unc mathematics contest 2015 2016 first round problems with solutions
University of Northern Colorado Mathematics Contest
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.
7
6
1
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?
B
A
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.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
1
University of Northern Colorado Mathematics Contest
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.
Fold
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.”
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
2
University of Northern Colorado Mathematics Contest
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.
B
A
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
3
University of Northern Colorado Mathematics Contest
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
#100 !
also reason as follows: There are #
! = 20 numbers which are multiples of 5. Among them
$ 5 "
# 100 !
there are #
! = 10 numbers which are multiples of 2.
$5⋅ 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.
7
6
1
Answer: 10
Solution:
Let the zig-zag path be ABCD as shown.
E
C
7
E
6
A
1
D
B
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
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
4
University of Northern Colorado Mathematics Contest
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?
B
A
Answer: 35
Solution 1:
We mark the numbers:
4
1
1
1
A
1
10
20
3
6
10
2
3
4
1
1
1
35
B
15
5
1
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
&7#
$$ !! ways to choose 3 blocks to be vertical.
% 3"
&7#
Therefore, the answer is $$ !! = 35.
3
% "
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
5
University of Northern Colorado Mathematics Contest
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
Solution:
The region is shaded below:
y
x+y=3
–x + y = 3
3
x
–3
3
–3
x – y = 3 –x – y = 3
The area is
6.
1
⋅ 6 ⋅ 6 = 18 .
2
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
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
6
University of Northern Colorado Mathematics Contest
Solution:
Any line passing through the center of the regular hexagon divides this hexagon into two pieces
of equal area.
L
(2, 5)
(4, 5)
(6, 5)
The center of the hexagon is (4, 5) . The line L passes (0, 0) and (4, 5) . The slope is
7.
5
= 1.25 .
4
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.
Fold
Answer: 138
Solution:
D
D
A
E
C
Fold
B
C
B
A
Let ABCD be the rectangle. The folding is along diagonal BC. After folding, let AC and BD
intersect at E.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
7
University of Northern Colorado Mathematics Contest
Let AE = DE = x . Then CE = 18 − x .
In right triangle CDE we have (18 − x )2 = x 2 + 122 . Solving for x, we obtain x = 5 .
The area of triangle BAC is
1
1
⋅ 12 ⋅ 18 = 108 , and the area of triangle BAC is ⋅ 12 ⋅ 5 = 30 .
2
2
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.”
What are R2-D2’s and BB-8’s numbers?
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
8
University of Northern Colorado Mathematics Contest
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
5
2
6
2
6
4
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, R2D2 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.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
9
University of Northern Colorado Mathematics Contest
Answer:
32
91
Solution:
Case 1: the first touch is copper
4
. Then we will have 3 copper, 5 silver, and 6 gold coins.
13
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
3+ 5
4
for this to happen is
= .
3+ 5+ 6 7
The probability for this to happen is
In this case, the probability that the gold are still more than either of the other two colors is
4 4 16
⋅ = .
13 7 91
Case 2: the first touch is silver
Because of the symmetry between copper and silver, the probability in this case is
16
as well.
91
Case 3: the first touch is gold
5
. Then we will have 5 copper, 5 silver, and 4 gold coins.
13
For whatever the second touch is, the gold cannot be more than either of the other two colors.
The probability for this to happen is
Therefore, the answer is
16 16 32
.
+ =
91 91 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
13
the gold are still more than either of the other two colors.
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
4
as well.
13
Case 3: the first touch is gold
The probability for this to happen is
5
. Then we will have 5 copper, 5 silver, and 4 gold coins.
13
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
5 10 25
⋅ = .
13 14 91
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
10
University of Northern Colorado Mathematics Contest
Therefore, the answer is
11.
4
4 25 81
+ +
= .
13 13 91 91
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.
B
A
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:
B
D
A
C
Pay attention to C and D.
& 5#
The number of routes which don’t cross or touch the red route is $$ !! = 10 .
