Difference between revisions of "2005 AMC 10B Problems/Problem 25"

 
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== Problem ==
 
== Problem ==
 
== Solution ==
 
== Solution ==
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The question asks for the maximum possible. The integers from 1~24 can be included because you cannot make 125 with integers from 1~24 without the other number being greater than 100. The integers 25~100 are left. They can be paired so the sum is 125. 25+100, 26+99, 27+98, ...... 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 = 62 --> C.
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== See Also ==
 
== See Also ==
 
*[[2005 AMC 10B Problems]]
 
*[[2005 AMC 10B Problems]]

Revision as of 20:02, 19 January 2010

Problem

Solution

The question asks for the maximum possible. The integers from 1~24 can be included because you cannot make 125 with integers from 1~24 without the other number being greater than 100. The integers 25~100 are left. They can be paired so the sum is 125. 25+100, 26+99, 27+98, ...... 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 = 62 --> C.

See Also

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