unc mathematics contest 2015 2016 first round problems with solutions

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University Of Northern Colorado Mathematics Contest 2015-2016

Problems of First Round

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

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

(

x,y

)

!for which

3 ≤

+ y

x .

The notation x _{stands for the “absolute value” of x. That is} x =x if x≥0_and x =−x if 0

≤

x _.

A

B 1

7

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6. The points

(

2,5

)

and

(

6,5

)

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

{

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 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.”

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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

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Problems with Solutions of First Round

1. How many positive integers less than 100 are multiples of 5 but not multiples of 2? 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 20 5 100

= !" ! #$ #

numbers which are multiples of 5. Among them

there are 10 2 5 100

= !" ! #$ #

⋅ numbers which are multiples of 2.

Therefore, the answer is 20₋10₌10_.

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 AE_⊥DC_{intersecting line}_{DC at E. Then}_AE₌6_and_DE₌1₊7₌8_{. Triangle}_{AED is a} 3-4-5 triangle.

Therefore, _AD₌10

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

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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

!! " # $$ % &

3 7

ways to choose 3 blocks to be vertical.

Therefore, the answer is 35 3 7

= !! " # $$ % &

. A

B

1 1 1 1 1

1 1 1

2 3 4

3 6 10

4 10 20

5 15 35 A

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5. Find the area of the region in the x-y!plane that consists of the points

(

x,y

)

!for which

3 ≤

+ y

x .

The notation x _{stands for the “absolute value” of x. That is} x =x if x≥0_and x =−x if 0

≤

x _.

Answer: 18

Solution:

The region is shaded below:

The area is 6 6 18 2

1 = ⋅

⋅ .

6. The points

(

2,5

)

and

(

6,5

)

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

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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

(

4,5

)

. The line L passes

(

0,0

)

and

(

4,5

)

. The slope is 1.25 4 5

= .

7. A rectangular sheet of paper whose dimensions are 12_ʺ _× 18_ʺ 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) (4, 5) (6, 5)

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Let AE=DE=x. Then CE=18−x_.

In right triangle CDE we have

(

18₋x

)

2₌x2₊122_{. Solving for x, we obtain}x₌5_.

The area of triangle BAC is 12 18 108 2

1

= ⋅

⋅ , and the area of triangle BAC is 12 5 30 2

1

= ⋅

⋅ .

The total area of the shaded region is 108+30=138. 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 4⋅

(

1+3+5+!+23

)

+2⋅232 =4⋅122 +2⋅232₌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 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.”

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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.

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?

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Answer: 91 32

Solution:

Case 1: the first touch is copper

The probability for this to happen is 13

4

. 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

Case 2: the first touch is silver

Because of the symmetry between copper and silver, the probability in this case is 91 16

as well.

Case 3: the first touch is gold

5

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

Therefore, the answer is

91

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

4

. 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 13

4

as well.

Case 3: the first touch is gold

5

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

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Therefore, the answer is

91 81

91 25

13 4

= +

+ .

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 10 2 5

=

!! " # $$ % &

.

Case 2:

A

B

C

D A

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The number of routes which don’t cross or touch the red route is 9

The number of routes which don’t cross or touch the red route is 7 2

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Case 5:

Pay attention to E and D.

=

!! " # $$ % &

.

Case 6:

Pay attention to F and D.

= !! " # $$ % &

.

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

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Case 8:

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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

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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.

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

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

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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.

Mark the point where the two paths first 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

A

B

D

C

F

E

X Y

A

B

D

C

F

E U

V A

B

D

C

F

E U

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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.

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That is, there are _!!

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

50