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Difference between revisions of "2000 AMC 10 Problems"

(Start page, most answers are wrong)
 
(Tired, continuing tomorrow)
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Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was <math>\</math><math>12.48</math>, but in January her bill was <math>\</math><math>17.54</math> because she used twice as much connect time as in December. What is the fixed monthly fee?
 
Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was <math>\</math><math>12.48</math>, but in January her bill was <math>\</math><math>17.54</math> because she used twice as much connect time as in December. What is the fixed monthly fee?
  
−
<math>\mathrm{(A)}\ \<cmath>2.53 \qquad\mathrm{(B)}\ \</cmath>5.06 \qquad\mathrm{(C)}\ \<cmath>6.24 \qquad\mathrm{(D)}\ \</cmath>7.42 \qquad\mathrm{(E)}\ \</math><math>8.77</math>
+
<math>\mathrm{(A)}\ \</math> <math>2.53 \qquad\mathrm{(B)}\ \</math> <math>5.06 \qquad\mathrm{(C)}\ \</math> <math>6.24 \qquad\mathrm{(D)}\ \</math> <math>7.42 \qquad\mathrm{(E)}\ \</math> <math>8.77</math>
  
 
[[2000 AMC 10 Problems/Problem 4|Solution]]
 
[[2000 AMC 10 Problems/Problem 4|Solution]]
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== Problem 6 ==
 
== Problem 6 ==
  
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The Fibonacci sequence <math>1</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, <math>8</math>, <math>13</math>, <math>21</math>, <math>\ldots</math> starts with two <math>1</math>s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?
+
The Fibonacci sequence <math>1, 1, 2, 3, 5, 8, 13, 21, \ldots</math> starts with two <math>1</math>s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?
  
 
<math>\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 9</math>
 
<math>\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 9</math>
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== Problem 7 ==
 
== Problem 7 ==
−
 
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<math>\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10</math>
 
  
 
[[2000 AMC 10 Problems/Problem 7|Solution]]
 
[[2000 AMC 10 Problems/Problem 7|Solution]]
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== Problem 10 ==
 
== Problem 10 ==
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If <math>\abs{x − 2} = p</math>, where <math>x < 2</math>, then <math>x − p =</math>
 
−
 
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<math>\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 2 - 2p \qquad\mathrm{(D)}\ 2p - 2 \qquad\mathrm{(E)}\ \abs{2p - 2}</math>
 
  
 
[[2000 AMC 10 Problems/Problem 10|Solution]]
 
[[2000 AMC 10 Problems/Problem 10|Solution]]
  
 
== Problem 11 ==
 
== Problem 11 ==
−
 
−
<math>\mathrm{(A)}\ \text{the empty set}\qquad\mathrm{(B)}\ \text{one point}\qquad\mathrm{(C)}\ \text{two lines}\qquad\mathrm{(D)}\ \text{a circle}\qquad\mathrm{(E)}\ \text{the entire plane}</math>
 
  
 
[[2000 AMC 10 Problems/Problem 11|Solution]]
 
[[2000 AMC 10 Problems/Problem 11|Solution]]
  
 
== Problem 12 ==
 
== Problem 12 ==
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−
<math>\mathrm{(A)}\ \text{I, by}\ 8\pi\qquad\mathrm{(B)}\ \text{I, by}\ 6\pi\qquad\mathrm{(C)}\ \text{II, by}\ 4\pi\qquad\mathrm{(D)} \text{II, by}\ 8\pi\qquad\mathrm{(E)}\ \text{II, by}\ 10\pi</math>
 
  
 
[[2000 AMC 10 Problems/Problem 12|Solution]]
 
[[2000 AMC 10 Problems/Problem 12|Solution]]
  
 
== Problem 13 ==
 
== Problem 13 ==
−
 
−
<math>\mathrm{(A)}\ 12\qquad\mathrm{(B)}\  30\qquad\mathrm{(C)}\  50\qquad\mathrm{(D)}\  60\qquad\mathrm{(E)}\  100</math>
 
