Twenty-fourth Annual UNC Math Contest First Round November ... · 10/15/2011 · Twenty-fourth...

Twenty-fourth Annual UNC Math Contest First Round November, 2015

Rules: 90 minutes; no electronic devices.

The positive integers are 1, 2, 3, 4, . . .

1. How many positive integers less than 100 are multiples of 5 but not multiples of 2?

1

6

72. A zig-zag path has three straight segments thatmeet at right angles and have lengths 1, 6, and 7, asshown in the diagram. What is the distance betweenthe endpoints of the path? That is, find the length ofthe dashed segment.

3. Three barrels currently contain 60 lbs, 50 lbs, and 10 lbs of sand. Sandy wants to equalizethe weight of sand in the barrels by redistributing the sand among the barrels. What is theleast total weight Sandy must move between barrels?

A

B

4. A spider has a web in the shape of the grid shownin the diagram. How many different ways can the spi-der move from corner A to corner B by traveling alongexactly seven segments?

5. Find the area of the region in the x-y plane that consists of the points (x, y) for which

|x|+ |y| 3.

The notation |x| stands for the “absolute value” of x. That is, |x| = x if x � 0 and |x| = �x if x 0.

!

!

!

6. The points (2,5) and (6,5) are two of the vertices ofa regular hexagon of side length two on a coordinateplane. There is a line L that goes through the point(0,0) and cuts the hexagon into two pieces of equal area.What is the slope of line L? Express that slope as a dec-imal number. A regular hexagon is a hexagon whose sides

have equal length and whose angles are congruent.

TURN PAGE OVER

Fold7. A rectangular sheet of paper whose dimensions are12” ⇥ 18” is folded along a diagonal, which creates theM- shaped region drawn below. Find the area of theshaded region.

8. A pyramid is built from solid unit cubes that are stacked in square layers which are centeredover one another. The top layer has one cube. The layer below it has 3 ⇥ 3 = 9 cubes. Thelayer below that has 5 ⇥ 5 = 25 cubes, and so on, with each layer having two more cubes ona side than the layer above it. The pyramid has a total of 12 layers. Find the exposed surfacearea of this solid pyramid, including the bottom.

9. Yoda chooses two integers from {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}.He whispers one number to R2-D2. He whispers the other number to BB-8. He then an-nounces to both R2-D2 and BB-8, "R2-D2’s number is smaller than BB-8’s number. But, theyhave the same number of divisors."R2-D2 says, "I don’t know BB-8’s number."Then BB-8 says, "With the information Yoda gave us, I did not know R2-D2’s number, either.However, I now know R2-D2’s number." What are R2-D2’s and BB-8’s numbers?A divisor of an integer is an integer that divides into the integer with no remainder. For example, the

divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

10. A treasure chest starts with 4 copper coins, 4 silver coins, and 5 gold coins. When Midasrandomly touches any colored coin, it magically disappears, and is replaced by two new coinsthat are of the complementary colors. For example, if Midas touches a silver coin, it transformsinto one copper coin and one gold coin. After two consecutive random Midas touches, what isthe probability that the gold coins are still more numerous than either of the other two colors?

A

B

11. A spider has a web in the shape of the grid shown inthe diagram. How many different ways can the spidermove in a loop from corner A to corner B and back tocorner A by traveling along exactly fourteen distinctsegments, if her path must never cross or touch itselfuntil it arrives back at corner A? The spider may not

move along any segment more than one time, and the spider’s path may not touch any inter-section it has previously visited until it returns to corner A. Count a clockwise loop as differentfrom its counter-clockwise counterpart.

END OF CONTEST

Twenty-fourth Annual UNC Math Contest First Round with Solutions Nov 2015

Rules: 90 minutes; no electronic devices.

The positive integers are 1, 2, 3, 4, . . .

1. How many positive integers less than 100 are multiples of 5 but not multiples of 2?ANSWER: 10SOLUTION: 5, 15, 25, 35, 45, 55, 65, 75, 85, and 95. There are ten of these.

1

6

72. A zig-zag path has three straight segments thatmeet at right angles and have lengths 1, 6, and 7, asshown in the diagram. What is the distance betweenthe endpoints of the path? That is, find the length ofthe dashed segment.

ANSWER: 10SOLUTION: Fit a rectangle that is 8 units wide and 6 units tall around the figure. Apply thePythagorean Theorem to find the length of the diagonal..

3. Three barrels currently contain 60 lbs, 50 lbs, and 10 lbs of sand. Sandy wants to equalizethe weight of sand in the barrels by redistributing the sand among the barrels. What is theleast total weight Sandy must move between barrels?ANSWER: 30 poundsSOLUTION: There are 50+60+10=120 pounds of sand in all. This makes 120 ÷ 3 = 40 poundsfor each bucket. Move 20 pounds from the heaviest bucket and 10 pounds from the middlebucket to the lightest bucket.

A

B

4. A spider has a web in the shape of the grid shownin the diagram. How many different ways can the spi-der move from corner A to corner B by traveling alongexactly seven segments?

ANSWER: 35SOLUTION: The spider must move up three times and down four times. Choose 3 slots outof 7 for the up moves and the answer is 7 choose 3, or 35.

5. Find the area of the region in the x-y plane that consists of the points (x, y) for which

|x|+ |y| 3.

The notation |x| stands for the “absolute value” of x. That is, |x| = x if x � 0 and |x| = �x if x 0.

ANSWER: 18 square unitsSOLUTION: The region is the square whose corners are the points (3,0), (0,3), (�3, 0) and (0,�3). You can find the length of a side but it is simpler to use the formula for the area of atriangle on the top and bottom halves of the figure.

!

!

!

6. The points (2,5) and (6,5) are two of the vertices ofa regular hexagon of side length two on a coordinateplane. There is a line L that goes through the point(0,0) and cuts the hexagon into two pieces of equal area.What is the slope of line L? Express that slope as a dec-imal number. A regular hexagon is a hexagon whose sides

have equal length and whose angles are congruent.

ANSWER: 1.25SOLUTION: By symmetry, any line that goes through the center of the hexagon will bisect thearea. We want the line through that center (4,5) and the point (0,0). That line has slope 5/4 =1.25.

Fold7. A rectangular sheet of paper whose dimensions are12” ⇥ 18” is folded along a diagonal, which creates theM- shaped region drawn below. Find the area of theshaded region.

ANSWER: 138 square inchesSOLUTION: Use similar triangles & the Pythagorean theorem

8. A pyramid is built from solid unit cubes that are stacked in square layers which are centeredover one another. The top layer has one cube. The layer below it has 3 ⇥ 3 = 9 cubes. Thelayer below that has 5 ⇥ 5 = 25 cubes, and so on, with each layer having two more cubes ona side than the layer above it. The pyramid has a total of 12 layers. Find the exposed surfacearea of this solid pyramid, including the bottom.ANSWER: 1634 square unitsSOLUTION: The bottom layer is a square 23 units on a side. Look straight down on the topto see that the top surface area = bottom surface area = 232 = 529. The side surface areafacing in each of four directions is 1 + 3 + 5 + ... + 23 = 144. Total area is 2 ⇥ 529 + 4 ⇥ 144 =1058 + 576 = 1634.

9. Yoda chooses two integers from {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}.He whispers one number to R2-D2. He whispers the other number to BB-8. He then an-nounces to both R2-D2 and BB-8, "R2-D2’s number is smaller than BB-8’s number. But, theyhave the same number of divisors."R2-D2 says, "I don’t know BB-8’s number."Then BB-8 says, "With the information Yoda gave us, I did not know R2-D2’s number, either.However, I now know R2-D2’s number." What are R2-D2’s and BB-8’s numbers?A divisor of an integer is an integer that divides into the integer with no remainder. For example, the

divisors of 12 are 1, 2, 3, 4, 6, and 12. ANSWER: R2-D2’s number is 12 and BB-8’s number is 20.

SOLUTION: Each prime has exactly two divisors and there are many primes to choose from,so neither number is a prime. List the divisors of the other possible numbers and count them:1 {1 } 1 divisor4 {1, 2, 4 } 3 divisors6 {1, 2, 3, 6 } 4 divisors8 {1, 2, 4, 8 } 4 divisors9 {1, 3, 9 } 3 divisors10 {1, 2, 5, 10 } 4 divisors12 {1, 2, 3, 4, 6, 12 } 6 divisors14 {1, 2, 7, 14 } 4 divisors16 {1, 2, 4, 8, 16 } 5 divisors18 {1, 2, 3, 6, 9, 18 } 6 divisors20 {1, 2, 4, 5, 10, 20 } 6 divisors12, 18, 20 each have 6 divisors. R2-D2’s number is 12 and BB-8’s number is 20.

10. A treasure chest starts with 4 copper coins, 4 silver coins, and 5 gold coins. When Midasrandomly touches any colored coin, it magically disappears, and is replaced by two new coinsthat are of the complementary colors. For example, if Midas touches a silver coin, it transformsinto one copper coin and one gold coin. After two consecutive random Midas touches, what isthe probability that the gold coins are still more numerous than either of the other two colors?

ANSWER: 32/91SOLUTION: 2 ⇥ 4

13 ⇥ ( 314 +

514) =

3291

A

B

11. A spider has a web in the shape of the grid shown inthe diagram. How many different ways can the spidermove in a loop from corner A to corner B and back tocorner A by traveling along exactly fourteen distinctsegments, if her path must never cross or touch itselfuntil it arrives back at corner A? The spider may not

move along any segment more than one time, and the spider’s path may not touch any inter-section it has previously visited until it returns to corner A. Count a clockwise loop as differentfrom its counter-clockwise counterpart.

ANSWER: 100SOLUTION: To count the outgoing paths, choose 3 spots out of the 7 steps to be the up steps.There are 7 choose 3 or 35 seven step paths from A to B.Count the permissible return paths for each outgoing path.Permissible loops come in pairs, a clockwise loop and the counterclockwise loop that tracesthe same segments. Therefore, it will suffice to count the clockwise paths and double theresult.The clockwise paths are the ones that begin with an "up" step.An outgoing path that does not arrive at B with a "down" step will not have any permissiblereturn paths. All return paths would have to cross the outbound path. Therefore, we needconsider only paths that begin at A with an "up" and arrive at B with a "down." These arepaths that go from corner to corner on a 2 ⇥ 3 grid and there are 5 choose 2 =10 of these. Seethe scans for a count for these ten cases.

END OF CONTEST

University of Northern Colorado Mathematics Contest

Problems are duplicated and solved by Ming Song ([email protected]) 1

University Of Northern Colorado Mathematics Contest 2015-2016 Problems of First Round


2. A zig-zag path has three straight segments that meet at right angles and have lengths 1, 6, and 7, as shown in the diagram. What is the distance between the endpoints of the path? That is, find the length of the dashed segment.

3. Three barrels currently contain 60 lbs, 50 lbs, and 10 lbs of sand. Sandy wants to equalize the weight of sand in the barrels by redistributing the sand among the barrels. What is the least total weight Sandy must move between barrels?

4. A spider has a web in the shape of the grid shown in the diagram. How many different ways can the spider move from corner A to corner B by traveling along exactly seven segments?

5. Find the area of the region in the x-y!plane that consists of the points ( )yx, !for which

3≤+ yx .

