https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Kiseki&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-22T06:36:26Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_25&diff=33216 2005 AMC 10B Problems/Problem 25 2010-01-20T00:02:15Z <p>Kiseki: /* Solution */</p> <hr /> <div>== Problem ==<br /> == Solution ==<br /> <br /> 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 --&gt; C.<br /> <br /> == See Also ==<br /> *[[2005 AMC 10B Problems]]</div> Kiseki https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_16&diff=33215 2005 AMC 10B Problems/Problem 16 2010-01-19T21:20:27Z <p>Kiseki: /* Solution */</p> <hr /> <div>== Problem ==<br /> == Solution ==<br /> <br /> == See Also ==<br /> *[[2005 AMC 10B Problems]]</div> Kiseki https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_16&diff=33214 2005 AMC 10B Problems/Problem 16 2010-01-19T21:17:34Z <p>Kiseki: /* Solution */</p> <hr /> <div>== Problem ==<br /> == Solution ==<br /> <br /> x^2+mx+n=0<br /> Roots: 2a, 2b<br /> 2a + 2b = 2(a+b) = -m<br /> 2a • 2b = 4ab = n<br /> <br /> x^2+px+m=0<br /> Roots: a, b<br /> a + b = -p<br /> ab = m<br /> <br /> Use substitution:<br /> <br /> /frac{n}{p}<br /> = /frac{4ab}{-(a+b)}<br /> = /frac{8ab}{-2(a+b)}<br /> = /frac{8m}{m}<br /> = 8<br /> <br /> Our answer is 8.<br /> <br /> == See Also ==<br /> *[[2005 AMC 10B Problems]]</div> Kiseki