2005 AMC 10B Problems/Problem 25: Difference between revisions
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== Problem == | == Problem == | ||
== Solution == | == 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 == | == See Also == | ||
*[[2005 AMC 10B Problems]] | *[[2005 AMC 10B Problems]] | ||
Revision as of 19:02, 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.
