DISINI 06-07
UNIVERSITY OF NORTHERN COLORADO
MATHEMATICS CONTEST
First Round
For all Colorado Students Grades 7-12
November 4, 2006
•
•
•
•
An arithmetic progression is a sequence of the form a, a+d, a+2d, …
n factorial is computed as n!= n(n − 1)(n − 2) K 3 ⋅ 2 ⋅ 1 .
An isosceles triangle has two sides with equal length.
The positive odd integers are 1, 3, 5, 7, 9, 11, 13, … .
1. In the 4 by 4 square the boxes can be filled
in with each of the numbers 1, 2, 3, 4 used
exactly once in each row and column. What is A?
2. Determine the sum of the first 500 digits of the unending decimal expansion for
3
= .2307692 K .
13
3. The length of each leg of an isosceles triangle
is 2 x − 1 . If the base is 5 x − 7 determine
all possible integer values of x .
4. The odd number 7 can be expressed as 16 − 9 = 4 2 − 3 2 , a difference of two squares. Express each
of the following odd integers as the difference of two squares:
(a) 17
(b) 83
5. The three roots of the cubic equation 9 x 3 − 36 x 2 + cx − 16 = 0 are in arithmetic progression.
Determine the value of c and the three roots.
Over
6. The perimeter of the triangle is 24 in., and
its area is 8 in.2 . What is the exact
area of the inscribed circle? [That is, express
the area as a fractional multiple of π ].
7. Let f ( x) = 5 x 4 − 6 x 3 − 3 x 2 + 8 x + 2 . Determine coefficients a, b, c, d and e so that
f (x ) = a + b( x − 2) + c( x − 2) 2 + d ( x − 2) 3 + e( x − 2) 4 .
8. Determine all n for which x n + y n factors. [As a reminder, x 3 + y 3 factors as
( x + y )( x 2 − xy + y 2 ) , and x 4 − y 4 factors as ( x 2 − y 2 )( x 2 + y 2 ) ].
9. An International Conference on Global Warming has 5 diplomats from the US, 3 diplomats
from Russia and 4 diplomats from China. These 12 diplomats are to be seated at the head
table in a single row.
Determine the number of possible seating arrangements if the diplomats from each country must be
seated together as a group. Express your answer using the n! notation.
10. A square P1 P2 P3 P4 is drawn in the coordinate plane with P1
at (1,0) and P3 at (3,0) . Let Pn denote the point ( x n , y n ) .
Compute the numerical value of the following product
of complex numbers: ( x1 + iy1 )( x 2 + iy 2 )( x3 + iy 3 )( x 4 + iy 4 ) .
11. A quaternary “number” is an arrangement of digits, each of which is 0, 1, 2, 3.
Some examples: 12203, 01130, 22222, 00031
(a) How many 5-digit quaternary numbers are there?
(b) How many 5-digit quaternary numbers are there in which the digit 3 appears at least once?
In each case express your answer using exponents; you need not multiply out your answers.
BRIEF SOLUTIONS TO FIRST ROUND
NOVEMBER 2006
1. A = 3; by trial and error.
2. Sum = 2246 ; There are 83 blocks of 2 + 3 + 0 + 7 + 6 + 9 = 27 , with 2 + 3 left over.
83 × 27 + 5 = 2246 .
3. x = 2, 3, or 4; 2 x − 1 + 2 x − 1 > 5 x − 7 gives 5 > x . 2 x − 1 + 5 x − 7 > 2 x − 1 gives x > 1.
4. (a) 17 = 9 2 − 8 2
(b) 83 = 42 2 − 412
5. 2 3, 4 3, 6 3 and c = 44; 36 9 = 4 = (a + d ) + a + (a − d ) = 3a and a = 4 3 is one root.
(
)
(
)
− 16 9 = (a + d )a(a − d ) = a a 2 − d 2 = 4 3 16 9 − d 2 and d = 2 3 . So the roots are a+d, a, a-d, or
6 3 , 4 3 , 2 3 . Substituting any root into the cubic gives c = 44 .
4
1
1
1
6. A = π ; label the sides as a, b, c. Then ar + br + cr = 8 , where r is the radius. Using a + b + c = 24 , you
2
2
2
9
2
r
2
4
2
have (a + b + c) = 8 or 24r = 16 , r = . Then A = π = π .
