2025 AMC 12A Problems/Problem 13

Problem

Let $C = \{1, 2, 3, \dots, 13\}$ . Let $N$ be the greatest integer such that there exists a subset of $C$ with $N$ elements that does not contain five consecutive integers. Suppose $N$ integers are chosen at random from $C$ without replacement. What is the probability that the chosen elements do not include five consecutive integers?

$\textbf{(A)}~\frac{3}{130} \qquad \textbf{(B)}~\frac{3}{143} \qquad \textbf{(C)}~\frac{5}{143} \qquad \textbf{(D)}~\frac{1}{26} \qquad \textbf{(E)}~\frac{5}{78}$

Solution 1

We first find what $N$ is by figuring out how many numbers we need to take out of the set so that the set does not contain $5$ consecutive integers. Since $N$ must be maximized, we must minimize what numbers are removed, and we quickly find that taking two numbers out works. Consider taking out $5$ and $10$ . You are left with $\{1,2,3,4,6,7,8,9,11,12,13\}$ , 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 $9$ ), and ${\binom{13}{2}}=78$ total ways to choose such integers. Therefore, the probability is $\boxed{\dfrac{1}{26}}$ .

~Kevin Wang

Minor edits ~aashrithm29

Very minor LaTeX edit ~ PerseverePlayer

Note: An alternative way to come to the conclusion that there are only three possible subsets of length twelve is as follows:

Let the two numbers removed from $C$ be $a$ and $b$ such that $a<b$ . Then, because we cannot have any sequence of five consecutive numbers, $a \leq 5$ , $b-a \leq 5$ , and $13 - b +1\leq 5$ . Then, graphing this system in the $ab$ plane, we easily see that there are only three solutions for $(a,b)$ : $(5,10), (4,9)$ , and $(5,9)$ . Then, simply proceed as described above.

~ 526

Solution 2

Trying to find a subset that satisfies the condition, we get $\{1,2,3,4,6,7,8,9,11,12,13\}$ , which has $N=11$ elements. The subsets $\{1,2,3,5,6,7,8,10,11,12,13\}$ and $\{1,2,3,4,6,7,8,10,11,12,13\}$ also work. In total, we have $3$ subsets and $\binom{13}{11}$ ways to choose $11$ elements from $C$ , so the probability is $\dfrac{3}{\binom{13}{11}} = \dfrac{1}{26}$ . Thus, the answer is $\boxed{\textbf{(D)}}$ ~anzhuPro

Solution 3

First, let's find the number $N$ : Consider if:

$N=13$ :

It is impossible - It doesn't remove any integer, violating the condition "not contain five consecutive integers".

$N=12$ :

It is also impossible, and still does not satisfy the condition. An example could be $\{1,2,3,4,6,7,8,9,10,11,12,13\}$ . $1$ , $2$ , $3$ , and $4$ doesn't violate the condition, but $6-13$ are all consecutive, hence $N$ can't be $12$ .

$N=11$ :

This is possible, because $\dfrac{11}{3} = 3$ remainder $2$ . This means that we can have one string of consecutive length $3$ and two strings of consecutive length $4$ . Hence, it could be either:

$\{1,2,3,5,6,7,8,10,11,12,13\}$ with consecutive integer lengths of $(3, 4, 4)$ .
$\{1,2,3,4,6,7,8,10,11,12,13\}$ with consecutive integer lengths of $(4, 3, 4)$ .
$\{1,2,3,4,6,7,8,9,11,12,13\}$ with consecutive integer lengths of $(4, 4, 3)$ .

Thus giving $3$ cases. The total ways of picking out $2$ integers from a set of $13$ is $\binom{13}{2} = 78$ . Hence, the probability is $\dfrac{3}{78} = \boxed{\dfrac{1}{26}}$ .

Note: You can also consider 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 in the same answer.

~Mitsuihisashi14

Formatting and minor edits ~Logibyte

Video Solution by Power Solve

https://www.youtube.com/watch?v=bpx-AaReYI8

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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