Difference between revisions of "2018 UNCO Math Contest II Problems"

(Problem 3)
(Problem 3)
 
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==Problem 2==
 
==Problem 2==
 +
<asy>
 +
pair D=(0,0),B=(0,12),C=(16,0),A=(-9,0);
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draw(A--B--C--A,dot);
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draw(D--B,dot);
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MP("D",D,S);MP("A",A,S);MP("C",C,S);MP("B",B,N);
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MP("16",(8,0),S);MP("15",(A+B)/2,NW);
 +
</asy>
  
 
Segment AB is perpendicular to segment BC and
 
Segment AB is perpendicular to segment BC and
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==Problem 3==  
 
==Problem 3==  
  
Find all values of B that have the property that if (x, y) lies on the hyperbola 2y^2-x^2 = 1,
+
Find all values of B that have the property that if <math>(x, y</math>) lies on the hyperbola <math>2y^2-x^2 = 1</math>,
 
then so does the point (3x + 4y, 2x + By).
 
then so does the point (3x + 4y, 2x + By).
  
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==Problem 4==
 
==Problem 4==
  
 +
How many positive integer factors of <math>36,000,000</math> are not perfect squares?
  
 
[[2018 UNCO Math Contest II Problems/Problem 4|Solution]]
 
[[2018 UNCO Math Contest II Problems/Problem 4|Solution]]
  
 
==Problem 5==
 
==Problem 5==
+
 
 +
<asy>
 +
pair A=(3,0),B=((6+sqrt(1536))/50,(144-sqrt(1536))/(50*sqrt(24))),C=((6-sqrt(1536))/50,(144+sqrt(1536))/(50*sqrt(24))),D=(-1.8,sqrt(24)/5);
 +
filldraw(circle((2,0),1),white);
 +
filldraw(circle((0,0),1),white);
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filldraw(circle((-2,0),1),white);
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draw(A--D,black);
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draw((3.5,0)--(-3.5,0),black);
 +
 
 +
dot(A,black+0.25cm);dot(B,black+0.25cm);dot(C,black+0.25cm);
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MP("A",A,NE);MP("B",B,N);MP("C",C,N);
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</asy>
 +
 
 +
Find the length of segment BC formed in the middle circle by a line that goes through point A and is
 +
tangent to the leftmost circle. The three circles in the
 +
figure all have radius one and their centers lie on the
 +
horizontal line. The leftmost and rightmost circles are
 +
tangent to the circle in the middle. Point A is at
 +
the rightmost intersection of the rightmost circle and the horizontal line.
  
 
[[2018 UNCO Math Contest II Problems/Problem 5|Solution]]
 
[[2018 UNCO Math Contest II Problems/Problem 5|Solution]]
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==Problem 6==
 
==Problem 6==
  
 +
Circling the square. Exactly one of these polynomials is a perfect square; that is, can be
 +
written as <math>(p(x))^2</math> where <math>p(x)</math> is also a polynomial. Circle the choice that is a perfect square,
 +
and for that choice, find the square root, the polynomial <math>p(x)</math>.
 +
 +
(A) <math>36-49x^2 + 14x^4 </math>
  
 +
(B) <math>36-48x^2 + 14x^4-x^6</math>
 +
 +
(C) <math>9-12x + 4x^2 + 12x^3-8x^4 + 4x^6 </math>
 +
 +
(D) <math>36-49x^2 + 15x^4-x^6</math>
  
 
[[2018 UNCO Math Contest II Problems/Problem 6|Solution]]
 
[[2018 UNCO Math Contest II Problems/Problem 6|Solution]]
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==Problem 7==
 
==Problem 7==
  
 
+
Let <math>x = 2A + 10B</math> where <math>A</math> and <math>B</math> are randomly chosen with replacement from among the
 
+
positive integers less than or equal to twelve. What is the probability that <math>x</math> is a multiple of
 +
<math>12</math>?
  
 
[[2018 UNCO Math Contest II Problems/Problem 7|Solution]]
 
[[2018 UNCO Math Contest II Problems/Problem 7|Solution]]
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==Problem 8==
 
==Problem 8==
  
 
+
Let <math>p(x) = x^{2018} + x^{1776}-3x^4-3</math>. Find the remainder when you divide <math>p(x)</math> by <math>x^3-x</math>
  
 
[[2018 UNCO Math Contest II Problems/Problem 8|Solution]]
 
[[2018 UNCO Math Contest II Problems/Problem 8|Solution]]
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==Problem 9==
 
==Problem 9==
  
 +
Call a set of integers Grassilian if each of its elements is at least as large as the number of
 +
elements in the set. For example, the three-element set <math>\{2, 48, 100\}</math> is not Grassilian, but the
 +
six-element set <math>\{6, 10, 11, 20, 33, 39\}</math> is Grassilian. Let <math>G(n)</math> be the number of Grassilian subsets of <math>\{1, 2, 3, ..., n\}</math>. (By definition, the empty set is a subset of every set and is Grassilian.)
 +
(a) Find <math>G(3)</math>, <math>G(4)</math>, and <math>G(5)</math>.
 +
(b) Find a recursion formula for <math>G(n + 1)</math>. That is, find a formula that expresses <math>G(n + 1)</math> in
 +
terms of <math>G(n), G(n  1),\;dots</math>
 +
(c) Give an explanation that shows that the formula you give is correct.
  
