Team Level 5 | 07/10/2010(1) Rectangle ABCD is inscribed in triangle EFG such that side AD of the

rectangle is on side EG of the triangle, as shown. The triangle's altitude from F to side

EG is 7 inches, and EG = 10 inches. The length of segment AB is equal to half the length

of segment AD. What is the area of rectangle ABCD? Express your answer as a common

fraction.

A

B C

DE

F

G

(2) What is the sum of the 2009 fractions of the form 2n(n+2)

if the values of n

are the positive integers from 1 through 2009? Express your answer as a decimal to the

nearest thousandth.

(3) In the game Tic-Tac-Toe, the �rst player writes an \X" in one of the nine

locations of the grid, and then the second player writes an \O" in one of the remaining

eight places. To analyze the game, Melissa decided to study all of the possible

\�rst-two-moves" combinations. She realized that she should consider symmetric patterns

like the three to the right as the same \�rst-two-moves" combination, and therefore only

count it once. And she decided that she should count patterns like the two to the left as

two di�erent \�rst-two-moves" combinations. Using Melissa's counting technique, how

many di�erent \�rst-two-moves" combinations are there?

(4) The fourteen digits of an account number are to be written, one per unit

square, in the string of unit squares below. The sum of any three consecutive digits is 15.

What is the value of x?

6 x 8

(5) In numbering the pages of a book from 1 to 576, how many times is the

digit 3 used?

(6) A regular hexagon is truncated to form a regular dodecagon (12-gon) by

removing identical isosceles triangles from its six corners. What percent of the area of the

original hexagon was removed? Express your answer to the nearest tenth.

(7) Within a 4� 4 grid of unit squares, there are squares of four various sizes

(1� 1 through 4� 4) each of which is comprised entirely of whole unit squares. In each of

the 16 unit squares, write the total number of squares (1� 1 through 4� 4) that contain

that square. One square has already been �lled in to show that six squares (one 1� 1, two

2� 2, two 3� 3 and one 4� 4) contain that square. What is the sum of the 16 numbers

shown when the grid is completely �lled in?

(8) A sequence of positive integers is formed by following two rules:

� If a term of the sequence is a multiple of 3, the following term is generated by dividing

the current term by 3.

� If a term of the sequence is not a multiple of 3, the following term is generated by

subtracting 1 from the current term.

If the �rst term of the sequence is a three-digit integer and the 6th term is 1, what is the

sum of all possible �rst terms?

(9) A right circular cone sits on a table, pointing up. The cross-section triangle,

perpendicular to the base, has a vertex angle of 60 degrees. The diameter of the cone's

base is 12p3 inches. A sphere is placed inside the cone so that it is tangent to the sides of

the cone and sits on the table. What is the volume, in cubic inches, of the sphere? Express

your answer in terms of �.

(10) Yesterday, Teddy played a game repeatedly on his computer and won exactly

85% of the games he played. Today, Teddy played the same game repeatedly, winning

every game he played, until his over-all winning percentage for yesterday and today was

exactly 94%. What is the minimum number of games that he played today?

(11) The sequence of integers in each of the two rows of squares and in each of

the two columns of square form four distinct arithmetic sequences. What is the least

positive integer value for N?

87

75

N

26

(12) Emma plays with her square unit tiles by arranging all of them into di�erent

shaped rectangular �gures. (For example, a 5 by 7 rectangle would use 35 tiles and would

be considered the same rectangle as a 7 by 5 rectangle). Emma can form exactly ten

di�erent such rectangular �gures that each use all of her tiles. What is the least number of

tiles Emma could have?

(13) For some integers that are not palindromes, like 91, a person can create a

palindrome by repeatedly reversing the number and adding the original number to its

reverse. For example, 91 + 19 = 110. Then 110 + 011 = 121, which is a palindrome, so 91

takes two steps to become a palindrome. Of all positive integers between 10 and 100, what

is the sum of the non-palindrome integers that take exactly six steps to become

palindromes?

(14) The math department at a small high school has six classes with 25 students

each, four classes with 20 students each, and two classes with 35 students each. No

student takes more than one math class. If each student correctly �lls out a questionnaire

asking for the total number of students in his/her math class, what is the average of all of

the numbers turned in on the questionnaires?

(15) Joseph, Pedro, Samuel and David leave their backpacks by the classroom

door. When the �re alarm rings they rush to leave, each randomly grabbing a backpack on

the way out. What is the probability that each boy takes a backpack that is not his own?

Express your answer as a common fraction.

(16) A math conference is presenting a lecture series with six di�erent lecturers.

