2011 AMC 10A Problems/Problem 6
Contents
[hide]Problem 6
Set has elements, and set has elements. What is the smallest possible number of elements in ?
Solution 1
will be smallest if is completely contained in , in which case all the elements in would be counted for in . So the total would be the number of elements in , which is .
Solution 2
Assume WLOG that , and . Then, all the integers through would be redundant in , so .
~MrThinker
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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