2025 AMC 12A Problems/Problem 13: Difference between revisions
No edit summary |
|||
| Line 15: | Line 15: | ||
Trying to find a subset that satisfies the condition, we get <imath>\{1,2,3,4,6,7,8,9,11,12,13\}</imath>, which has <imath>N=11</imath> elements. The subsets <imath>\{1,2,3,5,6,7,8,10,11,12,13\}</imath> and <imath>\{1,2,3,4,6,7,8,10,11,12,13\}</imath> also work. In total, we have <imath>3</imath> subsets and <imath>\binom{13}{11}</imath> ways to choose <imath>11</imath> elements from <imath>C</imath>, so the probability is <imath>\dfrac{3}{\binom{13}{11}} = \dfrac{1}{26}</imath>. Thus, the answer is <imath>\boxed{\textbf{(D)}}</imath> | Trying to find a subset that satisfies the condition, we get <imath>\{1,2,3,4,6,7,8,9,11,12,13\}</imath>, which has <imath>N=11</imath> elements. The subsets <imath>\{1,2,3,5,6,7,8,10,11,12,13\}</imath> and <imath>\{1,2,3,4,6,7,8,10,11,12,13\}</imath> also work. In total, we have <imath>3</imath> subsets and <imath>\binom{13}{11}</imath> ways to choose <imath>11</imath> elements from <imath>C</imath>, so the probability is <imath>\dfrac{3}{\binom{13}{11}} = \dfrac{1}{26}</imath>. Thus, the answer is <imath>\boxed{\textbf{(D)}}</imath> | ||
-anzhuPro | -anzhuPro | ||
==Solution 3== | |||
First lets find the number N: | |||
Consider if: | |||
1. N=13: It is impossible - It doesn't remove any integer, violating the condition "not contain five consecutive integers". | |||
2. N=12: It is also impossible, because removing 1 integer doesn't satisfy "not contain five consecutive integers." An example could be <imath>\{1,2,3,4,6,7,8,9,10,11,12,13\}</imath>. 1, 2, 3, 4 has a break, but 6-13 are all consecutive, hence N can't be 12. | |||
3. N=11: This is possible, because 11/3 = 3 remainding 2. This means that we can have 1 string of consecutive length 3 and 2 strings of consecutive length 4. Hence, it could be either: | |||
- <imath>\{1,2,3,5,6,7,8,10,11,12,13\}</imath> (3, 4, 4) lengths | |||
- <imath>\{1,2,3,4,6,7,8,10,11,12,13\}</imath> (4, 3, 4) lengths | |||
- <imath>\{1,2,3,4,6,7,8,9,11,12,13\}</imath> (4, 4, 3) lengths | |||
Thus giving 3 cases. The total ways of by picking out 2 integers from a set of 13 is 13C2 = 78. Hence, the probability is 3/78 = 1/26 (D). | |||
Note: You can also consider by the combinations of 1 string of consecutive length 3 and 2 strings of consecutive length 4. The length of 3 has 3 places to go, filling the rest 2 with 4 places each, giving 3/78 as well, resulting answer D. | |||
(Feel free to correct the LaTex, thanks.) | |||
~Mitsuihisashi14 | |||
Revision as of 20:45, 6 November 2025
Problem 13
Let 
. Let 
be the greatest integer such that there exists a subset of 
with elements that does not contain five consecutive integers. Suppose integers are chosen at random from without replacement. What is the probability that the chosen elements do not include five consecutive integers?
Solution 1
We first find what N is by figuring out how much numbers we need to take out of the set so that the set does not contain 5 consecutive integers. Taking two numbers out works. Consider taking out 5 and 10. You are left with 
, which does not have a string of 5 consecutive integers.
There are only 3 ways to take out two integers such that the resulting set meets our condition (5 and 10, 5 and 9, or 4 and 9). Therefore, the probability is 
.
~Kevin Wang
Solution 2
Trying to find a subset that satisfies the condition, we get 
, which has 
elements. The subsets 
and 
also work. In total, we have 
subsets and 
ways to choose 
elements from , so the probability is 
. Thus, the answer is 
-anzhuPro
Solution 3
First lets find the number N:
Consider if:
1. N=13: It is impossible - It doesn't remove any integer, violating the condition "not contain five consecutive integers".
2. N=12: It is also impossible, because removing 1 integer doesn't satisfy "not contain five consecutive integers." An example could be 
. 1, 2, 3, 4 has a break, but 6-13 are all consecutive, hence N can't be 12.
3. N=11: This is possible, because 11/3 = 3 remainding 2. This means that we can have 1 string of consecutive length 3 and 2 strings of consecutive length 4. Hence, it could be either:
- (3, 4, 4) lengths
- (4, 3, 4) lengths
- (4, 4, 3) lengths
Thus giving 3 cases. The total ways of by picking out 2 integers from a set of 13 is 13C2 = 78. Hence, the probability is 3/78 = 1/26 (D).
Note: You can also consider by the combinations of 1 string of consecutive length 3 and 2 strings of consecutive length 4. The length of 3 has 3 places to go, filling the rest 2 with 4 places each, giving 3/78 as well, resulting answer D.
(Feel free to correct the LaTex, thanks.) ~Mitsuihisashi14
