CMC 2000 2006

Community College of Philadelphia
Colonial Maths Challenge
Spring 2000
Individual Contest
The individual winner of the Spring 2000 edition of the Colonial Maths Challenge was Azeem Ansar of West
Catholic, with 12 correct answers (out of 20).
Peter Baratta, William Clee, Joanne Darken, Dot French, Elena Koublanova, Margaret Hitczenko, David Santos
helped with the competition.
You have 50 minutes to complete this exam. No consultation of any kind is allowed. Write your numerical answers on the
answer-sheet provided. No partial credit is given.
1. Evaluate
1
2
1
2
1
2 1
2
2. If 333 + 333 + 333 = 3x , find x.

3. How many integers between 1 and 21 (excluding 1 and 21) do not have a common factor (i.e, are relatively
prime) with 21?
4. A palindrome is an integer whose decimal expansion is symmetric. For example, 1, 11, 121, 344565443, are all
palindromes. The sequence of palindromes, starting with 1 is written in ascending order
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, . . .
Find the 1984-th positive palindrome.
5. A sequence of isosceles triangles is constructed starting with AB = BC, then BC = CD, and so on. If the
angle BAC = 17 , how many such triangles can be drawn?
D
B
A
6. If (x 1)2 + (y 2)2 + (z 3)2 + (w 4)2 = 0, what is x + y + z + w?

7. The sum of two numbers is 7 and their product 21. What is the sum of their reciprocals?
8. If x2 + x 1 = 0, find x4 + 2x3 + x2 .
9. A car with 5 tires (four road tires and a spare) travelled 30 000 miles. If all five tires were used equally, how
many miles wear did each tire receive?
10. In the diagram, what
A is the angle sum a + b + c + d between the two parallel lines?
B
C
D
11. In a river with a steady current, it takes a person 6 minutes to swim a certain distance upstream, but it takes
him only 3 minutes to swim back. How many minutes would it take a piece of wood to float this same
distance downstream?
1
Spring 2000
Individual Contest
12. If (8x 5)5 = Ax5 + Bx4 + Cx3 + Dx2 + Ex + F, then A + B + C + D + E + F =

13. What is the angle between the hands of a clock at a quarter to five?
14. Leo the clockmaker has two antique clocks. One gains ten seconds every hour, while the other loses twenty
seconds every hour. He set both clocks to show the correct time at 9 AM on 4 February 2001. On what date
will they next show the correct times simultaneously?
15. A squeezable toothpaste tube is originally in the form of a cylinder 12 cm long, with diameter 4 cm. The
short cylindrical nozzle has diameter 0.5 cm. What length of toothpaste can the tube produce?
16. Given that
1
1
1
1
1
+
+
+ +
+
1+ 2
2+ 3
3+ 4
1022 + 1023
1023 + 1024
is an integer, find it.

17. Solve the system
x+y+u
y+u+v
u+v+x
v+x+y
= 4,
= 5,
= 0,
= 8.
18. If eggs had cost x cents less per dozen, it would have cost 3 cents less for x + 3 eggs than if they had cost x
cents more per dozen. What is x?
19. A quiz has 25 questions with four points awarded for each correct answer and one point deducted for each
incorrect answer, with zero for each question omitted. Anna scores 77 points. How many questions did she
omit?
20. At a classroom party, the average age of b boys is g, and the average age of g girls is b. If the average age of
everyone at the party (all boys and girls and their 42-year-old teacher) is b + g, what is the value of b + g?
Spring 2000
Individual Contest Solutions
1. Proceeding from the innermost fraction one easily sees that

1
2
1
2
1
2 1
2
1
2
1
2 2
3
1
2
3
4
4
.
5
2. 333 + 333 + 333 = 3 333 = 334 , whence x = 34.

3. Of the 21 numbers, one eliminates the seven which are multiples of 3 and the three which are multiples of 7.
But one has included 21 twice, so the total number is 21 7 3 + 1 = 12.
4. It is easy to see that there are 9 palindromes of 1 digit, 9 palindromes with two digits, 90 with three digits, 90
with 4 digits, 900 with 5 digits and 900 with 6 digits. The last palindrome with 6 digits, 999999, constitutes
the 9 + 9 + 90 + 90 + 900 + 900 = 1998th palindrome. Hence, the 1997th palindrome is 998899, the 1996th
palindrome is 997799, the 1995th palindrome is 996699, the 1994th is 995599, etc., until we find the 1984th
palindrome to be 985589.
5. Consider the diagram below.
b2 c3
a1
c1
a2
b1
c2 a3
b3
Let a1 = x = 17 . As the triangles formed are isosceles, b1 = a1 = x. This entails that c1 = 180 a1 b1 and
so
a2 = 180 c1 = 2x,
b2 = a2 = 2x,
c2 = 180 a2 b2 = 180 4x,
a3 = 180 b1 c2 = 3x,
b3 = 3x,
..
.
an = nx
bn = nx.
In other words n-th triangle has two angles equal to nx. We can construct such triangles as long as 2nx < 180
. Since x = 17 , that means 34n < 180. The biggest possible n is 5. We can construct 5 triangles.
6. Each summand is non-negative so if the sum is to equal 0, each summand must be zero. This entails
x = 1, y = 2, z = 3, w = 4, whence x + y + z + w = 10.
7. Let x, y be the numbers. One has x + y = 7, xy = 21 whence
1 1
y+x
7
1
+ =
=
= .
x y
xy
21
3
8. Observe that x2 + x = 1. Hence x4 + x3 = x2 and x3 + x2 = x. Thus

x4 + 2x3 + x2 = x4 + x3 + x3 + x2 = x2 + x = 1.
9. Travelling 30000 miles with 4 tires is as travelling 120000 miles on one tire. The average wear of each of the 5
tires is thus 120000 5 = 24000 miles.
Spring 2000
10. Draw a straight line perpendicular to both parallel lines, as shown below. This closed figure forms a
hexagon, with interior angle measure 4 180 = 720 . Subtract both right angles from the closed figure to
obtain the angle sum

A a + b + c + d = 720 180 = 540 .
B
C
D
11. Let w be the speed of the swimmer and let c be the speed of the current. From d = st where d is distance, s
d
d
d d
d
speed, and t time, one has w c = and w + c = . Whence 2c = (w + c) (w c) = =
. This
6
3
3
6
12
d
yields 12 = . The piece of wood therefore needs 12 minutes to float downstream.
c
12. Letting x = 1 one gathers that
243 = 35 = (8(1) 5)5 = A + B + C + D + E + F.
13. Measure angles clockwise, with origin (0 ), at 12:00. Each minute ran by the minute hands accounts for
360
= 6 , and so, each, when jumping from hour to hour, the hands travel 5(6 ) = 30 . When the minute
60
2
40
hand is on the 8, it has travelled
= of a circumference, that is
60
3
2
360 = 240
3
and the hour hand has moved
2
of the way from 4 to 5, travelling
3
2
4+
(30 ) = 140 .
3
Hence, angle between the hands is 240 140 = 100 .

14. The clocks will both show the right time again on 3 August 2001, the 180th day after 4 February 2001. The
first clock will gain 1 minute every 6 hours, which is 1 hour every 360 hours, i.e., 1 hour every 15 days. The
second clock will lose 2 hours every 15 days. After 15d days, the hour on the first clock is 9 + d mod 12 and
the hour on the second clock is 9 2d mod 12. Since we want 9 + d 9 2d mod 12, one must have 3d 0
mod 12. Thus d must be a multiple of 4. Since one also wants the clocks to show the exact hour d must be a
multiple of 12. The smallest such d is clearly d = 12. Thus the clocks agree on the correct time for the first
time after 15d = 180 days.
15. The volume of the toothpaste is V = r2 h = (2)2 (12) = 48 cm 3 . After coming out of the nozzle, a cylinder
of radius = 0.25 cm and height l cm is formed. Its volume is (.25)2 l = 0.0625l cm 3 . Thus
l = 48 0.0625 = 768 cm.
16. Observe that ( x + 1 x)( x + 1 + x) = 1 and so
= x + 1 + x.
x+1 x
Thus
becomes
1
1
1
1
1
+
+
+ +
+
1+ 2
2+ 3
3+ 4
1022 + 1023
1023 + 1024
( 2 1) + ( 4 3) + + ( 1024 1021) = 1024 1 = 32 1 = 31.

4
Spring 2000
17. Adding the equations 3(x + y + u + v) = 9, and so x + y + u + v = 3. Thus

4 + v = 3,
x 5 = 3,
y + 0 = 3,
u 8 = 3.
This yields v = 7, x = 2, y = 3, u = 5.
x
18. If eggs had cost x cents less per dozen, I would have saved
cents per egg. If eggs had cost x cents more
12
x
x
per dozen, I would have lost
cents per egg. The difference between these two prices is 2( ) per egg.
12
12
Thus if I buy x + 3 eggs and the total difference is 3 cents, I can write
(x + 3)2
x
12
= 3,
which is to say x2 3x 18 = (x 3)(x + 6) = 0. Since x has to be a positive number, x = 3.