The notation x stands for the “absolute value” of x. That is xx = if 0≥x and xx −= if 0≤x .

A

B

1

7

6



6. The points ( )5,2 and ( )5,6 are two of the vertices of a regular hexagon of side length two on a coordinate plane. There is a line L that goes through the point ( )0,0 and cuts the hexagon into two pieces of equal area. What is the slope of line L? Express that slope as a decimal number. A regular hexagon is a hexagon whose sides have equal length and whose angles are congruent.

7. A rectangular sheet of paper whose dimensions are 12ʺ × 18ʺ is folded along a diagonal, which creates the M- shaped region drawn at the right. Find the area of the shaded region.

8. A pyramid is built from solid unit cubes that are stacked in square layers which 3 × 3 = 9 cubes. The layer below that has 5 × 5 = 25 cubes, and so on, with each layer having two more cubes on a side than the layer above it. The pyramid has a total of 12 layers. Find the exposed surface area of this solid pyramid, including the bottom.

9. Yoda chooses two integers from

{ }20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1 .

He whispers one number to R2-D2. He whispers the other number to BB-8. He then announces to both R2-D2 and BB-8, “R2-D2’s number is smaller than BB-8’s number. But, they have the same number of divisors.”

R2-D2 says, “I don’t know BB-8’s number.”

Then BB-8 says, “With the information Yoda gave us, I did not know R2-D2’s number, either. However, I now know R2-D2’s number.”

Fold



What are R2-D2’s and BB-8’s numbers?

A divisor of an integer is an integer that divides into the integer with no remainder. For example, the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

10. A treasure chest starts with 4 copper coins, 4 silver coins, and 5 gold coins. When Midas randomly touches any colored coin, it magically disappears, and is replaced by two new coins that are of the complementary colors. For example, if Midas touches a silver coin, it transforms into one copper coin and one gold coin. After two consecutive random Midas touches, what is the probability that the gold coins are still more numerous than either of the other two colors?

11. A spider has a web in the shape of the grid shown in the diagram. How many different ways can the spider move in a loop from corner A to corner B and back to corner A by traveling along exactly fourteen distinct segments, if her path must never cross or touch itself until it arrives back at corner A? The spider may not move along any segment more than one time, and the spider’s path may not touch any intersection it has previously visited until it returns to corner A. Count a clockwise loop as different from its counter-clockwise counterpart.

A

B



Problems with Solutions of First Round


Answer: 10

Solution:

These are easy to list: 5, 15, 25, 35, 45, 55, 65, 75, 85, and 95. There are ten of them. One can

also reason as follows: There are 205100

=!"

!#$

# numbers which are multiples of 5. Among them

there are 1025

100=!"

!#$

#⋅

numbers which are multiples of 2.

Therefore, the answer is 101020 =− .

2. A zig-zag path has three straight segments that meet at right angles and have lengths 1, 6, and 7, as shown in the diagram. What is the distance between the endpoints of the path? That is, find the length of the dashed segment.

Answer: 10

Solution:

Let the zig-zag path be ABCD as shown.

Draw DCAE ⊥ intersecting line DC at E. Then 6=AE and 871 =+=DE . Triangle AED is a 3-4-5 triangle.

Therefore, 10=AD

3. Three barrels currently contain 60 lbs, 50 lbs, and 10 lbs of sand. Sandy wants to equalize the weight of sand in the barrels by redistributing the sand among the barrels. What is the least total weight Sandy must move between barrels?

Answer: 30

1

7

6

A B

C D

E

E

1

7

6



Solution:

Since one of three barrels contains 10 lbs, at least 30 lbs must be moved into this barrel.

30 lbs is attainable: move 20 lbs from the first barrel into the third barrel and move 10 lbs from the second barrel into the third barrel. Then all barrels have 40 lbs each.

The answer is 30.

4. A spider has a web in the shape of the grid shown in the diagram. How many different ways can the spider move from corner A to corner B by traveling along exactly seven segments?

Answer: 35

Solution 1:

We mark the numbers:

A number at a cross-section indicates the number of shortest routes (with exactly 7 segments) from A to the cross-section.

The answer is 35.

Solution 2:

From A to B there are 7 blocks in a row: four horizontal and three vertical. In 7 blocks there are

!!"

#$$%

&

37

ways to choose 3 blocks to be vertical.

Therefore, the answer is 3537

=!!"

#$$%

&.

A

B

1 1 1 1 1

1

1

1

2

3

4

3

6

10

4

10

20

5

15

35

A

B



5. Find the area of the region in the x-y!plane that consists of the points ( )yx, !for which

3≤+ yx .

The notation x stands for the “absolute value” of x. That is xx = if 0≥x and xx −= if 0≤x .

Answer: 18

Solution:

The region is shaded below:

The area is 186621

=⋅⋅ .

6. The points ( )5,2 and ( )5,6 are two of the vertices of a regular hexagon of side length two on a coordinate plane. There is a line L that goes through the point ( )0,0 and cuts the hexagon into two pieces of equal area. What is the slope of line L? Express that slope as a decimal number. A regular hexagon is a hexagon whose sides have equal length and whose angles are congruent.

Answer: 1.25

x

y

x + y = 3 –x + y = 3

x – y = 3 –x – y = 3

3

–3

–3

3



Solution:

Any line passing through the center of the regular hexagon divides this hexagon into two pieces of equal area.

The center of the hexagon is ( )5,4 . The line L passes ( )0,0 and ( )5,4 . The slope is 25.145= .

7. A rectangular sheet of paper whose dimensions are 12ʺ × 18ʺ is folded along a diagonal, which creates the M- shaped region drawn at the right. Find the area of the shaded region.

Answer: 138

Solution:

Let ABCD be the rectangle. The folding is along diagonal BC. After folding, let AC and BD intersect at E.

Fold

A

B C

D

Fold

A

B C

D

E

(2, 5) (6, 5) (4, 5)

L



Let xDEAE == . Then xCE −=18 .

In right triangle CDE we have ( ) 222 1218 +=− xx . Solving for x, we obtain 5=x .

The area of triangle BAC is 108181221

=⋅⋅ , and the area of triangle BAC is 3051221

=⋅⋅ .

The total area of the shaded region is 13830108 =+ .

A quick student may find the 5-12-13 triangle immediately.

8. A pyramid is built from solid unit cubes that are stacked in square layers which 3 × 3 = 9 cubes. The layer below that has 5 × 5 = 25 cubes, and so on, with each layer having two more cubes on a side than the layer above it. The pyramid has a total of 12 layers. Find the exposed surface area of this solid pyramid, including the bottom.

Answer: 1634

Solution:

In the 12th layer there are 23 × 23 cubes.

The pyramid looks like

From one lateral side we see

From the top or bottom we see the 23 × 23 square grid. Therefore, the total exposed surface area is ( ) 1634232124232235314 222 =⋅+⋅=⋅+++++⋅ ! .

9. Yoda chooses two integers from

{ }20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1 .

He whispers one number to R2-D2. He whispers the other number to BB-8. He then announces to both R2-D2 and BB-8, “R2-D2’s number is smaller than BB-8’s number. But, they have the same number of divisors.”

R2-D2 says, “I don’t know BB-8’s number.”

Then BB-8 says, “With the information Yoda gave us, I did not know R2-D2’s number, either. However, I now know R2-D2’s number.”

What are R2-D2’s and BB-8’s numbers?



A divisor of an integer is an integer that divides into the integer with no remainder. For example, the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

Answer: 12 and 20

Solution:

Let us list the number f of factors for each number n from 1 to 20:

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

f 1 2 2 3 2 4 2 4 3 4 2 6 2 4 4 5 2 6 2 6

There is only one number having one factor, which is 1. Obviously 1 is not a candidate.

There is only one number having 5 factors, which is 16. 16 is not a candidate.

There are only two numbers having 3 factors, which are 4 and 9. If one holds 4 or 9, the other can immediately know what the component holds. So 4 and 9 are not candidates.

All prime numbers each have 2 factors.

Assume that both R2-D2 and BB-8 hold two primes.

R2-D2 cannot hold 19 because BB-8’s number is larger.

R2-D2 cannot hold 17 because R2-D2 can claim BB-8’s number immediately.

After R2-D2 says “I don’t know BB-8’s number”, BB-8 knows that R2-D2’s number is one of 2, 3, 5, 7, 11, 13. BB-8 cannot conclude what R2-D2’s number is.

So they don’t hold two prime numbers.

They are five numbers having 4 factors, which are 6, 8, 10, 14, 15. By the same reasoning, these are not candidates.

Now only 12, 18, and 20 are remaining, which each have 6 factors. R2-D2 and BB-8 hold two of them.

If one holds 18, he can claim the component’s number immediately. It is not the case.

Therefore, R2-D2’s number is 12, and BB-8’s number is 20.

R2-D2 holds 12. Since there are two larger numbers 18 and 20 having the same factor as 12, R2-D2 cannot conclude what BB-8’s number is.

BB-8 holds 20. Since there are two smaller numbers 12 and 18 having the same factor as 20, at the beginning BB-8 cannot conclude what R2-D2’s number is.

After R2-D2 says “I don’t know BB-8’s number”, BB-8 knows that R2-D2’s number is 12.

Therefore, the answers are 12 and 20.

10. A treasure chest starts with 4 copper coins, 4 silver coins, and 5 gold coins. When Midas randomly touches any colored coin, it magically disappears, and is replaced by two new coins that are of the complementary colors. For example, if Midas touches a silver coin, it transforms into one copper coin and one gold coin. After two consecutive random Midas touches, what is the probability that the gold coins are still more numerous than either of the other two colors?

If we understand “more than either of the other two colors” as “more than any of the other two colors”, we have the following solution.



Answer: 9132

Solution:

Case 1: the first touch is copper

The probability for this to happen is 134 . Then we will have 3 copper, 5 silver, and 6 gold coins.

If the second touch is gold, then the gold cannot be more than silver coins. For the gold to be more than any of the two other colors, the second touch can be copper and silver. The probability

for this to happen is 74

65353

=++

+ .

In this case, the probability that the gold are still more than either of the other two colors is

9116

74

134

=⋅ .

Case 2: the first touch is silver

Because of the symmetry between copper and silver, the probability in this case is 9116 as well.

Case 3: the first touch is gold


For whatever the second touch is, the gold cannot be more than either of the other two colors.


9116

9116

=+ .

If we understand “more than either of the other two colors” as “more than one of the other two colors”, we have the following solution.

Answer: 81/91

Solution:

Case 1: the first touch is copper

The probability for this to happen is 134 . Then the second touch can be any, which leads to that

the gold are still more than either of the other two colors.

Case 2: the first touch is silver

Because of the symmetry between copper and silver, the probability in this case is 134 as well.

Case 3: the first touch is gold


If the second touch is copper or silver, the gold are more than either of the other two colors.

In this case, the probability is 9125

1410

135

=⋅ .




9125

134

134

=++ .

11. A spider has a web in the shape of the grid shown in the diagram. How many different ways can the spider move in a loop from corner A to corner B and back to corner A by traveling along exactly fourteen distinct segments, if her path must never cross or touch itself until it arrives back at corner A? The spider may not move along any segment more than one time, and the spider’s path may not touch any intersection it has previously visited until it returns to corner A. Count a clockwise loop as different from its counter-clockwise counterpart.

