Difference between revisions of "2021 AIME II Problems/Problem 11"
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Revision as of 07:56, 6 April 2021
Problem
A teacher was leading a class of four perfectly logical students. The teacher chose a set of four integers and gave a different number in to each student. Then the teacher announced to the class that the numbers in were four consecutive two-digit positive integers, that some number in was divisible by , and a different number in was divisible by . The teacher then asked if any of the students could deduce what is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of . Find the sum of all possible values of the greatest element of .
Solution 1 (Text)
Let's start by listing the multiples of 6 and 7: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96; 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, and 98. First of all, the multiples of 6 and 7 wouldn't work, so we can cross out 42 and 84. Since the students said they didn't know the answer, it must be that the set has no distinct integers from the sets. Let's list out the possible sets. Since they are in lists of 4, we need multiples of 6 or 7. Let's list them out. 11, 12, 13, 14 12, 13, 14, 15 But if a student has 11 or 15, they would know. So that wouldn't work. The second set would be this: They couldn't know since the multiples of 7 is 35, and if you list it out, they wouldn't know about it since there are other sets such as 35, 36, 37, 38, and 33, 34, 35, 36. The people with these sets would say no, so the second set would WORK. The other sets are only 2 numbers, and since 2 numbers are sufficiently smaller than our 2nd set, we have concluded the answer is 37+50+79+92 = . ~Arcticturn
Solution 2 (Illustrations)
Note that It is clear that and otherwise the three other elements in are divisible by neither nor
In the table below, the multiples of are colored in yellow, and the multiples of are colored in green. By the least common multiple, we obtain cycles: If is a possible maximum value of then must be another possible maximum value of and vice versa. By observations, we write all possible maximum values of in red.
From the second row of the table above, we perform casework on the maximum value of
Finally, all possibilities for are and from which the answer is
Remarks
We can reconstruct the second table in this solution as follows, where Y and N stand for replies of "yes" and "no", respectively. Notice that this table has some kind of symmetry!
~MRENTHUSIASM
Video Solution
https://www.youtube.com/watch?v=7jKjilTRhs4
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.