 
[[2018 UNCO Math Contest II Problems/Problem 9|Solution]]
 
[[2018 UNCO Math Contest II Problems/Problem 9|Solution]]
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==Problem 10==
 
==Problem 10==
  
 +
The Seripians have seen the error of their ways and
 +
issued new pit-coins in 2-pit and 3-pit denominations,
 +
containing 2 and 3 serigrams of gold. One of the new
 +
coins is in the shape of a domino (two adjoining squares) and the other two are in the shape of
 +
triominoes (three adjoining squares), shown above. To celebrate the new coins, the Seripians
 +
have announced a contest. Seripian students can win fame and glory and 100 of each of the
 +
new Seripian pit-coins by successfully completing quests (a)-(d) below.
 +
Call a tiling by pit-coins prime if there is no vertical
 +
line that splits the tiling into tilings of two smaller
 +
shapes without cutting across any of the coins. The
 +
2x5 tiling above on the left is prime and the 2x5 tiling on the right is not prime.
 +
Define P(n) to be the number of distinct prime tilings
 +
of a horizontal 2xn grid. For example, P(4) = 6, and
 +
the six distinct prime 2x4 tilings are shown at left.
 +
Define Q(n) to be the number of distinct prime tilings of the two 2xn grids with one unit
 +
corner square missing at the right end. Q(3) = 4 and
 +
the four prime tilings are shown to the left. We wish
 +
you success on the Seripian Quests. Show your work.
 +
(a) Determine P(6).
 +
(b) Determine formulas for P(n) and Q(n) in terms of Q(n  1), Q(n  2), and/or Q(n  3)
 +
that are valid for n  4.
 +
(c) Determine a formula for P(n) that does not use Q. You may use P(n  1), P(n  2),
 +
P(n  3), . . .. Specify how large n must be for your formula to work.
 +
(d) Determine explicitly P(11) and P(13).
  
 
[[2018 UNCO Math Contest II Problems/Problem 10|Solution]]
 
[[2018 UNCO Math Contest II Problems/Problem 10|Solution]]
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==Problem 11==
 
==Problem 11==
  
 +
(a) Find an integer <math>n > 1</math> for which <math>1 + 2 + \ldots + n^2</math> is a perfect square.
 +
(b) Show that there are infinitely many integers <math>n > 1</math> that have the property that
 +
<math>1 + 2 + \ldots + n^2</math> is a perfect square, and determine at least three more examples of such <math>n</math>.
 +
Hint: There is one approach that uses the result of a previous problem on this contest.
  
 
[[2018 UNCO Math Contest II Problems/Problem 11|Solution]]
 
[[2018 UNCO Math Contest II Problems/Problem 11|Solution]]
 +
 +
 +
== See Also ==
 +
{{UNCO Math Contest box|year=2018|n=II|before=[[2017 UNCO Math Contest II]]|after=[[2019 UNCO Math Contest II]]}}

Latest revision as of 11:37, 30 July 2021

Twenty-sixth Annual UNC Math Contest Final Round January 20, 2018

Rules: Three hours; no electronic devices. The positive integers are 1, 2, 3, 4, . . . A prime is an integer strictly greater than one that is evenly divisible by no integers other than itself and 1. The primes are 2, 3, 5, 7, 11, 13, 17, . . .

Problem 1

A printer used 1890 digits to number all the pages in the Seripian Puzzle Book. How many pages are in the book? (For example, to number the pages in a book with twelve pages, the printer would use fifteen digits.)

Solution

Problem 2

[asy] pair D=(0,0),B=(0,12),C=(16,0),A=(-9,0); draw(A--B--C--A,dot); draw(D--B,dot); MP("D",D,S);MP("A",A,S);MP("C",C,S);MP("B",B,N); MP("16",(8,0),S);MP("15",(A+B)/2,NW); [/asy]

Segment AB is perpendicular to segment BC and segment AC is perpendicular to segment BD. If segment AB has length 15 and segment DC has length 16, then what is the area of triangle ABC?