Answer: 100

Solution 1:

In a short time, we may not able to find an elegant solution. I would like to find the answer by listing since 3 and 4 are not large numbers.

Let us list.

In the following diagrams, the back routes from B to A are red, and the green shaded rectangles help in counting.

Case 1:

Pay attention to C and D.

The number of routes which don’t cross or touch the red route is 1025=!!

"

#$$%

&.

Case 2:

A

B

C

D

A

B



The number of routes which don’t cross or touch the red route is 922

25

=!!"

#$$%

&−!!"

#$$%

&.

Case 3:


25

=!!"

#$$%

&−!!"

#$$%

&.

Case 4:


25

=!!"

#$$%

&−!!"

#$$%

&.

A

B

C

D

A

B

C

D

A

B

C

D



Case 5:

Pay attention to E and D.


=!!"

#$$%

&.

Case 6:

Pay attention to F and D.

The number of routes which don’t cross or touch the red route is 323=!!

"

#$$%

&.

Case 7:

Pay attention to F and D.

There is only 1 route which doesn’t cross or touch the red route.

A

B

A

B

F

D

A

B

E

D



Case 8:


24

=!!"

#$$%

&−!!"

#$$%

&.

Case 9:


24

=!!"

#$$%

&−!!"

#$$%

&.

Case 10:


=!!"

#$$%

&.

The total number of the routes which don’t cross or touch a given back route is

5023513647910 =+++++++++ .

Consider that the forward route and back route can be switched.

The answer is 100250 =⋅ .

A

B

F

G

A

B

E

D

A

B

E

D



Solution 2:

I am not satisfied with a solution by listing. It take me quite much time to find the following solution.

We assume two spiders moving from A to B. We count the pairs of their paths which don’t cross or touch each other.

Look at the figure:

One spider must be from C to D, and the other must be from E to F such that their paths don’t cross or touch each other.

We will count the pairs in each of which two paths cross or touch each other.

The following figure shows a pair of path X (green) from C to D and path Y (blue) from E to F which cross each other.

Mark the point where the two paths first meet (the black point).

A

B

D

C

F

E

A

B

D

C

F

E X

Y

A

B

D

C

F

E X

Y



Switch the parts of path X and path Y after the black point.

We have two paths: one (red) is from C to F and the other (pink) is from E to D.

Red path U = the part of path X before the black point + the part of path Y after the black point

Pink path V = the part of path Y before the black point + the part of path X after the black point

Let us look at another example: a pair of path X (green) from C to D and path Y (blue) from E to F which touch each other.


Switch paths:

A

B

D

C

F

E U

V

A

B

D

C

F

E X

Y

A

B

D

C

F

E X

Y

A

B

D

C

F

E U

V



We have two paths: one (red) is from C to F and the other (pink) is from E to D.

Red path U = the part of path X before the black point + the part of path Y after the black point

Pink path V = the part of path Y before the black point + the part of path X after the black point

We see that for any pair of paths: one from C to D and the other one from E to F, which cross or touch each other, we have a pair of paths: one from C to F and the other from E to D.

Now look at a pair of path U (red) from C to F and path V (pink) from E to D. The two paths must cross each other.


Switch paths:

Green path X = the part of path U before the black point + the part of path V after the black point

Blue path Y = the part of path V before the black point + the part of path U after the black point

A

B

D

C

F

E X

Y

A

B

D

C

F

E

U

V

A

B

D

C

F

E

U

V



Take a look at another pair of path U (red) from C to F and path V (pink) from E to D.

Mark the first point where the two paths meet (the black point).

Switch paths:

Green path X = the part of path U before the black point + the part of path V after the black point

Blue path Y = the part of path V before the black point + the part of path U after the black point

We see that for any pair of paths: one from C to F and the other one from E to D, we have a pair of paths: one from C to D and the other from E to F, which cross or touch each other.

We find the one-to-one correspondence.

From C to D there are !!"

#$$%

&

25

paths, and from E to F there are !!"

#$$%

&

25

paths as well. So there are

!!"

#$$%

&⋅!!"

#$$%

&

25

25

pairs.

From C to F there are !!"

#$$%

&

35

paths, and from E to D there are !!"

#$$%

&

15

paths. So there are !!"

#$$%

&⋅!!"

#$$%

&

15

35

pairs.

A

B

D

C

F

E U

V

A

B

D

C

F

E X

Y

A

B

D

C

F

E U

V



That is, there are !!"

#$$%

&⋅!!"

#$$%

&

15

35

pairs of paths: one from C to D and the other from E to F, which cross

or touch each other.

Therefore, the number of pairs of paths: one from C to D and the other from E to F, which don’t cross or touch each other is

50510101015

35

25

25

=⋅−⋅=##$

%&&'

(⋅##$

%&&'

(−##$

%&&'

(⋅##$

%&&'

( .

Consider that two spiders can switch their paths. The answer is

100502 =⋅ .

In general, for the nm× grid, the answer is

!!"

#

$$%

&''(

)**+

,

−

−+−⋅''(

)**+

, −+−−''

(

)**+

,

−

−+−

21111

111

22

mnm

mnm

mnm

.

24th Annual UNC MATH CONTEST: SUMMARY OF FIRST ROUND RESULTS A total of 1398 students participated. The highest possible score was 110. Seven participants scored 110. Eighteen students scored 100. The table below shows the minimum qualifying scores required for invitation to the Final Round for each grade. It also shows, by grade, the number of students who participated and the number invited to the Final Round.

Grade 5, 6 7 8 9 10 11 12 Total Qualifying

score 50 50 50 60 60 60 60 Number of qualifiers 7 31 51 37 33 38 39 236

Number of participants 85 307 317 130 140 213 206 1398

Points 110 100 90 80 70 60 50 40 <40 Number of students 7 18 24 45 52 55 103 142 942

The next table shows the number of students who answered each of the questions correctly.

1 2 3 4 5 6 7 8 9 9b 10 11

ANSWER 10 10 30 35 18 1.25 138 1634 12 20 32/91 100

Points 10 10 10 10 10 10 10 10 10 10 10 # CORRECT 1222 709 961 183 306 344 93 198 403 372 140 27 % CORRECT 87% 51% 69% 13% 22% 25% 7% 14% 29% 27% 10% 2%

Scoring procedures. Each numbered question is worth 10 points. A student had to correctly produce both numbers in question 9 to receive credit. No deduction was made for including units when asked not to. The answers are checked for that kind of irregularity and credit is assigned as appropriate. Congratulations! Thanks to all the teachers, parents, and students who participated. A complete current set of problems and solutions is now available on the contest website: www.uncmathcontest.wordpress.com

Twenty-fourth Annual UNC Math Contest Final Round January 23, 2016

Rules: Three hours; no electronic devices. The positive integers are 1, 2, 3, 4, . . .

1. Pythagorean Triplet The sum of the lengths of the three sides of a right triangle is 56. Thesum of the squares of the lengths of the three sides of the same right triangle is 1250. What isthe area of the triangle?

2. Fearsome Foursome Factorial Find the complete prime factorization of

(4!)![(3!)!]4

(The answer will be a product of powers of eight distinct primes.)

3. Polyhedral Die A cube that is one inch wide hashad its eight corners shaved off. The cube’s verticeshave been replaced by eight congruent equilateral tri-angles, and the square faces have been replaced by sixcongruent octagons. If the combined area of the eighttriangles equals the area of one of the octagons, what isthat area? (Each octagonal face has two different edgelengths that occur in alternating order.)

4. Number Sieve How many positive integers less than 100 are divisible by exactly two of thenumbers 2, 3, 4, 5, 6, 7, 8, 9? For example, 75 is such a number: it is divisible by 3 and by 5, butis not divisible by any of the others on the list. (If you show the integers you find, then youmay be assigned partial credit if you have accurately found most of them, even if you do notfind all of them.)

5. Rock and Roll. Zeus has decreed that Sisyphus must spend each day removing all the rocksin a certain valley and transferring them to Mount Olympus. Each night, each rock Sisyphusplaces on Mount Olympus is subject to the whims of Zeus: it will either be vaporized (withprobability 10%), be rolled back down into the valley (with probability 50% ), or be split by athunderbolt into two rocks that are both rolled down into the valley (with probability 40%).When the sun rises, Sisyphus returns to work, delivering rocks to Olympus. At sunrise onthe first day of his punishment, there is only one rock in the valley and there are no rockson Mount Olympus. What is the probability that there are exactly two rocks in the valley atsunrise on the third day? (If a rock is vaporized, it is gone.)

6. Rock and Roll Forever? (a) Given the situation in Question 5, what is the probability thatSisyphus must labor forever? That is, if Sisyphus begins with one rock in the valley on hisfirst morning, what is the probability that the Olympian rocks are never all vaporized?(b) Suppose that the whims of Zeus obey the following rules instead: a rock will either bevaporized (with probability 10%), be rolled back down into the valley (with probability 20%),be split by a thunderbolt into two rocks that are both rolled down into the valley (with prob-ability 30%), or be split by two thunderbolts into three rocks that are all rolled down into thevalley (with probability 40%). Now what is the probability that Sisyphus must labor forever?

TURN PAGE OVER

7. Evaluate

S =•

Ân=2

4n

(n2 � 1)2

!

A

B C

D E

8. Tree Each circle in this tree diagram is to be as-signed a value, chosen from a set S, in such a waythat along every pathway down the tree, the assignedvalues never increase. That is, A � B, A � C,C � D, C � E, and A, B, C, D, E 2 S. (It is permis-sible for a value in S to appear more than once.)(a) How many ways can the tree be so numbered, usingonly values chosen from the set S = {1, . . . , 6}?

(b) Generalize to the case in which S = {1, . . . , n}. Find a formula for the number of ways thetree can be numbered.For maximal credit, express your answer in closed form as an explicit algebraic expression in n.

9. Chess Masters Four identical white pawns and fouridentical black pawns are to be placed on a standard8 ⇥ 8, two-colored chessboard. How many distinct ar-rangements of the colored pawns on the colored boardare possible?No two pawns occupy the same square. The color of a pawnneed not match the color of the square it occupies, but it might.You may give your answer as a formula involving factorials orcombinations: you are not asked to compute the number.

10. Chess Wallpaper How many distinct plane wallpa-per patterns can be created by cloning the chessboardarrangements described in Question 9?Each periodic wallpaper pattern is generated by this method:starting with a chessboard arrangement from Question 9 (themaster tile), use copies of that tile to fill the plane seamlessly,placing the copies edge-to-edge and corner-to-corner. Note thatthe resulting wallpaper pattern repeats with period 8, horizon-tally and vertically.When the tiling is done, the chessboard edges and corners van-ish, leaving only an infinite periodic pattern of black and whitepawns visible on the wallpaper.

Regard two of the infinite wallpaper patterns as the same if and only if there is a plane translation thatslides one wallpaper pattern onto an exact copy of the other one. You may slide vertically, horizontally, or acombination of both, any number of squares. (Rotations and reflections are not allowed, just translations.)Note that the wallpaper pattern depicted above can be generated by many different master tiles (by regard-ing any square 8 ⇥ 8 portion of the wallpaper as the master tile chessboard). The challenge is to account for

such duplication. Remember that each master tile has exactly four pawns of each color.You may give your answer as an expression using factorials and/or combinations (binomial coefficients).You are not asked to compute the numeric answer.

