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 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 (If you have no time)

"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)
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2005 AMC 10B Problems/Problem 25

Contents

Problem

Solution 1

Solution 2 (If you have no time)

See Also