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Difference between revisions of "2007 UNCO Math Contest II Problems"

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draw(shift(9,-4)*Card,black);
 
draw(shift(9,-4)*Card,black);
 
draw(shift(12,-4)*Card,black);
 
draw(shift(12,-4)*Card,black);
−
MP("1",(.3,2),N);MP("1",(1.7,0),N);
+
MP("1",(.25,2.1),N);MP("1",(1.75,-.1),N);
−
MP("2",(3.3,2),N);MP("2",(4.7,0),N);
+
MP("1",(.25,2.1-4),N);MP("1",(1.75,-4.1),N);
−
MP("3",(6.3,2),N);MP("3",(7.7,0),N);
+
MP("\spadesuit",(1,1),N);MP("\clubsuit",(1,1-4),N);
−
MP("4",(9.3,2),N);MP("4",(10.7,0),N);
+
 
−
MP("5",(12.3,2),N);MP("5",(13.7,0),N);
+
MP("2",(3.25,2.1),N);MP("2",(4.75,-.1),N);
 +
MP("2",(3.25,2.1-4),N);MP("2",(4.75,-4.1),N);
 +
MP("\spadesuit",(4,1.75),N);MP("\clubsuit",(4,1.75-4),N);
 +
MP(180,"\spadesuit",(4,.25),N);MP(180,"\clubsuit",(4,.25-4),N);
 +
 
 +
 
 +
MP("3",(6.25,2.1),N);MP("3",(7.75,-.1),N);
 +
MP("3",(6.25,2.1-4),N);MP("3",(7.75,-4.1),N);
 +
MP("\spadesuit",(7,1.75),N);MP("\clubsuit",(7,1.85-4),N);
 +
MP("\spadesuit",(7,1),N);MP("\clubsuit",(7,1-4),N);
 +
MP(180,"\spadesuit",(7,.25),N);MP(180,"\clubsuit",(7,.25-4),N);
 +
 
 +
MP("4",(9.25,2.1),N);MP("4",(10.75,-.1),N);
 +
MP("4",(9.25,2.1-4),N);MP("4",(10.75,-4.1),N);
 +
MP("\spadesuit",(10-.3,1.75),N);MP("\clubsuit",(10-.3,1.75-4),N);
 +
MP(180,"\spadesuit",(10-.3,.25),N);MP(180,"\clubsuit",(10-.3,.25-4),N);
 +
MP("\spadesuit",(10+.3,1.75),N);MP("\clubsuit",(10+.3,1.75-4),N);
 +
MP(180,"\spadesuit",(10+.3,.25),N);MP(180,"\clubsuit",(10+.3,.25-4),N);
 +
 
 +
MP("5",(12.25,2.1),N);MP("5",(13.75,-.1),N);
 +
MP("5",(12.25,2.1-4),N);MP("5",(13.75,-4.1),N);
 +
MP("\spadesuit",(13-.3,1.75),N);MP("\clubsuit",(13-.3,1.75-4),N);
 +
MP("\spadesuit",(13,1),N);MP("\clubsuit",(13,1-4),N);
 +
MP(180,"\spadesuit",(13-.3,.25),N);MP(180,"\clubsuit",(13-.3,.25-4),N);
 +
MP("\spadesuit",(13+.3,1.75),N);MP("\clubsuit",(13+.3,1.75-4),N);
 +
MP(180,"\spadesuit",(13+.3,.25),N);MP(180,"\clubsuit",(13+.3,.25-4),N);
  
−
MP("1",(.3,2-4),N);MP("1",(1.7,-4),N);
 
−
MP("2",(3.3,2-4),N);MP("2",(4.7,-4),N);
 
−
MP("3",(6.3,2-4),N);MP("3",(7.7,-4),N);
 
−
MP("4",(9.3,2-4),N);MP("4",(10.7,-4),N);
 
−
MP("5",(12.3,2-4),N);MP("5",(13.7,-4),N);
 
 
</asy>
 
</asy>
  
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[[2007 UNCO Math Contest II Problems/Problem 10|Solution]]
 
[[2007 UNCO Math Contest II Problems/Problem 10|Solution]]
 +
 +
== See Also ==
 +
{{UNCO Math Contest box|year=2007|n=II|before=[[2006 UNCO Math Contest II]]|after=[[2008 UNCO Math Contest II]]}}

Latest revision as of 21:14, 7 November 2014

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST FINAL ROUND February 3,2007.

For Colorado Students Grades 7-12.

• The sequence of Fibonacci numbers is $1, 1, 2, 3, 5, 8, 13, 21, \ldots$

• The positive odd integers are $1, 3, 5, 7, 9, 11, 13,\ldots$

• A regular decagon is a $10$-sided figure all of whose sides are congruent.

Problem 1

Express the following sum as a whole number:

$1+ 2 - 3 + 4 + 5 - 6 + 7 + 8 - 9 +10 +11-12 +\cdots + 2005 + 2006 - 2007.$

Solution

Problem 2

In Grants Pass, Oregon $\frac{4}{5}$ of the men are married to $\frac{3}{7}$ of the women. What fraction of the adult population is married? Give a possible generalization.

Solution

Problem 3

State the general rule illustrated here and prove it:

$1 ,\quad \begin{tabular}{cc} 1&1\\1&2\end{tabular} ,\quad \begin{tabular}{ccc} 1&1&1\\1&2&2\\1&2&3\end{tabular},\quad \begin{tabular}{cccc} 1&1&1&1\\1&2&2&2\\1&2&3&3\\1&2&3&4 \end{tabular} ,\quad \begin{tabular}{ccccc} 1&1&1&1&1\\1&2&2&2&2\\1&2&3&3&3\\1&2&3&4&4\\1&2&3&4&5 \end{tabular} ,\quad \cdots$

Solution

Problem 4

If $x$ is a primitive cube root of one (this means that $x^3 =1$ but $x \ne 1$) compute the value of \[x^{2006}+\frac{1}{x^{2006}}+x^{2007}+\frac{1}{x^{2007}}.\]


Solution

Problem 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 $7$?