END OF CONTEST

Twenty-fourth Annual UNC Math Contest Final Round Solutions Jan 2016

Rules: Three hours; no electronic devices. The positive integers are 1, 2, 3, 4, . . .

1. Pythagorean Triplet The sum of the lengths of the three sides of a right triangle is 56. Thesum of the squares of the lengths of the three sides of the same right triangle is 1250. What isthe area of the triangle?

Answer: The area is 84.

Solution Call the lengths of the three sides a, b, and c, with c the hypotenuse. Then a + b + c =56, a2 + b2 + c2 = 1250, and a2 + b2 = c2. Conclude 2c2 = 1250, c2 = 625, and c = 25.Subtract c from both sides of the equation a + b + c = 54 to conclude a + b = 56 � 25 = 31.Subtract c2 from both sides of a2 + b2 + c2 = 1250 to conclude a2 + b2 = 1250� 625 = 650. Youcan solve these equations for a, b, and c, but what we are asked for is the area of the triangle,and that is 1

2 ab. We can take a small shortcut. Compute (a + b)2 = a2 + 2ab + b2. Use the factsthat a + b = 31 and a2 + b2 = 625 to conclude 312 = 625 + 2ab. Thus, 2ab = 312 � 625 =961 � 625 = 336. The area is 1

2 ab = 14336 = 84.

2. Fearsome Foursome Factorial Find the complete prime factorization of

(4!)![(3!)!]4

(The answer will be a product of powers of eight distinct primes.)

Answer: 26 ⇥ 32 ⇥ 73 ⇥ 112 ⇥ 13 ⇥ 17 ⇥ 19 ⇥ 23.Solution One can write the expressions out carefully and cancel common factors systemati-cally. However, this can also be solved quickly using Legendre’s simple but useful formulafor finding the prime factorization of a factorial. The formula makes use of the greatest integerfunction [[ ]] that rounds down to the nearest integer. As an example, the power of 2 in theprime factorization of 24!, computed by Legendre’s method, is[[24/2]] + [[24/22]] + [[24/23]] + [[24/24]] + . . . = 12 + 6 + 3 + 1 = 22

3. Polyhedral Die A cube that is one inch wide hashad its eight corners shaved off. The cube’s verticeshave been replaced by eight congruent equilateral tri-angles, and the square faces have been replaced by sixcongruent octagons. If the combined area of the eighttriangles equals the area of one of the octagons, what isthat area? (Each octagonal face has two different edgelengths that occur in alternating order.)

Answer: 2p

31+2

p3= 12�2

p3

11 .

Solution Let x be the amount that is cut off of each edge of the original square at the corners,so that one side length on the resulting octahedron is 1 � 2x and the other side length is

p2x.

Find the area of each octagon by subtracting the area of the four cut off corners from area ofthe square. Each cut off corner is a right isosceles triangle with legs of length x. The area of

each of these cut off corners is 12 x2. Thus the area of one of the octagons is 1� 2x2. Each of the

new triangular faces is an equilateral triangle with side lengthp

2x and heightp

32p

2x. Thearea of each equilateral corner face is 1

2

p3

2p

2xp

2x = (p

32 )x2. Set eight of these equal to the

area of one octagon:8p

32

x2 = 1 � 2x2.

Solve the equation for x. Since what we are asked for is the area 1 � 2x2, we might as wellsolve for x2.

4. Number Sieve How many positive integers less than 100 are divisible by exactly two of thenumbers 2, 3, 4, 5, 6, 7, 8, 9? For example, 75 is such a number: it is divisible by 3 and by 5, butis not divisible by any of the others on the list. (If you show the integers you find, then youmay be assigned partial credit if you have accurately found most of them, even if you do notfind all of them.)

Answer. There are eighteen such numbers:4, 9, 10, 14, 15, 21, 27, 35, 44, 50, 52, 68, 75, 76, 81, 92, 98, 99

Solution A reasonable way to do this one is to check each integer. Many of the integers willbe rejected quickly- the multiples of 4, perfect squares, and the primes, for instance.5. Rock and Roll. Zeus has decreed that Sisyphus must spend each day removing all the rocksin a certain valley and transferring them to Mount Olympus. Each night, each rock Sisyphusplaces on Mount Olympus is subject to the whims of Zeus: it will either be vaporized (withprobability 10%), be rolled back down into the valley (with probability 50% ), or be split by athunderbolt into two rocks that are both rolled down into the valley (with probability 40%).When the sun rises, Sisyphus returns to work, delivering rocks to Olympus. At sunrise onthe first day of his punishment, there is only one rock in the valley and there are no rockson Mount Olympus. What is the probability that there are exactly two rocks in the valley atsunrise on the third day? (If a rock is vaporized, it is gone.)

Answer: 0.332

Solution. A tree diagram shows that the possibilities are(i) the first rock is preserved the first night and split the second night (.5)(.4) = .2;(ii) the first rock is split into two the first night, and the second night both rocks are preserved(.4)(.5)(.5) = .1;(iii) the first rock is split into two the first night, and the second night one of those two is splitand the other is vaporized: 2(.4)(.1)(.4) = .032.In total, .2 + .1 + .032 = 0.332

6. Rock and Roll Forever? (a) Given the situation in Question 5, what is the probability thatSisyphus must labor forever? That is, if Sisyphus begins with one rock in the valley on hisfirst morning, what is the probability that the Olympian rocks are never all vaporized?(b) Suppose that the whims of Zeus obey the following rules instead: a rock will either bevaporized (with probability 10%), be rolled back down into the valley (with probability 20%),be split by a thunderbolt into two rocks that are both rolled down into the valley (with prob-ability 30%), or be split by two thunderbolts into three rocks that are all rolled down into thevalley (with probability 40%). Now what is the probability that Sisyphus must labor forever?

Answer: (a) 34 .

Solution Let P be the probability that starting with one rock, the subsequent rocks are even-tually all vaporized. Observe that if Zeus gives you two rocks, the probability that the stringsof successors of both rocks disappear eventually is P2, because these are independent events.Now use recursiveness to compute the probability from the viewpoint of the second day. NoteP = .1 (1/10 of the time that first rock is gone on the second day)+.5P (half the time you start over with one rock on the second day and P is the probabilitythat it and all its successors are eventually gone )+.4P2 (4/10 of the time you get two rocks and the probability that the strings of succes-sors of both rocks disappear eventually is P2) .Solve this quadratic for P. You get two solutions: P = 1 and P = .25.Another approach: The polynomial p(x) = .1 + .5x + .4x2 encodes the whims of Zeus: thecoefficient of xk in p(x) gives the probability of there being k rocks in the valley at sunrise onthe second day. The coefficient of xk in the composite function p(p(x)) gives the probabilityof there being k rocks in the valley at sunrise on the third day. The coefficient of xk in thethree-fold composition p(p(p(x))) gives the probability of there being k rocks in the valley atsunrise on the fourth day, and so on.The constant term of the polynomial is telling the probability that there are zero rocks in thevalley, that is, the probability that all the rocks have been vaporized.Look at the (very simple) graph of y = p(x) and y = x to track down p(0) and then p(p(0))and so on: use a cobweb picture to find that p(p(p(. . . (0) . . .)))) is the fixed point x wherex = p(x). Solve to find x = 1

4 . Conclude that the complementary event that the labor neverends is 3

4 .

Out[2]=

Figure 1: The curves cross at the fixed points x = p(x) where x = .25 and x = 1

(b) Answer: 15�p

658 .

Solution This time you solve P = .1 + .2P + .3P2 + .4P3. Observe that P = 1 is one solu-tion, so divide by the known linear factor P � 1 and solve the remaining quadratic, which is4P2 + 7P � 1 = 0. The solutions are P = �7±

p65

8 . Choose the positive P = �7+p

658 . The

complementary probability desired is 1 � P = 15�p

658 .

7. Evaluate

S =•

Ân=2

4n(n2 � 1)2

Answer: 54

Solution Using the partial fraction decomposition 4n(n2�1)2 = 1

(n�1)2 � 1(n+1)2 and then expand-

ing out the sum, we see a telescoping pattern of many canceling paired terms.( 1

12 � 132 ) + ( 1

22 � 142 ) + ( 1

32 � 152 ) + ( 1

42 � 162 ) + . . .

The only terms that do not cancel are the 112 and the 1

22 . The final infinite sum is ( 112 +

122 ) =

1 + 14 = 5

4 .

!

A

B C

D E

8. Tree Each circle in this tree diagram is to be as-signed a value, chosen from a set S, in such a waythat along every pathway down the tree, the assignedvalues never increase. That is, A � B, A � C,C � D, C � E, and A, B, C, D, E 2 S. (It is permis-sible for a value in S to appear more than once in thetree.)(a) How many ways can the tree be so numbered, usingonly values chosen from the set S = {1, . . . , 6}?

(b) Generalize to the case in which S = {1, . . . , n}. Find a formula for the number of ways thetree can be numbered.For maximal credit, express your answer in closed form as an explicit algebraic expression in n.Answers.A. 994B. 1

120 n(n + 1)(n + 2)�8n2 + 11n + 1

�= n5

15 + 7n4

24 + 5n3

12 + 5n2

24 + n60

Solution Once values for A and C are selected, the restrictions on the remaining values implythat they can be chosen in this many ways:(i) B can be chosen A ways;(ii) D and E can be chosen independently, each in C ways.Thus the total number of choices is T = ÂA=n

A=1(A ÂC=AC=1 C2).

The inner sum simplifies to 16 A(1 + A)(1 + 2A) and the total sum is T = Ân

A=0 p(A) where

p(A) =16

A2(1 + A)(1 + 2A)

This polynomial can be summed by a variety of standard methods, such as the method of un-determined coefficients, Bernoulli number expansions, or the Newton interpolation formulabased on forward differences.

Below we outline a method based on the idea of expressing p(A) as a linear combination ofbinomial coefficients and then applying the Hockey Stick Summation Lemma to each term.The Hockey Stick Lemma states that the entries in Pascal’s Triangle satisfyfor any fixed k � 0,

N

ÂA=0

✓A + k

k

◆=

✓N + k + 1

k + 1

◆

This is useful for summation of polynomials. As an example, observe that the binomial coeffi-cient (A+2

3 ) = (A)(A+1)(A+2)3! has a numerator expressible as a product of linear factors whose

roots are consecutive integers. Incidentally, this product formula allows us to extend the binomialcoefficients in a meaningful way, even to cases where A is negative.This example illustrates a basic principle: Every binomial coefficient is a product of linear factorsthat are shifted by consecutive integers. Conversely, any such product of linear factors is expressible asa binomial, hence can be summed by the Hockey Stick Lemma.With this in mind we rewrite the factorizationp(A) = 1

6 [(A � 1) + 1]A(1 + A)(1 + 2A)= 1

6 [(A � 1)A(A + 1)(1 + 2A)] + 16 A(1 + A)(1 + 2A)

= 16 [(A � 1)A(A + 1)(2(A + 2)� 3)] + 1

6 A(1 + A)(2(A + 2)� 3)so that it has been expressed in terms of products of consecutive linear factors. Some furthermultiplication givesp(a) = � 1

2(a � 1)a(a + 1) � 12 a(a + 1) + 1

3(a � 1)a(a + 2)(a + 1) + 13 a(a + 2)(a + 1) which

preserves the structure of the products of consecutive factors.Now for example, the first term involving Ân

a=212(a� 1)a(a+ 1) can be evaluated by changing

the summation index to k = a � 2 and then using the Hockey Stick:Ân�2

k=033! (k + 1)(k + 2)(k + 3) = 3 Ân�2

k=0 (k+3

3 ) = 3(n�2+44 ) . The same method works for the

other three terms to be summed.