If Dr. Smith's lecture depends on Dr. Jones' lecture, so that Dr. Smith must be scheduled

at some time after Dr. Jones, in how many orders can the six lecturers be scheduled?

(17) In the �gure, what is the area of triangle ABD? Express your answer as a

common fraction.

A

2

E

2

C 3 F 4 B

D

(18) Lamant has a box containing only red marbles, blue marbles and green

marbles. He needs to select at least 17 marbles without replacement to be sure at least

one of them is green. He needs to select at least 18 marbles without replacement to be

sure at least 1 of them is red. He needs to select at least 20 marbles without replacement

to be sure all three colors appear among the marbles selected. How many marbles are in

the box?

(19) A collection of nickels, dimes and pennies has an average value of 7 cents

per coin. If a nickel were replaced by �ve pennies, the average would drop to 6 cents per

coin. What is the number of dimes in the collection?

(20) Homewood Middle School has 1200 students, and 730 of these students

attend a summer picnic. If two-thirds of the girls in the school and one-half of the boys in

the school attend the picnic, how many girls attend the picnic?

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Answer Sheet

Number Answer Problem ID

1 1225/72 sq inches DDC1

2 1.499 03522

3 12 combinations BB02

4 1 D4C1

5 218 times D2522

6 7.2 percent C1D51

7 104 51D51

8 513 33522

9 288� cu inches 23522

10 30 games D1D51

11 4 0C02

12 240 tiles 4DC

13 176 4B5

14 26 students 3B5

15 3/8 43522

16 360 orders 5B5

17 56/11 12D51

18 26 marbles A1312

19 12 dimes DB02

20 520 girls BA5

Copyright MATHCOUNTS Inc. All rights reserved

Solutions

(1) 1225/72 sq inches ID: [DDC1]

Suppose that the altitude from F to EG intersects EG at point H. Then

4EAB � 4EHF , and we have that HE

HF= AE

AB. Also, 4GDC � GHF , and HG

HF= DG

DC.

Adding these equalities, we �nd that HE+HGHF

= AE+DGAB

, since AB = DC. But

HE +HG = EG = 10, HF = 7, and �nally AE +DG = EG � AD = 10� 2AB. Plugging

in, we �nd that 107= 10�2AB

AB, or AB = 35

12. Thus the area of ABCD is 35

12� 35

6=

1225

72.

(2) 1.499 ID: [03522]

We are asked to �nd2

1 � 3 +2

2 � 4 +2

3 � 5 +2

4 � 6 + � � �+ 2

2009 � 2011 :Observe that 2

n(n+2)may be written as 1

n� 1

n+2. Applying this identity, our sum becomes

1

1� 1

3+

1

2� 1

4+

1

3� 1

5+

1

4� 1

6+ � � �+ 1

2009� 1

2011:

Every negative term cancels with the term three places to the right. The only terms which

remain are

1 +1

2� 1

2010� 1

2011:

To the nearest thousandth, the sum is 1:499 .

(3) 12 combinations ID: [BB02]

The �rst player can place an X in one of three distinct positions: the center, an edge, or a

corner.

Case 1: An X in the center. The second player can then place their O in either an edge

or a corner, for 2 possibilities.

Case 2: An X on an edge. The second player can then place their O in the center, an

adjacent corner, an opposite corner, an adjacent edge, or the opposite edge, for 5

possibilities.

Case 3: An X in a corner. The second player can then place their O in the center, an

adjacent edge, an opposite edge, an adjacent corner, or the opposite corner, for 5

possibilities.

So there are 2 + 5 + 5 = 12 possibilities total.

(4) 1 ID: [D4C1]

Consider any block of 4 consecutive digits, say a; b; c; d respectively. Then we are given

that a + b + c = 15 = b + c + d , so a = d . Thus we see that the digit to the right of 6

must be x , and applying this logic twice the digit to the right of that must be 8. So we

have that 8 + x + 6 = 15, or x = 1 .

(5) 218 times ID: [D2522]

Three is used as a units digit in the integers 3 + 10n, where n ranges from 0 to 57.

Therefore, 3 appears as a units digit 58 times. Similarly, 3 is used as a tens digit in the

integers of the form a + 100n where n ranges from 30 to 39 and n ranges from 0 to 5.

Thus 3 appears as a tens digit 10 � 6 = 60 times. Finally, 3 is used as a hundreds digit in

100 numbers from 300 to 399. Altogether, we have 58 + 60 + 100 = 218 apperances of

the digit 3.

