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Difference between revisions of "2000 AIME I Problems"

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m (→‎Problem 12: updating '<center>$...$</center>' to '$$...$$')
 
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The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
 
The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
  
−
<center><asy>draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36));
+
<asy>defaultpen(linewidth(0.7));
 +
draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36));
 
draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61));
 
draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61));
 
draw((34,36)--(34,45)--(25,45));
 
draw((34,36)--(34,45)--(25,45));
 
draw((36,36)--(36,38)--(34,38));
 
draw((36,36)--(36,38)--(34,38));
 
draw((36,38)--(41,38));
 
draw((36,38)--(41,38));
−
draw((34,45)--(41,45));</asy></center>
+
draw((34,45)--(41,45));</asy>
  
 
[[2000 AIME I Problems/Problem 4|Solution]]
 
[[2000 AIME I Problems/Problem 4|Solution]]
  
 
== Problem 5 ==
 
== Problem 5 ==
−
Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is <math>25.</math> One marble is taken out of each box randomly. The probability that both marbles are black is <math>27/50,</math> and the probability that both marbles are white is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?
+
Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is <math>25.</math> One marble is taken out of each box randomly. The probability that both marbles are black is <math>\frac{27}{50},</math> and the probability that both marbles are white is <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?
  
 
[[2000 AIME I Problems/Problem 5|Solution]]
 
[[2000 AIME I Problems/Problem 5|Solution]]
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== Problem 8 ==
 
== Problem 8 ==
−
A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the liquid is <math>m - n\sqrt [3]{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the cube of any prime number. Find <math>m + n + p</math>.
+
A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is <math>m - n\sqrt [3]{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the cube of any prime number. Find <math>m + n + p</math>.
  
 
[[2000 AIME I Problems/Problem 8|Solution]]
 
[[2000 AIME I Problems/Problem 8|Solution]]
Line 50: Line 51:
 
== Problem 9 ==
 
== Problem 9 ==
 
The system of equations
 
The system of equations
−
<center><math>\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\
+
<cmath>\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\
 
\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\
 
\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\
 
\log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\
 
\log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\
−
\end{eqnarray*}</math></center>
+
\end{eqnarray*}</cmath>
  
 
has two solutions <math>(x_{1},y_{1},z_{1})</math> and <math>(x_{2},y_{2},z_{2})</math>. Find <math>y_{1} + y_{2}</math>.
 
has two solutions <math>(x_{1},y_{1},z_{1})</math> and <math>(x_{2},y_{2},z_{2})</math>. Find <math>y_{1} + y_{2}</math>.
Line 60: Line 61:
  
 
== Problem 10 ==
 
== Problem 10 ==
−
A sequence of numbers <math>x_{1},x_{2},x_{3},\ldots,x_{100}</math> has the property that, for every integer <math>k</math> between <math>1</math> and <math>100,</math> inclusive, the number <math>x_{k}</math> is <math>k</math> less than the sum of the other <math>99</math> numbers. Given that <math>x_{50} = m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m + n</math>.
+
A sequence of numbers <math>x_{1},x_{2},x_{3},\ldots,x_{100}</math> has the property that, for every integer <math>k</math> between <math>1</math> and <math>100,</math> inclusive, the number <math>x_{k}</math> is <math>k</math> less than the sum of the other <math>99</math> numbers. Given that <math>x_{50} = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m + n</math>.
  
 
[[2000 AIME I Problems/Problem 10|Solution]]
 
[[2000 AIME I Problems/Problem 10|Solution]]
  
 
== Problem 11 ==
 
== Problem 11 ==
−
Let <math>S</math> be the sum of all numbers of the form <math>a/b,</math> where <math>a</math> and <math>b</math> are relatively prime positive divisors of <math>1000.</math> What is the greatest integer that does not exceed <math>S/10</math>?
+
Let <math>S</math> be the sum of all numbers of the form <math>\frac{a}{b}</math>, where <math>a</math> and <math>b</math> are relatively prime positive divisors of <math>1000.</math> What is the greatest integer that does not exceed <math>\frac{S}{10}</math>?
  
