2005 AMC 10B Problem 25

Which numbers can be part of a pair summing to $125$? A number $k$ needs a partner $125 - k$ that also lies in $1$ to $100$, which requires $25 \le k \le 100$. So the numbers $1$ through $24$ are completely safe: all $24$ of them can go into $B$.

The numbers $25$ through $100$ ($76$ numbers) split into the $38$ disjoint pairs
$$\{25, 100\},\ \{26, 99\},\ \ldots,\ \{62, 63\},$$
each summing to $125$. $B$ may contain at most one number from each pair, so it has at most $38$ of these numbers.

Upper bound: $|B| \le 24 + 38 = 62$.

This is achievable: take $B = \{1, 2, \ldots, 62\}$. Its two largest elements sum to $62 + 61 = 123 < 125$, so no two elements reach $125$.

The answer is $\boxed{\textbf{(C)}\ 62}$.

"No two elements sum to $S$" is handled by pairing each number with its complement $S - k$: the forbidden pairs are disjoint, so one element per pair is the ceiling (pigeonhole), and numbers whose complement lies outside the set are free. A maximum problem needs both the bound and a construction; here the smallest $62$ numbers give the construction immediately.

Greedy from the bottom: keep adding $1, 2, 3, \ldots$ as long as no two chosen numbers sum to $125$. The first conflict is $62 + 63 = 125$, so you can take everything up to $62$, and any larger number would pair with something already chosen.

• Forgetting that 1 through 24 are unconstrained and answering 50 (one from each of fifty imagined pairs).
• Counting the pairs from $25$ to $100$ as $37$ or $39$ instead of $38$ (there are $76$ numbers, so $38$ pairs).