Difference between revisions of "2019 AIME I Problems/Problem 2"
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==Solution 4== | ==Solution 4== | ||
| − | If we | + | We can see that if B chooses 20, J has options 1-18, such that <math>B-J\geq 2</math>. 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 <math>18+17+\cdots+1</math> as the total number of solutions. The total amount of choices is <math>20\times19</math> (B and J must choose different numbers), so the probability is <math>\frac{18*19/2}{20*19}=\frac{9}{20}</math>. Therefore, the answer is <math>9+20=\boxed{029}</math> |
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=1|num-a=3}} | {{AIME box|year=2019|n=I|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:28, 15 March 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
Realize that by symmetry, the desired probability is equal to the probability that 
is at most 
, which is 
where 
is the probability that and differ by 1 (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 basically asking how many ways there are to choose 2 distinct elements from a 20 element set such that no 2 elements are adjacent. Using the well-known formula 
, there are 
ways. Dividing 171 by 380, our desired probability is 
. Thus, our answer is 
.
-Fidgetboss_4000
Solution 3
Simply 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 we know that 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
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. 