19. Anna answered 20 questions correctly (She could not answer less than 20, because then her score would have
been less than 19 4 = 76 < 77; She could not answer more than 20, because her score would have been at
least 21 4 3 = 81.77). To get exactly 77 points Anna had to answer exactly 3 questions wrong, which
means she omitted 2 questions.
20. The total age of boys is bg. The total age of girls is bg. The total number of people at a classroom party
isb + g + 1. Thus the average age is
bg + bg + 42
= b + g,
b+g+1
whence 42 = b2 + g2 + b + g.
Since b and g are whole numbers (they represent the number of boys and girls respectively), the only
numbers that satisfy the above equation are 5 and 8. Thus b + g = 8.
Spring 2000
Group Contest
a
2
1. (4 minutes, 4 marks) If = , find 900a2 400b2 .
b
3
a
2
Solution: From = , one gathers 3a 2b = 0. Thus
b
3
900a2 400b2 = 100(9a2 4b2 ) = 100(3a 2b)(3a + 2b) = 0.
2. (4 minutes, 4 marks) A car with five tires (four road tires and a spare) travelled 30000 miles. All five tires
were used equally. How many miles wear did each tire received?
Solution: Travelling 30000 miles with 4 tires is as travelling 120000 miles on one tire. The average wear of
each of the 5 tires is thus 120000 5 = 24000 miles.
3. (4 minutes, 6 marks) If a is 50 % larger than c, and b is 25 % larger than c, then a is what percent larger than
b?
1.5b
= 1.2b. Therefore a is 20% larger than b.
Solution: One has a = 1.5c and b = 1.25c Thus a =
1.25
4. (6 minutes, 6 marks) Let
1
x=6+
,
6 + 6+ 1 1
6+
1
6+ 1
..
.
where there are an infinite number of fractions. Write x in the form a + b, with a, b integers. Justify your
answer. (Hint: Some animals, when cut, grow anew, the new animal resembling the old!)
1
Solution: One has x = 6 + or x2 6x 1 = 0. Solving for x one has x = 3 10. Since x must be positive,
x
x = 3 + 10.
5. (7 minutes, 8 marks) Find the sum of all the integers from 1 to 1000 inclusive, which are not multiples of 3 or
n(n + 1)
5. You may use the formula 1 + 2 + 3 + + n =
.
2
Solution: One computes the sum of all integers from 1 to 1000 and weeds out the sum of the multiples of 3
and the sum of the multiples of 5, but puts back the multiples of 15, which one has counted twice. The
desired sum is
(1 + 2 + 3 + + 1000) (3 + 6 + 9 + + 999)
= (1 + 2 + 3 + + 1000) 3(1 + 2 + 3 + + 333)
(5 + 10 + 15 + + 1000)
+(15 + 30 + 45 + + 990)
5(1 + 2 + 3 + + 200)
+15(1 + 2 + 3 + + 66)
= 500500 3 55611
5 20100 + 15 2211
= 266332
6. (7 minutes, 6 marks) Three circles of radius R are mutually tangential as shown in the figure below. What is
the area of the region surrounded by the three circles?
Solution: The area sought is the area of the equilateral triangle formed joining the centers
of the three circles
minus the three angular sectors. Since the equilateral triangle has side 2R, it has area R2 3. Since 60 is 1/6
6
of the circumference, each of the angular sectors has area
Spring 2000
Group Contest
R2
R2
. Thus the area sought is R2 3
.
6
2
7. (4 minutes, 4 marks) In the following diagram, AE = 3, DE = 4, and AD = 5. What is the area of the
rectangle ABCD?
B
Solution: AED is a 3-4-5 right triangle. Drop a perpendicular from E meeting AD at F. Then the area of
1
1
AED is both given by (3 4) = 6 and by (AD)(EF) = 2.5(EF). This entails that EF = 2.4 Since EF = AB,
2
2
the area of the rectangle is given by (AB)(AD) = 2.4 5 = 12.
B
8. (4 minutes, 6 marks) At a certain college 99% of the 100 students are female, but only 98% of the students
living on campus are female. If some females live on campus, how many students live off campus?
Solution: Of the 100 students, only one is male. He is 2% of the on-campus population. Thus the whole
on-campus population consists of 50 students, so there are 100-50=50 off-campus students.
9. (4 minutes, 6 marks) DEFG is a square which is drawn on the outside of a regular pentagon ABCDE. How
big is the angle EAF in degrees?
Solution: The sum of the internal angles of a pentagon is 3 180 = 540 . Each internal angle measures thus
540 5 = 108 . Thus AEF = 360 AED EDF = 360 108 90 = 162 . As AEF is isosceles, we
180 162
gather that EAF =
= 9 .
2
10. (TIE BREAKER, 5 minutes) Solve for all real values of x
xx
x
(x)( x3 ) =
.
x
Solution: Clearly one solution is x = 1. Now x1+3/x = xx1 entails 1 +
solution set is thus {1, 1, 3}.
3
= x 1 or x = 1, 3. The real
x
Fall 2000
Individual Contest
The individual winners of the Fall 2000 edition of the Colonial Maths Challenge were: Rich Anderson, senior,
George Washington HS (first). Tie for second: Sasha Ovetsky, junior, Central HS (15) Maksim Rapoport, senior,
George Washington. Tie for third: Bryan Ferguson, junior, George Washington (13) Feng-Yen Li, senior, Central
Stephanie Moyerman, junior, Central Dmitry Pavolotsky, senior, Central HS. The Team Competition was won by
Central HS.
Peter Baratta, William Clee, Joanne Darken, Celina Evans, Dot French, Elena Koublanova, Margaret Hitczenko,
David Santos helped with the competition.
Instructions: You have 1 hour to complete this exam. Scrap paper, graph paper, ruler, compass, and calculators are
permitted. Write your answers on the answer sheet provided. No credit will be given for procedure. No partial credit
whatsoever.
1. N is a prime number and 90 < N < 99. What is 2N?

2. Let the binary operation be defined as a b = a2 b2 + ab. What is (3) (2)?
3. Iblis entered an elevator in a tall building. She went up 4 floors, down 6 floors, up 8 floors and down 10
floors. She then found herself on the 23rd floor. In what floor did she enter the elevator?
4. A number of birds are resting on two branches of a tree, one branch vertically above the other. A bird on the
lower branch says to the birds on the upper branch: If one of you will come down here, we will have an
equal number of birds on each branch. A bird on the upper branch replies, If one of you will come up
here, we will have twice as many up here as down there. How many birds are on each branch?
5. If the figure shown is folded to form a cube, then three faces meet at every vertex. If for each vertex we take
the product of the numbers on the three faces that meet there, what is the largest product we get?
1
4
6. At the annual Fourth of July picnic, the big event of the day is always the giant Tug-of-War! In the first round,
four weightlifters (all of equal strength) tugged against five grandmothers (all of equal strength). The result
was a tie! In the second round, Yogi Bear tugged against two grandmothers and one weightlifter. Again, the
result was a tie! In the third and final round, Yogi Bear and two grandmothers tugged against four of the
weightlifters. What was the outcome? Enter Y if Yogi and the grandmothers won, W if the weightlifters won,
or T if it was a tie.
7. To traverse the St. Laurent river, first I arrived at Isle of St. Therese, and then I went to the Isle of St.
Catherine, as shown in the diagram. Assume that the river banks are parallel. What is the angle x, in degrees?
110
St. Catherine
35
x
St. Therese
Fall 2000
Individual Contest
8. Suppose the product
1
1 2
2
1
1 2
3
1
1 2
4
1
1 2
14
1
1 2
15
a
,
b
where a and b are positive integers with no prime factor in common. What is a2 + b2 ?
9. If
45 + 45 + 45 + 45 65 + 65 + 65 + 65 + 65 + 65
= 2n ,
35 + 35 + 35
25 + 25
find n.
10. Find the area, in square units, of the cross-shaped region enclosed in the 5 5 square.
2
2
11. Four singers take part in a musical round of 4 equal lines, each finishing after singing the round through 3
times. The second singer begins when the first singer begins the second line, the third singer begins when the
first singer begins the third line, the fourth singer begins when the first singer begins the fourth line. What is
the fraction of the total singing time when all the singers are singing simultaneously?
12. Three identical squares are drawn side to side, as shown in the figure below. Compute the angle sum
ACH + ADH in degrees.
B
C
D
A
13. Find the smallest positive integer n such that

(perfect)fifth power.
n
n
n
is a (perfect) square, is a (perfect) cube, and is a
2
3
5
14. If
(x2 x + 1)6 (x5 x + 1)7 = a47 x47 + a46 x46 + a45 x45 + + a3 x3 + a2 x2 + a1 x + a0 ,
find the value of the sum of the coefficients of every even power:
a46 + a44 + a42 + a2 + a0 .
9
Fall 2000
Individual Contest
15. The circle with centre at O is tangent to the line AG at D, to the line AH at E, and to the line at BC at F. Find
BOC in degrees, if HAG measures 40 .
G
D
B
O
F
H
A
C
E
16. Point P is interior to an equilateral triangle ABC of side 3 units. The distance of P to AB is a, the distance of
P to AC is 2a and the distance of P to CB is 3a. Find the length of a.
17. Five people are waiting in line. Starting with the second one, it is noticed that the age of a person added to
three times the age of the person immediately in front is always equal to 110. What is the age of the first
person in line? Assume that the ages are integral numbers.
18. Find all positive solutions of the system
xy = 2,
yz = 3,
zx = 4.
19. One Price Shoes sell all their shoes for the same price, which is an integral number of dollars. Anacleta and
Sinforosa go shoe shopping to One Price Shoes. Anacleta has $200 and she buys as many pairs of shoes as
possible, and there remain $32. Sinforosa has $150 and she buys as many pairs of shoes as possible, and there
remain $24. What is the price of a pair of shoes?
20. A triangular banner is coloured as shown below, the area of each region given in cm2 . How many cm2 of
blue are there? How many cm2 of yellow?
yellow
252
105
green
orange
90
blue
red
120
violet
10
Fall 2000
1.
194
N = 97 and 2N = 194.
2.
(3) (2) = (3)2 22 + (3)(2) = 9 4 6 = 1.
3.
27
If she entered at floor F, then F + 4 6 + 8 10 = 23, whence F = 27.
4.
7 and 5
Let x be the original number of birds in the upper branch and let y be the original numbers of
birds in the lower branch. From the given data,

x 1 = y + 1,
x + 1 = 2(y 1).
This system solves to (x, y) = (7, 5).
5.
90
6.
Observe that 4 and 5 are on opposite sides, so they never meet. The largest product is 3 5 6 = 90.
Denote by y the strength of Yogi Bear, by w the strength of a weightlifter and by g that of a
grandmother. All these strengths are positive numbers. We are given that 4w = 5g and that y = 2g + w. Now
y + 2g = 2g + w + 2g = 4g + w =
16
21
w+w=
w > 4w,
5
5
and so Yogi and the grandmothers won.

7.
52.5 or 52 30
Draw a line perpendicular to the shores to complete a hexagon as shown. The sum of
110
35
the interior angles of a hexagon is 180 (6 2) = 720 . Thus

2x + 2 90 + 110 + (360 35 ) = 720 ,
whence x = 52.5 .
8.
289
We use the fact that 1 x2 = (1 x)(1 + x). First
1
1
2
1
1
1
1
3
14
11
1
1
15
1 2
13 14
1

=
.
2 3
14 15
15
Second
1
1+
2
Hence
1
1
1+
1 +
3
14
1
1
22
1
32
1
1+
15
Fall 2000
1
142
3 4
15 16
16

=
= 8.
2 3
14 15
2
1
152
8
,
15
and a2 + b2 = 289.
9.
12
The product equals

4(45 ) 6(65 )
= 4(45 ) = 212 .
3(35 ) 2(25 )
10.
290
17
Let A be the area of one of the four smaller corner right triangles (the ones with hypotenuse 3).
Then the area sought is 25 4A. Now, each of these smaller triangles is similar
to a large triangle (with legs 5
and 3). Since the hypotenuse of these triangles must be in proportion, 3k = 34, for some constant k. Now,
15
15
135
the areas keep the square proportionality, hence Ak2 =
. This gives A = 2 =
. Hence
2
2k
68
135
290
25 4A = 25
=
.
17
17
11.
3
5
Three lines have been sung before the fourth singer starts, and after that he sings 12 more lines. So
the total span of lines is 15. They start singing simultaneously from line 4, and the first singer is the first to
9
3
end, in line 12. Thus 12 4 + 1 = 9 lines out of 15 are sung simultaneously and the fraction sought is
= .
15
5
12.
45
Construct the diagram

shown below:
T
Observe that ADH = ACT and that THC is an isosceles right triangle. Hence ADH + ACH = 45 .
13.
30233088000000
The integer must be of the form n = 2a 3b 5c , with a, b, c positive integers. The
conditions imply that a 1 is even and that a is divisible by 15; that b 1 is divisible by 3 and that b is
divisible by 10; and that c 1 is divisible by 5 and c divisible by 6. Clearly, the smallest positive integers
satisfying those conditions are a = 15, b = 10, c = 6. Hence the integer sought is
n = 215 310 56 = 30233088000000.
12
14.
365
Fall 2000
Put
p(x) = (x2 x + 1)6 (x5 x + 1)7 = a47 x47 + a46 x46 + a45 x45 + + a3 x3 + a2 x2 + a1 x + a0 .
Then
a46 + a44 + + a0 =
15.
70
p(1) + p(1)
1 + 36
=
= 365.
2
2
First observe that ADO and AEO are right angles. DOA = 90 20 = 70 . As BD and BF are
tangent to the circle, OB bisects DOF. Hence

