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Difference between revisions of "2011 UNCO Math Contest II Problems"

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== See Also ==
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{{UNCO Math Contest box|year=2011|n=II|before=[[2010 UNCO Math Contest II]]|after=[[2012 UNCO Math Contest II]]}}

Latest revision as of 21:10, 7 November 2014

University of Northern Colorado MATHEMATICS CONTEST FINAL ROUND January 29, 2011 For Colorado Students Grades 7-12

• $n!$, read as n factorial, is computed as $n! = 1 \cdot 2 \cdot 3 \cdot 4 \cdots n$

• The factorials are $1, 2, 6, 24, 120, 720,\ldots$

• The square integers are $1, 4, 9, 16, 25, 36, 49, 64, 81,\ldots$

Problem 1

The largest integer $n$ so that $3^n$ evenly divides $9! = 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9$ is $n = 4$. Determine the largest integer $n$ so that $3^n$ evenly divides $85! = 1\cdot 2\cdot 3\cdot 4\cdots 84\cdot 85$.

Solution

Problem 2

Let $m$ and $n$ be positive integers. List all the integers in the set $\left\{ 20 ,21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31\right\}$ that $\underline{cannot}$ be written in the form $m+n+m \cdot n$. As an example, $20$ $\underline{can}$ be so expressed since $20 = 2 + 6 + 2\cdot 6$.

Solution

Problem 3

The two congruent rectangles shown have dimensions $5$ in. by $25$ in. What is the area of the shaded overlap region? [asy] filldraw((0,0)--(13,0)--(25,5)--(12,5)--cycle,blue); pair A1=(0,0),B1=(25,0),C1=(25,5),D1=(0,5); draw(A1--B1--C1--D1--cycle,black); pair P,R; P=unit(A1-D1)+D1; R=unit(C1-D1)+D1; draw(P--(P+R-D1)--R,black); pair A2=(0,0),B2=(25/13,-60/13),C2=(25,5),D2=(25-25/13,5+60/13); draw(A2--B2--C2--D2--cycle,black); P=unit(A2-D2)+D2; R=unit(C2-D2)+D2; draw(P--(P+R-D2)--R,black); [/asy] Solution

Problem 4

Let $A = \left\{ 2,5,10,17,\cdots,n^2+1,\cdots\right\}$ be the set of all positive squares plus $1$ and $B = \left\{101, 104, 109, 116,\cdots,m^2 + 100,\cdots\right\}$ be the set of all positive squares plus $100$.

(a) What is the smallest number in both $A$ and $B$?

(b) Find all numbers that are in both $A$ and $B$.

Solution

Problem 5

Determine the area of the square $ABCD$, with the given lengths along a zigzag line connecting $B$ and $D$. [asy] pair A,B,C,D,E,F,P,R; A=(0,4*sqrt(10)); B=(4*sqrt(10),4*sqrt(10)); C=(4*sqrt(10),0); D=(0,0); E=((9/5)*sqrt(10),(3/5)*sqrt(10)); F=(sqrt(10),3*sqrt(10)); draw(A--B--C--D--cycle,black); draw(D--E--F--B,black); P=unit(D-E)+E; R=unit(F-E)+E; draw(P--(P+R-E)--R,black); P=unit(B-F)+F; R=unit(E-F)+F; draw(P--(P+R-F)--R,black); MP("A",A,NW);MP("B",B,NE);MP("C",C,SE);MP("D",D,SW); MP("6",(D/2+E/2),NW);MP("8",(E/2+F/2),NE);MP("10",(F/2+B/2),S); [/asy]

Solution

Problem 6

What is the remainder when $1! + 2! + 3! + ?+ 2011!$ is divided by $18$?

Solution

Problem 7

What is the $\underline{sum}$ of the first $999$ terms of the sequence $1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63,\cdots$ that appeared on the First Round? Recall that a term in an even numbered position is twice the previous term, while a term in an odd numbered position is one more that the previous term.

Solution

Problem 8

The integer $45$ can be expressed as a sum of two squares as $45 = 3^2 + 6^2$.

(a) Express $74$ as the sum of two squares.

(b) Express the product $45\cdot 74$ as the sum of two squares.

(c) Prove that the product of two sums of two squares, $(a^2+b^2)(c^2+d^2)$ , can be represented as the sum of two squares.

Solution

Problem 9

Let $T(n)$ be the number of ways of selecting three distinct numbers from $\left\{1, 2, 3,\cdots ,n\right\}$ so that they are the lengths of the sides of a triangle. As an example, $T(5) = 3$; the only possibilities are $\{2-3-4\},\{ 2-4-5\}$, and $\{3-4-5\}$.

(a) Determine a recursion for T(n).

(b) Determine a closed formula for T(n).

Solution

Problem 10

The integers $1, 2, 3,\cdots , 50$ are written on the blackboard. Select any two, call them $m$ and $n$ and replace these two with the one number $m+n+mn$. Continue doing this until only one number remains and explain, with proof, what happens. Also explain with proof what happens in general as you replace $50$ with $N$. As an example, if you select $3$ and $17$ you replace them with $3 + 17 + 51 = 71$. If you select $5$ and $7$, replace them with $47$. You now have two $47$’s in this case but that’s OK.

Solution

Problem 11

Tie breaker – Generalize problem #2, and prove your statement.

Solution

See Also

2011 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
2010 UNCO Math Contest II 		Followed by
2012 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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