※ 引述《TOMOHISA (YAMASHITA)》之銘言：
: 2 3
: 自1~n中任取八個數，若其中均至少有兩個數的比值介於[─,─]，
: 3 2
: 則n的最大值為何？
The answer is 39
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If take 3 arbitary numbers from 1~n, max n = 3,
because when n=4, we take 1,2,4
then 1/2=2/4<2/3
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Take 4 numbers a1,...,a4, and 3≦a3<a4≦n
if a3=3 then a4 has no constraint
if a3≧4 then a4≦3/2*a3 hence a4≦6 hence max n=6
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Similarly,
Take 5 numbers a5≦3/2*7=21/2, max n=10
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Take 6 numbers, a6≦3/2*11=33/2, max n=16
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Take 7 numbers, a7≦3/2*17=51/2, max n=25
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Take 8 numbers, a8≦3/2*26=39, max n=39
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◆ From: 112.104.171.53
※ 編輯: JohnMash 來自: 112.104.171.53 (06/05 00:46)