DOA
70
=
= 35 .
2
2
G
BOF =
Therefore, the angle sought is 2(BOF) = 70 .
D
B
O
F
A
16.
3
a=
4
The area of ABC = Area APB+ Area APC+ Area CPA. Now,
1
3
2
2
32
3
2
1
1
1
3 a + 3 3a + 3 2a,
2
2
2
3
which is to say a =
.
4
P
A
17.
28
Let xk be the age of the k-th person. Then

x2 + 3x1 = 110,
x3 + 3x2 = 110,
x4 + 3x3 = 110,
x5 + 3x4 = 110.
These equations imply that each of x2 , x3 , x4 , x5 is an integer leaving remainder 2 upon division by 3. Also
x5
=
=
=
=
110 3x4
110 3(110 3x3 )
110 3(110 3(110 3x2 ))
110 3(110 3(110 3(110 3x1 ))),
13
Fall 2000
from where it follows that

x5 = 110(1 3 + 9 27) + 81x1 = 2200 + 81x1 .
Hence,
x5 + 13
.
81
Since x1 is an integer, x5 + 13 must be divisible by 81. The least positive x5 for which this occurs is x5 = 68.
By successively solving the equations we gather that x5 = 68, x4 = 14, x3 = 32, x2 = 26, x1 = 28.
x1 = 27 +
2 6
6
18.
z = 6, x =
,y =
Multiplying all the equations together, we obtain x2 y2 z2 = 24, whence
3
2
xyz = 2 6. Dividing xyz successively by each of the equations, we find,
2 6
6
z = 6, x =
,y =
.
3
2
19.
$42
Let A be the number of pairs of shoes Anacleta buys and let S be the number of pairs of shoes
Sinforosa buys. Let P be the price, in dollars, of each pair of shoes. Observe that P > 32. Then
200 = AP + 32, 150 = SP + 24, whence
AP = 168 = 23 3 7, SP = 126 = 2 32 7.
The price of each pair of shoes is a common divisor of 168 and 126, but it must exceed 32. The only such
divisor is 42.
20.
y = 210, b = 168
Observe from the figure below that ABE and OBE share the same base BE. Thus
the ratio of their areas is proportional to the ratio of their heights, which we will name h and h respectively.
Therefore
b + 372
h
= .
120
h
Also, AEC shares the same base EC with OEC and their heights are also h and h , giving
h
y + 195
=
.
h
90
Therefore
b + 372
y + 195
=
.
120
90
Again, ACD and AOD share the same base AD. The ratio of their areas is proportional to the ratio of
their heights h1 and h2 . This gives
y + 357
h1
=
.
252
h2
Also, CDB shares the same base DB with ODB and their heights are also h1 and h2 , giving
h1
b + 210
=
.
h2
b
Therefore
y + 357
b + 210
=
.
252
b
Upon solving the equations

b + 372
y + 195
=
,
120
90
14
and
Fall 2000
y + 357
b + 210
=
252
b
we deduce that y = 210, b = 168.

A
F
C
O
D
E
B
15
Fall 2000
Group Contest
1. (4 minutes, 4 marks) Junk Pizzas sell pizzas with diameters of 4 inches, 6 inches, 8 inches and 10 inches for $2,
$4, $8, and $12 respectively. If the pizzas have all the same thickness, which size will give the most pizza per
dollar?
Solution:
22
32
= 2 square inches per dollar, a $4 gives
= 2.25 square inches per
2
4
42
52
= 2 square inches per dollar, and a $12 gives
2.08 square inches per dollar.
dollar, an $8 gives
8
12
The best deal is the $4 pizza, with 6 inches of diameter.
6 inches
A $2 pizza gives
2. (4 minutes, 5 marks) The average of six numbers is 4. A seventh number is added and the new average
increases to 5. What was the seventh number?
Solution:
11
The sum of the original 6 numbers is S = 6 4 = 24. If the 7th number is x, then
24 + x
= 5,
7
whence x = 11.
3. (4 minutes, 6 marks) Assume that there is a positive real number x such that
.
x.
xx
= 2,
where there is an infinite number of xs. What is the value of x?

Solution:
x=
We have x2 = 2, and since x is positive, x =
2.
4. (7 minutes, 6 marks) The year is 2001. Dwaynes father gives Dwayne $N dollars on the Nth day of the year,
but he uses only $1 and $10 bills, using as many $10 bills as possible. Thus, on February 1st he gives Dwayne
$32 using three $10s and two $1s. On February 1st Dwayne has 1 + 2 + 3 + ... + 32 = 528 dollars saved, of
these 138 are $1s and 39 are $10s. Dwayne will leave on vacation the day he has exactly as many $10s as $1s.
When will Dwayne leave on vacation? You may use the formula
1 + 2 + 3 + + N =
Solution:
8 April 2001
N(N + 1)
.
2
Let x, y be, respectively, the number of $1s he uses and the number of $10s he uses. On day
number 10n + m, (n 0, 0 m 9), we have

x = 45n +
m(m + 1)
2
and
y = 5n(n 1) + n(m + 1).
If follows that
45n x 45(n + 1)
and
5n2 4n y 5n2 + 5n.
From this we deduce that if n 7, 5n2 + 5n < 45n and thus x < y, and if n 11, then 5n2 4n > 45(n + 1),
hence x > y. We must examine one by one the thirty values corresponding for n {8, 9, 10}. The only
possibilities giving x = y are n = m = 8 or n = m 1 = 8. Dwayne will leave after saving up money for 98
days, on 8 April.
16
Fall 2000
Group Contest
5. (7 minutes, 8 marks) The square has side 1. The shaded region below is formed by overlapping
quarter-circles each centred at a corner of the square. Diagonally opposite quarter-circles are tangent. Find
the area of the cross-shaped figure shaded.
Solution:
2 21
2
The area of the figure is twice that of the figure shaded below.
r
Now,
(2r)2 = 12 + 12 ,
2
. The figure above has as area that of the square minus two quarter circles and two isosceles
2
right triangles. That is, its area is
whence r =
12
4
The required area is thus
2
2
2
1
2
2
1
2
1
= 2 .
2
2 4
2 21 .
2
6. (6 minutes, 6 marks) Betty and Wilma can paint a building in two hours, Wilma and Pebbles can paint the
same building in three hours and Betty and Pebbles can paint the very same building in 5 hours. How long
would it take the three of them working together to paint the building?
Solution:
60
hours
31
Let b, w, p be the fractional effort put by Betty, Wilma, and Pebbles, respectively, for one hour.
We are given that

1
,
2
1
w+p= ,
3
1
p+b= .
5
b+w=
Adding these equations,

2(b + w + p) =
17
31
,
30
Fall 2000
Group Contest
31
31
hence b + w + p =
. In one hour they can do
of the building, hence they can do the whole building in
60
60
60
hours.
31
7. (6 minutes, 6 marks) The diagram is a sketch road map of roads linking Joannescity to Petersborough,
passing through Dotsville, Elenopolis, Billstowne, Margaretsburgh, and Davidherri. The number on each
section of road is the distance in miles from one town to another. How long is the shortest route from
Joannescity to Petersborough?
Joannescity
4
Dotsville
Margaretsburgh
11 miles
Elenopolis
Billstowne
5
Petersborough
Solution:
Davidherri
Starting from Joannescity, go to the nearest town. The shortest path is shown below.
Joannescity
Dotsville
Margaretsburgh
Elenopolis
Billstowne
Petersborough
8. (4 minutes, 5 marks) List all the triangles with integral sides and perimeter 10. Lists that differ only in order
are considered identical.
Solution:
{2, 4, 4} and {3, 3, 4}
Let a b c be the sides of the triangle. Then a, b, c, are integers satisfying

a + b + c = 10, a + b > c.
This entails
10 c > c,
whence c < 5. Thus the only such lists are 2, 4, 4, and 3, 3, 4.
Tie Breaker I (4 minutes, 3 marks.)
Find the units digit of
Solution:
72000 .
We have 74 = 2401 = 10N + 1, with N = 240. Hence

72000 = (74 )500 = (10N + 1)400 = 10Q + 1,
for some integer Q, and so the units digit is 1.
18
Fall 2000
Group Contest
Tie Breaker II (6 minutes, 6 marks)

Find positive integers a, b, c, d such that
x80 + x79 + x78 + + x3 + x2 + x + 1 = (x2a + xa + 1)(x2b + xb + 1)(x2c + xc + 1)(x2d + xd + 1).
You may use the identity T 3 1 = (T 1)(T 2 + T + 1).
Solution:
(x54 + x27 + 1)(x18 + x9 + 1)(x6 + x3 + 1)(x2 + x + 1)
Put S = x80 + x79 + + x + 1. Then
xS = x81 + x80 + + x = x81 + S 1.

Solving for S,
x80 + x79 + + x + 1 = S =
Now,
x81 1
.
x1
x81 1
x81 1 x27 1 x9 1 x3 1
= 27
.
x1
x 1 x9 1 x3 1 x 1
Using the identity T 3 1 = (T 1)(T 2 + T + 1) we deduce

x81 1
= (x54 + x27 + 1)(x18 + x9 + 1)(x6 + x3 + 1)(x2 + x + 1).
x1
19
Spring 2002
Individual Contest
The individual winners of the Spring 2002 edition of the Colonial Maths Challenge were: Sasha Ovetsky, senior,
Central HS (first); Michael Segal, freshman Central HS (second); George Nesterenko, junior, Central HS (third).
The Team Competition was won by Central HS.
Peter Baratta, William Clee, Joanne Darken, Dot French, Dan Jacobson, Elena Koublanova, Margaret Hitczenko, Mi
Mi Lian, Clark Loveridge, David Santos, Benjamin Wong helped with the competition.
whatsoever.
1. 20 (Forming Squares) If any four different dots are chosen, how many squares can be formed from the
design in Figure 1?
Figure 1: Forming Squares

Solution:
There are
9 squares of side of length 1, 4 of side of length 2, 1 of side of length 3, 4 of side of length

2, and 2 of side 5. Thus the total number of squares is 9 + 4 + 1 + 4 + 2 = 20.
2. 61 If
1
1
1+
5
where the fraction

Solution:
a
,
b
a
is in least terms, find a2 + b2 .
b
1
1+
1
5
1
5
a
= = ,
6
6
b
5
whence a2 + b2 = 52 + 62 = 61.
3. 6 An isosceles triangle has one side measuring 3 units and another side measuring 6 units. What is the length
of the third side?
Solution: Another side of length 3 is impossible, since the triangle inequality is not satisfied. Thus the third
side has length 6.
4. 21 A binary operation is defined by
ab =
a+b
.
2
If
x(x21) = x,
find x.
Solution: We have
x(x21) = x
20
x + 21
x+
2
=x
2
x + 21
x+
= 2x
2
2x + x + 21 = 4x
x = 21.
Spring 2002
Individual Contest
5. 28 How many digits does 416 525 have?

