Difference between revisions of "2002 AMC 8 Problems/Problem 3"
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==Solution== | ==Solution== | ||
| − | In order to get the smallest possible average, we want the 4 even numbers to be as small as possible. The first 4 positive even numbers are | + | In order to get the smallest possible average, we want the 4 even numbers to be as small as possible. The first 4 positive even numbers are 2, 4, 6, and 8. Their average is <math>\frac{2+4+6+8}{4}=\frac{20}{4}=\boxed{\text{(C)}\ 5}</math>. |
| − | + | ||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/JN5-YpUPeso | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=2|num-a=4}} | {{AMC8 box|year=2002|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:26, 29 October 2024
Problem
What is the smallest possible average of four distinct positive even integers?
Solution
In order to get the smallest possible average, we want the 4 even numbers to be as small as possible. The first 4 positive even numbers are 2, 4, 6, and 8. Their average is 
.
Video Solution by WhyMath
See Also
| 2002 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. 
