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Difference between revisions of "2006 Alabama ARML TST Problems"

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[[2006 Alabama ARML TST Problems/Problem 4 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 4 | Solution]]
 
==Problem 5==
 
==Problem 5==
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There exist positive integers <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>with no common factor greater than 1, such that
+
There exist positive integers <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> with no common factor greater than 1, such that
  
 
<center><math>A\log_{1200} 2+B\log_{1200} 3+C\log_{1200} 5=D.</math></center>
 
<center><math>A\log_{1200} 2+B\log_{1200} 3+C\log_{1200} 5=D.</math></center>
Line 25: Line 25:
  
 
[[2006 Alabama ARML TST Problems/Problem 5 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 5 | Solution]]
 +
 
==Problem 6==
 
==Problem 6==
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Let<math>\lfloor a \rfloor</math> be the greatest integer less than or equal to <math>a</math> and let <math>\{a\}=a-\lfloor a \rfloor</math>. Find <math>10(x+y+z)</math> given that
+
Let <math>\lfloor a \rfloor</math> be the greatest integer less than or equal to <math>a</math> and let <math>\{a\}=a-\lfloor a \rfloor</math>. Find <math>10(x+y+z)</math> given that
  
−
<math>\begin{align}
+
<center><cmath>\begin{align*}
 
x+\lfloor y \rfloor +\{z\}=14.2,\\
 
x+\lfloor y \rfloor +\{z\}=14.2,\\
 
\lfloor x \rfloor+\{y\} +z=15.3,\\
 
\lfloor x \rfloor+\{y\} +z=15.3,\\
 
\{x\}+y +\lfloor z \rfloor=16.1.
 
\{x\}+y +\lfloor z \rfloor=16.1.
−
\end{align}</math>
+
\end{align*}</cmath></center>
  
 
[[2006 Alabama ARML TST Problems/Problem 6 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 6 | Solution]]
 +
 
==Problem 7==
 
==Problem 7==
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Four equilateral triangles are drawn such that each one shares a different side with a square
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Four equilateral triangles are drawn such that each one shares a different side with a square of side length 10. None of the areas of the triangles overlap with the area of the square. The four vertices of the triangles that aren’t vertices of the square are connected to form a larger square. Find the area of this larger square.
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of side length 10. None of the areas of the triangles overlap with the area of the square. The
 
−
four vertices of the triangles that aren’t vertices of the square are connected to form a larger
 
−
square. Find the area of this larger square.
 
  
 
[[2006 Alabama ARML TST Problems/Problem 7 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 7 | Solution]]
 +
 
==Problem 8==
 
==Problem 8==
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A bored mathematician has his computer calculate 1000 consecutive terms in the Fibonacci
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A bored mathematician has his computer calculate 1000 consecutive terms in the Fibonacci sequence. He notes that the smallest of the numbers is a multiple of 7. How many of the other 999 Fibonacci numbers are multiples of 7?
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sequence. He notes that the smallest of the numbers is a multiple of 7. How many of the
 
−
other 999 Fibonacci numbers are multiples of 7?
 
  
 
[[2006 Alabama ARML TST Problems/Problem 8 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 8 | Solution]]
 
==Problem 9==
 
==Problem 9==
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Amanda ordered a dozen donuts. She said she wanted only chocolate, glazed, and powdered
+
Amanda ordered a dozen donuts. She said she wanted only chocolate, glazed, and powdered donuts, and at least one of each kind. Let <math>a</math>, <math>b</math>, and <math>c</math> be the number of chocolate, glazed, and powdered donuts she wound up with. Find the number of possible ordered triples <math>(a, b, c)</math>.
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donuts, and at least one of each kind. Let <math>a</math>, <math>b</math>, and <math>c</math> be the number of chocolate, glazed, and
 
−
powdered donuts she wound up with. Find the number of possible ordered triples <math>(a, b, c)</math>.
 