Solution: There are 28 digits, since
416 525 = 232 525 = 27 225 525 = 128 1025 ,
which is the 3 digits of 128 followed by 25 0s.
6. 10 In how many different ways can one change 50 cents using nickels, dimes or quarters?
Solution: We want the number of solutions of
5x + 10y + 25z = 50,
that is, of
x + 2y + 5z = 10,
with integer 0 x 10, 0 y 5, 0 z 2. The table below exhausts all ten possibilities.
z y
2 0
1 2
1 1
1 0
0 5
0 4
0 3
0 2
0 1
0 0
x
0
1
3
5
0
2
4
6
8
10
7. 400 A palindrome is an integer whose decimal expansion is symmetric and does not end in 0. How many
palindromes of 5 digits are even?
Solution: A five digit even palindrome has the form ABCBA, where A {2, 4, 6, 8}, and
(B, C) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}2 . Thus there are 4 choices for the first digit, 10 for the second, and 10 for the
third. Once these digits are chosen, the palindrome is completely determined. Therefore, there are
4 10 10 = 400 even palindromes of 5 digits.
8.
3
Find the exact value of the product
101
1
Solution: We have
2
5
2
7
2
5
2
7
2
2
1
9
99
2
2
1
9
99
2
101
=
=
9. 13 Let t = 32001 . If
find a.
Solution: We have
2
.
101
3 5 7 9
97 99

5 7 9 11
99 101
3
.
101
32001 + 32002 + 32003 = at,
32001 + 32002 + 32003 = 32001 (1 + 3 + 32 ) = (13)32001 ,
whence a = 13.
10. 120 As a publicity stunt, a camel merchant has decided to pose the following problem: If one gathers all of
my camels into groups of 4, 5 or 6, there will be no remainder. But if one gathers them into groups of 7
camels, there will be 1 camel left in one group. The number of camels is the smallest positive integer
satisfying these properties. How many camels are there?
Solution: The least common multiple of 4, 5 and 6 is 60, hence we want the smallest positive multiple of 60
leaving remainder 1 upon division by 7. This is easily seen to be 120.
21
Spring 2002
Individual Contest
11. 36 (Area of Quadrilateral) Figure 2 is composed of a grid of identical squares. The vertices of the
quadrilateral ABCD are corners of some of these squares. Find the exact area, in square units, of the
quadrilateral ABCD.
Cb
b
Figure 2: Area of Quadrilateral

Solution: Enclose the quadrilateral as shewn below.
The area sought is that of the enclosing square minus the area of the small rectangle and the four
right-triangles formed, whence it is
1
1
1
1
102 (3)(2) (3)(6) (2)(2) (8)(10) (7)(2) = 36.
2
2
2
2
12. 79 You are given that x + y = 3 and xy = 8. Find
x4 + y4 .
Solution: We have
and
x2 + y2 = (x + y)2 2xy = 9 16 = 7,
x4 + y4 = (x2 + y2 )2 2x2 y2 = 49 128 = 79.
22
Spring 2002
Individual Contest
13. 28 (Seven Circles) Each of six circles of radii 1 is tangent to its neighbour and toa seventh larger circle, as in
figure 3. The exact area that these six circles bound can be written in the form a 3 + b, where a and b are
integers. Determine the value of a2 + b2 + ab.
Figure 3: Seven Circles

Solution: The desired area is the area of the regular hexagon of side 2 with vertices at the centres of the
smaller circles minus the six third-of-circles. The apothem of the hexagon is, by the Pythagorean Theorem
p
1
3(2)) = 6 3. The area of the six thirds-of-circles is
= 22 1 = 3. The area of the hexagon is thus 6(
2
1
6( ) = 2. The area wanted is thus
3
6 3 2.
Hence a = 6, b = 2, from where it follows that a2 + b2 + ab = 36 + 4 12 = 28.
14. 13501 The integers from 1 to 1000 are written in succession. Find the sum of all the digits.
Solution: When writing the integers from 000 to 999 (with three digits), 3 1000 = 3000 digits are used. Each
of the 10 digits is used an equal number of times, so each digit is used 300 times. The the sum of the digits in
the interval 000 to 999 is thus
(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)(300) = 13500.
Therefore, the sum of the digits when writing the integers from 000 to 1000 is 13500 + 1 = 13501.
15. 96 How many 4-digit integers can be formed with the set of digits {0, 1, 2, 3, 4, 5} such that no digit is repeated
and the resulting integer is a multiple of 3?
Solution: The integers desired have the form D1 D2 D3 D4 with D1 6= 0. Under the stipulated constraints, we
must have
D1 + D2 + D3 + D4 {6, 9, 12}.
We thus consider three cases.
Case I: D1 + D2 + D3 + D4 = 6. Here we have {D1 , D2 , D3 , D4 } = {0, 1, 2, 3}, D1 6= 0. There are then 3 choices
for D1 . After D1 is chosen, D2 can be chosen in 3 ways, D3 in 2 ways, and D1 in 1 way. There are thus
3 3 2 1 = 3 3! = 18 integers satisfying case I.
23
Spring 2002
Individual Contest
Case II: D1 + D2 + D3 + D4 = 9. Here we have {D1 , D2 , D3 , D4 } = {0, 2, 3, 4}, D1 6= 0 or

{D1 , D2 , D3 , D4 } = {0, 1, 3, 5}, D1 6= 0. Like before, there are 3 3! = 18 numbers in each possibility, thus we
have 2 18 = 36 numbers in case II.
Case III: D1 + D2 + D3 + D4 = 12. Here we have {D1 , D2 , D3 , D4 } = {0, 3, 4, 5}, D1 6= 0 or
{D1 , D2 , D3 , D4 } = {1, 2, 4, 5}. In the first possibility there are 3 3! = 18 numbers, and in the second there are
4! = 24. Thus we have 18 + 24 = 42 numbers in case III.
The desired number is finally 18 + 36 + 42 = 96.
16. 3666 4 Determine the largest number which is the product of positive integers whose sum is 2002.
Solution: We are given some positive integers a1 , a2 , . . . , an with a1 + a2 + + an = 2002. To maximise
a1 a2 an , none of the ak s can be 1. Let us shew that to maximise this product, we make as many possible
ak = 3 and at most two aj = 2.
Suppose that aj > 4. Substituting aj by the two terms aj 3 and 3 the sum is not changed, but the product
increases since aj < 3(aj 3). Thus the ak s must equal 2, 3 or 4. But 2 + 2 + 2 = 3 + 3 and 2 2 2 < 3 3,
thus if there are more than two 2s we may substitute them by 3s. As 2002 = 3(667) + 1 = 3(666) + 4, the
maximum product sought is 3666 4.
17. 117856 Let S be the set of all natural numbers whose digits are chosen from the set {1, 3, 5, 7} such that no
digits are repeated. Find the sum of the elements of S.
Solution: First observe that 1 + 7 = 3 + 5 = 8. The numbers formed have either one, two, three or four digits.
The sum of the numbers of 1 digit is clearly 1 + 7 + 3 + 5 = 16.
There are 4 3 = 12 numbers formed using 2 digits, and hence 6 pairs adding to 8 in the units and the tens.
The sum of the 2 digits formed is 6((8)(10) + 8) = 6 88 = 528.
There are 4 3 2 = 24 numbers formed using 3 digits, and hence 12 pairs adding to 8 in the units, the tens,
and the hundreds. The sum of the 3 digits formed is 12(8(100) + (8)(10) + 8) = 12 888 = 10656.
There are 4 3 2 1 = 24 numbers formed using 4 digits, and hence 12 pairs adding to 8 in the units, the
tens the hundreds, and the thousands. The sum of the 4 digits formed is
12(8(1000) + 8(100) + (8)(10) + 8) = 12 8888 = 106656.
The desired sum is finally
16 + 528 + 10656 + 106656 = 117856.
18. 25 Fifty pounds of caviar are for sale at a store. Several customers are lined up to buy this caviar. Having sold
the demanded portion to a customer, the cashier calculates the average weight of the portions already sold
and tells the number of customers for which there is enough caviar left provided these customers each buy
exactly the average pounds of caviar that she has declared. It turns out that for each of the first 10 customers
the cashier is able to announce, after each of these customers has bought his caviar, that there is still enough
caviar for the next 10 customers. How much caviar will there be left after the first 10 customers have made
their purchases, provided the stated conditions are met?
Solution: If ak is the amount of caviar sold to the k-th customer, then for each k, 1 k 10 we are given that
50 (a1 + a2 + + ak ) = 10
whence
a1 + a2 + + ak
,
k
50k
= a1 + a2 + + ak .
10 + k
If the stipulated conditions are met then

25 =
50 10
= a1 + a2 + + a10 .
10 + 10
Hence there are 50 25 = 25 pounds of caviar left.

24
Spring 2002
Individual Contest
Remark: Since this is valid for each k, 1 k 10, subtracting the k-th equation from the k 1-th equation,
we deduce that
50k
50(k 1)
500
ak =
=
.
10 + k
9+k
(10 + k)(9 + k)
As long the k-th customer buys exactly
500
pounds, the conditions of the problem will be met.
(10 + k)(9 + k)
19. 2000 Let x denote the greatest integer less than or equal to x. For example, 3.6 = 3, = 4. Find the
largest integer N verifying the following two conditions.
First condition:
N
has three digits, each identical.
3
Second condition: There is a natural number n such that
N
= 1 + 2 + 3 + n.
3
Solution: From the first condition
N
= 111k
3
for some k {1, 2, 3, 4, 5, 6, 7, 8, 9}. From the second condition
n(n + 1)
1 +
= 111k n2 + n 222k = 0 n =
2
1 + 888k
.
2
We test each k {1, 2, 3, 4, 5, 6, 7, 8, 9} and see that 1 + 888k is a square only when k = 6, whence n = 36 and
1 + 2 + + 36 = 666.
From the first condition
666
N
< 667,
3
whence
1998 N < 2001.
The largest integer meeting both conditions is thus N = 2000.
20. 4913 and 5832 Find all the natural numbers of 4 digits that are equal to the cube of the sum of their digits.
Solution: Let n = 1000a + 100b + 10c + d be the integer sought and let s = a + b + c + d be the sum of its digits.
From 1000 n 9999 and n = s3 it follows that 11 s 21. Observe that
(s 1)s(s + 1) = s3 s = 999a + 99b + 9c = 9(111a + 11b + c),
whence 9 divides one of the three consecutive integers s 1, s or s + 1. Since 11 s 21, we have three cases to
consider: s3 s = 16 17 18, s3 s = 17 18 18 or s3 s = 18 19 20.
If s3 s = 16 17 18, then 111a + 11b + c = 544 whence a = 4, b = 9, c = 1, s = 17 and so, d = 3. The number is
n = 4913.
If s3 s = 17 18 19, then 111a + 11b + c = 646 whence a = 5, b = 8, c = 3, s = 18 and so, d = 2. The number is
n = 5832.
If s3 s = 18 19 20, then 111a + 11b + c = 760 whence a = 6, b = 8, c = 6, s = 19 and no such digit d exists.
The only two numbers are thus 4913 and 5832.
25
Spring 2002
Group Contest
1. 2004 (4 minutes) Find the value of n in