9. Chess Masters Four identical white pawns and fouridentical black pawns are to be placed on a standard8 ⇥ 8, two-colored chessboard. How many distinct ar-rangements of the colored pawns on the colored boardare possible?No two pawns occupy the same square. The color of a pawnneed not match the color of the square it occupies, but it might.You may give your answer as a formula involving factorials orcombinations: you are not asked to compute the number.

Answer: There are 64!56!4!4! arrangements of the colored pawns on the standard board.

Solution First choose four squares from the 64 on the board for the black pawns. There are 64!60!4!

possibilities. Then choose four squares from the 60 remaining squares for the white pawns.This gives 64!

60!4!60!

56!4! =64!

56!4!4! arrangements.

10. Chess Wallpaper How many distinct plane wallpa-per patterns can be created by cloning the chessboardarrangements described in Question 9?Each periodic wallpaper pattern is generated by this method:starting with a chessboard arrangement from Question 9 (themaster tile), use copies of that tile to fill the plane seamlessly,placing the copies edge-to-edge and corner-to-corner. Note thatthe resulting wallpaper pattern repeats with period 8, horizon-tally and vertically.When the tiling is done, the chessboard edges and corners van-ish, leaving only an infinite periodic pattern of black and whitepawns visible on the wallpaper.

Regard two of the infinite wallpaper patterns as the same if and only if there is a plane translation thatslides one wallpaper pattern onto an exact copy of the other one. You may slide vertically, horizontally, or acombination of both, any number of squares. (Rotations and reflections are not allowed, just translations.)Note that the wallpaper pattern depicted above can be generated by many different master tiles (by regard-ing any square 8 ⇥ 8 portion of the wallpaper as the master tile chessboard). The challenge is to account forsuch duplication. Remember that each master tile has exactly four pawns of each color.You may give your answer as an expression using factorials and/or combinations (binomial coefficients).You are not asked to compute the numerical answer.

Answer: There are1

32

h 64!56!4!4!

+ 3⇣ 32!

28!2!2!

⌘+ 12

⇣ 16!14!1!1!

⌘i=

132

h(64

4 )(604 )� 3

⇣(32

2 )(302 )� 3(16

1 )(151 )

⌘� 7(16

1 )(151 )

i+ 1

16

h3⇣(32

2 )(302 )� 3(16

1 )(151 )

⌘i+ 1

8

h7(16

1 )(151 )

i

= 9, 682, 216, 530 different wallpaper patterns.

Solution By "8 ⇥ 8 pattern" we will mean a pattern as described in the question, that is, astandard 8 ⇥ 8 chessboard that has a light square in the upper left corner and that containsfour black pawns and four white pawns.The difficulty in counting the wallpaper patterns generated by these 8⇥ 8 patterns is that twodifferent 8 ⇥ 8 patterns sometimes produce the same wallpaper pattern. Moreover, the num-ber of distinct 8 ⇥ 8 patterns that produce a particular wallpaper depends on the subpatternsof the pattern. We cannot simply divide the number of 8⇥ 8 patterns by a single integer to ac-count for the duplication. We must sort the 8⇥ 8 patterns according to how much duplicationthey produce.Begin with any 8 ⇥ 8 pattern and extend it to its wallpaper. The other 8 ⇥ 8 patterns thatproduce the same wallpaper are exactly the patterns obtained by placing an 8 ⇥ 8 windowover the wallpaper so that the upper left corner of the window is one of the light coloredsquares on the original board. There are 32 possible 8 ⇥ 8 patterns obtained this way, butsome of these 32 may be duplicates. This duplication occurs precisely when the 8 ⇥ 8 boardcomes from a wallpaper that has sub patterns. This duplication is what we must account for.We show two different solutions. The first is a general system known as the Burnside method.The second solution directly counts the number of different patterns.Examine the possibilities, as follows. Consider any two squares in an 8 ⇥ 8 pattern and lookat the two new 8 ⇥ 8 patterns obtained by placing an 8 ⇥ 8 window so that each of the twoselected

squares is in the upper left corner. If the resulting8⇥ 8 patterns are identical then we say the two squares"match" for these boards and wallpapers. This meansthe two squares are functionally the same for this pat-tern. For instance, in the example shown here (theone shown in Question 9) all of the squares with blackpawns match each other. All of the squares with white

pawns match each other. All of the squares just to the right of a white pawn match each other,and so on. (It is patterns like this, the ones with subpatterns, that produce duplications.)Again consider a particular 8 ⇥ 8 pattern. If exactly k squares match one of the squares in thepattern, then exactly k squares match each of the other squares in the pattern. In the example,k = 4. The 64 squares in the 8 ⇥ 8 pattern are partitioned into families of matching squareswith k members each. Thus k is a factor of 64. Because there are just four pawns of each color,k cannot be more than four. We conclude that k = 1, 2, or 4. Call the number of squares in eachmatching family the "symmetry number" of the 8 ⇥ 8 pattern. For the example, k = 4.Matching squares are the same color, so each family is either light or dark, and there are 32

kfamilies of light squares and 32

k families of dark squares.The families are constructed so that putting the upper left corner of the 8 ⇥ 8 window at onesquare in the family always matches the 8 ⇥ 8 pattern that is obtained by putting the windowat another square of the family and never matches the pattern obtained by putting the upperleft corner of the window at a square from a different family. Each family produces one 8 ⇥ 8pattern. Since the upper left corner is always light and since there are 32

k families of lightsquares, there are exactly 32

k different 8 ⇥ 8 patterns that generate the wallpaper pattern. Forthe pattern in the example, there are 32

4 = 8 different 8⇥ 8 patterns that generate the wallpaperpattern: put any of the light squares in the top two rows in the upper left corner. Each choicemakes a different 8 ⇥ 8 pattern and all will produce the same wallpaper pattern. If you putany other square in the upper left corner, your 8 ⇥ 8 pattern will be the same as the patternyou get by choosing the square in the top two rows that matches your square.Rewrite the fact that there are 32

k different patterns for the wallpaper for a pattern with sym-metry number k using multiplication instead of division,

[symmetry number] ⇥ [number of different 8 ⇥ 8 patterns that generate the wallpaper] = 32

The 8⇥ 8 patterns can be grouped by their associated wallpaper patterns. There is a set of 8⇥ 8patterns for each wallpaper pattern. If you combine all the sets for all the distinct wallpaperpatterns you will get all the 8 ⇥ 8 patterns.Here is the first nice trick: The product

[symmetry number] ⇥ [number of different 8 ⇥ 8 patterns that generate the wallpaper]can be viewed as the sum of one copy of the symmetry number for each pattern that generatesthe wall paper pattern. It is what you get if you add up all the symmetry numbers of all the8 ⇥ 8 patterns for a single wallpaper pattern. That is, in symbols

[symmetry number] ⇥ [number of different 8 ⇥ 8 patterns that generate the wallpaper] = Â k

where the sum is taken over all the 8 ⇥ 8 patterns that generate the wallpaper pattern.For a single wallpaper pattern the symmetry numbers are all the same. The sum is a compli-cated way to write a simple quantity.

So far we have obtained that for a single wallpaper pattern

Â k = 32

where the sum is taken over all the 8 ⇥ 8 patterns that generate the wallpaper and k is thesymmetry number of each 8 ⇥ 8 pattern.Now add up all the symmetry numbers for all the 8 ⇥ 8 patterns - not just for one singlewallpaper, but for all the wallpaper patterns. That is, sum over all wallpaper patterns. On theright hand side, you will get 32 times the number of different wallpaper patterns!!!!!!!!!!!! Thatis, calling the number of different wallpaper patterns W,

Â k = 32W.where the sum is now taken over all of the different 8 ⇥ 8 patterns for all of the wallpaperpatterns and k is symmetry number.The plan is to find the sum of all the symmetry numbers of all the patterns and divide by 32.Remember that the symmetry number of a particular 8 ⇥ 8 pattern is the number of squaresin that pattern that "match" the square in the upper left corner (include the corner itself). Thesum of all the symmetry numbers is the sum of the number of squares that match the upperleft corner, taken over all 8⇥ 8 patterns. We need to count all the squares that match the upperleft corners in all the patterns. We could do this by counting those squares, 8 ⇥ 8 pattern by8 ⇥ 8 pattern. Think of having a stack of cards, each with one of the patterns, and goingthrough them, one by one, counting the squares that match the upper left corner.Here is the second nice trick: Turn this around and instead of counting pattern by pattern,look square by square. For each square in the 8 ⇥ 8 chessboard, count the number of 8 ⇥ 8patterns for which that square matches the upper left corner. That is, instead of going card bycard, go square by square. Pick a square and go through all the cards, counting the number ofcards for which this square matches the upper left corner. Go through the stack once for eachof the 64 squares. (This is similar to adding up all the entries in a matrix by either going rowby row or going column by column.)It turns out that it is not hard to count square by square:None of the dark squares ever match the upper left corner, so the contributions from thosesquares are all zero. Now we must look at the 32 light squares.No square in any even numbered column or even numbered row can match the upper leftcorner. There are several ways to see this. Suppose the square diagonally below the upperleft corner, for instance, matches the corner. Then so also does the next one diagonally down.And so on: there will have to be 8 matching squares. However, there are never more thanfour matching squares. We conclude that the square diagonally below the upper left cornerwill never match the corner. The reasoning is similar for all squares in even numbered rowsor columns.There remain 16 squares to look at.The upper left square matches itself for all 8 ⇥ 8 patterns, so the number for this square is

64!56!4!4!Suppose the third square in the top row matches the corner. Then so also do the fifth andseventh. We find that the pattern must consist of repeats of the first two columns and thatthere must be exactly one white pawn and one black pawn in those first two columns. Also,any such pattern will make this third square in the top row match the corner. Choose one ofthe 16 squares for the black pawn and one of the remaining 15 squares for the white pawn.There are 16!

14! such patterns.

Similar analysis produces the numbers for the other squares. For instance, count how manypatterns have the fifth square in the top row matching the corner. In this case we find thepattern consists of a repeat of the first four columns and that any pattern that consists ofrepeats of its first four columns will do. There will be two pawns of each color in those firstfour columns. Some of the patterns will have subpatterns in those first four columns, e.g. thesecond pair of columns could be a repeat of the first pair. The beauty of this method is thatthis does not matter here! We have 32 squares in which to place 2 black pawns and 2 whitepawns. Choose 2 squares out of 32 and then 2 more out of the remaining 30. The number ofpatterns is 32!

28!2!2!Continuing, we obtain that the upper left corner square matches itself for all 64!

