2005 AMC 10B Problems/Problem 25

Problem

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 $B$ ?

$\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68$

Solution

Solution 1

The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$ . The integers from $25$ to $100$ are left. They can be paired so the sum is $125$ : $25+100$ , $26+99$ , $27+98$ , $\ldots$ , $62+63$ . That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = \boxed{\mathrm{(C)}\ 62}$ . Also, it is possible to see that since the numbers $1$ to $24$ are in the set there are only the numbers $25$ to $100$ to consider. As $62+63$ gives $125$ , the numbers $25$ to $62$ can be put in subset $B$ without having two numbers add up to $125$ . In this way, subset $B$ will have the numbers $1$ to $62$ , and so $\boxed{\mathrm{(C)}\ 62}$ .

Solution 2

"Cut" $125$ into half. The maximum integer value in the smaller half is $62$ . Thus the answer is $\boxed{\mathrm{(C)}\ 62}$ .

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2005 AMC 10B Problems/Problem 25

Contents

Problem

Solution

Solution 1

Solution 2

See Also