奧林匹克數學問題
回答 (2)
2011-01-16 8:02 pm
2011-01-16 8:37 pm
因 {x} > x - 1 , 有
{yz - {y}{z}} < {yz - (y-1)(z-1)} = {y + z - 1} < {10 + 10 - 1} = {19} = 19
故 {yz-{y}{z}}的最大可能值 = 18
{yz - {y}{z}} < {yz - (y-1)(z-1)} = {y + z - 1} < {10 + 10 - 1} = {19} = 19
故 {yz-{y}{z}}的最大可能值 = 18
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110116000051KK00284