56!4!4! patterns,three of the squares (the one in the fifth column and first row, the one in the fifth row and firstcolumn, and the one in the fifth row and fifth column) match the corner for 32!

28!2!2! patterns,and the remaining twelve squares each match the corner for 16!

14! patterns. Summarizing, thecount over all the 8 ⇥ 8 patterns of all the squares matching the upper left corner is

64!56!!4!4!

+ 3(32!

28!2!2!) + 12(

16!14!1!1!

).

Divide by 32 to get the number of different wallpaper patterns.

A second way to solve the problem: Direct enumerationFollow the first solution until you come to the "first nice trick." Instead of using the trick, wewill attempt to enumerate the number of different patterns directly by counting the numberof patterns with each symmetry number.Denote by Wk the number of different wallpaper patterns that have symmetry number kand denote by Bk the number of different 8 ⇥ 8 patterns that have symmetry number k.The statement above says that 32

k ⇥ Wk = Bk. The thing we have been asked to count isW = W1 + W2 + W4, the total number of distinct wallpaper patterns. To obtain W, we cancount the Bk and compute W = ( 1

32 ⇥ B1) + ( 116 ⇥ B2) + ( 1

8 ⇥ B4)

First we find B4, the number of distinct 8 ⇥ 8 patterns with symmetry number four. For eachsuch pattern, the 64 squares fall into 16 families of 4 matching squares. From each family selectthe square that is in the highest row. If there are several in the same row, then take the one ofthose that is farthest left. This will partition the 8 ⇥ 8 pattern into four blocks of 16 squares.Each block will contain one black and one white pawn. The block containing the upper leftsquare can be either the top two rows, the left two columns, or a 4x4 square. The pawns canoccupy any two squares in the block, so there are 16 ⇥ 15 different possible blocks of eachshape. Beginning with a two row block, one pattern results from repeating the block fourtimes down the 8 ⇥ 8 pattern. Another results from cloning the first two rows and shiftingthem right two squares for the second two rows, to the right another two squares for the thirdpair of rows, and right another two squares for the last pair of rows. A third pattern resultsfrom cloning and shifting right four squares each time. A fourth pattern results from cloningand shifting right six squares. After that, the we come back to the original cloning with noshift. We conclude that four different 8 ⇥ 8 patterns can be obtained from a single choice oftwo rows that contains one pawn of each color. We get 4 ⇥ 16 ⇥ 15 patterns. Beginning witha square block, we obtain patterns in three ways. Cloning and tiling the 8 ⇥ 8 pattern withfour of the clones gives one pattern. Clone the square block and put one copy in the upperleft corner and one in the lower left corner. Put the other two in the right side of the square,

but offset them so that they are shifted down two squares relative to the left side. This turnsout to produce a pattern that will have as its assigned block its top two rows, so we do notobtain a new pattern. Lastly, we can put two of the 4x4 squares in the top half of the 8 ⇥ 8pattern and put the other two below, offset by two squares. We find that each of the 16 ⇥ 15different 4x4 blocks produces two new 8 ⇥ 8 patterns. Beginning with a two column block,one pattern results from repeating the block four times across the 8 ⇥ 8 pattern. Anotherresults from cloning the first two columns and shifting them down two squares for the secondtwo columns, down another two squares for the third pair of columns, and down anothertwo squares for the last pair of columns. A third pattern results from cloning and shiftingdown four squares each time. A fourth pattern results from cloning and shifting down sixsquares. After that, the we come back to the original cloning with no shift. The second andfourth possibilities turn out to have already been counted among the patterns with the toptwo rows for blocks. The third possibility was already counted with the patterns whose blockis 4x4. Thus only one of the ways to use the two column block produces new patterns. Weconclude that in all there are 7 ⇥ 16 ⇥ 15 different 8 ⇥ 8 blocks with symmetry number 4.B4 = 7 ⇥ 16 ⇥ 15 and W4 = 4

32 ⇥ B4 = 18 ⇥ B4 = 1

8 ⇥ 7 ⇥ 16 ⇥ 15.

Now we find B2, the number of distinct 8⇥ 8 patterns with symmetry number two. This time,the 64 squares fall into 32 families of 2 matching squares. Choose one square from each familyjust as in the case above. This time we obtain two possible shapes for the blocks and each blockcontains two pawns of each color. This time the block that contains the upper left corner canbe either the top four rows or the left four columns. The pawns can occupy any four squaresin the block, so there are (32

2 )⇥ (302 ) different ways to distribute the pawns in each block. In

the case of B2, however, not all of the choices for filling the 32 squares produce a pattern withsymmetry number 2. Blocks that have a subsymmetry will produce patterns with symmetrynumber 4. Reasoning as we did for finding B4, we find that if there is a subpattern, thenthe pattern is made in one of three ways. For the case of the block with four rows and eightcolumns either the right half of the rectangle is a clone of the left half, the bottom half of therectangle is a clone of the top half, or the bottom half of the rectangle is the same as the tophalf shifted half way across the rectangle. The structure is analogous for the rectangular blockwith eight rows and four columns. In all these cases, we can obtain a pattern by placing onepawn of each color in a single one of the 16-square halves of the pattern. Thus the number ofsuch blocks that produce patterns with symmetry number 4 instead of 2 is 3 ⇥ 16 ⇥ 15. Weconclude that there are (32

2 ) ⇥ (302 ) � 3 ⇥ 16 ⇥ 15 useful basic blocks of 32 that are four rows

by eight columns and the same number that are eight columns by four rows. There are twodifferent 8 ⇥ 8 patterns that can be produced from each of these basic blocks- either clone thebasic block to the other half of the 8 ⇥ 8 square or clone it and shift it half way. Cloning avertical one and shifting half way will also be obtained by cloning a horizontal pattern andshifting half way. Thus, B2 = 3 ⇥ ((32

2 )⇥ (302 )� 3 ⇥ 16 ⇥ 15)

and W2 = 232 ⇥ B2 = 1

16 ⇥ 3 ⇥ ((322 )⇥ (30

2 )� 3 ⇥ 16 ⇥ 15).

Finding B1 is now easy, because each 8 ⇥ 8 pattern that does not have symmetry number 2 or4 must have symmetry number 1. That is,

B1 = total number of 8 ⇥ 8 patterns (computed in problem 9) �B2 � B4

= 64!56!4!4! � 3 ⇥ ((32

2 )⇥ (302 )� 3 ⇥ 16 ⇥ 15)� 7 ⇥ 16 ⇥ 15

and W1 = 132 B1

Now we compute

W = ( 132 ⇥ B1) + ( 1

16 ⇥ B2) + ( 18 ⇥ B4)

= 132

⇣64!

56!4!4! � 3((322 )(

302 )� 3⇥ 16⇥ 15)� 7⇥ 16⇥ 15

⌘+ 1

16

⇣3�(32

2 )(302 )� 3⇥ 16⇥ 15

�⌘+ 1

8(7⇥16 ⇥ 15)For neatness, we can write 16 ⇥ 15 as (16

1 )(151 ) and 64!

56!4!4! as (644 )(

604 ). This gives

W = 132

h(64

4 )(604 )� 3

⇣(32

2 )(302 )� 3(16

1 )(151 )

⌘� 7(16

1 )(151 )

i+ 1

16

h3⇣(32

2 )(302 )� 3(16

1 )(151 )

⌘i+ 1

8

h7(16

1 )(151 )

i

END OF CONTEST

Twenty-fourth Annual UNC Math Contest Final Round January 2016 Page 1 Solutions to questions 1 through 6(a), by Rocke Verser February 1, 2016

1. Even on the easiest problem on a contest, I often make a sketch and/or copy the information from the question. I find it helps me to internalize the problem, and to reduce "dumb" mistakes. Since this is a right triangle, I immediately think of the Pythagorean Theorem.

222

222 125056

cbacbacba

=+

=++

=++

Solution 1: Pythagorean Triplets. (A nice hint from the name of the question.) Combine the last two equations written above.

25625

12502

22

=

=

=+

cccc

The most well-known Pythagorean Triplet is the 3-4-5 triangle. If we scale this triangle by a factor of 5, we will have the correct hypotenuse. That would give us a 15-20-25 triangle. But the sum of sides is 15+20+25=60. And that doesn't meet the other specification of the problem. The first series of Pythagorean Triplets continues with odd sides: 3-4-5, 5-12-13, 7-24-25, 9-40-41, ... . The 7-24-25 triangle has the correct hypotenuse. Check the sum of the sides, 7+24+25=56. This meets all the criteria of the question. Now, we can answer the question: "What is the area?" Using the formula for the area of a triangle, we get the final answer:

8471272421

21 =×=××=×= heightbaseA

Solution 2: Algebra. First, we solve for c. Then we solve for a+b. A technique that sometimes helps with right triangle problems that don't seem to give enough information is to square the sum or difference of two sides.

25625

12502

22

=

=

=+

cccc

( )9612

31

315625

56

22

22

=++

=+

=+

=++

=++

bababa

baba

cba

We already found 625222 ==+ cba . Subtracting the last two equations leaves us with the area of 4 congruent triangles. 84 3366259612 2

121 ==×⇒=−= abheightbaseab .


2. The factorial of n is the product of all of the positive integers from n down to 1. The first step is to break the expression down starting from the innermost parentheses.

( )

( )[ ] ( )44 123456!6

123456!6!!36123!3

123222324!24!!4241234!4

×××××=

×××××==

=××=

××××××==

=×××=

L

We needed to multiply out 4! and 3!. But we must resist the urge to multiply out 24! and 6!. Why? First, quotients involving factorials are often easier to divide when they are already factored. Second, this question is specifically asking for the "prime factorization". So, anything we multiply will have to be factored, later. Probably the most straightforward way to solve this is to write out all the factors in the numerator and all the factors in the denominator and "cancel" out factors.

123456123456123456123456123456789101112131415161718192021222324

×××××××××××××××××××××××

×××××××××××××××××××××××

This is a perfectly fine solution method. A subtle enhancement is to group the numerator into 4 sets of 6 consecutive integer factor. Each group of 6 factors in the numerator is evenly divisible by 6!. (Think about why that's true, as we cancel factors.)

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

123456123456

123456789101112

123456131415161718

123456192021222324

123456123456123456123456123456789101112131415161718192021222324

×××××

××××××

×××××

××××××

×××××

××××××

×××××

×××××=

×××××××××××××××××××××××

×××××××××××××××××××××××

The rightmost fraction completely cancels. Rearranging the factors in the other denominators:

194711231532146

192021222324×××××=

×××××

×××××

13734171254136

131415161718×××××=

×××××

×××××

7232111435126

789101112×××××=

×××××

×××××

Now, everything is completely factored, except 224 = , so we can directly write the final answer: 6232 2371113171923 ×××××××


3. First, if the question had not provided a reasonably nice sketch, that is the very first thing I would draw. Even with the given perspective sketch, I still need to draw and label a face of the cube and an equilateral triangle that replaced a corner.

