2012 AMC 8 Problems/Problem 11
Problem
The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and 
are all equal. What is the value of ?
Solution
Since there must be an unique mode, and 
is already repeated twice, cannot be any of the numbers already listed (3, 4, 5, 7). (If it were, the mode would not be unique.) So must be , or a new number.
Solution 1: Guess & Check
We can eliminate answer choices 
and 
, because of the above statement. Now we need to test the remaining answer choices.
Case 1: 
Mode:
Median:
Mean: 
Since the mean does not equal the median or mode, 
can also be eliminated.
Case 2: 
Mode:
Median:
Mean:
We are done with this problem, because we have found when , the condition is satisfied. Therefore, the answer is 
.
Solution 2: Algebra
Notice that the mean of this set of numbers, in terms of , is:
Because we know that the mode must be (it can't be any of the numbers already listed, as shown above, and no matter what is, either or a new number, it will not affect being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and equal:
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
