Difference between revisions of "2003 AMC 10B Problems/Problem 3"
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Now we solve: | Now we solve: | ||
| − | <cmath>5n+24=64 | + | <cmath>5n+24=64</cmath> |
| − | 5n=40 | + | <cmath>5n=40</cmath> |
| − | n=8</cmath> | + | <cmath>n=8</cmath> |
Thus, the smallest integer is <math>\boxed{\textbf{(B) } 8}</math>. | Thus, the smallest integer is <math>\boxed{\textbf{(B) } 8}</math>. | ||
| Line 23: | Line 23: | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=2|num-a=4}} | {{AMC10 box|year=2003|ab=B|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:51, 4 January 2017
Problem
The sum of 
consecutive even integers is 
less than the sum of the first 
consecutive odd counting numbers. What is the smallest of the even integers?
Solution
It is a well-known fact that the sum of the first 
odd numbers is 
. This makes the sum of the first odd numbers equal to 
.
Let be equal to the smallest of the even integers. Then 
is the next highest, 
even higher, and so on.
This sets up the equation 
Now we solve:
![\[5n+24=64\]](http://latex.artofproblemsolving.com/8/4/2/842502c8003c93ed53dca618860b689d6544279e.png)
![\[5n=40\]](http://latex.artofproblemsolving.com/d/e/3/de3e5e0f25b320e50496de28ad34bd574e697841.png)
![\[n=8\]](http://latex.artofproblemsolving.com/2/b/a/2ba83bf8758acefe099680e230795f472acecdae.png)
Thus, the smallest integer is 
.
See Also
| 2003 AMC 10B (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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