(6) 7.2 percent ID: [C1D51]

Without loss of generality, let the side length of the hexagon be 1 unit. Also let u be the

length of each of the equal sides in the removed isosceles triangles. De�ne points A, B, C,

D, E, and F as shown in the diagram. Triangle CDB is a 30-60-90 triangle, so CD = u=2

and DB = up3=2. Also, AB = 1� 2u because CF = 1 and CB = AF = u. For the

resulting dodecagon to be regular, we must have AB = 2 � BD. We �nd

1� 2u = up3 =)

1 = 2u + up3 =)

1 = u(2 +p3) =)

1

2 +p3= u:

Multiplying numerator and denominator by 2�p3 to rationalize the denominator, we get

u = 2�p3. The area of a regular hexagon with side length s is 3s2p3=2 so the area of the

hexagon is 3p3=2. The removed area is 6� 1

2(CD)(2 � BD) = 3u2

p3=2. Therefore, the

fraction of area removed is u2, which to the nearest tenth of a percent is 0:072 = 7:2% .

A

B

C

D

EF

u

1− 2u

(7) 104 ID: [51D51]

There are sixteen 1� 1 squares, nine 2� 2 squares, four 3� 3 squares and one 4� 4

square. A square of area A causes A unit squares to increase the number written in them

by 1. Therefore, rather than summing the numbers written in the squares, we may sum the

areas of the squares. The sum of the areas of the 1� 1 squares is 16, the sum of the areas

of the 2� 2 squares is 36, the sum of the areas of the 3� 3 squares is 36, and the 4� 4

square has area 16. The sum of the areas of all the squares is 16 + 36 + 36 + 16 = 104 .

(8) 513 ID: [33522]

Work backwards one term at a time from the sixth term to �nd the �rst term. At each

step we either add 1 or multiply by 3, making sure that we never add 1 to produce a

multiple of 3. Before checking cases, we make an observation that will aid our search. Let

n be the fourth term and a the �rst term. The largest value of a is 27n since we may triple

n at each of the three steps from the fourth term to the �rst term. The second largest

value of a is 9(n + 1). Since a is a three-digit integer, we �nd that a fourth term of n gives

rise to only one value of a when

9(n + 1)240

.

(3) k = p91 � p12 for distinct primes p1 and p2. The smallest such k is attained when

p1 = 2 and p2 = 3, which gives k = 29 � 3 > 240.

(4) k = p19 for some prime p. The smallest such k is attained when p = 2, which gives

k = 219 > 240.

Therefore, the least number of tiles Emma could have is 240 tiles.

(9) 176 ID: [4B5]

Say our two-digit non-palindrome is n = ab = 10a + b, with a and b digits. Reversing n

and adding it to itself is 10a + b + 10b + a = 11(a + b). This operation only depends on

a + b, so 57 and 48 for example yield the same result. When a + b � 9, the resulting

number is just a number in f11; 22; : : : ; 99g, all of which are palindromes, so numbers with

a + b � 9 take one step. We can now check how many times the operation needs to be

applied on each remaining value of a + b. Since a; b � 9, a + b � 18.

a + b = 10! 110! 121

a+b=11 ! 121

a+b=12 ! 132! 363

a+b=13 ! 143! 484

a+b=14 ! 154! 605! 1111

a+b=15 ! 165! 726! 1353! 4884

a+b=16 ! 176! 847! 1595! 7546! 14003! 44044

a+b=17 ! 187! 968! 1837! 9218! 17347! 91718! : : :

a+b=18 ! 189! 1089! 10890! 20691! 40293! 79497The only two values of

a + b which require exactly six steps are a + b = 16 and a + b = 18. However, the only n

for which a + b = 18 is n = 99, a palindrome. We are left with 97 + 79 = 176 , as we

exclude the palindrome n = 88.

(10) 26 students ID: [3B5]

While it's tempting to just calculate that average number of students in each class, that is

not what the question is asking for. Instead, we need to calculate the average of the

numbers on the questionaires. To do this, we can calculate the sum of all of the numbers

written on the questionaires and divide by the total number of questionaires. There are

6 � 25 = 150 students in the 6 classes of 25, each of whom will write a 25 on their

questionaire, 4 � 20 = 80 students in the 4 classes of 20, each of whom will write a 20 on

their questionaire, and 2 � 35 = 70 students in the 2 classes of 35, each of whom will write

a 35 on their questionaire. So the total sum of all the numbers on all the questionaires is

150 � 25 + 80 � 20 + 70 � 35 = 7800 and the total number of questionaires will be

150 + 80 + 70 = 300, so the average number on each questionaire is7800

300= 26 .

