2012 AMC 8 Problems/Problem 11
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[hide]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 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:
Solution 3: Balance scale
We know the unique mode must be , so the mean must be the same number . Let's imagine a scale. exactly stands the mid-point of the scale. Numbers of represent the left side "weights" of the scale. Numbers of represent the right side "weights" of the scale. On the left side, the difference of the three "weights" between are , respectively. It gives us the total difference is . In order to allow the scale to keep balance, on the right side, the total difference must be . Because we have already known the difference of the right side "weights" between is , partially, so the difference between and unknown must be . It exactly gives us the answer:. ---LarryFlora
Solution 4: Must be an integer(Super fast)
Since all the answer choices are integers, we know that , which simplifies to , must be an integer. The only answer choice that satisfies this is .
~NXC
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=1946
~ pi_is_3.14
Video Solution
https://youtu.be/dloxxgDBm88 ~savannahsolver
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 |
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