[asy] size(200,100); path Card=(arc((.2,.2),.2,180,270)--(1.8,0)--arc((1.8,.2),.2,270,360)--(2,2.8)--arc((1.8,2.8),.2,0,90)--(.2,3)--arc((.2,2.8),.2,90,180)--(0,.2)); draw(Card,black); draw(shift(3,0)*Card,black); draw(shift(6,0)*Card,black); draw(shift(9,0)*Card,black); draw(shift(12,0)*Card,black); draw(shift(0,-4)*Card,black); draw(shift(3,-4)*Card,black); draw(shift(6,-4)*Card,black); draw(shift(9,-4)*Card,black); draw(shift(12,-4)*Card,black); MP("1",(.25,2.1),N);MP("1",(1.75,-.1),N); MP("1",(.25,2.1-4),N);MP("1",(1.75,-4.1),N); MP("\spadesuit",(1,1),N);MP("\clubsuit",(1,1-4),N); MP("2",(3.25,2.1),N);MP("2",(4.75,-.1),N); MP("2",(3.25,2.1-4),N);MP("2",(4.75,-4.1),N); MP("\spadesuit",(4,1.75),N);MP("\clubsuit",(4,1.75-4),N); MP(180,"\spadesuit",(4,.25),N);MP(180,"\clubsuit",(4,.25-4),N); MP("3",(6.25,2.1),N);MP("3",(7.75,-.1),N); MP("3",(6.25,2.1-4),N);MP("3",(7.75,-4.1),N); MP("\spadesuit",(7,1.75),N);MP("\clubsuit",(7,1.85-4),N); MP("\spadesuit",(7,1),N);MP("\clubsuit",(7,1-4),N); MP(180,"\spadesuit",(7,.25),N);MP(180,"\clubsuit",(7,.25-4),N); MP("4",(9.25,2.1),N);MP("4",(10.75,-.1),N); MP("4",(9.25,2.1-4),N);MP("4",(10.75,-4.1),N); MP("\spadesuit",(10-.3,1.75),N);MP("\clubsuit",(10-.3,1.75-4),N); MP(180,"\spadesuit",(10-.3,.25),N);MP(180,"\clubsuit",(10-.3,.25-4),N); MP("\spadesuit",(10+.3,1.75),N);MP("\clubsuit",(10+.3,1.75-4),N); MP(180,"\spadesuit",(10+.3,.25),N);MP(180,"\clubsuit",(10+.3,.25-4),N); MP("5",(12.25,2.1),N);MP("5",(13.75,-.1),N); MP("5",(12.25,2.1-4),N);MP("5",(13.75,-4.1),N); MP("\spadesuit",(13-.3,1.75),N);MP("\clubsuit",(13-.3,1.75-4),N); MP("\spadesuit",(13,1),N);MP("\clubsuit",(13,1-4),N); MP(180,"\spadesuit",(13-.3,.25),N);MP(180,"\clubsuit",(13-.3,.25-4),N); MP("\spadesuit",(13+.3,1.75),N);MP("\clubsuit",(13+.3,1.75-4),N); MP(180,"\spadesuit",(13+.3,.25),N);MP(180,"\clubsuit",(13+.3,.25-4),N); [/asy]

Solution

Problem 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.

Solution

Problem 7

(a) Express the infinite sum $S= 1+ \frac{1}{3}+\frac{1}{3^2}+ \frac{1}{3^3}+ \cdots$ as a reduced fraction.

(b) Express the infinite sum $T=\frac{1}{5}+ \frac{1}{25}+ \frac{2}{125}+ \frac{3}{625}+ \frac{5}{3125}+ \cdots$ as a reduced fraction. Here the denominators are powers of $5$ and the numerators $1, 1, 2, 3, 5, \ldots$ are the Fibonacci numbers $F_n$ where $F_n=F_{n-1}+F_{n-2}$.

Solution

Problem 8

A regular decagon $P_1P_2P_3\cdots P_{10}$ is drawn in the coordinate plane with $P_1$ at $(2,0)$ and $P_6$ at $(8,0)$. If $P_n$ denotes the point $(x_n ,y_n )$, compute the numerical value of the following product of complex numbers: $( x_1+iy_1)( x_2+iy_2)( x_3+iy_3) \cdots (x_{10} + iy_{10})$ where $i^2 = -1$ as usual.

[asy] draw(polygon(10),dot); draw((-2,0)--(2,0),black); draw((-5/3,-2)--(-5/3,2),black); MP("P_1",(-1,0),NW);MP("(2,0)",(-.9,0),SW); MP("P_6",(1,0),NE);MP("(8,0)",(.9,0),SE); [/asy]

Solution

Problem 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 $\pi$ (and not as a decimal approximation).

[asy] path T=polygon(3); draw(unitcircle,black); draw(scale(2)*T,black); draw(shift(2/sqrt(3),-2/3)*scale(1/3)*unitcircle,black); draw(shift(2/sqrt(3)/3,-2/9)*shift(2/sqrt(3),-2/3)*scale(1/9)*unitcircle,black); [/asy]

Solution

Problem 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.

Solution

See Also

2007 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
2006 UNCO Math Contest II 		Followed by
2008 UNCO Math Contest II
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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