Difference between revisions of "2002 AMC 8 Problems/Problem 3"
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| + | ==Problem 3== | ||
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| + | What is the smallest possible average of four distinct positive even integers? | ||
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| + | <math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7</math> | ||
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| + | ==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 <math>\frac{2+4+6+8}{4}=\boxed{5}</math>. | 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}=\boxed{5}</math>. | ||
Revision as of 12:14, 17 June 2011
Problem 3
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 
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