Solution

Problem 3

Find all values of B that have the property that if $(x, y$) lies on the hyperbola $2y^2-x^2 = 1$, then so does the point (3x + 4y, 2x + By).

Solution

Problem 4

How many positive integer factors of $36,000,000$ are not perfect squares?

Solution

Problem 5

[asy] pair A=(3,0),B=((6+sqrt(1536))/50,(144-sqrt(1536))/(50*sqrt(24))),C=((6-sqrt(1536))/50,(144+sqrt(1536))/(50*sqrt(24))),D=(-1.8,sqrt(24)/5); filldraw(circle((2,0),1),white); filldraw(circle((0,0),1),white); filldraw(circle((-2,0),1),white); draw(A--D,black); draw((3.5,0)--(-3.5,0),black);  dot(A,black+0.25cm);dot(B,black+0.25cm);dot(C,black+0.25cm); MP("A",A,NE);MP("B",B,N);MP("C",C,N); [/asy]

Find the length of segment BC formed in the middle circle by a line that goes through point A and is tangent to the leftmost circle. The three circles in the figure all have radius one and their centers lie on the horizontal line. The leftmost and rightmost circles are tangent to the circle in the middle. Point A is at the rightmost intersection of the rightmost circle and the horizontal line.

Solution

Problem 6

Circling the square. Exactly one of these polynomials is a perfect square; that is, can be written as $(p(x))^2$ where $p(x)$ is also a polynomial. Circle the choice that is a perfect square, and for that choice, find the square root, the polynomial $p(x)$.

(A) $36-49x^2 + 14x^4$

(B) $36-48x^2 + 14x^4-x^6$

(C) $9-12x + 4x^2 + 12x^3-8x^4 + 4x^6$

(D) $36-49x^2 + 15x^4-x^6$

Solution

Problem 7

Let $x = 2A + 10B$ where $A$ and $B$ are randomly chosen with replacement from among the positive integers less than or equal to twelve. What is the probability that $x$ is a multiple of $12$?

Solution

Problem 8

Let $p(x) = x^{2018} + x^{1776}-3x^4-3$. Find the remainder when you divide $p(x)$ by $x^3-x$

Solution

Problem 9

Call a set of integers Grassilian if each of its elements is at least as large as the number of elements in the set. For example, the three-element set $\{2, 48, 100\}$ is not Grassilian, but the six-element set $\{6, 10, 11, 20, 33, 39\}$ is Grassilian. Let $G(n)$ be the number of Grassilian subsets of $\{1, 2, 3, ..., n\}$. (By definition, the empty set is a subset of every set and is Grassilian.) (a) Find $G(3)$, $G(4)$, and $G(5)$. (b) Find a recursion formula for $G(n + 1)$. That is, find a formula that expresses $G(n + 1)$ in terms of $G(n), G(n  1),\;dots$ (c) Give an explanation that shows that the formula you give is correct.

Solution

Problem 10

The Seripians have seen the error of their ways and issued new pit-coins in 2-pit and 3-pit denominations, containing 2 and 3 serigrams of gold. One of the new coins is in the shape of a domino (two adjoining squares) and the other two are in the shape of triominoes (three adjoining squares), shown above. To celebrate the new coins, the Seripians have announced a contest. Seripian students can win fame and glory and 100 of each of the new Seripian pit-coins by successfully completing quests (a)-(d) below. Call a tiling by pit-coins prime if there is no vertical line that splits the tiling into tilings of two smaller shapes without cutting across any of the coins. The 2x5 tiling above on the left is prime and the 2x5 tiling on the right is not prime. Define P(n) to be the number of distinct prime tilings of a horizontal 2xn grid. For example, P(4) = 6, and the six distinct prime 2x4 tilings are shown at left. Define Q(n) to be the number of distinct prime tilings of the two 2xn grids with one unit corner square missing at the right end. Q(3) = 4 and the four prime tilings are shown to the left. We wish you success on the Seripian Quests. Show your work. (a) Determine P(6). (b) Determine formulas for P(n) and Q(n) in terms of Q(n 1), Q(n 2), and/or Q(n 3) that are valid for n 4. (c) Determine a formula for P(n) that does not use Q. You may use P(n 1), P(n 2), P(n 3), . . .. Specify how large n must be for your formula to work. (d) Determine explicitly P(11) and P(13).

Solution

Problem 11

(a) Find an integer $n > 1$ for which $1 + 2 + \ldots + n^2$ is a perfect square. (b) Show that there are infinitely many integers $n > 1$ that have the property that $1 + 2 + \ldots + n^2$ is a perfect square, and determine at least three more examples of such $n$. Hint: There is one approach that uses the result of a previous problem on this contest.

Solution


See Also

2018 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
2017 UNCO Math Contest II
Followed by
2019 UNCO Math Contest II
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions