Difference between revisions of "2000 AMC 8 Problems/Problem 3"
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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>. | 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>. | ||
==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> | ||
Latest revision as of 11:21, 25 October 2016
Contents
Problem
How many whole numbers lie in the interval between 
and 
?
Solution
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 
.
Solution 2
We can approximate to . Now we approximate to . Now we list the integers between and including and :
![\[2,3,4,5,6\]](http://latex.artofproblemsolving.com/0/2/8/028f420f716f02a3a83db473b542f4c31c759daf.png)
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. 
