DISINI 08-09
UNIVERSITY OF NORTHERN COLORADO
MATHEMATICS CONTEST
First Round
For all Colorado Students Grades 7-12
November 1, 2008
•
•
•
The positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ….
The Pythagorean Theorem says that a 2 + b 2 = c 2 where a, b, and c are side lengths of a right
triangle and c is the hypotenuse.
A scalene triangle has three sides of unequal length.
1. A unit fraction is a proper fraction of the form 1/n where is an integer greater than 1. The
numerator is always 1. Examples: 1/3, 1/29, 1/100 Find two ways to write 4/5 as the sum of three
different unit fractions.
2. Insert all 11 integers 1, 2, 3, , 10, 11 into
this shape so that the sum of all the vertical
squares is 43 and the sum of all the horizontal
squares is 28. What number is in the corner?
3. Find a set of three different positive integers given that their product is 72 and that their sum is a
multiple of 7.
4. The area of the scalene triangle shown is 84 sq. units. Two side lengths are given as AB=10 and
AC=21. Determine the length of the third side BC.
OVER
5. Determine the area of the trapezoid.
6. An army of ants is organizing a peace march across a room. If they form columns of 8 ants there
are 4 left over. If they form columns of either 3 or 5 ants there are 2 left over. What is the smallest
number of ants that could be in this army?
7. Let
. How many subsets of consisting of 8 (eight) different
elements are such that the sum of the eight elements is a multiple of 5?
8. Let
denote the maximum number of points of intersection strictly between lines
formed by joining the m points on to the points on
in all possible ways.
shown in the diagram.
and
as
(a) Compute
(b) Compute
(c) Give a formula for
9. (a) How many subsets
no two elements of
of
have the property that
differ by 1? As an example,
does not.
(b) Repeat with the set {1, 2, 3, 4, 5, 6}.
contains at least 2 elements and
satisfies these two properties but
Brief Solutions
First Round 2008
1.
=
+
2. n = 5; Let n be in the corner, x the sum of the non corner horizontal squares, and y the corner
+ 11= 66; solving gives
vertical square. Then x + n = 28, y + n = 43; x + y + n = 1 + 2 +
x = 23, n = 5.
3. Trial and error yields 1, 3, 24.
4. BC = 17; Let h be the altitude from B. Then
and h = 8. Since 10 – 8 – 6 is a
Pythagorean triplet, the base AC could be expressed as the sum 21 = 6 + 15 but then BC is the
hypotenuse of a right triangle whose side lengths are 8 and 15. BC = 17.
5. 244; Drop altitudes to form a center rectangle and two triangles. The base could be expressed as
the sum 41 = 15 + 20 + 6, noting the potential for 17 –15 – 8 and a 10 – 8 – 6 right triangle. The
area is then 60 + 160 + 24 = 244.
6. 92; Let n be the number of ants. Then n = 8a + 4 = 3b + 2 = 5c + 2. Then 8a + 2 is both a multiple
of 3 and 5, hence of 15. Trying even k for 8a + 2 = 15k gives us k = 6, a = 11.
7. 9; Since 1 + 2 + 3 +
+ 10 = 55, the sum of all the integers in S is a multiples of 5. Then the
complement of the subset consisting of 8 elements must also be a multiple of 5. These doubletons
are easier to count: 1,4 1,9 2,3 2,8 3,7 4,6 5,10 6,9 and 7,8.
8. (a) f (3, 3) = 9
(b) f (3, 4) = 18
(c) f (2, n) = n (n –1) / 2
A point of intersection is determined by the intersection of two lines. Any choice of 2 points
will produce such an intersection point. Hence for (c) there are
on along with 2 points on
points.
9.
(a) 7; They are 13, 14, 15, 24, 25, 35 and 135 (without using set notation).
(b) 14; 10 doubletons and the 4 triplets 135, 136, 146, 246
University of Northern Colorado
MATHEMATICS CONTEST
FINAL ROUND 2009
For Colorado Students Grades 7-12
1.
How many positive 3-digit numbers abc are there such that
have this property but 245 and 317 do not.
2.
(a) Let
? For example, 202 and 178
. For how many n between 1 and 100 inclusive is
(b) For how many n between 1 and 100 inclusive is
a multiple of 5?
a multiple of 5?
3.
