Difference between revisions of "2013 AMC 12A Problems/Problem 20"
m (→Solution 1) |
|||
Line 40: | Line 40: | ||
{{AMC12 box|year=2013|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2013|ab=A|num-b=19|num-a=21}} | ||
− | [Category: Introductory Set Theory Problems]] | + | [[Category: Introductory Set Theory Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:21, 4 May 2020
Contents
Problem 20
Let be the set . For , define to mean that either or . How many ordered triples of elements of have the property that , , and ?
Solution 1
Imagine that the 19 numbers are just 19 persons sitting evenly around a circle ; each of them is facing to the center.
One may check that if and only if is one of the 9 persons on the left of , and if and only if is one of the 9 persons on the right of . Therefore, " and and " implies that cuts the circumference of into three arcs, each of which has no more than numbers sitting on it (inclusive).
We count the complement: where the cut generated by has ONE arc that has more than persons sitting on. Note that there can only be one such arc because there are only persons in total.
Suppose the number of persons on the longest arc is . Then two places of are just chosen from the two end-points of the arc, and there are possible places for the third person. Once the three places of are chosen, there are three possible ways to put into them clockwise. Also, note that for any , there are ways to choose an arc of length . Therefore the total number of ways (of the complement) is
So the answer is
NOTE: this multiple-choice problem can be done even faster -- after we realized the fact that each choice of the three places of corresponds to possible ways to put them in, and that each arc of length has equitable positions, it is evident that the answer should be divisible by , which can only be from the five choices.
Solution 2
First, we can find out that the only that satisfy the conditions in the problem are , , and .
Consider the 1st set of conditions for . We get that there are
cases for the first set of conditions.
Since the 2nd and 3rd set of conditions are simply rotations of the 1st set, the total number of cases is
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.