University of Northern Colorado Mathematics Contest 2019-2020 · 2019-11-20 · University of Northern Colorado Mathematics Contest Problems are duplicated and solved by Ming Song

University of Northern Colorado Mathematics Contest

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

University of Northern Colorado Mathematics Contest 2019-2020 Solutions of First Round

1. This floor plan of the portal to the Addams castle shows twin chambers and a wedge-shaped pit (shaded) where a trap door will be placed. Find the degree measure of the acute angle at the tip of the wedge. The chambers are congruent regular ten-sided polygons.

Answer: 72

Solution 1:

One exterior angle of a regular n-sided polygon measures

.

One exterior angle of a regular decagon (10-sided polygon) measures

.

The marked angle contains two exterior angles of regular decagons as shown.

The marked angle in the original diagram measures

.

Solution 2:

One interior angle of a regular n-sided polygon measures

.

360n°

360 3610°= °

2 36 72× ° = °

( )2 180nn

- × °



One interior angle of a regular decagon measures

.

The marked angle in the original diagram measures

.

2. Find the value of for the function F defined by for and for .

Answer: 0.666

Solution:

.

3. Grandmama starts with identical cauldrons, one on her left that is full of liquid and one on her right that is empty. She pours three-quarters of the liquid from the left one into the right one and casts a spell. Then she pours one third of the liquid in the right one back into the left one. If she now pours one third of the contents of the left cauldron back into the right cauldron, what fraction of the liquid is now in the cauldron on the left?

Answer:

Solution 1:

Let us do the operations step by step:

Left Right

1 0

8 180 14410× °

= °

360 2 144 72° - × ° = °

( )13.333F ( ) 2F x x= 1x <

( ) ( )1F x F x= - 1x ³

( ) ( ) ( ) ( )13.333 12.333 1.333 0.333 2 0.333 0.666F F F F= = = = = × =!

13

14

34

1 112 2

- =3 2 14 3 2× =

1 2 12 3 3× =



The answer is .

Solution 2:

We may assume a number for the capacity of the left cauldron. We use 12.

Now we do the operations step by step:

Left Right

12 0

3 9

6 6

4

The desired fraction is .

4. Find the smallest number of nightshade seedlings Uncle Fester could have if when he arranges them with 8 in each row, he has three left over and when he puts 9 in each row, he has 4 left over.

Answer: 67

Solution 1:

We are searching the number n:

mod 8,

mod 9.

The smallest positive value of n is

.

Solution 2:

We are searching the number n:

mod 8,

mod 9.

If you don’t see the same “negative” remainders in solution 1, we may do the following.

Let for some nonnegative integer k. We have

mod 9.

Do mod calculations:

.

The smallest nonnegative value of k is 8.

The smallest positive value of n is .

The answer is 67.

13

4 112 3

=

3 5n = = -

4 5n = = -

( )lcm 8, 9 5 72 5 67- = - =

3n =

4n =

8 3n k= +

8 3 4k + =

8 1mod 9 1mod 9 1 8mod 9k k k= ® - = ® = - =

8 8 3 67n = × + =



Solution 3:

If you don’t know the “mod” sign, we may think this way.

Let Uncle Fester borrow 5 nightshade seedlings.

When he arranges the original nightshade seedlings + 5 borrowed with 8 in each row, he has none left over and when he puts 9 in each row, he has none left over.

Then the least number of nightshade seedlings is the least common multiple of 8 and 9. It is .

Now ask Uncle Fester to return 5 nightshade seedlings.

The answer is .

Solution 4:

We list the numbers that yield remainder 3 upon division by 8 until we hit a number that yields remainder 4 upon division by 9:

3, 11, 19, 27, 35, 43, 51, 59, 67, .

We see that 67 yields remainder 4 upon division by 9.

The answer is 67.

5. Pugsley tries to fit a round circle into a square hole, but part of the circle sticks out. What is the total area of the two excess regions that Pugsley will carve from the circle? The circle has radius 10, and Pugsley has it placed so that it passes through one corner of the square and is tangent to two sides of the square.

Answer:

Solution:

Let ABCD be the square with A on the circle.

Let O be the center of the circle. Let AB intersect the circle at E.

8 9 72× =

72 5 67- =

!

50 100p -

A B

C D

O

E



Draw the diagonal AC of the square. Obviously, O is on AC.

With and , is an isosceles right triangle.

Then we have .

The area of the shaded circular segment AE is

.

The total area of the shaded regions is

.