  
 
[[2006 Alabama ARML TST Problems/Problem 9 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 9 | Solution]]
 +
 
==Problem 10==
 
==Problem 10==
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Let <math>p</math> be the probability that Scooby Doo solves any given mystery. The probability that
+
Let <math>p</math> be the probability that Scooby Doo solves any given mystery. The probability that Scooby Doo solves 1800 out of 2006 given mysteries is the same as the probability that he solves 1801 of them. Find the probability that Scooby Doo solves the mystery of why Eddie Murphy decided to stop being funny.
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Scooby Doo solves 1800 out of 2006 given mysteries is the same as the probability that he
 
−
solves 1801 of them. Find the probability that Scooby Doo solves the mystery of why Eddie
 
−
Murphy decided to stop being funny.
 
  
 
[[2006 Alabama ARML TST Problems/Problem 10 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 10 | Solution]]
 
==Problem 11==
 
==Problem 11==
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The integer <math>5^{2006}</math> has 1403 digits, and 1 is its first digit (farthest to the left). For how many
+
The integer <math>5^{2006}</math> has 1403 digits, and 1 is its first digit (farthest to the left). For how many integers <math>0\leq k \leq 2005</math> does <math>5^k</math> begin with the digit 1?
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integers <math>0\leq� k \leq� 2005</math> does <math>5k</math> begin with the digit 1?
 
  
 
[[2006 Alabama ARML TST Problems/Problem 11 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 11 | Solution]]
  
 
==Problem 12==
 
==Problem 12==
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Yoda begins writing the positive integers starting from 1 and continuing consecutively as he
+
Yoda begins writing the positive integers starting from 1 and continuing consecutively as he writes. When he stops, he realizes that there is no set of 5 composite integers among the ones he wrote such that each pair of those 5 is relatively prime. What’s the largest possible number Yoda could have stopped on?
−
writes. When he stops, he realizes that there is no set of 5 composite integers among the
 
−
ones he wrote such that each pair of those 5 is relatively prime. What’s the largest possible
 
−
number Yoda could have stopped on?
 
  
 
[[2006 Alabama ARML TST Problems/Problem 12 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 12 | Solution]]
Line 77: Line 67:
 
Find the sum of the solutions to the equation
 
Find the sum of the solutions to the equation
  
−
<center><math>\sqrt[4]{x+27}+\sqrt[4]{55-x}.</math></center>
+
<center><math>\sqrt[4]{x+27}+\sqrt[4]{55-x}=4.</math></center>
  
 
[[2006 Alabama ARML TST Problems/Problem 13 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 13 | Solution]]
 +
 
==Problem 14==
 
==Problem 14==
 
Find the real solution <math>(x, y)</math> to the system of equations
 
Find the real solution <math>(x, y)</math> to the system of equations
  
−
<center><math>x^3−3xy^2=−610,</math></center>
+
<center><math>x^3-3xy^2=-610,</math></center>
−
<center><math>3x^2y−y^3=182.</math></center>
+
<center><math>3x^2y-y^3=182.</math></center>
  
 
[[2006 Alabama ARML TST Problems/Problem 14 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 14 | Solution]]
  
 
==Problem 15==
 
==Problem 15==
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Ying lives on Strangeland, a tiny planet with 4 little cities that are each 100 miles apart
+
Ying lives on Strangeland, a tiny planet with 4 little cities that are each 100 miles apart from each other. One day, Ying begins driving from her home city of Viavesta to the city of Havennew, which takes her about an hour. When she gets to Havennew, she decides she wants to go straight to another city on Strangeland, so she randomly chooses one of the other three cities (possibly Viavesta), and starts driving there. Ying drives like this for most of the day, making 8 total trips between cities on Strangeland, choosing randomly where to drive to next from each stop. She then stops at her final city of destination, digs a hole, and buries her car.
−
from each other. One day, Ying begins driving from her home city of Viavesta to the city
 
−
of Havennew, which takes her about an hour. When she gets to Havennew, she decides she
 
−
wants to go straight to another city on Strangeland, so she randomly chooses one of the other
 
−
three cities (possibly Viavesta), and starts driving there. Ying drives like this for most of the
 
−
day, making 8 total trips between cities on Strangeland, choosing randomly where to drive to
 
−
next from each stop. She then stops at her final city of destination, digs a hole, and buries
 
−
her car.
 
  
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Let <math>p</math> be the probability Ying buried her car in Viavesta and let q be the probability she
+
Let <math>p</math> be the probability Ying buried her car in Viavesta and let q be the probability she buried it in Havennew. Find the value of <math>p + q</math>.
−
buried it in Havennew. Find the value of <math>p + q</math>.
 