  
 
[[2000 AMC 10 Problems/Problem 13|Solution]]
 
[[2000 AMC 10 Problems/Problem 13|Solution]]
  
 
== Problem 14 ==
 
== Problem 14 ==
−
 
−
<math>\mathrm{(A)}\ 171\qquad\mathrm{(B)}\ 173\qquad\mathrm{(C)}\ 182\qquad\mathrm{(D)}\ 188\qquad\mathrm{(E)}\ 210</math>
 
  
 
[[2000 AMC 10 Problems/Problem 14|Solution]]
 
[[2000 AMC 10 Problems/Problem 14|Solution]]
  
 
== Problem 15 ==
 
== Problem 15 ==
−
 
−
<math>\mathrm{(A)}\ 29\qquad\mathrm{(B)}\ 42\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 47\qquad\mathrm{(E)}\ 50</math>
 
  
 
[[2000 AMC 10 Problems/Problem 15|Solution]]
 
[[2000 AMC 10 Problems/Problem 15|Solution]]
  
 
== Problem 16 ==
 
== Problem 16 ==
−
 
−
<math>\mathrm{(A)}\ \frac{35}{2}\qquad\mathrm{(B)}\ 15\sqrt{2}\qquad\mathrm{(C)}\ \frac{64}{3}\qquad\mathrm{(D)}\ 16\sqrt{2}\qquad\mathrm{(E)}\ 24</math>
 
  
 
[[2000 AMC 10 Problems/Problem 16|Solution]]
 
[[2000 AMC 10 Problems/Problem 16|Solution]]
  
 
== Problem 17 ==
 
== Problem 17 ==
−
 
−
<math>\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ \frac{\sqrt{2}}{2}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2}\qquad\mathrm{(D)}\ \frac{2\sqrt{2}}{2}\qquad\mathrm{(E)}\ \frac{2\sqrt{3}}{3}</math>
 
  
 
[[2000 AMC 10 Problems/Problem 17|Solution]]
 
[[2000 AMC 10 Problems/Problem 17|Solution]]
  
 
== Problem 18 ==
 
== Problem 18 ==
−
 
−
<math>\mathrm{(A)}\ 10^4\times26^2\qquad\mathrm{(B)}\ 10^3\times26^3\qquad\mathrm{(C)}\ 5\times10^4\times26^2\qquad\mathrm{(D)}\ 10^2\times26^4\qquad\mathrm{(E)}\ 5\times10^3\times26^3</math>
 
  
 
[[2000 AMC 10 Problems/Problem 18|Solution]]
 
[[2000 AMC 10 Problems/Problem 18|Solution]]
  
 
== Problem 19 ==
 
== Problem 19 ==
−
 
−
<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 59\qquad\mathrm{(D)}\ 89\qquad\mathrm{(E)}\ 178</math>
 
  
 
[[2000 AMC 10 Problems/Problem 19|Solution]]
 
[[2000 AMC 10 Problems/Problem 19|Solution]]
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== Problem 20 ==
 
== Problem 20 ==
 
   
 
   
−
<math>\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ \frac{3}{5}\qquad\mathrm{(C)}\ \frac{2}{3}\qquad\mathrm{(D)}\ \frac{4}{5}\qquad\mathrm{(E)}\ 1</math>
 
−
 
 
[[2000 AMC 10 Problems/Problem 20|Solution]]
 
[[2000 AMC 10 Problems/Problem 20|Solution]]
  
 
== Problem 21 ==
 
== Problem 21 ==
−
 
−
<math>\mathrm{(A)}\ 2439\qquad\mathrm{(B)}\ 4096\qquad\mathrm{(C)}\ 4903\qquad\mathrm{(D)}\ 4904\qquad\mathrm{(E)}\ 5416</math>
 
  
 
[[2000 AMC 10 Problems/Problem 21|Solution]]
 
