The 2013 Colonial Mathematics Challenge was held on 6 May 2013.

Eighty-five students from fifteen schools

participated. Helping to run the contest were: professors Reid Huntsinger, Dan Jacobson, Elena Koublanova,
Clark Loveridge, Wimayra Luy, Eric Neumann, Deivy Petrescu, Geoff Schulz, Sanda Shwe, Mohammed
Teymour and Brent Webber; staff members Deb Coley and Ruth Al-Ameen; students Ryan Ivey, Khylil
Robinson and Reynaldo Noel Jeune; and former student Gianni Ferrara.
The individual contest was won by Michael Leggerie from Girard Academic Music Program (GAMP), with 15
problems out of 20 correct. In second place was Alexander Palmer also from GAMP, with 11 out of 20 correct.
And in third place was Marinos Rrapaj also from GAMP, with 10 out of 20 correct
First Place

Second Place

Third Place

Michael Leggerie

Alexander Palmer

Marinos Rrapaj

The team contest was won by the team from Girard Academic Music Program with team members Michael
Leggerie, Matthew Mutter, Alexander Palmer, Marinos Rrapaj and Harnail Sembhi.

l to r: Teja Smith (teacher), Marinos Rrapaj, Alexander Palmer, Harnail Sembhi, Michael Leggerie, Matthew Mutter

2013 Colonial Mathematics Challenge

Mathematics Department

6 May 2013

Community College of Philadelphia

Name:

School:

Sex (Optional):

Age (Optional):

1:

11:

2:

12:

3:

13:

4:

14:

5:

15:

6:

16:

7:

17:

8:

18:

9:

19:

10:

20:

Community College of Philadelphia

Colonial Math Challenge

Spring 2013

Individual Contest

Instructions: This is a closed-book individual examination. Calculators, rulers, compasses, and square-grid paper are allowed. Cell
phones and other communication devices must be turned off during the examination. Contestants consulting with other
contestants, teachers or any one else during the examination will be disqualified.
Time: 90 minutes

Point P is outside circle C in the plane. At most how many points on C are a distance 3 units from P?

If log ( log 4( log 5x ) ) = 0 , then find the value of x 1 2 .

Judys birthday is May 6. In 2013, this falls on a Monday. In what year will her birthday next fall on a Monday?

Find the number of digits in 4 5

Line segment AB is both a diameter of a circle of radius 2 and a side of an equilateral triangle ABC. The circle also intersects
AC and BC at points D and E, respectively. Find the exact length of AE.

If the sequence { a n } is defined by

16 25

, written in the usual base 10 form.

a1 = 2
a n = a n 1 + 2n

for n > 1

then find a 100 .

b
c
a+b
If --- = 2 and --- = 3 , what is the ratio ------------ ?
a
b
b+c

Community College of Philadelphia

Spring 2013

Individual Contest

The odd positive integers, 1, 3, 5, 7, . . . , are arranged in five columns (A, B, C, D, E) continuing with the pattern shown below. In
which column (A, B, C, D, or E) will the number 2013 occur?
A
15
31
47
.
.

Colonial Math Challenge

B
1
13
17
29
33
45
49
.
.
.
.

C
3
11
19
27
35
43
51
.
.
.
.

D
5
9
21
25
37
41
53
.
.
.
.

E
7
23
39
55
.
.

A cryptographer devises the following method for encoding positive integers. First the integer is expressed in base 5. Second, a
one-to-one correspondence is established between the sets {0, 1, 2, 3, 4} and {V, W, X, Y, Z}. Using this correspondence, the
cryptographer finds that three consecutive positive integers in increasing order are coded as VYZ, VYX, VVW, respectively. What is
the base-10 expression for the integer coded as XYZ?

10

Let x = .123456789101112. . . 998999 , where the digits are obtained by writing the integers 1 through 999 in order. Find the
2013th digit to the right of the decimal point.

11

There are two natural ways to inscribe a square in a given isosceles right triangle. If it is done as in Figure 1 below, then the area of
the square is 441 cm2. What is the area (in cm2) of the square inscribed in the same triangle as shown in Figure 2 below?

Figure 1

12

Find the product of all real roots of the equation x

Figure 2

log x

= 10 .

( log x = log10 x )

Community College of Philadelphia

Colonial Math Challenge

Spring 2013

Individual Contest

13

How many distinguishable rearrangements of the letters in CONTEST have both vowels first?

14

In ABC, AB = 8, BC = 7 and CA = 6 . If side BC is extended to a point D so that DAB is similar to DCA , find the
length of CD.
D
C
7

6
A

15

Find the exact sum of all real solutions of the equation x 2x + 1 = 3 .

16

a
The first four terms of an arithmetic sequence are a, x, b, 2x. Find the ratio --- .
b

17

A wooden cube with edge length n units (where n is an integer > 2) is painted black on all sides. By slices parallel to its faces,
the cube is cut into n 3 smaller cubes of unit edge length. If the number of smaller cubes with just one face painted black is
equal to the number of smaller cubes with no paint, what is n?

18

The 120 permutations of CRAZY are listed in alphabetical order as if each were an ordinary 5-letter word. What is the last letter in
the 94th word in this list?

19

A drawer in a darkened room contains 100 red socks, 80 green socks, 60 blue socks and 40 yellow socks. John selects socks
one at a time from the drawer but is unable to see the colors of the socks drawn. What is the smallest number of socks that must
be selected to guarantee that the selection contains at least 10 pairs? (A pair of socks is two socks of the same color. No sock
may be counted in more than one pair.)

20

A ball was floating in a lake when the lake froze. The ball was removed, without breaking the ice, leaving a hole 24 cm across
at the top and 8 cm deep. What is the radius of the ball in centimeters?

Community College of Philadelphia

Colonial Math Challenge

Team:

Spring 2013

Team Contest

School:

21 (3 minutes)
Answer:
In the hexagon ABCDEF, sides AF and CD are parallel, as are sides AB and FE, and sides BC and ED. Each side has length 1. Also,
AFE = CDE = 60 . Find the exact area of the hexagon.
F

D
60

60

A
B

Community College of Philadelphia

Colonial Math Challenge

Team:

Spring 2013

School:

22 (4 minutes)
Answer:
Find the units digit of 3

1001 1002

13

1003

Team Contest

Community College of Philadelphia

Team:

Colonial Math Challenge

Spring 2013

School:

23 (5 minutes)
Answer:
2

1000
Find the value of ----------------------------- .
2
2
252 248

Team Contest

Community College of Philadelphia

Colonial Math Challenge

Team:

Spring 2013

School:

24 (4 minutes)
Answer:
A rectangle intersects a circle as shown. AF = 4, FE = 9 and BC = 6 . Find CD.

Team Contest

Community College of Philadelphia

Colonial Math Challenge

Team:

Spring 2013

Team Contest

School:

25 (4 minutes)
Answer:
Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1. The distance from the top of the
block on the floor to the top of the block on the table is 32 inches. After rearranging the blocks as in Figure 2, the distance from
the top of the block on the floor to the top of the block on the table is 20 inches. How high is the table in inches?

32

Figure 1

20

Figure 2

Community College of Philadelphia

Colonial Math Challenge

Team:

Spring 2013

Team Contest

School:

26 (3 minutes)
Answer:
Let p, q and r be distinct prime numbers and let a, b and c be positive integers. If p a q b r c is the smallest positive perfect cube
having pq 2 r 4 as a divisor, then find a + b + c .

Community College of Philadelphia

Colonial Math Challenge

Team:

Spring 2013

Team Contest

School:

27 (4 minutes)
Answer:
Triangle ABC is a right triangle with legs AB = 12 and BC = 5 . Two arcs of circles are drawn, one with center A and radius
12, the other with center C and radius 5. The arcs intersect the hypotenuse, AC, at points D and E. Find the length DE.
D

E
5

12

Community College of Philadelphia

Colonial Math Challenge

Team:

Spring 2013

School:

28 (6 minutes)
Answer:
1
1
1
1
Compute the product 1 ----- 1 ----- . . . 1 ----- 1 -------- .