  
 
[[2006 Alabama ARML TST Problems/Problem 15 | Solution]]
 
[[2006 Alabama ARML TST Problems/Problem 15 | Solution]]

Latest revision as of 19:56, 24 March 2015

Problem 1

How many integers $x$ satisfy the inequality

$(x-2006)(x-2004)(x-2002)\cdots (x-4)<0?$

Solution

Problem 2

Compute $\sum_{k=1}^{5}\sum_{n=1}^{6}kn$.

Solution

Problem 3

River draws four cards from a standard 52 card deck of playing cards. Exactly 3 of them are 2’s. Find the probability River drew exactly one spade and one club from the deck.

Solution

Problem 4

Find the number of six-digit positive integers for which the digits are in increasing order.

Solution

Problem 5

There exist positive integers $A$, $B$, $C$, and $D$ with no common factor greater than 1, such that

$A\log_{1200} 2+B\log_{1200} 3+C\log_{1200} 5=D.$

Find $A+B+C+D$.

Solution

Problem 6

Let $\lfloor a \rfloor$ be the greatest integer less than or equal to $a$ and let $\{a\}=a-\lfloor a \rfloor$. Find $10(x+y+z)$ given that

\begin{align*} x+\lfloor y \rfloor +\{z\}=14.2,\\ \lfloor x \rfloor+\{y\} +z=15.3,\\ \{x\}+y +\lfloor z \rfloor=16.1. \end{align*}

Solution

Problem 7

Four equilateral triangles are drawn such that each one shares a different side with a square of side length 10. None of the areas of the triangles overlap with the area of the square. The four vertices of the triangles that aren’t vertices of the square are connected to form a larger square. Find the area of this larger square.

Solution

Problem 8

A bored mathematician has his computer calculate 1000 consecutive terms in the Fibonacci sequence. He notes that the smallest of the numbers is a multiple of 7. How many of the other 999 Fibonacci numbers are multiples of 7?

Solution

Problem 9

Amanda ordered a dozen donuts. She said she wanted only chocolate, glazed, and powdered donuts, and at least one of each kind. Let $a$, $b$, and $c$ be the number of chocolate, glazed, and powdered donuts she wound up with. Find the number of possible ordered triples $(a, b, c)$.

Solution

Problem 10

Let $p$ be the probability that Scooby Doo solves any given mystery. The probability that Scooby Doo solves 1800 out of 2006 given mysteries is the same as the probability that he solves 1801 of them. Find the probability that Scooby Doo solves the mystery of why Eddie Murphy decided to stop being funny.

Solution

Problem 11

The integer $5^{2006}$ has 1403 digits, and 1 is its first digit (farthest to the left). For how many integers $0\leq k \leq 2005$ does $5^k$ begin with the digit 1?

Solution

Problem 12

Yoda begins writing the positive integers starting from 1 and continuing consecutively as he writes. When he stops, he realizes that there is no set of 5 composite integers among the ones he wrote such that each pair of those 5 is relatively prime. What’s the largest possible number Yoda could have stopped on?

Solution

Problem 13

Find the sum of the solutions to the equation

$\sqrt[4]{x+27}+\sqrt[4]{55-x}=4.$

Solution

Problem 14

Find the real solution $(x, y)$ to the system of equations

$x^3-3xy^2=-610,$
$3x^2y-y^3=182.$

Solution

Problem 15

Ying lives on Strangeland, a tiny planet with 4 little cities that are each 100 miles apart from each other. One day, Ying begins driving from her home city of Viavesta to the city of Havennew, which takes her about an hour. When she gets to Havennew, she decides she wants to go straight to another city on Strangeland, so she randomly chooses one of the other three cities (possibly Viavesta), and starts driving there. Ying drives like this for most of the day, making 8 total trips between cities on Strangeland, choosing randomly where to drive to next from each stop. She then stops at her final city of destination, digs a hole, and buries her car.

Let $p$ be the probability Ying buried her car in Viavesta and let q be the probability she buried it in Havennew. Find the value of $p + q$.

Solution

See also

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