(11) 3/8 ID: [43522]

Let us refer to the four students by their �rst initials. We describe possible ways that the

backpacks could be picked up using four-letter strings in which the letters refer to the

students and the positions refer to the backpacks, in the order J, P, S, and D. For example,

JSPD means that Joseph and David picked up the correct backpack and Samuel and

Pedro's backpacks were switched. Let's call an arrangement in which everyone picks up the

wrong backpack a derangement.

We will count the number of derangements by checking di�erent possibilities for the �rst

letter. The �rst letter of a derangement cannot be J. By listing all 6 arrangements which

begin with P, we see that the only derangements which begin with P are PJDS, PDJS, and

PSDJ. Similarly, there are 3 derangements whose �rst letter is S and 3 derangements

whose �rst letter is D. Altogether, there are 3 + 3 + 3 = 9 derangements out of 4! = 24

arrangements. Since each arrangement is equally likely, the probability that no one picks up

the correct backpack is 9=24 = 3=8 .

(12) 360 orders ID: [5B5]

There are 6! = 720 ways to order the lecturers with no restrictions. By symmetry, exactly

half of these have Dr. Jones and Dr. Smith in the right order., for a total of 7202

= 360

orders.

(13) 56/11 ID: [12D51]

Let G and H be the feet of the perpendiculars from D to CB and AC, respectively. De�ne

x = DH and y = DG. By the similarity of triangles DGB and ECB, we have

y

2=

7� x

7;

and by the similarity of triangles ADH and AFC,

x

3=

4� y

4:

Clearing denominators in these two equations we obtain

2x + 7y = 14

4x + 3y = 12;

which we can solve by doubling the �rst equation and subtracting the two equations. We

�nd y = 16=11 and x = 21=11. The area of triangle ADC is 12(AC)(DH) = 42

11, the area of

triangle CDB is 12(CB)(DG) = 56

11, and the area of triangle ABC is 1

2(4)(7) = 14. The

area of triangle ADB is 14� 42=11� 56=11 = 56=11 square units.

A

E

C F B

D

G

H

(14) 26 marbles ID: [A1312]

Let the number of red marbles be r , the number of blue ones be b, and the number of

green ones be g. Look at the �rst condition: he needs to select at least 17 marbles to be

sure one is green. By the Pigeonhole Principle, this means that there are 16 red and blue

marbles, since if there were more then Lamant could select 17 red and blue marbles and

thus he could not be sure of getting a green. If there were fewer than 16 red and blue

marbles, Lamant would not need to take out at least 17 marbles to be sure of getting a

green. So r + b = 16.

Look at the second condition: he needs to select at least 18 marbles to make sure one is

red. Similarly, that means there are 17 blue and green marbles. So b + g = 17.

Look at the third condition: he needs to select at least 20 marbles to make sure of having

all three colors. Similarly, that means there are 19 red and green marbles. So r + g = 19.

Adding all three equations, we have

(r + b) + (b+ g) + (r + g) = 16+ 17+ 19) r + b+ g = 26. Thus, there are 26 marbles

in the box.

(15) 12 dimes ID: [DB02]

Let M be the total number of cents in the collection, and let N be the total number of

coins in the collection. Since the average is 7 cents per coin, MN= 7, or M = 7N. If a

nickel is replaced by �ve pennies, the amount of money doesn't change, but the number of

coins increases by 4. So M

N+4= 6. Substituting M = 7N we get 7N

N+4= 6, and solving gives

N = 24. Then M = 7(24) = 168.

Since we have a total of 168 cents in pennies, nickels, and dimes, the number of pennies

must be 3, 8, 13, 18, or so on (since nickels and dimes contribute multiples of 5 cents).

Suppose we have 8 or more pennies. We have a total of 24 coins, so there are at most 16

nickels and dimes. At least one of these is a nickel, since the problem says we can exchange

one nickel for �ve pennies. So, the most money we could have is 8 pennies, 1 nickel, and

15 dimes, a total of 8 + 5 + 15(10) = 163 cents. But we know we have more than this: we

have 168 cents. This is a contradiction, so we must have less than 8 pennies. Thus, we

have 3 pennies. This means we have total of 21 nickels and dimes, and these nickels and

dimes are worth 165 cents. Letting n and d be the number of nickels and dimes we have,

respectively, we get n + d = 21 and 5n + 10d = 165. Dividing the second equation by 5,

we have n + 2d = 33, and subtracting the �rst equation from this gives d = 12 .

(16) 520 girls ID: [BA5]

Let the number of girls at HMS be g and the number of boys be b. Thus, the total number

of students implies g + b = 1200 and the attendance �gures imply 23g + 1

2b = 730.

Multiplying the �rst equation by 3 and subtracting that from the second equation multiplied

by 6, we get g = 780. And, the number of girls who attended the picnic is 23� 780 = 520 .

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