22
32
92
10 2

Team Contest

Community College of Philadelphia

Colonial Math Challenge

Team:

Spring 2013

Team Contest

School:

29 (3 minutes)
Answer:
In the figure, ABC has a right angle at C and A = 20 . If BD is the bisector of ABC , then find the measure of BDC in
degrees.
B

20
D

Community College of Philadelphia

Team:

Colonial Math Challenge

Spring 2013

Team Contest

School:

31 (4 minutes)
Answer:
A lattice point is a point in the plane with integer coordinates. How many lattice points are on the line segment whose
endpoints are (3, 11) and (42, 275)? (Include both endpoints of the segment.)

Community College of Philadelphia

Team:

Colonial Math Challenge

Spring 2013

Team Contest

School:

32 (5 minutes)
Answer:
In ABC, AB = 9, BC = 10 and CA = 8 . Point D is the midpoint of side AB and E is the foot of the altitude from A to BC.
Find the length of DE.

Community College of Philadelphia

Team:

Colonial Math Challenge

Spring 2013

Team Contest

School:

34 (5 minutes)
Answer:
An 111111 wooden cube, formed by gluing together 113 unit cubes, is sitting on a table. You can take a photograph of the cube
from any point. What is the greatest number of unit cubes that can be seen in the photograph?

Community College of Philadelphia

Colonial Math Challenge

Spring 2013

Solutions

The points in question are the points of intersection of the original circle C with the circle of radius 3 centered at P. Two
circles with different centers intersect in at most two points.

1
-----25

2019

Note that 365 = 52 7 + 1 , so Judys birthday will advance one day in non-leap years and two days in leap years.
Since 2016 is a leap year, her birthday will fall on the following days:
2014 Tuesday
2015 Wednesday
2016 Friday
2017 Saturday
2018 Sunday
2019 Monday

28

4 5

2 3

4 1 2

log ( log 4( log 5x ) ) = 0 log 4( log 5x ) = 1 log 5x = 4 x = 5 x 1 2 = ( 5 )

7

16 25

32 25

= 2 5

= 2 10

25

= 128 10

25

= 5

1
= -----25

. So the 3 digits in 128 are followed by 25 zeros for a total of 28 digits.

ABE is inscribed in a semicircle, so AEB = 90 and AB = 4 since AB is a

3
diameter. Then AE is an altitude of equilateral ABC , so AE = 4 ------- = 2 3 .
2

10100 From the definition of the sequence a n a n 1 = 2n

( n > 1 ) . Then

a1 = 2 1
a2 a1 = 2 2
a3 a2 = 2 3
.
.
.
a 100 a 99 = 2 100
100 101
Adding these equations, a 100 = 2 ( 1 + 2 + 3 + . . . + 100 ) = 2 ---------------------- = 10100 .

3
--8

b
c
a+b
a + 2a
3
--- = 2 b = 2a . --- = 3 c = 3b = 6a . Then ------------ = ------------------- = --- .
a
b
b+c
2a + 6a
8

Community College of Philadelphia

Colonial Math Challenge

B The rows in pairs of two have the form

A
B
C
D
E
16k+1
16k+3
16k+5
16k+7
16k+15
16k+13
16k+11
16k+9
( k = 0, 1, 2, . . . ) . Since 2013 = 16 ( 125 ) + 13 , 2013 is in column B.

108

Spring 2013

Solutions

In base 5, VYZ + 1 = VYX , so X = Z + 1 . Also, VYX + 1 = VVW . This implies that X = 4 , W = 0 , Z = 3 and
V = Y + 1 . So Y = 1 and V = 2 . Then XYZ = 413 5 = 108 10 .

.1234567891011 . . . 99100101 . . .