6. The Mummy’s tomb is a regular pyramid whose base is square and whose four upright faces are congruent isosceles triangles. A scarab beetle crawls along the outer surface of the pyramid from one corner of the base to the midpoint of the opposite upright edge. Find the length of the shortest path the beetle can take. Each triangular face has top angle 30 degrees; its upright sides each have length 26.

Answer:

Solution:

Let PABCD be the pyramid as shown.

We display the lateral faces on a plane.

The shortest path is the straight segment AM on the plane.

With , , and , AMP is a 30°-60°-90° triangle.

Therefore, .

45EAOÐ = ° OA OE= AOE

90AOEÐ = °

2 21 110 10 25 504 2

p p× - × = -

( )2 25 50 50 100p p× - = -

13 3

26PA = 13PM = 2 30 60APMÐ = × ° = °

13 3AM =

30º

26

60º

26

A

B C

D

P

M

P

A B

C

M



7. Wednesday has a lie detector machine that is correct most of the time, but labels a truthful person as a “liar” with probability 0.06 and labels a liar as “truthful” with probability 0.08. She interrogates three children, who all claim they did not frighten Wednesday’s pet iguana. In actuality, exactly two are guilty, and one is innocent. She tests all three suspects. The machine labels two as liars and one as innocent. What is the probability that the machine has correctly identified the guilty pair?

Answer:

Solution:

The probability that the machine labels two as liars and one as innocent is

.

The probability that the machine correctly identifies the guilty pair is

.

The probability that the machine correctly identifies the guilty pair under that the machine labels two as liars and one as innocent is

.

This is the answer.

8. Morticia gives a number between 1 and 65 to Lurch, and then Pugsley tries to guess what it is by asking Lurch questions. Pugsley asks first “Is your number greater than 20?” and Lurch answers. Then Pugsley asks “Is your number even or odd?” and Lurch answers. “Hmm,” says Pugsley, “Is your number a perfect square?” When Lurch answers, Pugsley says “I think I almost have it. I will know if you tell me whether one of the digits in your number is a 4.” Lurch answers and Pugsley says “I have it!” Just then, Morticia looks up and says “Oh Pugsley-I forgot to tell you! Lurch is having a backwards day today. Every answer he gives is wrong.” Pugsley furrows his brow. After a minute he says, “Thanks. That is ok. I can still tell what his number is!” What does Pugsley think is Lurch’s number?

Answer: 14

Solution:

We analyze in two cases.

Case 1: the answer is “no” to “is your number greater than 20?”

The numbers are from 2 to 20 inclusive.

We divide the numbers into two groups: even and odd

Even: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Odd: 3, 5, 7, 9, 11, 13, 15, 17, 19

10811093

20.92 0.92 0.94 0.92 0.08 0.06

1æ ö

´ ´ + ´ ´ ´ç ÷è ø

0.92 0.92 0.94´ ´

0.92 0.92 0.94 92 94 23 47 10812 92 94 2 8 6 23 47 12 1093

0.92 0.92 0.94 0.92 0.08 0.061

´ ´ ´ ´= = =

´ + ´ ´ ´ +æ ö´ ´ + ´ ´ ´ç ÷

è ø



There is a digit 4 only in some even numbers. Hence the answer to “is your number even or odd?” is “even”.

In the even numbers there are 2 square numbers: 4 and 16. One number contains a digit 4 and the other doesn’t.

In the even numbers there are 8 non-square numbers: 2, 6, 8, 10, 12, 14, 18, 20. There are 7 numbers not containing a digit 4.

Since Pugsley can know the number by knowing whether of one of digits of the number is a 4, the answer to “is your number a perfect square?” is “yes”.

If the answer is “yes” to “whether one of the digits is a 4, Lurch’s number is 4.

If the answer is “no” to “whether one of the digits is a 4, Lurch’s number is 16.

Now we examine which situation with the opposite answers still allows Pugsley to determine Lurch’s number.

We use the opposite answers for all.

The numbers are greater than 20: 21 to 63 inclusive.

The numbers are odd: 21, 23, 25, 27, 29, , 47, 49, 51, 53, , 63.

The numbers are non-squares: 21, 23, 27, 29, , 47, 51, 53, , 63.

If the answer is “no” to “whether one of the digits is a 4, the number can be one of 21, 23, 27, 29, , 39, 51, 53, , 63. Pugsley cannot determine the number.

If the answer is “yes” to “whether one of the digits is a 4, the number can be one of 41, 43, 45, 47. Pugsley cannot determine the number either.

Therefore, there is no solution in this case.

Case 2: the answer is “yes” to “is your number greater than 20?”

