2013 AMC 12A Problems/Problem 20
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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
Solution 3 (Common sense)
The conditions are very strange. The major way can satisfy is if because at this point no matter what. This is the only case that works, but we can rotate the variables around, so our answer is a multiple of .
Suppose and where . That means and . WLOG . Start casework on what is, anywhere from to . There are 2 main factors: 1. the number of ways to choose , which is . 2. the number of ways to choose or start the sequence, which is The sum of all cases turns out to be the sum of squares from to , which is . Multiply that by and get
~dragnin
Solution 4 (Modular Arithmetic)
Note that if we take , the conditions are equivalent to , , and . Let . We have ways to pick , and cases in our sample space for . Of these cases, only those where will work. By stars and bars, this turns out to be , hence the answer is . ~sigmapie
Solution 5 (pretty braindead)
will denote the first inequality, and will denote the second.
We soon see that we cannot have by adding the inequalities we obtain from them together, as it results in .
Similarly, we can't have . is also not possible by observing the inequalities.
Thus, we only have to consider . Something like is just a rotation of this case.
Here, we start with , , and . We begin with casework on , since it is the smallest.
When , can range from . There are different values for as x ranges across .
When , can range from . There are different values for as x ranges across .
Continuing this pattern, we get the sum . Multiplying by three to count for rotations, we get .
-skibbysiggy
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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