2020 AIME I Problems/Problem 9
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Problem
Solution
First, prime factorize as
. Denote
as
,
as
, and
as
.
In order for to divide
, and for
to divide
,
, and
. We will consider each case separately. Note that the total amount of possibilities is
, as there are
choices for each factor.
We notice that if we add to
and
to
, then we can reach the stronger inequality
. Therefore, if we pick
integers from
to
, they will correspond to a unique solution, forming a 1-1 correspondence. The amount of solutions to this inequality is
.
The case for ,
, and
proceeds similarly for a result of
. Therefore, the probability of choosing three such factors is
Simplification gives
, and therefore the answer is
.
-molocyxu
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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