Difference between revisions of "2011 AMC 10A Problems/Problem 6"
m (→Solution) |
|||
Line 10: | Line 10: | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2011|ab=A|num-b=5|num-a=7}} | {{AMC10 box|year=2011|ab=A|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Revision as of 12:00, 4 July 2013
Problem 6
Set has
elements, and set
has
elements. What is the smallest possible number of elements in
?
Solution
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
.
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.