 
[[2000 AIME I Problems/Problem 11|Solution]]
 
[[2000 AIME I Problems/Problem 11|Solution]]
Line 71: Line 72:
 
== Problem 12 ==
 
== Problem 12 ==
 
Given a function <math>f</math> for which
 
Given a function <math>f</math> for which
−
<center><math>f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)</math></center>
+
<cmath>f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)</cmath>
 
holds for all real <math>x,</math> what is the largest number of different values that can appear in the list <math>f(0),f(1),f(2),\ldots,f(999)</math>?
 
holds for all real <math>x,</math> what is the largest number of different values that can appear in the list <math>f(0),f(1),f(2),\ldots,f(999)</math>?
  
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== Problem 13 ==
 
== Problem 13 ==
−
In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at <math>50</math> miles per hour along the highways and at <math>14</math> miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is <math>m/n</math> square miles, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
+
In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at <math>50</math> miles per hour along the highways and at <math>14</math> miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is <math>\frac{m}{n}</math> square miles, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  
 
[[2000 AIME I Problems/Problem 13|Solution]]
 
[[2000 AIME I Problems/Problem 13|Solution]]
  
 
== Problem 14 ==
 
== Problem 14 ==
−
In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> are congruent. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AC}</math> and <math>\overline{AB},</math> respectively, so that <math>AP = PQ = QB = BC.</math> Angle <math>ACB</math> is <math>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find the greatest integer that does not exceed <math>1000r</math>.
+
In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> are congruent. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AC}</math> and <math>\overline{AB},</math> respectively, so that <math>AP = PQ = QB = BC.</math> Angle <math>ACB</math> is <math>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find <math>\lfloor 1000r \rfloor</math>.
  
 
[[2000 AIME I Problems/Problem 14|Solution]]
 
[[2000 AIME I Problems/Problem 14|Solution]]
  
 
== Problem 15 ==
 
== Problem 15 ==
−
A stack of <math>2000</math> cards is labelled with the integers from <math>1</math> to <math>2000,</math> with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: <math>1,2,3,\ldots,1999,2000.</math> In the original stack of cards, how many cards were above the card labeled 1999?
+
A stack of <math>2000</math> cards is labelled with the integers from <math>1</math> to <math>2000,</math> with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: <math>1,2,3,\ldots,1999,2000.</math> In the original stack of cards, how many cards were above the card labeled <math>1999</math>?
  
 
[[2000 AIME I Problems/Problem 15|Solution]]
 
[[2000 AIME I Problems/Problem 15|Solution]]
  
 
== See also ==
 
== See also ==
 +
 +
{{AIME box|year = 2000|n=I|before=[[1999 AIME Problems]]|after=[[2000 AIME II Problems]]}}
 +
 
* [[American Invitational Mathematics Examination]]
 
* [[American Invitational Mathematics Examination]]
 
* [[AIME Problems and Solutions]]
 
* [[AIME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

Latest revision as of 02:40, 4 January 2023

2000 AIME I (Answer Key)
Printable version | AoPS Contest Collections • PDF

Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 1

Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0$.

Solution

Problem 2

Let $u$ and $v$ be integers satisfying $0 < v < u$. Let $A = (u,v)$, let $B$ be the reflection of $A$ across the line $y = x$, let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$. Find $u + v$.

Solution

Problem 3

In the expansion of $(ax + b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a + b$.

Solution

Problem 4

The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.

[asy]defaultpen(linewidth(0.7)); draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]

Solution

Problem 5

Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $\frac{27}{50},$ and the probability that both marbles are white is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$?

Solution

Problem 6

For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^{6}$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$?

Solution

Problem 7

Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Problem 8

A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$.

Solution

Problem 9

The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}

has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$. Find $y_{1} + y_{2}$.

Solution

Problem 10

A sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, find $m + n$.

Solution

Problem 11

Let $S$ be the sum of all numbers of the form $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $\frac{S}{10}$?

Solution

Problem 12

Given a function $f$ for which \[f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)\] holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)$?

Solution

Problem 13

In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is $\frac{m}{n}$ square miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Problem 14

In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$.

Solution

Problem 15

A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$?

Solution

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
1999 AIME Problems 		Followed by
2000 AIME II Problems
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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