2
3
9
3
7. a = 38, b = 84, c = 81, d = 34, e = 5; Let x − 2 = y ; Then x = y + 2 and
f (x ) = f (y + 2 ) = 5(y + 2 ) − 6(y + 2 ) − 3(y + 2 ) + 8(y + 2 )+ 2 =
4
3
2
5(x − 2) + 34(x − 2) + 81(x − 2) + 84(x − 2)+ 38 =
4
3
2
5 y 4 + 34 y 3 + 81 y 2 + 64 y + 38 =
5(x − 2 ) + 34(x − 2 ) + 81(x − 2 ) + 84(x − 2 )+ 38
4
3
2
8. For all n having an odd factor; or n ≠ 2 k ; For example x 6 + y 6 = (x 2 + y 2 )(x 4 − x 2 y 2 + y 4 )
9. 6 ⋅ 5 ! ⋅ 3! ⋅ 4! ; The US diplomats can be arranged in a group in 5 ! ways. Same for the other groups.
But then the three groups can be permuted around in 3 ! = 6 ways.
10. 15; Label the points as P1 = 1 , P2 = 2 + i , P3 = 3 , P4 = 2 − i and multiply:
P1 P2 P3 P4 = 1 ⋅ (2 + i )⋅ 3 ⋅ (2 − i ) = 3(4 + 1) = 15 .
11. (a) 4 5 ; There are four choices for each of five spots.
(b) 4 5 − 35 ; Take all from part (a) and subtract those where the digit 3 fails to appear.
If there is no 3, there are 35 such “numbers”.
UNIVERSITY OF NORTHERN COLORADO
MATHEMATICS CONTEST
FINAL ROUND
For Colorado Students Grades 7-12
February 3, 2007
•
•
•
The sequence of Fibonacci numbers is 1, 1, 2, 3, 5, 8, 13, 21, K .
The positive odd integers are 1, 3, 5, 7, 9, 11, 13, … .
A regular decagon is a 10-sided figure all of whose sides are congruent.
_________________________________________________________________________________
1.
Express the following sum as a whole number:
1 + 2 − 3 + 4 + 5 − 6 + 7 + 8 − 9 + 10 + 11 − 12 + L + 2005 + 2006 − 2007 .
2.
In Grants Pass, Oregon 4 5 of the men are married to 3 7 of the women. What fraction of the
adult population is married? Give a possible generalization.
3.
State the general rule illustrated here and prove it:
4.
If x is a primitive cube root of one (this means that x 3 = 1 but x ≠ 1 ) compute the value of
1
1
x 2006 + 2006 + x 2007 + 2007 .
x
x
5.
Ten different playing cards have the numbers
1, 1, 2, 2, 3, 3, 4, 4, 5, 5 written on them
as shown. Three cards are selected at random
without replacement. What is the
probability that the sum of the
numbers on the three cards is divisible by 7?
Over
6.
(a) Demonstrate that every odd number 2n + 1 can be expressed as a difference of two squares.
(b) Demonstrate which even numbers can be expressed as a difference of two squares.
7.
1 1
1
+ 2 + 3 + L as a reduced fraction.
3 3
3
1 1
2
3
5
+
+
+
+ L as a (reduced) fraction. Here
(b) Express the infinite sum T = +
5 25 125 625 3125
(a) Express the infinite sum S = 1 +
the denominators are powers of 5 and the numerators 1, 1, 2, 3, 5, K are the Fibonacci numbers
Fn where Fn = Fn −1 + Fn − 2 .
8.
A regular decagon P1 P2 P3 K P10 is drawn
in the coordinate plane with P1 at (2,0)
and P6 at (8,0). If Pn denotes the point
( x n , y n ), compute the numerical value of
the following product of complex numbers:
(x1 + i y1 )(x2 + iy 2 )(x3 + iy3 )L (x10 + i y10 )
where i = − 1 as usual.
9.
A circle is inscribed in an equilateral triangle whose side
length is 2. Then another circle is inscribed externally
tangent to the first circle but inside the triangle as shown.
And then another, and another. If this process continues
forever what is the total area of all the circles? Express
your answer as an exact multiple of π (and not as a
decimal approximation).
10.
A quaternary “number” is an arrangement of digits, each of which is 0, 1, 2, 3.
