Difference between revisions of "2019 AIME I Problems/Problem 2"
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==Solution 3== | ==Solution 3== | ||
| − | Create a grid using graph paper, with 20 columns for the values of J from 1 to 20 and 20 rows for the values of B from 1 to 20. Since <math>B</math> cannot equal <math>J</math>, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since <math>B - J</math> must be at least <math>2</math>, we can mark the line where <math>B - J = 2</math>. Now we sum the number of squares that are on this line and below it. We get <math>171</math>. Then we find the number of total squares, which is <math>400 - 20 = 380</math>. Finally, we take the ratio <math>\frac{171}{380}</math>, which simplifies to <math>\frac{9}{20}</math>. Our answer is <math>9+20=\boxed{029}</math>. | + | Create a grid using graph paper, with <math>20</math> columns for the values of <math>J</math> from <math>1</math> to <math>20</math> and <math>20</math> rows for the values of <math>B</math> from <math>1</math> to <math>20</math>. Since <math>B</math> cannot equal <math>J</math>, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since <math>B - J</math> must be at least <math>2</math>, we can mark the line where <math>B - J = 2</math>. Now we sum the number of squares that are on this line and below it. We get <math>171</math>. Then we find the number of total squares, which is <math>400 - 20 = 380</math>. Finally, we take the ratio <math>\frac{171}{380}</math>, which simplifies to <math>\frac{9}{20}</math>. Our answer is <math>9+20=\boxed{029}</math>. |
==Solution 4== | ==Solution 4== | ||
Revision as of 11:49, 26 December 2019
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 asks 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 B chooses 20, J has options 1-18, such that 
. If B chooses 19, J has choices 1-17. By continuing this pattern, B will choose 3 and J will have 1 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
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. 
