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PUMaC 2008 · 团队赛 · 第 1 题

PUMaC 2008 — Team Round — Problem 1

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

  1. (2 points) Calculate 6 + 6 + 6 + . . . + . 6 1+ 1+ ...
解析
  1. Calculate 6 + 6 + 6 + . . . + . 6 1+ 1+ ... √ √ √ 6 ( ANS: 5 Let X = 6 + 6 + 6 + . . . and Y = . We wish to find X + Y . Now we have 6 1+ 1+ ... √ 6 2 X = 6 + X , whence X = (1 + 1 + 24) / 2 = 3. Also, Y = , so Y (1 + Y ) = 6, thus 1+ Y √ 2 Y + Y − 6 = 0 whence Y = ( − 1 + 1 + 24) / 2 = 2. Thus we conclude that X + Y = 3 + 2 = 5 as claimed. CB: GBH?, ACH)