10

Consider the groupings above. Group A (the single digit numbers) has 9 digits. Group B (the 2-digit numbers) has 90 2 = 180
digits. Together they total 189 digits. To get to digit 2013 we need 2013 189 = 1824 more digits or 1824 3 = 608 more
three digit numbers. The 3-digit numbers start with 100 so the 608th 3-digit number is 707 and the 2013th digit is 7.
11

392 In the first triangle, the side of the square is 21 cm, so the side of the large triangle is 42 cm. In the second figure, let the
side of the square be x. The small triangle with hypotenuse x has side x 2 and the triangle with legs x has hypotenuse
x 2 . Then one leg of the large triangle is 42 = x 2 + 2x = 3x 2 . Then x = 14 2 and x 2 = 196 ( 2 ) = 392 .

21

x
21

12

Given x

log ( x

13

120

log x

log x

21

x
------2

x
x 2

= 10 , take the common log (base 10) of both sides:

2
1
1
) = log 10 ( log x ) = 1 log x = 1 x = 10 or x = ------ . Then the product of the roots is 10 ------ = 1 .
10
10

There are 2 ways to order the vowels. If the consonants were all distinct there would be 5! = 120 orderings of the

consonants. But since there are two Ts which are indistinguishable, there are 120 2 = 60 orderings of the consonants. Thus
there are 2 ( 60 ) = 120 rearrangements of CONTEST with the vowels first.

14

CD
AC
AD
x
6
AD
9 By similar triangles, -------- = -------- = -------- . Let CD = x . Then -------- = --- = ------------ . From the first part of this equation,
AD
AB
BD
AD
8
x+7
4x
3(x + 7)
AD = ------ , and from the last part, AD = -------------------- . Equating the two expressions for AD, and solving yields x = CD = 9 .
3
4

Community College of Philadelphia

15

2
--3

x 2x + 1 = 3

Case 1

Case 2

or

Colonial Math Challenge

Spring 2013

Solutions

x 2x + 1 = 3

1
2x + 1 0 x --- . Then x ( 2x + 1 ) = 3

or

x ( 2x + 1 ) = 3 .

x1 = 3

or

x 1 = 3

x = 4

or

x = 2

1
2x + 1 < 0 x < --- . Then x + ( 2x + 1 ) = 3

or

x + ( 2x + 1 ) = 3 .

2
x = --3

or

4
x = --3

1
But 4 < --- , so 4 is not a solution.
2

2
1
2
But --- > --- , so --- is not a solution.
3
2
3

2
4
So there are two real solutions and their sum is 2 + --- = --- .
3
3

16

1
--- Let k be the difference between successive terms. Then the difference between the 4th and 2nd terms is 2k.
3
--xx
x
x
x
3x
a
2
1
2k = 2x x and k = --- . Then a = x --- = --- and b = x + --- = ------ . So --- = ------ = --- .
2
2
2
2
2
b
3x
3
-----2

17

8 The large cube has 6 faces containing at least one painted face on each smaller cube. The unpainted smaller cubes come from
removing the 6 faces of the large cube, so there are ( n 2 ) 3 unpainted smaller cubes. The smaller cubes with one painted face
come from the 6 faces of the larger cube without the edges, so there are 6 ( n 2 ) 2 smaller cubes with one painted face. If
( n 2 ) 3 = 6 ( n 2 ) 2 , then n 2 = 6 and n = 8 .

18

A There are 4! = 24 permutations starting with each of the 5 letters - A, B, C, D, E. Listed in alphabetical order, the 96th
permutation is the last one starting with Y, namely YZRCA. Then the 95th is YZRAC and the 94th is YZCRA. So the last letter
in the 94th word is A.

19

23 For any selection, at most 4 socks (one sock of each color) will be left unpaired. This occurs when an odd number of socks
of each color is selected. So if 24 socks are selected, at least 20 socks are paired guaranteeing at least 10 pairs. But since 23 is not
the sum of four odd numbers, if 23 socks are selected there are at most 3 colors with an unpaired sock, so again, there are at least
10 pairs. Since 22 = 5 + 5 + 5 + 7 , there could be 4 unpaired socks and only 9 pairs. So 23 is the smallest number of socks that
must be selected to guarantee at least 10 pairs.