% 2"
Case 2:
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
11
University of Northern Colorado Mathematics Contest
B
D
A
C
& 5# & 2#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 9 .
% 2" % 2"
Case 3:
B
D
A
C
& 5# & 3#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 7 .
% 2" % 2"
Case 4:
B
D
A
C
& 5# & 4#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 4 .
% 2" % 2"
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
12
University of Northern Colorado Mathematics Contest
Case 5:
B
D
A
E
Pay attention to E and D.
& 4#
The number of routes which don’t cross or touch the red route is $$ !! = 6 .
% 2"
Case 6:
B
D
A
F
Pay attention to F and D.
& 3#
The number of routes which don’t cross or touch the red route is $$ !! = 3 .
% 2"
Case 7:
B
A
Pay attention to F and D.
There is only 1 route which doesn’t cross or touch the red route.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
13
University of Northern Colorado Mathematics Contest
Case 8:
B
D
A
E
& 4# & 2#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 5 .
% 2" % 2"
Case 9:
B
D
A
E
& 4# & 3#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 3.
% 2" % 2"
Case 10:
B
G
A
F
& 2#
The number of routes which don’t cross or touch the red route is $$ !! = 2 .
%1"
The total number of the routes which don’t cross or touch a given back route is
10 + 9 + 7 + 4 + 6 + 3 + 1 + 5 + 3 + 2 = 50 .
Consider that the forward route and back route can be switched.
The answer is 50 ⋅ 2 = 100 .
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
14
University of Northern Colorado Mathematics Contest
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:
F
B
D
E
A
C
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.
F
B
D
Y
E
X
A
C
Mark the point where the two paths first meet (the black point).
F
B
D
Y
E
X
A
C
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
15
University of Northern Colorado Mathematics Contest
Switch the parts of path X and path Y after the black point.
F
B
D
V
E
U
A
C
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.
F
Y
B
D
X
E
A
C
Mark the point where the two paths first meet (the black point).
F
Y
B
D
X
E
A
C
Switch paths:
F
V
B
D
E
U
A
C
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
16
University of Northern Colorado Mathematics Contest
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.
F
B
U
D
E
V
A
C
Mark the point where the two paths first meet (the black point).
F
B
U
D
E
V
A
C
Switch paths:
F
B
Y
D
E
X
A
C
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
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
17
University of Northern Colorado Mathematics Contest
Take a look at another pair of path U (red) from C to F and path V (pink) from E to D.
F
B
D
V
E
U
A
C
Mark the first point where the two paths meet (the black point).
F
B
D
V
E
U
A
C
Switch paths:
F
B
D
Y
E
X
A
C
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.
& 5#
& 5#
From C to D there are $$ !! paths, and from E to F there are $$ !! paths as well. So there are
%2"
%2"
& 5# & 5#
$$ !! ⋅ $$ !! pairs.
% 2" % 2"
& 5#
& 5#
& 5# & 5#
From C to F there are $$ !! paths, and from E to D there are $$ !! paths. So there are $$ !! ⋅ $$ !!
% 3"
% 1"
% 3" % 1 "
pairs.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
18
University of Northern Colorado Mathematics Contest
& 5# & 5#
That is, there are $$ !! ⋅ $$ !! pairs of paths: one from C to D and the other from E to F, which cross
% 3" % 1 "
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
( 5 % ( 5 % ( 5% ( 5%
&& ## ⋅ && ## − && ## ⋅ && ## = 10 ⋅10 − 10 ⋅ 5 = 50 .
' 2 $ ' 2 $ ' 3$ ' 1 $
Consider that two spiders can switch their paths. The answer is
2 ⋅ 50 = 100 .
In general, for the m × n grid, the answer is
&, m − 1 + n − 1) 2 , m − 1 + n − 1) , m − 1 + n − 1)#
'' − **
'' ⋅ **
''! .
2$**
m
$%+ m − 1 ( +
( + m − 2 (!"