(102002 + 25)2 (102002 25)2 = 10n .
Solution: We have
(102002 + 25)2 (102002 25)2
104004 + 2 25 102002 + 252 (104004 2 25 102002 + 252 )

4 25 102002
100 102002 = 102004 ,
=
=
=
whence n = 2004.
2. 13 (4 minutes) In a group of 100 camels, 46 eat wheat, 57 eat barley, and 10 eat neither. How many camels eat
both wheat and barley?
Solution: Let A be the set of camels eating wheat, and |A| its number, and B be the set of camels eating barley,
and |B| its number. Then
90 = 100 10 = |A B| = |A| + |B| |A B| = 46 + 57 |A B| = 103 |A B|,
whence |A B| = 13.
3. (6 minutes) Write positive integers in the empty boxes of the grid below so that the sum of any three
consecutive boxes be always the same and the sum of all the boxes be 217.
A B C
D
17
E F G
H
20
J K
L M
Solution: Since the sum of any three consecutive boxes is the same, we deduce from
A B C
a b c
D
17
E F G
H
20
J K
L M
H
20
J K
L M
H
20
I
c
J
17
that a = 17. If the conditions are met we must have

A
17
B C D
b c 17
E
b
F G
c 17
from where b = 20. This gives an arragement of the form

A
17
B
20
C D
c 17
E
20
F G
c 17
K
20
L M
c 17
from where
17 5 + 20 4 + 4c = 217.
This gives c = 13.

The required arrangement is finally
A
17
4.
B C
20 13
D
17
E
20
F
13
G
17
H
20
I
13
J
17
2002
(5 minutes) Find
2003
1
2
2
1
2
26
1
2
1
2
K
20
L
13
M
17
where the digit 2 is repeated 2002 times.

Solution: Put
x1 = 2, xn = 2
We see that
1
xn1
1
=
2
2
x3 = 2 =
3
x2 = 2
Spring 2002
Group Contest
, n 2.
3
,
2
4
,
3
and by induction we can prove that

xn =
Thus x2002 =
2003
and we want
2002
1
x2002
5.
n+1
.
n
2002
.
2003
1
(8 minutes) (Area of circular semi-segment) In Figure 4 each of the chords divides the larger circle into
2
1
two parts of areas in ratio . The intersection of the chords forms a square concentric with the larger circle.
3
Find the ratio of the area shaded to the area of the circle exscribed to the square (dashed circle). Solution:
Figure 4: Area of circular semi-segment.

The outer circle of area A can be divided into 1 square of area S, four regions of area Q and four smaller
regions of area P (the area shaded above).
Q
Notice that any two parallel chords described divide the outer circle into three bands, two outer bands
of identical area 2P + Q, and a central band of area 2Q + S. Any one of the given chords divides the outer
A
3A
A
circle into two parts, one of area and the other of area
. The area of the central band is 2Q + S =
is
4
4
2
A
twice the area of one of the outer bands, each of area 2P + Q = . This gives S = 4P. If x is the side of the
4
S
x2
x2
central square, then we have P = =
. The area of the inner circle is thus I = and thus finally
4
4
2
x2
P
1
= 4x2 =
.
I
2
2
27
Spring 2002
Group Contest
6. 100 (5 minutes) (Concentric Circles) In figure 5, the two circles are concentric, AB is tangent to the inner
circle, and AB = 20. Find the exact value of the area of the shaded annulus.
Figure 5: Concentric Circles

Solution: Let R and r be the radii of the outer and inner circle, respectively. The area sought is (R2 r2 ).
20
Now, by the Pythagorean Theorem, R2 r2 = ( )2 = 100. Hence the desired area is 100.
2
7. 2 : 1 (4 minutes) (Ratio of Areas) In figure 6, each side of the rectangle is divided into three equal segments.
The endpoints of the segments are joined to the center of the rectangle in order to form the banner shown.
Find the ratio of the areas of the white portion to the shaded portion.
Figure 6: Ratio of Areas.

Solution: Without loss of generality, assume that the rectangle is a (3a) (3b) rectangle of area 9ab. Four
3b
3a
isosceles triangles, two with base a and height
, and two with base b and height
are formed. The area
2
2
that these triangles comprise is thus
1 3b
2
a
2 2
1 3a
+2
b
2 2
= 3ab.
The area of the white portion is thus 9ab 3ab = 6ab and the desired ratio is
6ab : 3ab = 2 : 1.
8. 9901 (5 minutes) Given that
1, 000, 002, 000, 001
has a prime factor greater than 9000, find it.
28
Solution:
1, 000, 002, 000, 001
Spring 2002
=
1012 + 2 106 + 1
=
(106 + 1)2
=
((102 )3 + 1)2
2
= (10 + 1)2 ((102 )2 102 + 1)2
=
1012 99012 ,
whence the prime sought is 9901.
29
Group Contest
Spring 2003
Individual Contest
The individual winners of the Spring 2003 edition of the Colonial Maths Challenge were: 1st Place: David Heinz Central HS Wayne Whitfield - Central HS 2nd Place: George Nesterenko - Central HS Jacky Yuen - George
Washington 3rd Place: Lynn Haimowitz - Central Jarret Leiberman - George Washington. The Team Competition
was won by Central HS.
Baratta, William Clee, Joanne Darken, Dot French, Dan Jacobson, Margaret Hitczenko, Wimayra Luy, Clark
Loveridge, David Santos, Yun Yoo helped with the competition.
whatsoever.
1. Given that
10 + 102
1
1
+
10 100
is an integer, find it.

Solution: =
1000
10 + 102
103 + 104
11000
=
=
= 1000.
1
1
10 + 1
11
+
10 100
2. If today is Thursday, what day will it be 100 days from now?

Saturday
Solution: Since 100 = 7 14 + 2, 98 days from today will be a Thursday, and 100 days from
today will be a Saturday.

3. What is the closest (integer) perfect square to 2003?
2025 Observe that 2003 > 44. We see that 442 = 1936 and 452 = 2025. Thus the closest square is 2025.
= 150 . See figure 7. What is the measure of CD

in
4. In the circle with center O, COD = 2ABC and AB
degrees?
A
B
C
b
Figure 7: Problem 4.
1
= COD = 2ABC = 30 .
AC = 15 . Thus CD
2
5. The circle in figure 8 has radius 1. Side AB of ABC is a diameter of the circle,
and AC = 3. Find the area
inside the circle but outside the triangle. Leave your answer in the form a + b 3, where a and b are
rational numbers.
Solution: ABC is a right triangle at C since this C is inscribed in a semicircle. By the

2
3
3
Pythagorean Theorem BC = 4 3 = 1. Hence the area of ABC is
, and the area sought is
.
2
2
30
= 30 and hence ABC =

Solution: AC
30
Spring 2003
Individual Contest
6. Find all positive primes of the form n3 8, where n is a positive integer.

19
Solution: Since n3 8 = (n 2)(n2 + 2n + 2) and since n 2 < n2 + 2n + 4, we must have
n 2 = 1 = n = 3 and n3 8 = 19.
7. If k is a positive integer, the notation k! (read k factorial) indicates the product of the integers between 1
and k, e.g., 5! = 1 2 3 4 5 = 120. Find the last two digits of the integer
1! + 2! + 3! + + 100!.
13
For k 10, k! ends in two or more 0s. Thus the last two digits of the desired sum are the last two
digits of
1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! = 409113,
thus the desired sum ends in 13.
8. p and q are positive integers such that
7
p
11
< <
.
10
q
15
What is the least value of q?
7
Solution: If all a, b, c, d are positive, then

a
c
a
a+c
c
<
=
<
< .
b
d
b
b+d
d
Hence
7
11
7
18
11
7
25
18
11
<
=
<
<
=
<
<
<
.
10
15
10
25
15
10
35
25
15
25
5
5
11
4
7
Since
= , we have q 7. Could it be smaller? Observe that >
and that <
. Thus by
35
7
6
15
6
10
considering the cases with denominators q = 1, 2, 3, 4, 5, 6, we see that no such fraction lies in the desired
interval. The smallest denominator is thus 7.
9. Find the number of ordered pairs (x, y) of integers (positive, negative or zero) satisfying the equation
x + y + xy = 120.
Here (x, y) is different from (y, x) if x 6= y.
6
Solution: Observe that

x + y + xy + 1 = 121 = (x + 1)(y + 1) = 112 .
Thus x + 1 = 1, 11 or 121. Thus there are six solutions

(0, 120), (120, 0), (2, 122), (122, 2), (10, 10), (12, 12).
10. An urn has 900 chips, numbered 100 through 999. Chips are drawn at random and without replacement from
the urn, and the sum of their digits is noted. What is the smallest number of chips that must be drawn in
order to guarantee that at least three of these digital sums be equal?
53
Solution: There are 27 different sums. The sums 1 and 27 only appear once (in 100 and 999), each of
the other 25 sums appears thrice. Thus if 27 + 25 + 1 = 53 are drawn, at least 3 chips will have the same sum.
31
Spring 2003
Individual Contest
11. The quadratic equation ax2 + bx 3 = 0 has 1 as one of its roots. If a and b are positive primes, what is
a 2 + b2 ?
Solution: We have
29
a(1)2 + b(1) 3 = 0 = a b = 3.
The difference between odd primes is always even, so one of the primes must be 2. Thus a = 5 and b = 2.
Hence a2 + b2 = 29.
12. A circle is inscribed in right ABC, with a right angle at C. The circle is tangent to hypotenuse AB at P
where AP = 20 and BP = 6. Find the radius of the circle.
Solution: Tangents to a circle from the same point have equal lengths. Thus AC = 20 + r and
CB = 6 + r. Equality of areas reveals that
1
1
1
(20 + r)(6 + r) = r2 + 2
(20)(r) + 2
(6)(r)
2
2
2
= 120 26r r2 = 0 = (30 + r)(4 r) = 0.
Since r > 0 we must have r = 4.

