2024 AIME II Problems/Problem 14
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,
.
~Shen Kislay Kai and Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
https://youtu.be/0FCGY9xfEq0?si=9Fu4owVaSm-WWxFJ
~MathProblemSolvingSkills.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 |
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