The numbers are from 21 to 63 inclusive.

We divide the numbers into two groups: even and odd

Even: 22, 24, 26, 28, , 62

Odd: 21, 23, 25, 27, , 63

Subcase a: to “is your number even or odd?”, the answer is “even”

Since there is only one square number: 36, the answer to “is your number a perfect square?” is “no”.

Then Pugsley cannot determine Lurch’s number because there are two or more numbers containing a digit 4 and there are two or more numbers not containing a digit 4 in the non-squares.

Subcase b: to “is your number even or odd?”, the answer is “odd”

There are two squares 25 and 49. One contains a digit 4, and the other doesn’t. Therefore, the answer to “is your number a perfect square?” is “yes”.

If the answer is “yes” to “whether one of the digits is a 4?”, the number is 49.

If the answer is “no” to “whether one of the digits is a 4?”, the number is 25.

! !

! !

! !

!

!



Now we examine which situation with the opposite answers still allows Pugsley to determine Lurch’s number.

We use the opposite answers for all.

The numbers are less than or equal to 20: 2 to 20 inclusive.

The numbers are even: 2, 4, 6, 8, 10, 12, 14, 16, 18.

The numbers are non-squares: 2, 6, 8, 10, 12, 14, 18.

If the answer is “no” to “whether one of the digits is a 4?”, the number cannot be determined.

If the answer is “yes” to “whether one of the digits is a 4?”, the number is 14. Pugsley can still determine Lurch’s number.

Therefore, Lurch’s number is 14.

Let us make a summary:

Lurch’s number is 14.

To “is your number greater than 20?”, the liar’s answer is “yes” (the true answer is “no”).

To “is your number even or odd?”, the liar’s answer is “odd” (the true answer is “even”).

To “is your number a perfect square?”, the liar’s answer is “yes” (the true answer is “no”).

To “whether one of the digits is a 4?”, the liar’s answer is “no” (the true answer is “yes”).

The answer to the problem is 14.

9. Lurch plays Whack-a-Thing on a circular table that has holes numbered 1 through 19 counter-clockwise. If Thing is currently at hole numbered x, Thing will next travel steps counterclockwise to reach its new hole. Play begins with Thing selecting a starting hole, and a round of play ends when Thing returns to a hole previously visited in that round. What is the maximum number of distinct holes that Thing can occupy in a single round of play?

Example. If Thing starts at , it will then travel steps counterclockwise to

hole . Next Thing will travel steps to hole number . Next, from hole 16, Thing will travel steps, completing several circuits around the table and ultimately appearing at hole .

( )1x x -

2x = ( )2 2 1 2- =

2 2 4+ = ( )4 4 1 12- = 4 12 16+ =16 15×!



Answer: 7

Solution:

Note that

.

The next number is the square of the previous number in mod 19.

Let us list.

1 distinct hole

7 distinct holes

7 distinct holes

(see the 2nd line) 6 distinct holes

6 distinct holes

6 distinct holes

2 distinct holes

(see the 7th line) 3 distinct holes

(see the 3rd line) 6 distinct holes







( ) 21x x x x+ - =

1 1®

81 25 362 16 9 5 6 1

247

34® ® ® ® ® ® ®

- -

93 5 6 179 4 16® ® ® ® ® ® ®

4 ®

56 17 45 16 9® ® ® ® ® ®

617 4 16 6 9 5® ® ® ® ® ®

49 6417 718

® ®-

648 7®

9 5®

10 59

®-

11 78

®-

12 77

11 11® ® ®-

13 176

® ®-

14 65

® ®-

15 164

®-



(see the 3rd line) 6 distinct holes

(see the 2nd line) 6 distinct holes

2 distinct holes

1 distinct hole

We see the maximum number of distinct holes to be 7. It can be achieved when is 2, 3, 10, 13, 14, or 15.

The answer is 7.

The following is Professor Diaz’s approach:

You might find it helpful on this problem to organize information in a graph like this:

16 93

®-

17 42

®-

18 11

®-

190

19®

0x



10. Gomez, Morticia, Pugsley, Wednesday, Uncle Fester, and Cousin Itt sit at a round table. If each casts a hex on one of the others, chosen at random, what is the probability that everyone gets hexed?

Answer: 53/3125 When everyone gets hexed, you have each person matched to one of the others. This is a rearrangement of six items in which no item stays in its original place. This is called a “derangement.” There are many ways to compute the number of these. You may want to look up the term “derangement.”

Solution 1:

We first calculate the probability that at least one doesn’t get hexed.

