2005 AMC 10B Problem 25

Below is the professionally curated solution for Problem 25 of the 2005 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10B solutions, or check the answer key.

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Concepts:subsets pairing and grouping extremal argument

Difficulty rating: 1720

25.

A subset $B$ of the set of integers from $1$ to $100,$ inclusive, has the property that no two elements of $B$ sum to $125.$ What is the maximum possible number of elements in $B?$

$50$

$51$

$62$

$65$

$68$

Solution:

The pairs summing to $125$ are $(25, 100), (26, 99), \ldots, (62, 63),$ which is $62 - 25 + 1 = 38$ pairs. From each pair, $B$ may contain at most one element.

The numbers $1$ through $24$ cannot pair with anything in range to sum to $125,$ so all $24$ of them may be included.

Thus $B$ has at most $38 + 24 = 62$ elements, and the set $\{1, 2, \ldots, 62\}$ achieves this.

Thus, C is the correct answer.

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