Some examples: 001, 3220, 022113.
(a) How many 6-digit quaternary numbers are there in which each of 0, 1 appear at least once?
(b) How many n-digit quaternary numbers are there in which each of 0, 1, 2, appear at least
once? Test your answer with n = 3 .
(c) Generalize.
Brief Solutions Final Round
February 3, 2007
1. 670,338; (1 + 2 − 3) + (4 + 5 − 6) + (7 + 8 − 9) + L + (2005 + 2006 − 2007)
= 0 + 3 + 6 + L + 3 ⋅ 668 = 3(1 + 2 + 3 + L + 668) = 3(668)(369) / 2
12
12
24
; Restated,
of the men are married to
of the women. Then
of the adult population
43
15
28
43
is married. To generalize, if a b of the men are married to c d of the women, then ca cb of the
men are married to ca da of the women. The proportion that is married is 2ca (cb + da ).
2. 24
3. 12 + 2 2 + 32 + L + n 2 = 1 ⋅ n + 3(n − 1)+ 5(n − 2)+ L + (2n − 1)⋅1 . The picture tells the story. For
example, the fourth diagram shows one 4, three 3’s, five 2’s and seven 1’s. Stripping off layers of
1’s also gives 12 + 2 2 + 32 + 4 2 .
3
4. +1; Since x = 1 , x
2006
2
= x and x
2007
2
1
= 1 . The expression becomes x + 2 + 2 = x + =
x
x
2
1
2
1
x3
−x
2
2
2
2
= 1 since x + x + 1 = 0 . Or, x + 2 + 2 = x + 2 + 2 = x + x + 1 + 1 = 0 + 1 = 1 . Or,
x
x
x
1
1
1
x 2 + 2 + 2 = + x + 2 = −1 + 2 = 1 since the sum of the vectors x and is − 1 .
x
x
x
5. 2 15 ; There are five ways to achieve a sum divisible by 7; 115 (2 ways), 133 ( 2 ways), 124 (8
10
ways), 233 (2 ways), 455 (2 ways). Hence, there are 16 favorable ways out of 120 = total
3
choices.
6. (a) 2n + 1 = (n + 1)2 − n 2
(b) Multiples of 4; If x = 4n , 4n = (n + 1)2 − (n − 1)2 . The even numbers not divisible by 4, namely
2, 6, 10, 14, K cannot be expressed as a product of two even numbers and hence cannot be
expressed as (a + b )(a − b ) = a 2 − b 2 . If a, b are both even each of (a + b ), (a − b ) is even. If both
a, b are odd each of (a + b ), (a − b ) is even. If one is even, one odd both of (a + b ), (a − b ) are odd
and so is the product, handled in part (a).
1
3
1 1 1
1 1
7. (a) 3 2 ; S = 1 + + 2 + L = 1 + 1 + + 2 + L or S = 1+ S and S = .
3
2
3 3 3
3 3
(b) 5 19 ;
5T
= 1 +
T
=
4T
= 1
1
5
1
5
+
+
+
2
2
5
1
52
1
5
2
+
+
+
3
3
5
2
53
1
5
3
+
+
+
5
54
3
54
2
5
4
+ L
+ L
1
+ L = 1+ T
5
8. 9,706,576; Translate the center of the decagon to the origin. Now the vertices represent the roots
of f ( x) = x10 − 310 = 0 . Since the Pn are each 5 more than the roots of f ( x) = 0 , they would be
the roots of f (x − 5) = 0 or ( x − 5)10 − 310 = 0 . The product then is the constant term, or
510 − 310 = 9,706,576 .
9.
3
3π
; Let r1 , and r2 be radii of the first and second circles; r1 =
and the area of the first circle is
3
8
1
3
,
A1 = π 3 . From r1 + r2 = 2(r1 − r2 ), r2 = 3 9 and A2 = π 27 . Similarly r3 = r2 =
3
27
1
1 1
π 3
A3 = π 243 . Then the total area = π +
+
+ L =
= 3π 8 .
3 27 243
1− 1
9
()
()
10. (a) 4 6 − 2 ⋅ 36 + 2 6 ; from the total 4 6 subtract those that have no 0 36 or no 1 36 . Then add
()
back in those that have no 0 and no 1 2 6 .