An army of ants is organizing a march to the Obama inauguration. If they form columns of 10 ants
there are 8 left over. If they form columns of 7, 11 or 13 ants there are 2 left over. What is the
smallest number of ants that could be in the army?
4.
How many perfect squares are divisors of the product ! · ! · ! · ! · ! · ! · ! · ! ? (Here, for
example, ! means · · · .)
5.
The two large isosceles right triangles are congruent.
If the area of the inscribed square A is 225 square
units, what is the area of the inscribed square B?
6.
Let each of m distinct points on the positive
x-axis be joined to each of n distinct points on
the positive y-axis. Assume no three segments
are concurrent (except at the axes). Obtain
with proof a formula for the number of interior
intersection points. The diagram shows that
the answer is 3 when
and
.
OVER
7.
A polynomial
has a remainder of 4 when divided by
by
. What is the remainder when
is divided by
8.
Two diagonals are drawn in the trapezoid
forming four triangles. The areas of two of the
triangles are 9 and 25 as shown. What is the total
area of the trapezoid?
9.
A square is divided into three pieces of
equal area by two parallel lines as shown.
If the distance between the two parallel
lines is 8 what is the area of the square?
10.
Let
, , , … , . Determine the number of subsets A of such that A contains at least two
elements and such that no two elements of A differ by 1 when
(a)
11.
(b)
and a remainder of 14 when divided
?
(c) generalize for any
.
If the following triangular array of numbers is continued using the pattern established, how many
numbers (not how many digits) would there be in the 100th row? As an example, the 5th row has 11
numbers. Use exponent notation to express your answer.
1
2
3 4 5
6 7 8 9 10
11 12 13 14 15 16 17 18 19 20 21
22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42
Brief Solutions Final Round
January 31, 2009
1.
45; Count them as you let c range from 1 to 9. There are
2.
(a) 25; From the partial units digit table,
.
is a multiple of 5 only when
, where
ranges from 0 to 24.
There are 25.
n
2
3
4
5
6
7
8
1
1
1
1
1
1
1
1
2
4
8
6
2
4
8
6
,
1
2
3
4
5
, …,
6
7
8
1
1
1
1
1
1
1
1
2
4
8
6
2
4
8
6
3
9
7
1
3
9
7
1
4
6
4
6
4
6
4
6
10
20
20
14
10
20
20
14
is a multiple of 5 except when
(b) 75;
3.
1
6008; Ants =
then
. The smallest
that works is
.
. See table. There are 75 such
. Then
.
…
must be a multiple of 7, 11, and 13;
4.
360; ! · ! · ! · ! · ! · ! · ! · ! ·
· · · . A perfect square divisor must be of the form
· · ·
where a, b, c, d are even integers satisfying
,
,
,
. There are 12 choices for
a (they are 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22), 5 choices for b, 3 choices for c and 2 for d.
5.
200; If
6.
and
, the large right triangle has area 450. Then
; Select 2 of the m points and 2 of the n points; connect with two lines (twist them to make a point). Each
point of intersection corresponds to the crossing of the two lines.
7.
; Since the remainder must be linear
and
8.
.
gives
. Then
and
64; Since area ABD = area ACD, X
.
Y
and X
Since ABE and EBC have the same height h,
and hence X⁄
X
·
,X
⁄ . Similarly,
·
⁄X
. Total area
⁄
X,
Y.
and so
.
9.
832;
implies a
must equal
√
. So
√
. Then
and the area of the square is
10. (a) 133; Try this with
·
√
.
first, and organize by size of subset.
Size
2
3
4
#
15
10
1
Size
2
3
4
#
21
20
5
. For
. The total is
. For
for a total of
the chart is:
the total is
(b)
(c)
10. Alternate Solution
There is a one-to-one correspondence between the desired type of subsets of S consisting of k elements and the
number of ways of choosing k out of
objects. Each desired k-element subset matches with a way of
choosing k out of a sequence of 10 objects so that no two are adjacent. The following picture illustrates this with
: ○●○●○○○○●○
, ,
Delete
objects, one from each gap, to form ○●●○○○●○
In reverse ●●●○○○○○ corresponds to , , . Any of the
ways of selecting 3 objects from these 8 corresponds to
a desired subset.
⁄ ; The center element in row
11.
Row
# of elements
2
3
4
is
. The following chart helps establish the pattern:
5
6
Note that
So row 100 contains
⁄ numbers, by summing the geometric sum.
