Difference between revisions of "2005 AMC 10B Problems/Problem 25"
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== Problem == | == Problem == | ||
+ | A subset <math>B</math> of the set of integers from <math>1</math> to <math>100</math>, inclusive, has the property that no two elements of <math>B</math> sum to <math>125</math>. What is the maximum possible number of elements in <math>B</math>? | ||
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+ | <math>\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68 </math> | ||
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== Solution == | == Solution == | ||
− | + | The question asks for the maximum possible number of elements. The integers from <math>1</math> to <math>24</math> can be included because you cannot make <math>125</math> with integers from <math>1</math> to <math>24</math> without the other number being greater than <math>100</math>. The integers from <math>25</math> to <math>100</math> are left. They can be paired so the sum is <math>125</math>: <math>25+100</math>, <math>26+99</math>, <math>27+98</math>, <math>\ldots</math>, <math>62+63</math>. That is <math>38</math> pairs, and at most one number from each pair can be included in the set. The total is <math>24 + 38 = \boxed{\mathrm{(C)}\ 62}</math>. | |
− | The question asks for the maximum possible. The integers from 1 | ||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2005|ab=B|num-b=24|after=Last Problem}} |
Revision as of 16:44, 8 July 2011
Problem
A subset of the set of integers from to , inclusive, has the property that no two elements of sum to . What is the maximum possible number of elements in ?
Solution
The question asks for the maximum possible number of elements. The integers from to can be included because you cannot make with integers from to without the other number being greater than . The integers from to are left. They can be paired so the sum is : , , , , . That is pairs, and at most one number from each pair can be included in the set. The total is .
See Also
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 |