 44,100, and (1 + 2 + ... + 10)2 = 3025. What is the sum of the digits of
13 + 33 + 53 + ... + 193?
23 + 43 + 63 + ... + 203 = 23 • (13 + 23 + ... + 103) = 23 • (1 + 2 + ... + 10)2
= 8 • 3025 = 24,200.
13 + 33 + 53 + ... + 193 = (13 + 23 + ... + 203) - (23 + 43 + ... + 203)
= 44,100 - 24,200 = 19,900; the sum of its digits = 1 + 9 + 9 = 19.
2 π̇ i̇
k̇
20. The nth roots of 1 are r =e n where i2 = -1 and k is in {1, 2, ..., n}.
k
What is the sum of r1 + r2 + ... + rn-1?
Let x=e
2 π̇ i̇
n
; the problem now asks for
(1+ x+x 2+ x 3+...+ x(n−1))−1 , or
n
2 πi n
n
( ) =e
x =e
2π i
n
n
=e 2 πi=1, so
x+ x 2+x 3+...+x(n−1) , which is
n
x −1
−1 .
x −1
n
n
x −1=0, and the result is x −1 −1= 0 −1=−1
x −1
x−1
21. A sphere with radius 10 cm is dropped into a cone where the measure of the angle of the crosssection made by a plane containing its axis is 60°. How far along the side of the cone is the ring
of contact from the vertex?
If you look at the cross-section, you see two legs of a triangle with common vertex 60°, and a
circle of radius 10; the vertex of the triangle, the center of the circle, and the tangent point form
a 30-60-90 triangle with short side of length 10, and we are asked to find the length of the
longest side, which is 10  3 .
22. Two candles of equal length start burning at the same time. One candle will burn out in 3
hours, and the other will burn out in 7 hours. How long will it take before one candle is 3 times
as long as the other?
If both candles start with length 1, then, after x hours, the 3-hour candle's length is 1−
x
the 7-hour candle's length is 1− .
7
(7−x)
(3− x)
7
=3 •
→ 7 – x = 21 – 7x → x =
hours.
3
7
3
x
and
3
23. Layers of cubes (all the same size) are set up as follows: the bottom layer is 8 x 8; on top of this
is a 6 x 6 layer; on top of this is a 4 x 4 layer; on top of this is a 2 x 2 layer. Each layer, except
the bottom layer, is centered on top of the layer below it. Except for the bottom sides of the
bottom layer, all of the exposed faces of the cubes are painted red. How many cubes are there
with exactly one painted face?
Every cube on an edge will have at least two of its faces painted – and every cube is either on
an edge or has all of its faces (except, possibly, the bottom) covered, so the answer is 0.
24. What is the smallest positive two-digit number which, in base 10, equals the base 7 two-digit
number that has the same digits but in reverse order?
10a + b = 7b + a → 9a = 6b → 3a = 2b
The smallest base 10 number (10a + b) has the smallest a, which is 2; 23 base 10 = 32 base 7
The solution is 23
25. A collection of X insects (with six legs each), Y spiders (with eight legs each), and Z worms
(with 20 legs each) have a total of 504 legs. If the number of spiders equals the number of
insects plus twice the number of worms, and XYZ > 0, then how many creatures are in the
collection?
Y = X + 2Z
504 = 6X + 8Y + 20Z = 6X + 8X + 16Z + 20Z = 14X + 36Z
252 = 7X + 18Z
7X = 252 – 18Z = 18 (14 – Z); since 7 and 18 are relatively prime, X must be a multiple of 18
X >= 36 → 7X >= 252 → Z <= 0, so X = 18, Z = 7, and Y = 32, and the number of creatures is
18 + 7 + 32 = 57.
26. Square ABCD has area 1. Points E and F are on AB such that AB is trisected into AE, EF, and
FB; points G and H trisect BC into BG, GH, and HC; points J and K trisect CD into CJ, JK, and
KD. Line segments EH and GK intersect at X. What is the area of triangle GXH?
Since the area of ABCD is 1, each side has length 1.
Let L and M be points on DA such that DL = LM = MA.
Since G and H trisect BC, and L and M trisect AD, MG and LH are both parallel to both AB and
CD. Similarly, FJ and EK are both parallel to AD and BC.