MATHEMATICS CONTEST
First Round
For all Colorado Students Grades 7-12
November 1, 2008
•
•
•
The positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ….
The Pythagorean Theorem says that a 2 + b 2 = c 2 where a, b, and c are side lengths of a right
triangle and c is the hypotenuse.
A scalene triangle has three sides of unequal length.
1. A unit fraction is a proper fraction of the form 1/n where is an integer greater than 1. The
numerator is always 1. Examples: 1/3, 1/29, 1/100 Find two ways to write 4/5 as the sum of three
different unit fractions.
2. Insert all 11 integers 1, 2, 3, , 10, 11 into
this shape so that the sum of all the vertical
squares is 43 and the sum of all the horizontal
squares is 28. What number is in the corner?
3. Find a set of three different positive integers given that their product is 72 and that their sum is a
multiple of 7.
4. The area of the scalene triangle shown is 84 sq. units. Two side lengths are given as AB=10 and
AC=21. Determine the length of the third side BC.
OVER
5. Determine the area of the trapezoid.
6. An army of ants is organizing a peace march across a room. If they form columns of 8 ants there
are 4 left over. If they form columns of either 3 or 5 ants there are 2 left over. What is the smallest
number of ants that could be in this army?
7. Let
. How many subsets of consisting of 8 (eight) different
elements are such that the sum of the eight elements is a multiple of 5?
8. Let
denote the maximum number of points of intersection strictly between lines
formed by joining the m points on to the points on
in all possible ways.
shown in the diagram.
and
as
(a) Compute
(b) Compute
(c) Give a formula for
9. (a) How many subsets
no two elements of
of
have the property that
differ by 1? As an example,
does not.
(b) Repeat with the set {1, 2, 3, 4, 5, 6}.
contains at least 2 elements and
satisfies these two properties but
Brief Solutions
First Round 2008
1.
=
+
2. n = 5; Let n be in the corner, x the sum of the non corner horizontal squares, and y the corner
+ 11= 66; solving gives
vertical square. Then x + n = 28, y + n = 43; x + y + n = 1 + 2 +
x = 23, n = 5.
3. Trial and error yields 1, 3, 24.
4. BC = 17; Let h be the altitude from B. Then
and h = 8. Since 10 – 8 – 6 is a
Pythagorean triplet, the base AC could be expressed as the sum 21 = 6 + 15 but then BC is the
hypotenuse of a right triangle whose side lengths are 8 and 15. BC = 17.
5. 244; Drop altitudes to form a center rectangle and two triangles. The base could be expressed as
the sum 41 = 15 + 20 + 6, noting the potential for 17 –15 – 8 and a 10 – 8 – 6 right triangle. The
area is then 60 + 160 + 24 = 244.
6. 92; Let n be the number of ants. Then n = 8a + 4 = 3b + 2 = 5c + 2. Then 8a + 2 is both a multiple
of 3 and 5, hence of 15. Trying even k for 8a + 2 = 15k gives us k = 6, a = 11.
7. 9; Since 1 + 2 + 3 +
+ 10 = 55, the sum of all the integers in S is a multiples of 5. Then the
complement of the subset consisting of 8 elements must also be a multiple of 5. These doubletons
are easier to count: 1,4 1,9 2,3 2,8 3,7 4,6 5,10 6,9 and 7,8.
8. (a) f (3, 3) = 9
(b) f (3, 4) = 18
(c) f (2, n) = n (n –1) / 2
A point of intersection is determined by the intersection of two lines. Any choice of 2 points
will produce such an intersection point. Hence for (c) there are
on along with 2 points on
points.
9.
(a) 7; They are 13, 14, 15, 24, 25, 35 and 135 (without using set notation).
(b) 14; 10 doubletons and the 4 triplets 135, 136, 146, 246
University of Northern Colorado
MATHEMATICS CONTEST
FINAL ROUND 2009
For Colorado Students Grades 7-12
1.
How many positive 3-digit numbers abc are there such that
have this property but 245 and 317 do not.
2.
(a) Let
? For example, 202 and 178
. For how many n between 1 and 100 inclusive is
(b) For how many n between 1 and 100 inclusive is
a multiple of 5?
a multiple of 5?
3.