Denote the six people by G, M, P, W, F, and I.

What is the probability that one specified person (say G) doesn’t get hexed?

For G not to get hexed, any of M, P, W, F, and I can cast a hex on one of the four other people (not G), and G can cast a hex on one of M, P, W, F, and I.

The probability that one specified person doesn’t get hexed is .

What is the probability that two specified people (say G and M) don’t get hexed?

For G and M not to get hexed, any of P, W, F, and I can cast a hex on one of the three other people (not G and not M), and any of G and M can cast a hex on one of P, W, F, and I.

The probability two specified people don’t get hexed is .

5 14 55 5æ ö æ ö×ç ÷ ç ÷è ø è ø

4 23 45 5æ ö æ ö×ç ÷ ç ÷è ø è ø

G

M

P

W F

I

G

M P

W F

I



What is the probability that three specified people (say G, M, and P) don’t get hexed?

For G, M, and P not to get hexed, any of W, F, and I can cast a hex on one of the two other people (not G, not M, and not P), and any of G, M, and P can cast a hex on one of W, F, and I.

The probability that three specified people don’t get hexed is .

What is the probability that four specified people (say G, M, P, and W) don’t get hexed?

For G, M, P, and W not to get hexed, F and I must cast a hex on each other, and any of G, M, P, and W can cast a hex on one of F and I.

The probability that four specified people don’t get hexed is .

With the inclusive-exclusive principle, the probability that at least one doesn’t get hexed is

3 32 35 5æ ö æ ö×ç ÷ ç ÷è ø è ø

2 41 25 5æ ö æ ö×ç ÷ ç ÷è ø è ø

( )

( ) ( )

5 1 4 2 3 3 2 4

5 4 2 3 3 2 46

410 5 4 5 3 4 7 4 2

5 5

4

6 6 6 64 5 3 4 2 3 1 21 2 3 45 5 5 5 5 5 5 51 6 4 5 15 3 4 20 2 3 15 1 251 2 36 2 3 2 2 3 3 2 2 3 2 3 15 52

æ ö æ ö æ ö æ öæ ö æ ö æ ö æ ö æ ö æ ö æ ö æ ö× × - × × + × × - × ×ç ÷ ç ÷ ç ÷ ç ÷ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è ø è ø è ø è ø è øè ø è ø è ø è ø

= × × - × × + × × - × ×

×= × - × + × - × = - + × -

=10

5 53 2 3 307264

5 5 3125× ×× = =

G

P W F

I M

G

M

P

W

F

I



The probability that everyone gets hexed is

.

Solution 2:

Know that

the number of the circular permutations of n different items is .

The total number of ways that everyone casts a hex is .

For everyone to get hexed, there are four cases.

Case 1: 6 = 6 meaning that six people make one circle.

Let each one cast a hex on his/her left person.

In the diagram A, B, C, D, E, F are Gomez, Morticia, Pugsley, Wednesday, Uncle Fester, and Cousin Itt in some order

The number of ways for this to happen is that is the number of the circular permutations of 6 peoples.

Case 2: 6 = 2 + 4 meaning that six people make two circles with one circle of two people and the other of 4 people


There are ways to choose two people to the first circle. The remaining four people go to the

second circle.

There is only one arrangement in the first circle, and there are ways to arrange 4 people in the second circle.

3072 5313125 3125

- =

( )1 !n -

65

5! 120=

62æ öç ÷è ø

3!

A B

D

C

E

F

A

B

C

E

D F



The number of ways in this case is .

Case 3: 6 = 3 + 3 meaning that six people make two circles with each circle of three people


There are ways to divide 6 people into two groups with each group of three people.

There are ways to arrange 3 people in each circle.


Case 4: 6 = 2 + 2 + 2 meaning that six people make three circles with each circle of two people

Let two people in one circle cast a hex on each other.

There are ways to divide 6 people into three groups with each group of two

people.

There is only one way to arrange 2 people in each circle.


63! 15 6 90

2æ ö

× = × =ç ÷è ø

6 33 32!

æ ö æ ö×ç ÷ ç ÷

è ø è ø

2!

6 33 3 20 1 2 22! 2! 402! 2

æ ö æ ö×ç ÷ ç ÷ × × ×è ø è ø × × = =

6 4 22 2 2

3!

æ ö æ ö æ ö× ×ç ÷ ç ÷ ç ÷

è ø è ø è ø

6 4 22 2 2 15 6 1 15

3! 6

æ ö æ ö æ ö× ×ç ÷ ç ÷ ç ÷ × ×è ø è ø è ø = =

D

E

F

A

B

C

A

B

C

D

E F



The total number of ways that everyone gets hexed is

.

