Difference between revisions of "2000 AMC 8 Problems/Problem 3"
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− | ==Problem | + | ==Problem== |
How many whole numbers lie in the interval between <math>\frac{5}{3}</math> and <math>2\pi</math>? | How many whole numbers lie in the interval between <math>\frac{5}{3}</math> and <math>2\pi</math>? | ||
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<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}</math> | <math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}</math> | ||
− | ==Solution== | + | ==Solution 1 == |
+ | |||
+ | The smallest whole number in the interval is <math>2</math> because <math>5/3</math> is more than <math>1</math> but less than <math>2</math>. The largest whole number in the interval is <math>6</math> because <math>2\pi</math> is more than <math>6</math> but less than <math>7</math>. There are five whole numbers in the interval. They are <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math>, so the answer is <math>\boxed{\text{(D)}\ 5}</math>. slayy bestie | ||
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==Solution 2== | ==Solution 2== | ||
− | We can approximate <math>2\pi</math> to <math>6</math>. Now we approximate <math>5/3</math> to <math>2</math>. Now we list the integers between <math>2</math> and <math>6</math>: | + | We can approximate <math>2\pi</math> to <math>6</math>. Now we approximate <math>5/3</math> to <math>2</math>. Now we list the integers between <math>2</math> and <math>6</math> including <math>2</math> and <math>6</math>: |
<cmath>2,3,4,5,6</cmath> | <cmath>2,3,4,5,6</cmath> | ||
Hence, the answer is <math>\boxed{D}</math> | Hence, the answer is <math>\boxed{D}</math> |
Revision as of 12:31, 18 June 2024
Contents
Problem
How many whole numbers lie in the interval between and ?
Solution 1
The smallest whole number in the interval is because is more than but less than . The largest whole number in the interval is because is more than but less than . There are five whole numbers in the interval. They are , , , , and , so the answer is . slayy bestie
Solution 2
We can approximate to . Now we approximate to . Now we list the integers between and including and : Hence, the answer is
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.