20

13 The figure shows a cross-section of the ball still in the ice. In the right
triangle, r 2 = ( r 8 ) 2 + 12 2 . Solving for r, r = 13 .

r-8
8

r
12

Community College of Philadelphia

21

Solutions

3.

9 Any power of either 7 13 = 91 or 3 = 81 has a units digit of 1. So the units digit of

3

1001 1002

13

1003

4 250

= (3 )

( 7 13 )

23

Spring 2013

3 Draw segments AE, BE and CE. There are now 4 equilateral triangles with side length 1. So the total area is
1-
3-
=
4 -( 1 ) -----2
2

22

Colonial Math Challenge

1002

3 13 is the same as the units digit of 3 13 = 39 which is 9.

1000
1000 1000
1000
500 ----------------------------- = ---------------------------------------------------------- = ------------ ------------ = 2 250 = 500
2
2
(
252
+
248
)
(
252

248
)
500
4
252 248

24 5 Let G and H be points on segment FE so that CDHG is a

rectangle. AG = BC = 6 so FG = 2 . Similarly EH = 2 .
Then CD = GH = 9 2 2 = 5 .

25

26 Let the blocks be ( a inches ) ( b inches ) where a < b , and let the height of the table be h inches. From the first arrangement
h a + b = 32 and from the second h + a b = 20 . Adding these two equations, 2h = 52 and h = 26 .

26

12 If p a q b r c is the smallest positive perfect cube having pq 2 r 4 as a divisor, then a, b and c are all multiples of 3, and
a 1, b 2 and c 4 . The smallest such are a = 3, b = 3 and c = 6 , so a + b + c = 3 + 3 + 6 = 12 .

27

4 By the Pythagorean theorem, AC =

12 2 + 5 2 = 13 . AD = 12 so DC = 1 . CE = 5 so AE = 8 . Then

DE = AC DC AE = 13 1 8 = 4 .

28

11
-----20

1
1
1
1
n1 n+1
1
1
1
- 1 + --- = ------------ ------------ . So 1 ----- 1 ----- . . . 1 ----- 1 --------
1 ----2- = 1 -n
n
n n

n
2 2
3 2
9 2
10 2
1 11
11
1 3 2 4 3 5 4 6 5 7 6 8 7 9 8 10 9 11
= --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ------ ------ ------ = --- ------ = -----2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10
2 10
20

29

1
55 ABC = 90 20 = 70 . DBC = --- ABC = 35 . BDC = 90 35 = 55 .
2

31

275 11
88
4 The slope of the line segment is --------------------- = ------ . Since 88 and 15 are relatively prime, the next lattice point after ( 3, 11 ) is
42 + 3
15
( 3 + 15, 11 + 88) = ( 12, 99 ) . Continuing in this way we get lattice points ( 27, 187 ) and ( 42, 275 ) . So there are 4 lattice
points on the segment.

Community College of Philadelphia

32

Colonial Math Challenge

Spring 2013

9
--- DE is the median to the hypotenuse of right ABE , so it is half
2

Solutions

1
9
as long as the hypotenuse. DE = --- AB = --- .
2
2

34

331 At most three of the cubes six faces can be seen at one time. There is one unit cube that is on all three faces. The three edges
each contain 10 unit cubes not counting the one that is on all three faces. And the three faces each contain 102 unit cubes not
counting the ones on the three edges. So the greatest number of unit cubes that can be seen from a single point is
1 + 3 ( 10 ) + 3 ( 10 2 ) = 331 .
or
The cubes that cannot be seen form a cube that is 101010 unit cubes. So the number of cubes that can be seen is
3

11 10 = 331 .

Note: Problems numbered 30 and 33 were not used in the team contest.