Difference between revisions of "2019 AIME I Problems/Problem 2"
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==Solution 5== | ==Solution 5== | ||
− | Similar to solution 4, we can go through the possible values of <math>J</math> to find all the values of <math>B</math> that makes <math>B-J\geq 2</math>. If <math>J</math> chooses <math>1</math>, then <math>B</math> can choose anything from <math>3</math> to <math>20</math>. If <math>J</math> chooses <math>2</math>, then <math>B</math> can choose anything from <math>4</math> to <math>20</math>. By continuing this pattern, we can see that there is <math>18+17+\cdots+1</math> possible solutions. The amount of solutions is, therefore, <math>\frac{18*19 | + | Similar to solution 4, we can go through the possible values of <math>J</math> to find all the values of <math>B</math> that makes <math>B-J\geq 2</math>. If <math>J</math> chooses <math>1</math>, then <math>B</math> can choose anything from <math>3</math> to <math>20</math>. If <math>J</math> chooses <math>2</math>, then <math>B</math> can choose anything from <math>4</math> to <math>20</math>. By continuing this pattern, we can see that there is <math>18+17+\cdots+1</math> possible solutions. The amount of solutions is, therefore, <math>\frac{18*19}{2}=171</math>. Now, because <math>B</math> and <math>J</math> must be different, we have <math>20\times19=380</math> possible choices, so the probability is <math>\frac{171}{380}=\frac{9}{20}</math>. Therefore, the final answer is <math>9+20=\boxed{029}</math> |
-josephwidjaja | -josephwidjaja |
Revision as of 10:52, 7 June 2020
Contents
[hide]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
Solution 5
Similar to solution 4, we can go through the possible values of to find all the values of
that makes
. If
chooses
, then
can choose anything from
to
. If
chooses
, then
can choose anything from
to
. By continuing this pattern, we can see that there is
possible solutions. The amount of solutions is, therefore,
. Now, because
and
must be different, we have
possible choices, so the probability is
. Therefore, the final answer is
-josephwidjaja
Video Solution
https://www.youtube.com/watch?v=lh570eu8E0E
Video Solution 2
https://youtu.be/TSKcjht8Rfk?t=488
~IceMatrix
Video Solution 3
https://www.youtube.com/watch?v=nbtIBP6Auig&t=460s
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.