A
P
C

13. The vertices of a triangle have coordinates (1, 2), (1, 8), and (9, 5). There is a point (x, y) inside the triangle
with the following property: if line segments are drawn from each vertex to its opposite side and passing
through (x, y), the triangle is divided into six regions of equal area. What is the point (x, y)?
(3, 5)
Solution: The point (x, y) must be the centroid of the triangle, that is, (x, y) is the point of
intersection of the medians of the triangle. The centroid of the triangle with vertices at (a1 , b1 ), (a2 , b2 ), and
(a3 , b3 ) is located at ((a1 + a2 + a3 )/3, (b1 + b2 + b3 )/3), so the point required is (3, 5).
14. Little Dwayne has 100 cards where the integers from 1 through 100 are written. He also has an unlimited
supply of cards with the signs + and =. How many true equalities can he make, if he uses each card no more
than once?
33
The shortest equality under the stated conditions must involve 3 numbers, and hence a maximum of
32
Spring 2003
Individual Contest
33 equalities can be achieved. The 33 equalities below show that this maximum can be achieved.
1 + 75 = 76
3 + 74 = 77
5 + 73 = 78
7 + 72 = 79
9 + 71 = 80
11 + 70 = 81
13 + 69 = 82
15 + 68 = 83
17 + 67 = 84
19 + 66 = 85
21 + 65 = 86
23 + 64 = 87
25 + 63 = 88
27 + 62 = 89
29 + 61 = 90
31 + 60 = 91
33 + 59 = 92
35 + 58 = 93
37 + 57 = 94
39 + 56 = 95
41 + 55 = 96
43 + 54 = 97
45 + 53 = 98
47 + 52 = 99
49 + 51 = 100
24 + 26 = 50
20 + 28 = 48
16 + 30 = 46
12 + 32 = 44
8 + 34 = 42
2 + 38 = 40
4 + 6 = 10
14 + 22 = 36
15. In figure 10 DEC has area 4, AEB has area 10 and AB||CD. Find the area of the trapezoid ABCD.
D
14 + 4 10
Solution: By considering ADB and ACB we deduce that AED and BEC have the
same area, call it x. The area of the trapezoid is 14 + 2x. By considering similar triangles
x
10
=
= x = 2 10,
4
x
whence the area of the trapezoid is 14 + 4 10.

16. Find x if
13 + 3 = 2x +
2
1
x+
x+
x+
33
1
.
x + ..
Spring 2003
Individual Contest
Solution: We have
13 + 3
=x+
2
13 + 3
2
=x+
2
13 + 3
1
=
x+
x+
x+
1
.
x + ..
13 + 3
2
x=
2
13 + 3
6( 13 + 3)
x=
2( 13 + 3)
x = 3.
=
=
=
17. How many paths consisting of a sequence of horizontal and/or vertical line segments, each segment
connecting a pair of adjacent letters in figure 11 spell BIPOLAR?
B
B I
I P
B
B I
I P
P O
O L
B
I
P
O
L
A
B
I B
P I
O P
L O
A L
R A
B
I
P
O
L
B
I B
P I
O P
B
I B

127
Solution: Split the diagram, as in figure 12. Since every required path must use the R, we count
paths starting from R and reaching up to a B. Since there are six more rows that we can travel to, and since at
each stage we can go either up or left, we have 26 = 64 paths. The other half of the figure will provide 64
more paths. Since the middle column is shared by both halves, we have a total of 64 + 64 1 = 127 paths.
B
I
B
B I
I P
P O
B
I
P
O
L
B
B I
I P
P O
O L
L A
A R
34
Spring 2003
Individual Contest
18. A survey shows that 90% of high-schoolers in Philadelphia like at least one of the following activities: going
to the movies, playing sports, or reading. It is known that 45% like the movies, 48% like sports, and 35% like
reading. Also, it is known that 12% like both the movies and reading, 20% like only the movies, and 15%
only reading. What percent of high-schoolers like all three activities?
5%
Solution: We make the Venn diagram in as in figure 13. From it we gather the following system of
equations
x
x
x
x
x
+
+
+
+
y +
+
y
y
y +
+ 20
z
z +
+
t
t
+ u
z +
+ u
15
15
+ 20
=
=
=
=
=
45
48
35
12
90
The solution of this system is seen to be x = 5, y = 7, z = 13, t = 8, u = 22. Thus the percent wanted is 5%.
Sports
u
z
t
x
Movies
20
15
Reading

19. The numbers
1, 2, 3, , 2003
are written on a blackboard, in increasing order. Then the first, the fourth, the seventh, etc. are erased,
leaving the numbers
2, 3, 5, 6, 8, 9, 11, 12, 14,
on the board. This process is repeated, leaving the numbers
3, 5, 8, 9, .
The process continues until one number remains on the board and is finally erased. What is the last number
to be erased?
1598
Let Jn be the first number remaining after n erasures, so J0 = 1, J1 = 2, J3 = 3, J4 = 5, etc. We prove by
induction that
Jn+1 =
3
Jn if Jn is even,
2
and
3
(Jn + 1) 1 if Jn is odd.
2
Assume first that Jn = 2N. Consider the number 3N. There are initially N smaller numbers 1 mod 3. So
after the first erasure, it will lie in 2N-th place. Hence, it will lie in first place after n + 1 erasures. Assume
now that Jn = 2N + 1. Consider 3N + 2. There are initially N + 1 smaller numbers 1 mod 3. So after the
first erasure, it will lie in 2N + 1-st place. Hence, it will lie in first place after n + 1 erasures. That completes
the induction. We may now calculate successively the members of the sequence: 1, 2, 3, 5, 8, 12, 18, 27, 41, 62,
93, 140, 210, 315, 473, 710, 1065, 1598, 2397. Hence 1598 is the last surviving number from 1, 2, ..., 2003.
Jn+1 =
35
Spring 2003
Individual Contest
20. A is a set of one hundred distinct natural numbers such that any triplet a, b, c of A (repetitions are allowed in
a triplet) gives a non-obtuse triangle whose sides measure a, b, and c. Let S(A) be the sum of the perimeters
obtained by adding all the triplets in A. Find the smallest value of S(A). Note: we count repetitions in the
sum S(A), thus all permutations of a triplet (a, b, c) appear in S(A).
868500000
Solution: Let m be the largest member of the set and let n be its smallest member. Then
m n + 99 since there are 100 members in the set. If the triangle with sides n, n, m is non-obtuse then
m2 2n2 from where
(n + 99)2 2n2 n2 198n 992 0 n 99(1 + 2) n 240.
If n < 240 the stated condition is not met since m2 (n + 99)2 2n2 and the triangle with sides of length
n, n, m is not obtuse. Thus the set
A = {240, 241, 242, . . . , 339}
achieves the required minimum. There are 1003 = 1000000 triangles that can be formed with length in A and
so 3000000 sides to be added. Of these 3000000/100 = 30000 are 240, 30000 are 241, etc. Thus the value
required is
100(240 + 339)
30000(240 + 241 + + 339) = (30000)
= 868500000.
2
36
Spring 2003
Group Contest
1. (3 minutes) What is the difference between the sum of all odd positive integers up to 2003 and the sum of all
the even positive integers up to 2003?
We need
1002
(1 + 3 + + 2003) (2 + 4 + + 2002) = (1 2) + (3 4) + + (2001 2002) + 2003 = 1001 + 2003 = 1002.

2. (4 minutes) Evaluate the sum
2
3
1 2 4 + 2 4 8 + 3 6 12 + + 100 200 400

1 3 9 + 2 6 18 + 3 9 27 + + 100 300 900
1/3
Solution: Let
and
Then
a = 1 2 4 + 2 4 8 + 3 6 12 + + 100 200 400 = (13 + 23 + + 1003 )(1 2 4)

b = 1 3 9 + 2 6 18 + 3 9 27 + + 100 300 900 = (13 + 23 + + 1003 )(1 3 9).
a 1/3
b

=
(13 + 23 + + 1003 )(1 2 4)

13 + 23 + + 1003 )(1 3 9)
1/3
8
27
1/3
=
2
.
3
3. (6 minutes) Figure 14 is symmetric and is made up of three squares each of side 1. What is the radius of the
smallest circle that encloses the figure?
C
B

5 17
16
Solution: Let r = AB = AB be the radius sought, as in figure 15. From ABC we deduce
r2 =
2
1
2
From AB C we deduce
+ (2 C A)2 .
r2 = 12 + (C A)2 .
13
Equating both equalities, we deduce C A =
and putting this back into either of the above equalities we
16
5 17
deduce r =
16
37
Spring 2003
Group Contest
4. (3 minutes) Factor 111111 (six 1s) into primes.

(3)(7)(11)(13)(37)
Solution:
111111
=
=
=
=
(111)(1001)
(3)(37)(11)(91)
(3)(37)(11)(7)(13)
(3)(7)(11)(13)(37)
5. (4 minutes) Let r 6= 2003 be a real number. When y = x2 + rx + 2003 and y = x2 2003x + r are graphed on
the xy-plane, they intersect at a point (x0 , y0 ). Find the value of x0 .
r 2003
r + 2003
Solution: We have
x20 + x0 r + 2003 = x20 2003x0 + r = x0 (r + 2003) = r 2003 = x0 =
r 2003
.
r + 2003
6. (4 minutes) A brick in the form of a regular polygon is removed from a wall. It is observed that if the brick
suffered a rotation of either 40 or 60 about its center it would fit back again into its original place. What is
the smallest number of sides of this polygon?
18 sides
Solution: Let n be the required minimum of sides. Each exterior angle of the figure measures
360
. This exterior angle must be small enough so that when we turn the figure either 40 or 60 , we obtain
n
360
the same shape. This will be accomplished when
divides the greatest common divisor of 40 and 60 ,
n
360
which is 20 . Since
= 20 has the solution n = 18 this is the sought minimum.
n
7. (3 minutes) An urn contains 28 blue marbles, 20 red marbles, 12 white marbles, 10 yellow marbles, and 8
magenta marbles. How many marbles must be drawn from the urn in order to assure that there will be 15
marbles of the same color?
59
Solution: If all the magenta, all the yellow, all the white, 14 of the red and 14 of the blue marbles are
drawn, then in among these 8 + 10 + 12 + 14 + 14 = 58 there are no 15 marbles of the same color. Thus we
need 59 marbles in order to insure that there will be 15 marbles of the same color.
8. (4 minutes) Little Dwayne is back! He writes consecutive positive integers starting with 1 on a triangle, each
time adding two numbers to a new row, as in figure 16.
..
.
..
.
10
..
.
5
11
..
.
2
6
12
..
.
1
3
7
13
..
.
4
8 9
14 15
..
..
.
.
16
..
.
..
.
..
.