[[2000 AMC 10 Problems/Problem 21|Solution]]
  
 
== Problem 22 ==
 
== Problem 22 ==
−
 
−
<math>\mathrm{(A)}\ 5\qquad\mathrm{(B)}\ 10\qquad\mathrm{(C)}\ 30\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 210</math>
 
  
 
[[2000 AMC 10 Problems/Problem 22|Solution]]
 
[[2000 AMC 10 Problems/Problem 22|Solution]]
  
 
== Problem 23 ==
 
== Problem 23 ==
−
 
−
<math>\mathrm{(A)}\ 13\qquad\mathrm{(B)}\ \frac{44}{3}\qquad\mathrm{(C)}\ \sqrt{221}\qquad\mathrm{(D)}\ \sqrt{255}\qquad\mathrm{(E)}\ \frac{55}{3}</math>
 
  
 
[[2000 AMC 10 Problems/Problem 23|Solution]]
 
[[2000 AMC 10 Problems/Problem 23|Solution]]
  
 
== Problem 24 ==
 
== Problem 24 ==
−
 
−
<math>\mathrm{(A)}\ \frac{1}{8}\qquad\mathrm{(B)}\ \frac{1}{6}\qquad\mathrm{(C)}\ \frac{1}{4}\qquad\mathrm{(D)}\ \frac{1}{3}\qquad\mathrm{(E)}\ \frac{1}{2}</math>
 
  
 
[[2000 AMC 10 Problems/Problem 24|Solution]]
 
[[2000 AMC 10 Problems/Problem 24|Solution]]
  
 
== Problem 25 ==
 
== Problem 25 ==
−
 
−
<math>\mathrm{(A)}\ \frac{1}{2187}\qquad\mathrm{(B)}\ \frac{1}{729}\qquad\mathrm{(C)}\ \frac{2}{243}\qquad\mathrm{(D)}\ \frac{1}{81}\qquad\mathrm{(E)}\ \frac{5}{243}</math>
 
  
 
[[2000 AMC 10 Problems/Problem 25|Solution]]
 
[[2000 AMC 10 Problems/Problem 25|Solution]]

Revision as of 01:17, 2 January 2009

Problem 1

In the year 2001, the United States will host the International Mathematical Olympiad. Let $I$, $M$, and $O$ be distinct positive integers such that the product $I * M * O = 2001$. What is the largest possible value of the sum $I + M + O$?

$\mathrm{(A)}\ 23 \qquad\mathrm{(B)}\ 55 \qquad\mathrm{(C)}\ 99 \qquad\mathrm{(D)}\ 111 \qquad\mathrm{(E)}\ 671$

Solution

Problem 2

$2000({2000}^{2000}) =$

$\mathrm{(A)}\ {2000}^{2001} \qquad \mathrm{(B)}\ {4000}^{2000} \qquad \mathrm{(C)}\ {2000}^{4000} \qquad \mathrm{(D)}\ {4,000,000}^{2000} \qquad\mathrm{(E)}\ {2000}^{4,000,000}$

Solution

Problem 3

Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?

$\mathrm{(A)}\ 40 \qquad\mathrm{(B)}\ 50 \qquad\mathrm{(C)}\ 55 \qquad\mathrm{(D)}\ 60 \qquad\mathrm{(E)}\ 75$

Solution

Problem 4

Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was $$$12.48$, but in January her bill was $$$17.54$ because she used twice as much connect time as in December. What is the fixed monthly fee?

$\mathrm{(A)}\ $ $2.53 \qquad\mathrm{(B)}\ $ $5.06 \qquad\mathrm{(C)}\ $ $6.24 \qquad\mathrm{(D)}\ $ $7.42 \qquad\mathrm{(E)}\ $ $8.77$

Solution

Problem 5

Solution

Problem 6

The Fibonacci sequence $1, 1, 2, 3, 5, 8, 13, 21, \ldots$ starts with two $1$s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?

$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 9$

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

See also

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