Difference between revisions of "2015 AMC 8 Problems/Problem 13"
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-tryanotherangle | -tryanotherangle | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/ZbwdX6sZyQA | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 13:30, 31 March 2022
Problem
How many subsets of two elements can be removed from the set so that the mean (average) of the remaining numbers is 6?
Solutions
Solution 1
Since there will be elements after removal, and their mean is
, we know their sum is
. We also know that the sum of the set pre-removal is
. Thus, the sum of the
elements removed is
. There are only
subsets of
elements that sum to
:
.
Solution 2
We can simply remove subsets of
numbers, while leaving only
behind. The average of this one-number set is still
, so the answer is
.
-tryanotherangle
Video Solution
~savannahsolver
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.