If he stops when he writes the number 2003, in what row does he stop?
45-th
Solution: The last number of the n-th row is n2 . Since 442 = 1936 < 2003 < 2025 = 452 , 2003 lies in the 45-th
row.
38
Spring 2003
Group Contest
9. (6 minutes) ABC in figure 17 is equilateral, of side 2. Its vertex B is on the side AX of the square AXYZ of
side 4. If the triangle rotates clockwise in such a way that each of its sides successively touches a side of the
square traversing the whole square until the triangle returns to its original position (and original
orientation), what is the length of the path travelled by vertex C? Write your answer in the form a where a
is a rational number.
Z
Y
C
40
3
Solution: From figure 18 we need 8 iterations to turn ABC into CAB. Thus we need 24
iterations to return ABC to its original position and orientation. In 8 of the 24 iterations vertex C travels a
4
third of the circumference, that is
radians per iteration. In 8 of the 24 iterations C is the pivot and so
3
travels nothing. On the remaining 8 iterations vertex C travels one twelfth of the circle, that is, radians per
3

40
4
iteration. Thus vertex C has travelled 8
+8
=
radians.
3
3
3
A
B
C
A
A
C
B
B
A
C
39
Spring 2006
Individual Contest
The 2006 Colonial Math Contest, held on February 9, 2006, was a great success. Forty students on 9 teams
participated and although we were expecting 22 more students, we still had an exciting and challenging contest.
The Individual Contest in the morning was close. The winner was Ian Cassel from Lansdale Catholic HS. There
was a 4-way tie for 2nd place (just one point behind 1st place): Xiufen Chen from Lincoln HS, Steve Kroukowski,
David Martz and Charles Yocum all from Lansdale Catholic HS.
The Team Contest in the afternoon consisted of 12 problems. Each problem was scored and team scores were
posted before the next problem started, so teams could see the standings at all times. Lincoln HS jumped out to an
early lead on the first 5 problems, but a late surge by the Lansdale Catholic 1 team, won the event. Members of the
winning team were Ian Cassel, David Martz, Brianne Riviello and Jenna Wasylenko.
Members of the winning team and first and second place winners of the individual contest received copies of
MAPLE 10 software (provided by Maplesoft) as prizes. Contest problems and solutions will be posted on the Math
Dept website at http://faculty.ccp.edu/dept/math/index.html.
And now the thank yous: huge thanks to David Santos for preparing the problems and providing the solutions; to
Mark Saks and Eleonora Chertok for feedback on the problems; to Chris Lewis and the Admissions Office for
helping organize the contest and, of course, for lunch; to the AV staff for excellent support; to faculty Margaret
Hitczenko, Robert Smith, Mohamed Teymour, John Jernigan, Elena Koublanova, Wimayra Luy and David Santos
and to students Kyle Hofler and Joseph Heard for helping run the contest; a special thanks to Joanne Darken and
Maplesoft for the prizes; and to the HS faculty Jackie Burton (Lincoln HS), Roger Cazeau (Blair Christian
Academy), Jason File (MaST Community Charter),Casey Huckel (Franklintowne HS), Christine McCrane
(Lansdale Catholic HS) and Hung Phan (Paul Robeson HS) for bringing and encouraging your students and for
your help with the team contest. Kudos to all.
1 What is the greatest number of Mondays that can occur in 45 consecutive days?
2 A 4 4 4 wooden cube is painted red and then cut into sixty-four 1 1 1 cubes. How many of the 1 1 1
cubes have exactly two of their sides painted red?
3 What is
(2 + 4 + + 2006) (1 + 3 + + 2005),
that is, what is the difference between the sum of all the odd positive integers up to 2006 and the sum of all the
even positive integers up to (and including) 2006?
4 How many isosceles triangles are there with perimeter 11 and each side of integral length?
5 Consider the two infinite repeating decimals

a = 0.12345 = 0.1234512345 . . . ;
b = 0.98765 = 0.9876598765 . . . .
If the sum a + b is written as an improper fraction in lowest terms
40
p
, find p2 + q2 .
q
Spring 2006
Individual Contest
6 Two circles of radius 2 and centers O and P are tangent to one another as shown in figure 19. If AD and BD are
tangents, find the length of BD.
A
b
B
b
7 In figure 20 ABC, FDC, GEC are isosceles. Also AB = 3AC. The perimeter of ABC is 84. D is the
midpoint of BC; E is the midpoint of DC; F is the midpoint of AC and G is the midpoint of FC. Find the perimeter
of the shaded quadrilateral DEGF.
B
D
E
A
8 In ABC, E and F are on AB, with E between A and F and satisfying

AE : EF : FB = 1 : 2 : 3.
Points G and D are on CB with G between C and D and satisfying
CG : GD : DB = 4 : 3 : 2.
If FG intersects ED at H, find the ratio DH : HE.
9 Reduce to lowest terms:
116690151
.
427863887
10 The product of three consecutive even integers is 87 8, where the asterisks represent five missing digits.
What is the sum of the integers?
11 How many (unordered) triplets of real numbers {x, y, z} are there such that
x(x + y + z) = 26;
y(x + y + z) = 27;
41
z(x + y + z) = 28?
Spring 2006
Individual Contest
12 If a, b, c are positive integers such that a2 + b c = 100 and a + b2 c = 124, find a + b + c.
13 Let
p + q.
p
59
p
45
be the unique rational number in reduced form, such that
< <
and 0 < q < 200. Find the sum
q
80
q
61
14 A die consists of a cube which has a different color on each of 6 faces. How many distinguishably different
kinds of dice can be made?
15 How many integers in the set {100, 101, 102, . . . , 198, 199} of 100 consecutive integers are not the sum of four
consecutive integers?
16 Two squares are inscribed in a semicircle as in figure 21. If the area of the smaller square is a, what is the area
of the larger square?
42
Spring 2006
Individual Contest
17 Consider the sequence

1,
1, 2,
1, 2, 3,
1, 2, 3, 4,
1, 2, 3, 4, 5,
....
Find its 2005-th term. You may use the formula

1 + 2 + + n =
n(n + 1)
.
2
18 Consider 2005 hexagonal domino pieces with the numbers 1, 2, 3, 4, 5, 6 written on the edges in clockwise
fashion, as in figure 22. A chain is formed so that domino rules are observed. This means that two edges from
different pieces are joined only when the numbers on the edges agree. The numbers on the edges joining two
different domino pieces are now deleted and the remaining numbers are now added. What is the maximum value
of this sum?
6
5
2
3
19 Find the greatest common divisor of

a = 1| .{z
. . 1}
and
forty ones
b = 1| .{z
. . 1} .
twelve ones
20 The rectangle in figure 23 is decomposed into nine squares. If the shaded square has area 1, what is the area of
the rectangle?
c6
c8
c1
c5
c7
c4
Figure 23: Problem 20
43
c2
c3
Spring 2006
STUDENT:
SCHOOL:
(Optional) Age:
(Optional) Sex:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
44
Individual Contest
Spring 2006
Group Contest
21 3 minutes The positive non-squares are ordered in increasing order:

2,
3,
5,
6,
7,
8,
10, . . . .
Find the 100-th term.

22 4 minutes The odd positive integers are written in a triangular array as follows
.. ..
. .
13
..
.
7
15
..
.
3
9
17
..
.
1
5
11
19
..
.
where the n-th row has n numbers. What is the sum of the 20-th row?
23 4 minutes Find the largest integer k such that 5k divides
3 10! + 12 5! + 4 7!.
Here n! = 1 2 3 n.
24 6 minutes What is the sum of the positive integers less than ten million having exactly 77 divisors?
25 4 minutes An equilateral triangle and a regular hexagon have equal perimeters. If T is the area of the triangle
T
and H is the area of the hexagon, find the ratio .
H
26 3 minutes If
m3 + m3 + m3 + m3 + m3 = 25, find m.
27 5 minutes Find the sum

1
+
2
1 2
+
3 3
1 2 3
+ +
4 4 4
+ +
1
2
99
+
+ +
100 100
100
of all positive proper fractions (not necessarily in lowest terms) whose numerators are positive integers and whose
denominators are positive integers less than or equal to 100.
28 3 minutes How many zeros are there at the end of the product
156 285 557
29 4 minutes As a sales gimmick, a merchant began measuring with a yardstick that was x inches too long. If this
reduced his profit from 50% of his cost to 20% of his cost, find x.
30 5 minutes A seven-digit base ten number
N = 6666abc,
where a, b, and c represent digits, is a perfect square. Find a2 + b2 + c2 .
45
Spring 2006
Group Contest
31 3 minutes Fill in the numbers from 1 up to 5 in the five little circles, in such a way that in each triangle the
given number is the sum of the numbers on the vertices. Which number appears in the shaded circle?
10
12
32 4 minutes The mass of 200 kg cucumbers consists of 99% of water. The cucumbers are drying out due to the
sun, till the mass consists of 98% of water. Determine the weight of the cucumbers now.
33 4 minutes John, Bob and Tom are brothers of Jill and Ingrid. Ingrid is three years younger than Jill. The age of
Bob equals the average of the ages of John and Jill. John and Tom together, and also Bob and Ingrid together, are 1
year younger than twice the age of Jill. John and Ingrid together are 1 year older than twice the age of Jill. Who is
the eldest child?
34 5 minutes The lengths of the sides of a right triangle are respectively x y, x, and x + y, where 0 < y < x. Find
x
the ratio .
y
35 5 minutes A positive integer, written in decimal notation, has 4 digits. The first two digits are the same and the
last two digits are the same. The number is a perfect square. What is the sum of the digits of this number?
46
Community College of Philadelphia Colonial Maths Challenge Spring 2006 Solutions to the Individual and Group Contests
1 If one puts Monday at the very beginning of the period one obtains V
45
W = 7 Mondays.
7
2 For every edge, there are 2 such cubes. Since there are 12 edges, there required total is 12 2 = 24..
3 The desired quantity is
(2 1) + (4 3) + + (2006 2005) = 1003.
4 If the sides of the triangle measure a, a, b, then 2a + b = 11, from where b must be odd. The following triangle
inequalities must also be fulfilled
2a > b, a + b > a.
Thus one has the following possibilities:
b = 1, a = 5;
b = 3, a = 4;
b = 5, a = 3,
so there are three such triangles.

5 Observe that a + b = 1.1 =
10
p
= , hence p2 + q2 = 181
9
q
6 Let line AD be tangent to circle P at E. Then DE = DB and
AE2 = AP2 PE2 = 36 4 = 32 = AE = 4 2.