Let P be the midpoint of EH; PHG is similar to EHB, so PG is half the length of EB → PG =
FB → P is also the intersection point of GM and FJ.
Similarly, KG, FJ, and HL all intersect at Q.
GQ and HP divide GHPQ into four triangles of equal area, so the area of GHX = ¼ the area of
GHPQ.
1
1
1
GP= BF= , and GH = , so the area of GHPQ =
, and the area of GHX =
3
3
9
1 1 1
• =
9 4 36
27. James takes X days to do a job, John takes 2X days to do the same job, and together they can do
the job in B days. If John works faster so he could do the job by himself in 3/4 of his normal
time, then together it would take them A days. What is A / B?
1
1
of the job per day. John can do
of the job per day. Together, they
X
2X
1
1
3
+
=
can do
of the job per day, so together it would take them 2X days;
X 2X 2X
3
2
B= X.
3
4 1
2
•
, or
, of the job per day, so together they can do
“Faster” John can do
3 2X
3X
1
2
5
3
3X
+
=
of the job per day, and it would take them
days; A= X. .
X 3X 3X
5
5
3
X
A 5
9
=
=
B 2
10
X
3
James can do
28. Material is being dropped from a hopper to form a cone which remains in the same proportion
at all times. When the pile was a certain height, it took 2 hours to double the height of the pile.
How many hours will it take to double the height of the new pile?
In order to double the height of a pile, the volume must be increased to 8 times what it was. Let
the original pile have volume V; to double the height, the volume must be 8V, so the rate of
7V
7
added material is
, , or
V per hour. In order to double it again, the volume must
2 hours
2
increase from 8V to 64V, so 56V of material is needed; the number of hours needed =
56
=16
7
2
()
29. For what values X is
2
<− X true?
( X +3)
Assume X ≠ -3.
If X > -3, X + 3 > 0, so multiplying both sides by (X + 3) does not change the sign;
2 < -X (X + 3) → 2 < -X2 + 3X → X2 + 3X + 2 < 0 → (X+2) (X+1) < 0 →
-2 < X < -1; the entire range > -3
If X < -3, X + 3 < 0, so multiplying both sides by (X + 3) does change the sign;
2 > -X (X + 3) → 2 > -X2 + 3X → X2 + 3X + 2 > 0 → (X+2) (X+1) > 0 → X < -2 or X > -1; the
only values in that range < -3 are all X < -3.
Thus, the solution is {X < -3, -2 < X < -1}
30. January 1, 1979 was on a Monday. What was the next year in which January 1 was also on a
Monday?
In most cases, going from January 1 of a year to January 1 of the next year is 52 weeks and 1
day, so the day of the week would move forward one day. If the “from” year is a leap year,
February 29 is included in that jump, so the day of the week would move forward two days.
1980 – Tuesday
1981 – Thursday; 1982 – Friday; 1983 – Saturday; 1984 – Sunday
1985 – Tuesday; 1986 – Wednesday; 1987 – Thursday; 1988 – Friday
1989 – Sunday; 1990 – Monday – the solution is 1990
31. A game is played in which two players each roll a six-sided die; the winner is the person who
rolls the higher number. (If both players roll the same number, it is a tie, and neither player
wins.) What is the probability that a given player wins and/or ties (i.e. the other player does not
win) in three consecutive games?
There are 36 possible pairs of rolls; in 6 of them, it is a tie, and in the other 30, each player wins
21 7
= .
15, so the probability of winning or tying a particular game is
36 12
Each game is an independent event, so the probability of winning/tying three consecutive games
7 3 343
is
=
12 1728
(Note this is in lowest terms as the numerator is the power of an odd number which is not a
factor of the denominator)
32. The potential energy of a mass of M pounds at a height H feet is M • G • H, where G is 32.2
feet/sec2 . The kinetic energy of a mass of M pounds moving at V feet/sec is MV2/2. The sum
of potential energy and kinetic energy is constant. If a piledriver is dropped from 100 feet
above the ground, at what height is the kinetic energy equal to 5/4 of the potential energy?
At 100 feet, V = 0, so MGH + MV2/2 = 32.2 • M • 100 = 3220 M.