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
19
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.
7
6
1
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?
B
A
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.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
1
University of Northern Colorado Mathematics Contest
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.
Fold
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.”
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
2
University of Northern Colorado Mathematics Contest
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.
B
A
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
3
University of Northern Colorado Mathematics Contest
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
#100 !
also reason as follows: There are #
! = 20 numbers which are multiples of 5. Among them
$ 5 "
# 100 !
there are #
! = 10 numbers which are multiples of 2.
$5⋅ 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.
7
6
1
Answer: 10
Solution:
Let the zig-zag path be ABCD as shown.
E
C
7
E
6
A
1
D
B
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
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
4
University of Northern Colorado Mathematics Contest
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?
B
A
Answer: 35
Solution 1:
We mark the numbers:
4
1
1
1
A
1
10
20
3
6
10
2
3
4
1
1
1
35
B
15
5
1
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
&7#
$$ !! ways to choose 3 blocks to be vertical.
% 3"
&7#
Therefore, the answer is $$ !! = 35.
3
% "
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
5
University of Northern Colorado Mathematics Contest
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
Solution:
The region is shaded below:
y
x+y=3
–x + y = 3
3
x
–3
3
–3
x – y = 3 –x – y = 3
The area is
6.
1
⋅ 6 ⋅ 6 = 18 .
2
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
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
6
University of Northern Colorado Mathematics Contest
Solution:
Any line passing through the center of the regular hexagon divides this hexagon into two pieces
of equal area.
L
(2, 5)
(4, 5)
(6, 5)
The center of the hexagon is (4, 5) . The line L passes (0, 0) and (4, 5) . The slope is
7.
5
= 1.25 .
4
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.
Fold
Answer: 138
Solution:
D
D
A
E
C
Fold
B
C
B
A
Let ABCD be the rectangle. The folding is along diagonal BC. After folding, let AC and BD
intersect at E.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
7
University of Northern Colorado Mathematics Contest
Let AE = DE = x . Then CE = 18 − x .
In right triangle CDE we have (18 − x )2 = x 2 + 122 . Solving for x, we obtain x = 5 .
The area of triangle BAC is
1
1
⋅ 12 ⋅ 18 = 108 , and the area of triangle BAC is ⋅ 12 ⋅ 5 = 30 .
2
2
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.”
What are R2-D2’s and BB-8’s numbers?
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
8
University of Northern Colorado Mathematics Contest
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
5
2
6
2
6
4
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, R2D2 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.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
9
University of Northern Colorado Mathematics Contest
Answer:
32
91
Solution:
Case 1: the first touch is copper
4
. Then we will have 3 copper, 5 silver, and 6 gold coins.
13
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
3+ 5
4
for this to happen is
= .
3+ 5+ 6 7
The probability for this to happen is
In this case, the probability that the gold are still more than either of the other two colors is
4 4 16
⋅ = .
13 7 91
Case 2: the first touch is silver
Because of the symmetry between copper and silver, the probability in this case is
16
as well.
91
Case 3: the first touch is gold
5
. Then we will have 5 copper, 5 silver, and 4 gold coins.
13
For whatever the second touch is, the gold cannot be more than either of the other two colors.
The probability for this to happen is
Therefore, the answer is
16 16 32
.
+ =
91 91 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
13
the gold are still more than either of the other two colors.
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
4
as well.
13
Case 3: the first touch is gold
The probability for this to happen is
5
. Then we will have 5 copper, 5 silver, and 4 gold coins.
13
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
5 10 25
⋅ = .
13 14 91
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
10
University of Northern Colorado Mathematics Contest
Therefore, the answer is
11.
4
4 25 81
+ +
= .
13 13 91 91
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.
B
A
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:
B
D
A
C
Pay attention to C and D.
& 5#
The number of routes which don’t cross or touch the red route is $$ !! = 10 .
% 2"
Case 2:
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
11
University of Northern Colorado Mathematics Contest
B
D
A
C
& 5# & 2#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 9 .
