2025 AMC 12A Problem 13

Let

C = \{1, 2, 3, \ldots, 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?

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Solution

To avoid five consecutive integers, it suffices to remove two elements (for example

5

and

10

), and no single removal can hit both disjoint blocks

\{1,\ldots,5\}

and

\{9,\ldots,13\}.

Thus

N = 11.

Choosing

11

13

elements is the same as removing

2,

which can be done in

\binom{13}{2} = 78

ways. The chosen set avoids five consecutive integers exactly when the two removed elements together intersect every window

\{t, t+1, t+2, t+3, t+4\}

for

t = 1, \ldots, 9.

This forces one removed element in

\{1,\ldots,5\},

the other in

\{9,\ldots,13\},

and the two within

5

of each other. The valid removals are

\{4,9\},

\{5,9\},

and

\{5,10\},

giving

3

of them. The probability is

\dfrac{3}{78} = \dfrac{1}{26}.

Thus, the correct answer is D.