Difference between revisions of "2024 AIME II Problems/Problem 14"
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Revision as of 19:38, 9 February 2024
Contents
[hide]Problem
Let
Solution
We write the base- two-digit integer as . Thus, this number satisfies with and .
The above conditions imply . Thus, .
The above equation can be reorganized as
Denote and . Thus, we have where and .
Next, for each , we solve Equation (1).
We write in the prime factorization form as . Let be any ordered partition of (we allow one set to be empty). Denote and .
Because , there must exist such an ordered partition, such that and .
Next, we prove that for each ordered partition , if a solution of exists, then it must be unique.
Suppose there are two solutions of under partition : , , and , . W.L.O.G., assume . Hence, we have
Because and , there exists a positive integer , such that and .
Thus,
However, recall . We get a contradiction. Therefore, under each ordered partition for , the solution of is unique.
Note that if has distinct prime factors, the number of ordered partitions is . Therefore, to find a such that the number of solutions of is more than 10, the smallest is 4.
With , the smallest number is . Now, we set and check whether the number of solutions of under this is more than 10.
We can easily see that all ordered partitions (except ) guarantee feasible solutions of . Therefore, we have found a valid . Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.