(b) 4 n − 3 ⋅ 3n + 3 ⋅ 2 n − 1n
(c) Generalize
MATHEMATICS CONTEST
First Round
For all Colorado Students Grades 7-12
November 4, 2006
•
•
•
•
An arithmetic progression is a sequence of the form a, a+d, a+2d, …
n factorial is computed as n!= n(n − 1)(n − 2) K 3 ⋅ 2 ⋅ 1 .
An isosceles triangle has two sides with equal length.
The positive odd integers are 1, 3, 5, 7, 9, 11, 13, … .
1. In the 4 by 4 square the boxes can be filled
in with each of the numbers 1, 2, 3, 4 used
exactly once in each row and column. What is A?
2. Determine the sum of the first 500 digits of the unending decimal expansion for
3
= .2307692 K .
13
3. The length of each leg of an isosceles triangle
is 2 x − 1 . If the base is 5 x − 7 determine
all possible integer values of x .
4. The odd number 7 can be expressed as 16 − 9 = 4 2 − 3 2 , a difference of two squares. Express each
of the following odd integers as the difference of two squares:
(a) 17
(b) 83
5. The three roots of the cubic equation 9 x 3 − 36 x 2 + cx − 16 = 0 are in arithmetic progression.
Determine the value of c and the three roots.
Over
6. The perimeter of the triangle is 24 in., and
its area is 8 in.2 . What is the exact
area of the inscribed circle? [That is, express
the area as a fractional multiple of π ].
7. Let f ( x) = 5 x 4 − 6 x 3 − 3 x 2 + 8 x + 2 . Determine coefficients a, b, c, d and e so that
f (x ) = a + b( x − 2) + c( x − 2) 2 + d ( x − 2) 3 + e( x − 2) 4 .
8. Determine all n for which x n + y n factors. [As a reminder, x 3 + y 3 factors as
( x + y )( x 2 − xy + y 2 ) , and x 4 − y 4 factors as ( x 2 − y 2 )( x 2 + y 2 ) ].
9. An International Conference on Global Warming has 5 diplomats from the US, 3 diplomats
from Russia and 4 diplomats from China. These 12 diplomats are to be seated at the head
table in a single row.
Determine the number of possible seating arrangements if the diplomats from each country must be
seated together as a group. Express your answer using the n! notation.
10. A square P1 P2 P3 P4 is drawn in the coordinate plane with P1
at (1,0) and P3 at (3,0) . Let Pn denote the point ( x n , y n ) .
Compute the numerical value of the following product
of complex numbers: ( x1 + iy1 )( x 2 + iy 2 )( x3 + iy 3 )( x 4 + iy 4 ) .
11. A quaternary “number” is an arrangement of digits, each of which is 0, 1, 2, 3.
Some examples: 12203, 01130, 22222, 00031
(a) How many 5-digit quaternary numbers are there?
(b) How many 5-digit quaternary numbers are there in which the digit 3 appears at least once?
In each case express your answer using exponents; you need not multiply out your answers.
BRIEF SOLUTIONS TO FIRST ROUND
NOVEMBER 2006
1. A = 3; by trial and error.
2. Sum = 2246 ; There are 83 blocks of 2 + 3 + 0 + 7 + 6 + 9 = 27 , with 2 + 3 left over.
83 × 27 + 5 = 2246 .
3. x = 2, 3, or 4; 2 x − 1 + 2 x − 1 > 5 x − 7 gives 5 > x . 2 x − 1 + 5 x − 7 > 2 x − 1 gives x > 1.
4. (a) 17 = 9 2 − 8 2
(b) 83 = 42 2 − 412
5. 2 3, 4 3, 6 3 and c = 44; 36 9 = 4 = (a + d ) + a + (a − d ) = 3a and a = 4 3 is one root.
(
)
(
)
− 16 9 = (a + d )a(a − d ) = a a 2 − d 2 = 4 3 16 9 − d 2 and d = 2 3 . So the roots are a+d, a, a-d, or
6 3 , 4 3 , 2 3 . Substituting any root into the cubic gives c = 44 .
4
1
1
1
6. A = π ; label the sides as a, b, c. Then ar + br + cr = 8 , where r is the radius. Using a + b + c = 24 , you
2
2
2
9
2
r
2
4
2
have (a + b + c) = 8 or 24r = 16 , r = . Then A = π = π .