The probability that everyone gets hexed is

.

Solution 3:

We will build a recursion to solve the general problem with n people.

Let n people be A, B, C, .

Let be the number of ways that everyone gets hexed.

A casts a hex on somebody. There are ways to choose one on whom A casts a hex.

Without loss of generality let A cast a hex on B.

There are two cases now.

Case 1: B casts a hex on A.

Then there are ways for the rest people to cast hexes.

Case 2: B casts a hex not on A.

120 90 40 15 265+ + + =

6265 535 3125

=

!

na

( )1n -

2na - 2n -

A C B

D

E F

A

C B

D

E F

n–2 people

A C

B

E F

n–1 people

D



Now look at the people without A.

There are ways for the people to cast hexes.

Somebody ? casts a hex on B.

Now we place A back and let A cast a hex on B. And we let the hex on B by ? be moved to on A.

Therefore, we have

.

That is,

.

The total number of ways that everyone casts a hex is .

The probability that everyone gets hexed is .

In our problem the probability is .

Let us calculate by the recursion.

,

,

,

,

,

1n -

1na - 1n -

( ) ( )2 11n n na n a a- -= - +

( ) ( )1 21n n na n a a- -= - +

( )1 nn -

( )1n

n

an -

665a

6a

0 1a =

1 0a =

( ) ( )2 1 01 1 0 1 1a a a= × + = × + =

( ) ( )3 2 12 2 1 0 2a a a= × + = × + =

( ) ( )4 3 23 3 2 1 9a a a= × + = × + =

C B

E F

n–1 people

?

D

A C B

E F

n–1 people

?

D



.

The desired probability is

.

( ) ( )5 4 34 4 9 2 44a a a= × + = × + =

( ) ( )6 5 45 5 44 9 5 53a a a= × + = × + = ×

65 53 535 3125×

=

Solution to #10 contributed by Jon Meilstrup Problem: If 6 people cast a hex on one of the others, chosen at random, what is the probability that everyone gets hexed? Call Hex(n) a function that computes the number of ways n people can cast on others in the group, with everyone getting hexed, and nobody casting on themselves. We need to compute Hex(6). Choose a person at random to be number 1. They cast a hex on one of the remaining 5 people. Call the person they choose to be number 2. Case 1) person 2 casts on person 1. Number of possibilities here is Hex(4), computed below. Case 2) person 2 casts on person 3. There are 4 people to choose from to be person 3. Case 2 has two sub cases: Case 2.1) person 3 casts on person 1. Number of possibilities here is Hex(3). Case 2.2) person 3 casts on person 4. There are 3 people to choose from to be person 4. Case 2.2 has two sub cases: Case 2.2.1) person 4 casts on person 1. Number of possibilities here is Hex(2) Case 2.2.2) person 4 casts on person 5. There are 2 people to choose from to be person 5. After 5 is chosen, there is only one choice for person 6. Hex(2) has only 2 people, each casts on each other, so Hex(2) =1 Hex(3) has 3 people, the first person can choose one of 2 people, so Hex(3)=2 Hex(4) has 4 people. We can follow similar reasoning as above, or note that the possibilities are: (1-2-3-4) or (1-2)(3-4) For (1-2-3-4) there are 3 choices for the second person, and 2 for the third, so there are 6 possibilities. For (1-2)(3-4) there are 3 choices for person 2, and the rest are fixed, so only 3 possibilities. So, Hex(4) = 6+3 = 9 Now, to total up the answer, in reverse order, as in each step, we need to sum the cases, then multiply by the number of ways that case could be used: Case 2.2.2) 2 choices Case 2.2.1) Hex(2) = 1 choice Case 2.2) sum of 2.2.2 and 2.2.1, or 2+1, then times 3 (number of ways to choose person 4) = 3*(2+1) = 9 Case 2.1) Hex(3) = 2

Case 2) sum of 2.2 and 2.1, or 9+2, times 4 (number of ways to choose person 3) = 4*(9+2)=44 Case 1) Hex(4) = 9 So, Hex(6) is the sum of Case 1 and 2, times the number of ways to choose person 2 = (44+9)*5 The probability is this number, divided by the total possible, so (53*5)/5^6 or 53/5^5

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University of Northern Colorado Mathematics Contest 2019-2020 · 2019-11-20 · University of Northern Colorado Mathematics Contest Problems are duplicated and solved by Ming Song