Therefore, MGH + MV2/2 = 3220 M → 32.2 H + V2 / 2 = 3220 → 644 H + 102 = 64,400.
MV2 / 2 = 5/4 M G H → 2 V2 = 161 H → 10 V2 = 805 H →
64,400 = 644 H + 805 H → H = 64,400 / (644 + 805) → (161 • 400) / (161 • 9) = 400/9.
33. Five great circles are constructed on the surface of a sphere. If no three of these circles intersect
at the same point, into how many regions do they divide the sphere's surface?
The first circle divides the surface into two regions
Each subsequent circle will go through each existing circle twice, and each region between each
consecutive pair of intersections is divided into two, so the number of additional regions = 2 x
the number of existing circles
The total = 2 + (2 x 1) + (2 x 2) + (2 x 3) + (2 x 4) = 22.
˙
34. The nth hexagonal number Hn = n 2 ṅ−1
˙
 n1
2
If Hn = Tk, then express k in terms of n.
The nth triangular numner Tn = n
k2/2 + k/2 = 2n2 – n → k2 + k = 4n2 – 2n → k2 + k + ¼ = 4n2 – 2n + ¼
(k + ½)2 = 4n2 – 2n + ¼ = (2n – ½)2
Since both > 0, k + ½ = 2n - ½ → k = 2n – 1.
(Confirmation: Tk = T2n-1 = (2n-1) • 2n / 2 = n (2n-1) = Hn.)
35. A ten-question test is scored as follows: 10 points for a correct answer, -3 points for an incorrect
answer, and 0 points for no answer. How many test scores of 50 or higher are possible on this
test?
If there are 4 or fewer correct answers, the score < 50.
For 5 correct answers, the score = 50 – 3 x the number of incorrect answers, so 50 is the only
possible score
For 6, scores of 60, 57, 54, and 51 are possible
For 7, scores of 70, 67, 64, and 61 are possible (anything lower would require more than 3
incorrect answers, which means there were more than 10 questions)
For 8, scores of 80, 77, and 74 are possible
For 9, scores of 90 and 87 are possible
For 10, only 100 is possible
This is 15 different scores.
36. There are 10 cards numbered 1 to 10. Three cards are chosen at random one at a time without
replacement. What is the probability that the three cards will be chosen in ascending order?
There are six ways to draw three cards without replacement, and only one of them is in
ascending order, so the probability is 1/6.
(Note that the initial number of cards is irrelevant, as long as there are at least three and all of
them are different.)
37. For what positive integer X are both
 x8 and  x−15 positive integers?
In other words, x + 8 and x – 15 are perfect squares. Since (n + 1)2 – n2 = 2n+1, then, if 2n+1 =
23, the squares are 121 and 144, so x = 136.
38. If x = log3 5 and y = log5 3, then (expressed as a quotient of relatively prime positive integers)
−x
2y
3 5
what is
−2
−y ?
3 5
1
1
1
= log 5 =
x
5
3 3
2y
y 2
log 3 2
5 =(5 ) =(5 ) =32=9
1
−2
3 =
9
1
−y
5 =
3
46
1
+9
5
5
207
=
=
1 1
10
4
+
9 3
9
3− x =
3
5
( )
()
39. Solve x3 + y3 = 208 and x2 - xy + y2 = 52; express your answer as ordered pairs (x,y).
x3 + y3 = ( x2 - xy + y2) (x + y) → x + y = 208 / 52 = 4
x2 - xy + y2 = 52 = x2 – x (4-x) + (4-x)2 = x2 – 4x + x2 + 16 – 8x + x2 = 3x2 – 12x + 16
3x2 – 12x – 36 = 0 → x2 – 4x – 12 = 0 → x = {-2, 6} → y = {6, -2}
The solutions are (-2, 6) and (6, -2).
40. Given ten different prime numbers x1, x2,, x3,..., x10, how many different numbers of the form
xi4 xj2 xk2 can be formed where i, j, and k are distinct?
There are 10 different possibilities for X(i); for each one, there are 9 different possibilities for
X(j); for each of these 90 pairs, there are 8 different possibilities for X(k).
However, if X(j) and X(k) are switched, this a different possibility, but the result is the same, so
each number appears in two possibilities.
The number of different numbers = 10 • 9 • 8 / 2 = 360.