% 2" % 2"
Case 3:
B
D
A
C
& 5# & 3#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 7 .
% 2" % 2"
Case 4:
B
D
A
C
& 5# & 4#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 4 .
% 2" % 2"
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
12
University of Northern Colorado Mathematics Contest
Case 5:
B
D
A
E
Pay attention to E and D.
& 4#
The number of routes which don’t cross or touch the red route is $$ !! = 6 .
% 2"
Case 6:
B
D
A
F
Pay attention to F and D.
& 3#
The number of routes which don’t cross or touch the red route is $$ !! = 3 .
% 2"
Case 7:
B
A
Pay attention to F and D.
There is only 1 route which doesn’t cross or touch the red route.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
13
University of Northern Colorado Mathematics Contest
Case 8:
B
D
A
E
& 4# & 2#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 5 .
% 2" % 2"
Case 9:
B
D
A
E
& 4# & 3#
The number of routes which don’t cross or touch the red route is $$ !! − $$ !! = 3.
% 2" % 2"
Case 10:
B
G
A
F
& 2#
The number of routes which don’t cross or touch the red route is $$ !! = 2 .
%1"
The total number of the routes which don’t cross or touch a given back route is
10 + 9 + 7 + 4 + 6 + 3 + 1 + 5 + 3 + 2 = 50 .
Consider that the forward route and back route can be switched.
The answer is 50 ⋅ 2 = 100 .
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
14
University of Northern Colorado Mathematics Contest
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:
F
B
D
E
A
C
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.
F
B
D
Y
E
X
A
C
Mark the point where the two paths first meet (the black point).
F
B
D
Y
E
X
A
C
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
15
University of Northern Colorado Mathematics Contest
Switch the parts of path X and path Y after the black point.
F
B
D
V
E
U
A
C
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.
F
Y
B
D
X
E
A
C
Mark the point where the two paths first meet (the black point).
F
Y
B
D
X
E
A
C
Switch paths:
F
V
B
D
E
U
A
C
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
16
University of Northern Colorado Mathematics Contest
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.
F
B
U
D
E
V
A
C
Mark the point where the two paths first meet (the black point).
F
B
U
D
E
V
A
C
Switch paths:
F
B
Y
D
E
X
A
C
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
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
17
University of Northern Colorado Mathematics Contest
Take a look at another pair of path U (red) from C to F and path V (pink) from E to D.
F
B
D
V
E
U
A
C
Mark the first point where the two paths meet (the black point).
F
B
D
V
E
U
A
C
Switch paths:
F
B
D
Y
E
X
A
C
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.
& 5#
& 5#
From C to D there are $$ !! paths, and from E to F there are $$ !! paths as well. So there are
%2"
%2"
& 5# & 5#
$$ !! ⋅ $$ !! pairs.
% 2" % 2"
& 5#
& 5#
& 5# & 5#
From C to F there are $$ !! paths, and from E to D there are $$ !! paths. So there are $$ !! ⋅ $$ !!
% 3"
% 1"
% 3" % 1 "
pairs.
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
18
University of Northern Colorado Mathematics Contest
& 5# & 5#
That is, there are $$ !! ⋅ $$ !! pairs of paths: one from C to D and the other from E to F, which cross
% 3" % 1 "
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
( 5 % ( 5 % ( 5% ( 5%
&& ## ⋅ && ## − && ## ⋅ && ## = 10 ⋅10 − 10 ⋅ 5 = 50 .
' 2 $ ' 2 $ ' 3$ ' 1 $
Consider that two spiders can switch their paths. The answer is
2 ⋅ 50 = 100 .
In general, for the m × n grid, the answer is
&, m − 1 + n − 1) 2 , m − 1 + n − 1) , m − 1 + n − 1)#
'' − **
'' ⋅ **
''! .
2$**
m
$%+ m − 1 ( +
( + m − 2 (!"
Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)
19