Also
AD2 = BD2 AB2 = (AE + BD)2 = 64 + BD2 = (4 2 + BD)2 = 64 + BD2 ,
which in turn
= 32 + 8BD 2 + BD2 = 64 + BD2 = BD = 2 2.
7 Let x = AC. Then x + 3x + 3x = 84 = x = 12. Hence FC =

this the desired perimeter is 39.
BC
AC
= 6 and FG = 3. Also DC =
= 18. From
2
2
8 Choose K on line FG such that DK||EF. As GDK GBF, it follows that

DK
DG
9y
=
= DK =
.
BF
BG
5
As HFE HKD, it follows that
DH
DK
DK
3 DK
3 3x
9
=
=
=
=
=
.
HE
EF
2y
2 3y
2 5x
10
The required ratio is thus
9
.
10
B
b
F
E
A
b
b
b
b
H K
D
b
G
b
C
9
116690151
3 38896717
3
=
=
.
427863887
11 38896717
11
10 Let the integers be n 2, n, n + 2. Forcedly, they must end in 2, 4, 6, since their product ends in 8. The product
of the integers is n3 4n n3 . Now n3 87000008 = n 443. Inspection gives n = 444, and the sum is
442 + 444 + 446 = 1332.
11 Adding the equations (x + y + z)2 = 91 = x + y + z = 9. Thus there are two triplets.
47
12 Subtracting both equations,

a a2 + b2 b = 24 = (b a)(b + a 1) = 24,
and one of the factors must be odd and the other even. One tries either 3 8 or 1 24. The first factorization is
rejected, since it gives b = 6, a = 3, c = 85. The second factorization yields a = 12, b = 13, c = 57 and so
a + b + c = 82.
13 Since we are given that this is the unique fraction, we dont have to prove uniqueness (which is a lot harder to
prove). Observe that if a, b, c, d are positive integers then
a
c
a
a+c
c
<
=
<
< .
b
d
b
b+d
d
Hence
59
104
45
<
<
,
80
141
61
and p + q = 245.
14 Choose a specific colour for the upper face of the cube, say A. Then we have five choices for colouring the lower
face of the cube, say with colour B. Rotate the cube so that some colour C is facing us. Now the remaining sides are
fixed with respect to these three. We can distribute the three colours in 3 2 1 = 6 ways, giving 5 6 = 30
possibilities.
15 If n 1, n, n + 1, n + 2 are four consecutive integers, then their sum is 4n + 2, that is, the integers sought do not
leave remainder 2 upon division by 4. Since three quarters of the integers leave some other remainder, the answer
3
is 100 = 75.
4
16 Rotate the figure as in figure 24. The larger square is then decomposed into 4 of the smaller square. The area
sought is 4a.
17 The n-th group has n members and sum
n(n + 1)
n(n + 1)
. Upon writing the last term of the n-th group,
2
2
n2
2005 = n 63, that is, we should look around the 63rd group in order to
2
63(64)
find the 2005-th term. Now after the 63rd group has been written,
= 2016 terms have been written. Hence
2
the 2005-th term is the 52 in the group
1, 2, 3, . . . , 62, 63.
terms have been written. Now
Originally this problem asked for the sum of the first 2005 terms, but it was then considered too difficult. Lets see
how to find this sum. We have to sum
1(1 + 1) 2(2 + 1) 3(3 + 1)
62(62 + 1)
+
+
+ +
+ (1 + 2 + + 52).
2
2
2
2
48
This is
62
X
k2 + k
2
!
+ 1378.
k=1
Since
(12 + 22 + + n2 ) =
n(n + 1)(2n + 1)
6
and
1 + 2 + n =
n(n + 1)
,
2
the desired sum is
1
1 2
1 62 63 125 1 62 63 52 53
+
+
= 43042.
1 + 22 + + 622 + (1 + 2 + + 62) + (1 + 2 + + 52) =
2
2
2
6
2
2
2
18 In order to obtain the maximal sum, we must have a chain where every piece is glued to the smallest possible of
sides using the smallest numbers. The chain then must have 2003 pieces which are joined on two sides and the two
end pieces are joined at only one side. If the 1 and the 2 were shared, the first and second piece must be joined at 1
and the second and third at 2. But this is impossible because then the first and the third piece would be joined at 3
and 6, violating domino rules. Hence only 3s and 1s are shared. The middle 2003 contribute 2 + 4 + 5 + 6 = 17
each, one of the end pieces contributes 2 + 3 + 4 + 5 + 6 = 20, and the other end piece contributes
1 + 2 + 4 + 5 + 6 = 18 hence the maximum sum is (2003)(17) + 1(20) + 1(18) = 34089.
19 Let an = 1| .{z
. . 1}. Then using the Euclidean Algorithm,
n ones
gcd(a40 , a12 ) =
=
=
=
=
=
=
=
=
=
gcd(a40 a12 , a12 )

gcd(a28 1012 , a12 )
gcd(a28 , a12 )
gcd(a28 a12 , a12 )
gcd(a16 1012 , a12 )
gcd(a16 , a12 )
gcd(a16 a12 , a12 )
gcd(a4 1012 , a12 )
gcd(a4 , a12 )
a4 ,
whence the gcd sought is 1111.

20 Suppose c1 has side c. Then we see that c2 has side c + 1, c3 has side c + 2, and c4 has side c + 3. From this it
follows that c5 has side 4. Continuing this process, we deduce that c6 has side 2c + 1. The height of the initial
rectangle is thus
(2c + 1) + (c + 1) + (c + 2) = 4c + 4.
Square c7 has side
c + 3 + 4 = c + 7.
The width of the original rectangle is thus
(c + 7) + (c + 3) + (c + 2) = 3c + 12.
Square c8 has side of length
c + 7 + 4 = c + 11.
Finally, two opposite sides of the rectangle have dimensions 4c + 4 and
(c + 7) + (c + 11) = 2c + 18.
Since they must be equal we deduce 4c + 4 = 2c + 18 = c = 7. Conclusion: the original rectangle has area
33 32 = 1056.
21 Since there are 10 squares between 1 and 110 inclusive 110 is the 100-th term.
49
22 Upon writing the last element of the (n 1)th row,

1 + 2 + + n 1 =
(n 1)n
2
numbers have been written. Thus the first element of the n-th row is the
2(
(n 1)n
+ 1-st odd integer, namely
2
(n 1)n
+ 1) 1 = n2 n + 1. This means that the sum of 20-th row is
2
381 + 383 + + + 419 =
23 We have
3 10! + 12 5! + 4 7!
=
=
=
=
20(381 + 419)
= 8000.
2
6!(3 7 8 9 10 + 2 + 4 7)
6!(3 7 8 9 10 + 30)
30 6!(7 8 9 + 1)
30 6! 505,
whence k = 3.
24 First observe that 210 103 , and so 230 106 , which implies that 234 107 . This means that any exponent
present of any prime must be 34. If n has the prime factorization
ak
1
n = pa
1 pk
with primes p1 < p2 < < pk and ai 1, then the number of divisors of n is
(a1 + 1) (ak + 1).
If
(a1 + 1) (ak + 1) = 77,
then one must have k 2, and then either a1 = 6, a2 = 10, and one number is 26 310 = 3779136, or a1 = 10, a2 = 6,
and another number is 210 36 = 746496. Observe that 210 56 = 16000000 > 107 . The desired sum is thus
3779136 + 746496 = 4525632.
25 Let P = 6a = 3b, where a is the length of a side of the hexagon
This
andb is the length of a side of the triangle.
3
3
entails that 2a = b and 4a2 = b2 . The area of the hexagon is 6
a2 and the area of the triangle is
b2 . Thus
4
4
6
26 We have
3 2
b
4
3 2
4 a
b2
4a2
2
=
= .
2
2
6a
6a
3
5m3 = 25 = 5m3 = 54 = m3 = 53 = m = 5.
27 Recall from Gau trick that

1 + 2 + + n =
Hence the k-th sum, k 2, in parentheses is
Thus the desired sum is
28 The product equals
1 2
k1
+ + +
k k
k
n(n + 1)
.
2
(k 1)k
k1
=
.
2k
2
1
(1 + 2 + + 99) = 2475.
2
36 56 75 210 57 117 = 36 53 75 1010 117 ,
which ends in 10 zeros.

50
29 One has
1.5 = 1.2(1 +
x
) = x = 9.
36
30 We have 6666000 2581.9 > 2581. Now 25832 = 6671889 and 25822 = 6666724, and so the only answer is
6666724. Thus 72 + 22 + 42 = 69.
31 The figure below shows the correct arrangement.
3
6
2
10
4
12
5
32 Originally, we have 198 kg of water, and 2 kg of cucumber mass. After evaporation, the 2 kg of cucumber mass
make 2% of the current weight w and thus .02w = 2 = w = 100 kg.
33 Let j, b, t, l, i be the ages of John, Bob, Tom, Jill, and Ingrid, respectively. We are given
i + 3 = l;
b=
j+l
;
2
j + t = b + i = 2l 1;
j + i = 2l + 1.
We will express all other ages in terms of one, let us say, Jills age. We have i = l 3, so Ingrid is younger than Jill.
Also, b = 2l 1 i = l + 2, so Bob is older than Jill. Continuing j = 2l + 1 i = l + 4, so John is older than Bob.
Now t = 2l 1 j = l 5, so Tom is younger than Ingrid. Therefore, in increasing order of age we have Tom,
Ingrid, Jill, Bob, and John.
34 We have that
whence
(x y)2 + x2 = (x + y)2 = x2 = 4xy = x = 4y,
x
= 4.
y
35 The number is of the form aabb, where a 6= 0 and b are digits. Now, aabb = 11(a0b) clearly divisible by 11.
But since the number is a square, the number a0b must be divisible by 11. This means that a 0 + b must be
divisible by 11, and so the number could be either 4477 or 7744. The sum of the digits is thus 22.
51

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Community College of Philadelphia

Colonial Maths Challenge

2. If 333 + 333 + 333 = 3x , find x.

6. If (x 1)2 + (y 2)2 + (z 3)2 + (w 4)2 = 0, what is x + y + z + w?

Community College of Philadelphia

Colonial Maths Challenge

12. If (8x 5)5 = Ax5 + Bx4 + Cx3 + Dx2 + Ex + F, then A + B + C + D + E + F =

is an integer, find it.

Community College of Philadelphia

Colonial Maths Challenge

Individual Contest Solutions

1. Proceeding from the innermost fraction one easily sees that

2. 333 + 333 + 333 = 3 333 = 334 , whence x = 34.

8. Observe that x2 + x = 1. Hence x4 + x3 = x2 and x3 + x2 = x. Thus

Community College of Philadelphia

Colonial Maths Challenge

Individual Contest Solutions

obtain the angle sum

Hence, angle between the hands is 240 140 = 100 .

16. Observe that ( x + 1 x)( x + 1 + x) = 1 and so

( 2 1) + ( 4 3) + + ( 1024 1021) = 1024 1 = 32 1 = 31.

Community College of Philadelphia

Colonial Maths Challenge

Individual Contest Solutions

17. Adding the equations 3(x + y + u + v) = 9, and so x + y + u + v = 3. Thus

which is to say x2 3x 18 = (x 3)(x + 6) = 0. Since x has to be a positive number, x = 3.

Community College of Philadelphia

Colonial Maths Challenge

Community College of Philadelphia

Colonial Maths Challenge

of the circumference, each of the angular sectors has area

Community College of Philadelphia

Colonial Maths Challenge

1. N is a prime number and 90 < N < 99. What is 2N?

Community College of Philadelphia

Colonial Maths Challenge

8. Suppose the product

13. Find the smallest positive integer n such that

Community College of Philadelphia

Colonial Maths Challenge

Community College of Philadelphia

Colonial Maths Challenge

(3) (2) = (3)2 22 + (3)(2) = 9 4 6 = 1.

If she entered at floor F, then F + 4 6 + 8 10 = 23, whence F = 27.

Individual Contest Solutions

birds in the lower branch. From the given data,

and so Yogi and the grandmothers won.

the interior angles of a hexagon is 180 (6 2) = 720 . Thus

We use the fact that 1 x2 = (1 x)(1 + x). First

Community College of Philadelphia

Colonial Maths Challenge

Individual Contest Solutions

The product equals

Construct the diagram

The integer must be of the form n = 2a 3b 5c , with a, b, c positive integers. The

Community College of Philadelphia

Colonial Maths Challenge

Individual Contest Solutions

tangent to the circle, OB bisects DOF. Hence