2
3
9
3
7. a = 38, b = 84, c = 81, d = 34, e = 5; Let x − 2 = y ; Then x = y + 2 and
f (x ) = f (y + 2 ) = 5(y + 2 ) − 6(y + 2 ) − 3(y + 2 ) + 8(y + 2 )+ 2 =
4
3
2
5(x − 2) + 34(x − 2) + 81(x − 2) + 84(x − 2)+ 38 =
4
3
2
5 y 4 + 34 y 3 + 81 y 2 + 64 y + 38 =
5(x − 2 ) + 34(x − 2 ) + 81(x − 2 ) + 84(x − 2 )+ 38
4
3
2
8. For all n having an odd factor; or n ≠ 2 k ; For example x 6 + y 6 = (x 2 + y 2 )(x 4 − x 2 y 2 + y 4 )
9. 6 ⋅ 5 ! ⋅ 3! ⋅ 4! ; The US diplomats can be arranged in a group in 5 ! ways. Same for the other groups.
But then the three groups can be permuted around in 3 ! = 6 ways.
10. 15; Label the points as P1 = 1 , P2 = 2 + i , P3 = 3 , P4 = 2 − i and multiply:
P1 P2 P3 P4 = 1 ⋅ (2 + i )⋅ 3 ⋅ (2 − i ) = 3(4 + 1) = 15 .
11. (a) 4 5 ; There are four choices for each of five spots.
(b) 4 5 − 35 ; Take all from part (a) and subtract those where the digit 3 fails to appear.
If there is no 3, there are 35 such “numbers”.
UNIVERSITY OF NORTHERN COLORADO
MATHEMATICS CONTEST
FINAL ROUND
For Colorado Students Grades 7-12
February 3, 2007
•
•
•
The sequence of Fibonacci numbers is 1, 1, 2, 3, 5, 8, 13, 21, K .
The positive odd integers are 1, 3, 5, 7, 9, 11, 13, … .
A regular decagon is a 10-sided figure all of whose sides are congruent.
_________________________________________________________________________________
1.
Express the following sum as a whole number:
1 + 2 − 3 + 4 + 5 − 6 + 7 + 8 − 9 + 10 + 11 − 12 + L + 2005 + 2006 − 2007 .
2.
In Grants Pass, Oregon 4 5 of the men are married to 3 7 of the women. What fraction of the
adult population is married? Give a possible generalization.
3.
State the general rule illustrated here and prove it:
4.
If x is a primitive cube root of one (this means that x 3 = 1 but x ≠ 1 ) compute the value of
1
1
x 2006 + 2006 + x 2007 + 2007 .
x
x
5.
Ten different playing cards have the numbers
1, 1, 2, 2, 3, 3, 4, 4, 5, 5 written on them
as shown. Three cards are selected at random
without replacement. What is the
probability that the sum of the
numbers on the three cards is divisible by 7?
Over
6.
(a) Demonstrate that every odd number 2n + 1 can be expressed as a difference of two squares.
(b) Demonstrate which even numbers can be expressed as a difference of two squares.
7.
1 1
1
+ 2 + 3 + L as a reduced fraction.
3 3
3
1 1
2
3
5
+
+
+
+ L as a (reduced) fraction. Here
(b) Express the infinite sum T = +
5 25 125 625 3125
(a) Express the infinite sum S = 1 +
the denominators are powers of 5 and the numerators 1, 1, 2, 3, 5, K are the Fibonacci numbers
Fn where Fn = Fn −1 + Fn − 2 .
8.
A regular decagon P1 P2 P3 K P10 is drawn
in the coordinate plane with P1 at (2,0)
and P6 at (8,0). If Pn denotes the point
( x n , y n ), compute the numerical value of
the following product of complex numbers:
(x1 + i y1 )(x2 + iy 2 )(x3 + iy3 )L (x10 + i y10 )
where i = − 1 as usual.
9.
A circle is inscribed in an equilateral triangle whose side
length is 2. Then another circle is inscribed externally
tangent to the first circle but inside the triangle as shown.
And then another, and another. If this process continues
forever what is the total area of all the circles? Express
your answer as an exact multiple of π (and not as a
decimal approximation).
10.
A quaternary “number” is an arrangement of digits, each of which is 0, 1, 2, 3.
Some examples: 001, 3220, 022113.
