Difference between revisions of "2019 AIME I Problems/Problem 2"
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+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
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Revision as of 03:44, 6 March 2020
Contents
Problem 2
Jenn randomly chooses a number from
. Bela then randomly chooses a number
from
distinct from
. The value of
is at least
with a probability that can be expressed in the form
where
and
are relatively prime positive integers. Find
.
Solution
By symmetry, the desired probability is equal to the probability that is at most
, which is
where
is the probability that
and
differ by
(no zero, because the two numbers are distinct). There are
total possible combinations of
and
, and
ones that form
, so
. Therefore the answer is
.
Solution 2
This problem is essentially asking how many ways there are to choose distinct elements from a
element set such that no
elements are adjacent. Using the well-known formula
, there are
ways. Dividing
by
, our desired probability is
. Thus, our answer is
.
-Fidgetboss_4000
Solution 3
Create a grid using graph paper, with columns for the values of
from
to
and
rows for the values of
from
to
. Since
cannot equal
, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since
must be at least
, we can mark the line where
. Now we sum the number of squares that are on this line and below it. We get
. Then we find the number of total squares, which is
. Finally, we take the ratio
, which simplifies to
. Our answer is
.
Solution 4
We can see that if chooses
,
can choose from
through
such that
. If
chooses
,
has choices
~
. By continuing this pattern,
will choose
and
will have
option. Summing up the total, we get
as the total number of solutions. The total amount of choices is
(B and J must choose different numbers), so the probability is
. Therefore, the answer is
-eric2020
Video Solution
https://www.youtube.com/watch?v=lh570eu8E0E
Video Solution 2
~IceMatrix
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.