The area of the face of the cube was originally the area of a square of side-length 1. But we shaved off four isosceles right triangles, each with an unknown side length of x. So, the remaining area of the face of the square is 22

212 2141 xx −=×− . The hypotenuse of the isosceles right triangle has the same length as the side-length of

the equilateral triangles we created in place of the corners, xs 2= . A student may know the area of an equilateral triangle in terms of its side-length. If not, divide the equilateral triangle vertically into a 30-60-90 triangle. The ratio of the side-lengths of any 30-60-90 triangle is 2:3:1 . All of the side-lengths can be found by proportions if any of the three side-lengths are known. Or, if these proportions are not known, we can use the Pythagorean Theorem. The length of the hypotenuse is s. The length of the short side is s2

1 (because we just cut the equilateral triangle in half). The unknown side (the height) is found by the Pythagorean Theorem. With the height now known, we can find the area of the equilateral triangle.

( )( )

sh

sh

sh

shs

23

12

432

2412

22221

=

=

−=

=+

2

606060

43

23

21

21

sA

ssA

heightbaseA

=

××=

××=−−

Twenty-fourth Annual UNC Math Contest Final Round January 2016 Page 4 Solutions to questions 1 through 6(a), by Rocke Verser February 1, 2016 The requested area, A, is equal to the area of eight equilateral triangles. It is also equal to area of the octagonal face of the cube. Let's find both areas in terms of s2.

2

2

32438

sA

sA

=

×=

2

22

2

122

21

sAxsxs

xA

−=

=

=

−=

Finally, let's equate the two areas, solve for s2, and then substitute to find A. Although this problem didn't explicitly ask, some contests prefer that you rationalize the denominator.

( )

13232

32132

11132

132

2

2

2

22

+=

×=

+=

=+

−=

A

sA

s

s

ss

( )

113212

1343212

132

3234132132

13232

22

−=

−⋅

−=

−

−⋅=

−

−×

+=

A

A

A


4. Solution 1: Number Sieve. (A hint from the name of the question.) If a student has seen how the Sieve of Eratosthenes is used to quickly and reliably find the prime numbers less than 100, a similar strategy will work for this problem. First, we write the numbers from 1 to 99 or make a grid representing the numbers from 1 to 99. Then, for each factor given in the question, we put a tick mark on top of (or in the grid) for every number that is divisible by the specified factor.

0 1 2 3 4 5 6 7 8 9 0 / / // / /// / /// // 10 // //// // // /// //// 20 /// // / ///// / / // /// 30 //// /// / / // ///// / / 40 //// //// // /// / ///// / 50 // / // //// / //// / / 60 ///// / /// /// / /// // / 70 /// ////// / // // / /// 80 //// // / ///// / / / /// 90 ///// / // / / / ///// // //

Most of the tricks that make the Sieve of Eratosthenes fast (and clever) don't apply here. You have to laboriously mark each factor of each number. You have to be careful not to obscure one tick mark with another. And if you lose track of where you are, you may have to start from the beginning. Once the tick marks are done, find the numbers that are ticked exactly twice. I would suggest you write down those numbers (for partial credit). I found 18 such numbers. They are 4, 9, 10, 14, 15, 21, 27, 35, 44, 50, 52, 68, 75, 76, 81, 92, 98, and 99. Solution 2. Logical Reasoning. This method may be less tedious and more fun. But, I had a difficult time convincing myself I had not missed any categories.

• Powers of 3: (9, 27, 81) • Powers of 2: (4) • Odd prime (>9) multiples of 9: 9 × (11) • Odd prime (>9) multiples of 4: 4 × (11, 13, 17, 19, 23) • Odd multiples of 2: 2 × (5, 7, 52, 72) • Odd multiples of 3: 3 × (5, 7, 52) • Odd multiples of 5: 5 × (7)

All together, there are 18 numbers in my lists.


5. Solution 1. Probability tree. Probability trees are a safe and effective way to solve this kind of problem. However, if the problem only wants a tiny bit of information, we sometimes do a lot of extra work to complete the entire tree. Nevertheless, that's the method we will use for this solution. There are several different ways to draw and label a probability tree. In my sketch, we start at the top and work our way to the bottom. The top row represents the start of Day 1. The next row is the start of Day 2. And the bottom row is the start of Day 3. Horizontally, the left column represents zero rocks in the valley. Each subsequent column represents one more rock in the valley. The numbers in the nodes (circles) represent the probability that we are in that state at the start of the specified day. For example, we are guaranteed to have 1 rock in the valley on the start of Day 1, so we put "1" in the circle corresponding to "Day 1, 1 rock". Each edge (line segment) in our tree shows how the status can change from one day to the next. The numbers that label each edge represent the probability that the event will occur. So the edge that connects the node at Day 1 to the leftmost node at Day 2 indicates there is a 1/10 probability that we will go from 1 rock in the valley to zero rocks in the valley. The sum of the edges that leave any circle must always total 1. The numbers in each node can be found by summing the products of each probability along an incoming edge by the value in the immediately preceding node. If we haven't made a mistake, the sum of the probabilities leaving each node on a given day must total 1.

Twenty-fourth Annual UNC Math Contest Final Round January 2016 Page 7 Solutions to questions 1 through 6(a), by Rocke Verser February 1, 2016 The edge leaving "Start of Day 2 / 0 rocks" shows a probability of 1. Once there are zero rocks in the valley, there will always be zero rocks in the valley. The edges leaving "1 rock" are given directly by the problem. The edges leaving "2 rocks" require some additional thought. Each rock is independently vaporized, rolled back down into the valley, or split into two rocks. The probability that both rocks are vaporized is 0.12=0.01. The probability that both rocks are duplicated is 0.42=.16. One rock can be vaporized and one rock can be rolled back down into the valley (resulting in 1 rock in the valley) in 2 different ways. So, the probability this occurs is 2×0.1×0.5=0.10. One rock can be duplicated and one rock can be rolled back down into the valley (resulting in 3 rocks in the valley) in 2 different ways. So, the probability this occurs is 2×0.4×0.5=0.40. And finally, one rock can be vaporized and one rock can be duplicated in two different ways (2×0.1×0.4=0.08) or both rocks can be rolled back down into the valley (0.5×0.5=0.25). So the probability we go from 2 rocks to 2 rocks is 0.08+0.25=0.33. Let's check our probabilities of leaving "Start of Day 2 / 2 rocks". 0.01+0.10+0.33+0.40+0.16 = 1.00, which is the correct total. Next, we sum the probabilities entering the nodes along "Start of Day 3". For the node with zero rocks, we have 0.1×1+0.5×0.1+0.4×0.01=0.1+0.05+0.004=0.154. For the node with one rock, we have 0.5×0.5+0.4×0.10=0.25+0.04=0.29. For the node with two rocks, we have 0.5×0.4+0.4×0.33=0.2+0.132=0.332. For the node with three rocks, we have 0.4×0.40=0.16. And for the node with four rocks, we have 0.4×0.16=0.064. Let's check the total probabilities inside each node along "Start of Day 3". 0.154+0.290+0.332+.160+0.064=1.000, which is the correct total. The chance of having 2 rocks in the valley at the start of Day 3 is 0.332 or 33.2%. If you are confident in your skills with probability trees, we weren't required to compute the complete probability tree. In fact, we only needed to compute the edges and nodes leading to the node at the start of Day 3 with 2 rocks. (4 edges and 4 nodes.) Solution 2. Logical Reasoning. We can list all of the ways of ending up with 2 rocks: Roll the rock into the valley (5/10 probability). Then duplicate the rock (4/10 probability). 20/100. Duplicate the rock (4/10 probability). Then vaporize one, duplicate the other (4/10*1/10*4/10 * 2 ways) = 32/1000; or roll both rocks back down into the valley (4/10*5/10*5/10=100/1000). Tallying the above cases, 200/1000 + 32/1000 + 100/1000 = 332/1000 = 83/250.


6(a). We can simplify the problem by ignoring Sisyphus and the valley and the days. The question simply wants to know the probability that the number of rocks never reaches 0, when starting with 1 rock. Zeus picks up a rock at his leisure and vaporizes it (with probability 1/10), returns it to the pile (with probability 5/10), or splits the rock in two (with probability 4/10). If there are no rocks remaining, Zeus rolls over and goes to sleep. If Zeus returns the rock to the pile, nothing has changed, so we can further simplify the problem by ignoring that action, and considering only the relative probability of the other actions. Zeus picks up a rock at his liesure. He vaporizes it (with probability 1/5) or splits the rock in two (with probability 4/5). (The probabilities of vaporization:duplication = 1:4, as before the simplification.) Solution via State Diagram: We can represent the situation with a State Diagram. There is a single state for every possible number of rocks in the pile. Zero rocks is State S0. One rock is State S1. Etc. For each state, we have a variable nk which represents the total number of times the system has entered that state from the beginning to the end of time. See attached illustration.

We can write formulas to relate the variables, n, with each other. For example, if you are in State S1, Zeus will vaporize that one and only rock with probability 1/5. So, the number of times we enter State S0 is exactly equal to 1/5 the number of times we enter State S1. We enter State S1 once when the system is initialized. We also enter State S1 when we were in State S2 and Zeus vaporizes one of the two remaining rocks. So, the net

number of times we enter State S1 is exactly 5

1 21

nn += . At each other State, Sk we can enter the State when

Zeus splits a rock while we were in State Sk-1 or when Zeus vaporizes a rock while we were in State Sk+1. We have an infinite number of simple linear equations and an infinite number of unknowns. We could make an arbitrary guess and solve the system of equations iteratively. Or we could make a good guess and prove that our guess satisfies the system. First, note that each value of n is a weighted average of its two neighbors. Also note how perfectly regular the systems is for States S2 and higher.

Twenty-fourth Annual UNC Math Contest Final Round January 2016 Page 9 Solutions to questions 1 through 6(a), by Rocke Verser February 1, 2016 Let me redraw the State diagram ever so slightly:

Now, notice that state S1 looks almost the same as States S2 and higher. What if we guess that it *really* is the same as State S2? Let's set the input entering State S2 from the left equal to the input entering State S1 from the

left: 45

541 11 =⇒= nn . Let's solve for n2: 4

54

20-25n 5

145

51 2

221 ==⇒+=⇒+=

nnn . Now, it is a

simple matter to confirm 1 ,45

≥= knk . Then we solve for 41

51

0 ==nn . The final answer is

431 0 =− n .

6(a). Solution Estimate via Probability Tree See the adjoining probability tree. The leftmost column shows the probability the Zeus has vaporized the only remaining rock. The next column shows the probability that one rock remains. At each step, Zeus adds a rock or destroys a rock, so the number of rocks remaining alternates from odd to even. (Unless it was already zero.) The total probability at each stage must equal 1. So, before Zeus picks up the solitary rock, the probability that 1 rock is present is exactly 1. (Represented by the number in the circle at the top of the tree.) After Zeus handles a rock for the first time, we are on the next row. The total probability is 0.2+0.8=1. There is a 1/5 probability that Zeus destroyed the rock, and there is an 4/5 probability that there are now 2 rocks. At the next stage, Zeus has either destroyed one of the two rocks remaining or he has split one of the two rocks remaining, and now we have three rocks. The total probability is 0.2 (no rocks remained from the previous step) + 0.16 (one rock remains) + 0.64 (three rocks are in the pile) = 1.00. After the fourth step, the probabilities for 0, 1, 3, or 5 rocks are 0.232 + 0.0512 + 0.3072 + 0.4096 = 1.0000. After the ninth step, the probability that there are no rocks remaining is 0.248171008. Maybe we can guess the answer is converging towards 1–0.25, and hope the graders will award partial credit.