An army of ants is organizing a march to the Obama inauguration. If they form columns of 10 ants
there are 8 left over. If they form columns of 7, 11 or 13 ants there are 2 left over. What is the
smallest number of ants that could be in the army?
4.
How many perfect squares are divisors of the product ! · ! · ! · ! · ! · ! · ! · ! ? (Here, for
example, ! means · · · .)
5.
The two large isosceles right triangles are congruent.
If the area of the inscribed square A is 225 square
units, what is the area of the inscribed square B?
6.
Let each of m distinct points on the positive
x-axis be joined to each of n distinct points on
the positive y-axis. Assume no three segments
are concurrent (except at the axes). Obtain
with proof a formula for the number of interior
intersection points. The diagram shows that
the answer is 3 when
and
.
OVER
7.
A polynomial
has a remainder of 4 when divided by
by
. What is the remainder when
is divided by
8.
Two diagonals are drawn in the trapezoid
forming four triangles. The areas of two of the
triangles are 9 and 25 as shown. What is the total
area of the trapezoid?
9.
A square is divided into three pieces of
equal area by two parallel lines as shown.
If the distance between the two parallel
lines is 8 what is the area of the square?
10.
Let
, , , … , . Determine the number of subsets A of such that A contains at least two
elements and such that no two elements of A differ by 1 when
(a)
11.
(b)
and a remainder of 14 when divided
?
(c) generalize for any
.
If the following triangular array of numbers is continued using the pattern established, how many
numbers (not how many digits) would there be in the 100th row? As an example, the 5th row has 11
numbers. Use exponent notation to express your answer.
1
2
3 4 5
6 7 8 9 10
11 12 13 14 15 16 17 18 19 20 21
22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42
Brief Solutions Final Round
January 31, 2009
1.
45; Count them as you let c range from 1 to 9. There are
2.
(a) 25; From the partial units digit table,
.
is a multiple of 5 only when
, where
ranges from 0 to 24.
There are 25.
n
2
3
4
5
6
7
8
1
1
1
1
1
1
1
1
2
4
8
6
2
4
8
6
,
1
2
3
4
5
, …,
6
7
8
1
1
1
1
1
1
1
1
2
4
8
6
2
4
8
6
3
9
7
1
3
9
7
1
4
6
4
6
4
6
4
6
10
20
20
14
10
20
20
14
is a multiple of 5 except when
(b) 75;
3.
1
6008; Ants =
then
. The smallest
that works is
.
. See table. There are 75 such
. Then
.
…
must be a multiple of 7, 11, and 13;
4.
360; ! · ! · ! · ! · ! · ! · ! · ! ·
· · · . A perfect square divisor must be of the form
· · ·
where a, b, c, d are even integers satisfying
,
,
,
. There are 12 choices for
a (they are 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22), 5 choices for b, 3 choices for c and 2 for d.
5.
200; If
6.
and
, the large right triangle has area 450. Then
; Select 2 of the m points and 2 of the n points; connect with two lines (twist them to make a point). Each
point of intersection corresponds to the crossing of the two lines.
7.
; Since the remainder must be linear
and
8.
.
gives
. Then
and
64; Since area ABD = area ACD, X
.
Y
and X
Since ABE and EBC have the same height h,
and hence X⁄
X
·
,X
⁄ . Similarly,
·
⁄X
. Total area
⁄
X,
Y.
and so
.
9.
832;
implies a
must equal
√
. So
√
. Then
and the area of the square is
10. (a) 133; Try this with
·
√
.
first, and organize by size of subset.
Size
2
3
4
#
15
10
1
Size
2
3
4
#
21
20
5
. For
. The total is
. For
for a total of
the chart is:
the total is
(b)
(c)
10. Alternate Solution
There is a one-to-one correspondence between the desired type of subsets of S consisting of k elements and the
number of ways of choosing k out of
objects. Each desired k-element subset matches with a way of
choosing k out of a sequence of 10 objects so that no two are adjacent. The following picture illustrates this with
: ○●○●○○○○●○
, ,
Delete
objects, one from each gap, to form ○●●○○○●○
In reverse ●●●○○○○○ corresponds to , , . Any of the
ways of selecting 3 objects from these 8 corresponds to
a desired subset.
⁄ ; The center element in row
11.
Row
# of elements
2
3
4
is
. The following chart helps establish the pattern:
5
6
Note that
So row 100 contains
⁄ numbers, by summing the geometric sum.