(a) How many 6-digit quaternary numbers are there in which each of 0, 1 appear at least once?
(b) How many n-digit quaternary numbers are there in which each of 0, 1, 2, appear at least
once? Test your answer with n = 3 .
(c) Generalize.
Brief Solutions Final Round
February 3, 2007
1. 670,338; (1 + 2 − 3) + (4 + 5 − 6) + (7 + 8 − 9) + L + (2005 + 2006 − 2007)
= 0 + 3 + 6 + L + 3 ⋅ 668 = 3(1 + 2 + 3 + L + 668) = 3(668)(369) / 2
12
12
24
; Restated,
of the men are married to
of the women. Then
of the adult population
43
15
28
43
is married. To generalize, if a b of the men are married to c d of the women, then ca cb of the
men are married to ca da of the women. The proportion that is married is 2ca (cb + da ).
2. 24
3. 12 + 2 2 + 32 + L + n 2 = 1 ⋅ n + 3(n − 1)+ 5(n − 2)+ L + (2n − 1)⋅1 . The picture tells the story. For
example, the fourth diagram shows one 4, three 3’s, five 2’s and seven 1’s. Stripping off layers of
1’s also gives 12 + 2 2 + 32 + 4 2 .
3
4. +1; Since x = 1 , x
2006
2
= x and x
2007
2
1
= 1 . The expression becomes x + 2 + 2 = x + =
x
x
2
1
2
1
x3
−x
2
2
2
2
= 1 since x + x + 1 = 0 . Or, x + 2 + 2 = x + 2 + 2 = x + x + 1 + 1 = 0 + 1 = 1 . Or,
x
x
x
1
1
1
x 2 + 2 + 2 = + x + 2 = −1 + 2 = 1 since the sum of the vectors x and is − 1 .
x
x
x
5. 2 15 ; There are five ways to achieve a sum divisible by 7; 115 (2 ways), 133 ( 2 ways), 124 (8
10
ways), 233 (2 ways), 455 (2 ways). Hence, there are 16 favorable ways out of 120 = total
3
choices.
6. (a) 2n + 1 = (n + 1)2 − n 2
(b) Multiples of 4; If x = 4n , 4n = (n + 1)2 − (n − 1)2 . The even numbers not divisible by 4, namely
2, 6, 10, 14, K cannot be expressed as a product of two even numbers and hence cannot be
expressed as (a + b )(a − b ) = a 2 − b 2 . If a, b are both even each of (a + b ), (a − b ) is even. If both
a, b are odd each of (a + b ), (a − b ) is even. If one is even, one odd both of (a + b ), (a − b ) are odd
and so is the product, handled in part (a).
1
3
1 1 1
1 1
7. (a) 3 2 ; S = 1 + + 2 + L = 1 + 1 + + 2 + L or S = 1+ S and S = .
3
2
3 3 3
3 3
(b) 5 19 ;
5T
= 1 +
T
=
4T
= 1
1
5
1
5
+
+
+
2
2
5
1
52
1
5
2
+
+
+
3
3
5
2
53
1
5
3
+
+
+
5
54
3
54
2
5
4
+ L
+ L
1
+ L = 1+ T
5
8. 9,706,576; Translate the center of the decagon to the origin. Now the vertices represent the roots
of f ( x) = x10 − 310 = 0 . Since the Pn are each 5 more than the roots of f ( x) = 0 , they would be
the roots of f (x − 5) = 0 or ( x − 5)10 − 310 = 0 . The product then is the constant term, or
510 − 310 = 9,706,576 .
9.
3
3π
; Let r1 , and r2 be radii of the first and second circles; r1 =
and the area of the first circle is
3
8
1
3
,
A1 = π 3 . From r1 + r2 = 2(r1 − r2 ), r2 = 3 9 and A2 = π 27 . Similarly r3 = r2 =
3
27
1
1 1
π 3
A3 = π 243 . Then the total area = π +
+
+ L =
= 3π 8 .
3 27 243
1− 1
9
()
()
10. (a) 4 6 − 2 ⋅ 36 + 2 6 ; from the total 4 6 subtract those that have no 0 36 or no 1 36 . Then add
()
back in those that have no 0 and no 1 2 6 .
(b) 4 n − 3 ⋅ 3n + 3 ⋅ 2 n − 1n
(c) Generalize