Twenty-fourth Annual UNC Math Contest Final Round January 2016 Page 10 Solutions to questions 1 through 6(a), by Rocke Verser February 1, 2016 6(a). Solution via Probability Tree This is going to be an infinitely wide and deep probability tree. We need a formula. So, I'll begin by counting the number of paths that get us to each place where Zeus has destroyed the last rock. See attached diagram. The numbers in the circles represent the number of paths from the root of the tree to that particular node. This was performed using standard MATHCOUNTS path counting methods. Note the 14s in the circles near the bottom. That indicates there are 14 paths from the root of the tree to those particular nodes. The lower-left of the two nodes is where Zeus has just destroyed the final rock. No matter how we get from the root to this node, Zeus has created 4 additional rocks and he has vaporized 5 rocks. Since Zeus has 14 ways to do this, the formula for the probability

that we reach this node is 001835008.051

5414

54

=⎟

⎠

⎞⎜

⎝

⎛×⎟

⎠

⎞⎜

⎝

⎛× ,

which number you may recognize from the previous solution. The sequence we have found begins with 1, 1, 2, 5, 14. Not surprisingly, these are the Catalan Numbers. (Also known as mountain numbers. If we rotate the diagram counterclockwise 90 degrees, we can imagine climbing a mountain, possibly with ups and downs, and eventually descending back to earth. C(0)=1, C(1)=1, C(2)=2, C(3)=5, C(4)=14. Now we can write a general formula for each of the nodes where Zeus has just destroyed the last rock:

nnnnn nCnCnC ⎟

⎠

⎞⎜

⎝

⎛⋅=⎟

⎠

⎞⎜

⎝

⎛⎟

⎠

⎞⎜

⎝

⎛⎟

⎠

⎞⎜

⎝

⎛⋅=⎟

⎠

⎞⎜

⎝

⎛⎟

⎠

⎞⎜

⎝

⎛⋅+

254

5)(

51

51

54)(

51

54)(

11

The total probability that all rocks will eventually be destroyed, is given by the infinite sum:

∑

∞

=⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

⎟

⎠

⎞⎜

⎝

⎛⋅=0 25

45

)(n

n

evernCp , where ( )

112+

⋅⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=nn

nnC

In general, the contest awards more points for a closed-form solution. However, as this is an infinite sum and the problem asks for "the probability", I am skeptical that much credit would be awarded without a closed form solution. Although I am not a grader, I would recommend a student should spend some of their 3 hours attempting to reduce the infinite sum to a number. At the very least, I think a student should provide a convincing argument that the sum converges to a valid probability (a number between 0 and 1).

If we let n

nCnT ⎟

⎠

⎞⎜

⎝

⎛⋅⋅=254

51)()( represent each term in the summation, above, we can find the ratio between

consecutive terms. Using the formula for C(n), above, a little algebra quickly shows that ( ) ( )224)1()( +−=− nnnCnC . Then, we have:

2516

2525816

224

254

)1()(

<+−

=+−

⋅=− n

nnn

nTnT

Twenty-fourth Annual UNC Math Contest Final Round January 2016 Page 11 Solutions to questions 1 through 6(a), by Rocke Verser February 1, 2016 Since the ratios of every term in our sequence are less than the ratios in a geometric series, we can find the upper bound on our summation, using the formula for the sum of a geometric series. (Although this may look too scary to do without a calculator, if you know your powers of 2, only the final term we show really requires any hand calculation (a power of 2, decimal shifted, times 14, divided by 9).

( ) ( )( )

40.251433280.24817100

92580.001835000.0040960.010240.0320.2925

525614

5320

532

54

51

11

254

5114

254

515

254

512

254

5111

511

1)4()3()2()1()0(

9753

2516

432

325162

2516

2516

<<

⋅++++<

⋅⋅

++++<

−⋅⎟

⎠

⎞⎜

⎝

⎛⋅⋅+⎟

⎠

⎞⎜

⎝

⎛⋅⋅+⎟

⎠

⎞⎜

⎝

⎛⋅⋅+⋅⋅+⋅⋅⋅<

++++++++<

ever

ever

ever

ever

ever

P

P

P

P

TTTTTP L

No single step, above, is exceptionally hard. Putting it all together is rather brutal. I think most students who could put all this together will probably find an easier solution. However, with a few more (non-trivial) steps, the summation can be reduced to a closed form, but I will leave that as an exercise for the very dedicated reader.

41=everp . So, 4

3411 =−=neverp

24th%ANNUAL%UNC%STATE%MATHEMATICS%CONTEST%%SUMMARY%OF%RESULTS%2016%

!The!Final!Round!was!held!in!Greeley!at!the!University!of!Northern!Colorado!on!Saturday,!!January!23,!2016.!The!contest!is!open!to!all!students!in!Colorado!in!grades!7F12.!!

TOP%16%WINNERS%%

RANK NAME

SCHOOL GR TEACHERS/COACHES 1! Chen! Hongyi! Fairview!! 10! Del!Ebadi!2! Mazenko! Austen! Cherry!Creek!! 8! Dotty!Dady!3! Sowrirajan! Hari! Cherry!Creek!! 10! Dotty!Dady!4! Swartz! Avi! Cherry!Creek!! 11! Dotty!Dady!5! Leing! T.J.! Alexander!Dawson! 12! Brian!Hannen,!Silva!Chang!6! Lou! Aaron! Highlands!Ranch! 11! Anthony!Verbsky!7! Zhang! Casey! Fairview!! 12! Emily!Silverman!8! Soh! Christine! Fairview!! 12! Emily!Silverman!9!! Chu! Sonia! Rock!Canyon!! 9! Ming!Song,!Erynn!Albert!9! Wu! Derwin! Cherry!Creek! 9! Dotty!Dady!11! Wang! Amy! Pine!Creek! 11! Ming!Song,!April!Pierce!12! Nelson! Lucas! Loveland!! 9! Greg!Morrison,!Judy!Leatherman!13! Lawrence! Kathryn! Fairview!! 12! Emily!Silverman!14! Aparicio! Zachary! D’Evelyn! 11! Cory!Ryckman!15! Thomas! Rahul! Cherry!Creek!! 7! Dotty!Dady!16! Chen! Jackson! Fairview!! 12! Emily!Silverman,!Silva!Chang!

!%

ADDITIONAL%WINNERS%%

Listed!alphabetically!by!name!!

!Ahn! Jinyoung! Highlands!Ranch! 12! Anthony!Verbsky!

!Gardner! Jason! D'Evelyn! 11! Cannon!

!Golla! Anurag! Fairview!! 11! Silva!Chang,!Emily!Silverman!

!Lin! Andrea! Fairview!! 10! Del!Ebadi,!Silva!Chang!

!Mehta! Sanat! Arapahoe!! 11! Dan!Swomley!

!New! Stephen! Fort!Collins!! 11! Chris!Suppes,!Craig!Luckasen!

!Starr! Andrew! Grandview!! 11! John!Schultz,!Laura!Morales!

!Weygandt! Scott! Niwot!! 12! Adria!Repp,!Silva!Chang!

!Zhang! Hannah! Fairview!! 10! Del!Ebadi,!Brad!Arehart!

!Zhang! Luke! Rocky!Canyon! 10! Ming!Song!

!Zhang! Stephanie! Fairview!! 10! Del!Ebadi,!Brad!Arehart!

!The!winners!listed!above!are!invited,!along!with!their!families!and!teachers,!to!a!banquet!to!be!held!in!their!honor!on!Saturday,!April!2,!2016!in!the!Panorama!Room!at!the!UNC!Student!Center.!!

24th%ANNUAL%UNC%STATE%MATHEMATICS%CONTEST%%2016%HONORABLE%MENTIONS%

Listed!alphabetically!by!name!!!!!

NAME

SCHOOL GR TEACHERS/COACHES Chang! Min!Young! Challenge!School! 7! Melissa!Eggleston,!Matt!Bixby!Chen! Claudia! Pine!Creek!! 9! Amy!Verduft,!Ming!Song,!April!Pierce!Dieter! Justin! Pine!Creek!! 11! April!Pierce!Gao! Sarah! Cherry!Creek! 9! Dotty!Dady,!Ming!Song,!Mrsl!Hughes!Hu! Michael! Cherry!Creek!! 9! Ming!Song!Ijju! Siddarth! Challenge!School! 8! Matt!Bixby!Joshi! Arnav! Challenge!School! 8! Matt.Bixby!Kenigsberg! Matthew! Home!school! 11! Katie!Kenigsberg!Krarti! Taha! Niwot!HS! 10! Adria!Repp,!Megan!Schlagel!Kumar! Aditya! Cherry!Creek! 10! Dotty!Dady!Lambert! Shane! Steamboat!Springs! 6! Val!Gary,!Sally!Lambert!Li! Amanda! Fairview!! 10! Del!Ebadi,!Brad!Arehart!Lin! Cynthia! Fairview!! 9! Brad!Arehart!Liu! Amber! Rock!Canyon! 9! Ming!Song,!Erynn!Albert!Nambiar! Keshav! Rampart!! 9! Robyn!MacIvor!Park! Hannah! Poudre!! 9! Mr.!Sayers!Ratanapan! Wutikorn! Coral!Acad!of!Science! 10! Selcuk!Guncu!!Roy! Varun! Rampart!! 9! Mr.!Mestas!Rui! Daniel! Summit!Charter! 8! Tanya!DeNobrega!Singh! Isani! Cherry!Creek!! 10! Dotty!Dady!Sun! Orien! Palmer!Ridge!! 10! Ming!Song,!Leah!Phillips,!Scott!Obermeyer!Wang! Ashley! Fairview!! 11! Silva!Chang,!Del!Ebadi!Wang! Jonathan! Fairview!! 10! Brad!Arehart,!Emily!Silverman,!Silva!Chang!Younger! George! Colorado!Academy! 12! Charity!Smith!

%2016%JUNIOR%HONORABLE%MENTIONS%

Listed!by!grade!in!alphabetical!order!!!!

NAME SCHOOL GR TEACHERS / COACHES Guo! Brandon! Peak!to!Peak!! 8! Anita!ChakrabortyFSpotts!Joshi! Sinaya! Challenge! 6! Matt!Bixby!Larsen! Carter! Challenge!! 7! Matt!Bixby!Subramanian! Emhyr! Challenge!! 8! Matt!Bixby!Tang! Eric! West! 7! Joanne!Parkhouse!Yang! Kevin! Mountain!Ridge! 7! Christine!Draper!!!Competition!was!fierce!this!year.!Seven!students!correctly!determined!the!probability!that!Sisyphus!would!have!to!labor!forever!in!Question!6(a)!and!four!were!able!to!determine!that!probability!in!6(b).!There!were!several!different!approaches.!Three!students!obtained!a!perfect!closed!form!answer!for!Question!8(b)!and!several!more!came!very!close.!Again,!there!were!several!good!approaches!to!this!one.!Several!more!students!gave!correct!summation!formulae.!A!few!students!attempted!Question!10,!about!the!chess!wallpaper!patterns,!and!one!student!made!substantial!progress.!!

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