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1 证明∵(1·2·3…1984)~(1/1984)<1/1984 sum from k=1 to 1984 k=1/1984·(1984(1+1984))/2=1985/2, 上式两边1984次方,得 1984!<1985~(1984)·2~(-1984) 2 解∵ 1985能被5整除。又 1984~(1984)=(1985-1)~(1984)=1985~(1984)-C_(1984)~1·1985~(1983)+C_(1984)~2·1985~(198)~2+…-C_(1984)~(1983)·1985+1 ∴ 1984~(1984)除以5所得的余数是1。 3 证明由题设,得 l~2=a~2+b~2+c~2 且l>a l>b,l>c。∴l~(1984)=l~2、l~(1982)=(a~2+b~2+c~2)l~(1982)=a~2l~(1982)+b~2·l~(1982)+c~(2·1982)≥a~2·a~(1982)+b~2b~(1982)+c~2·c~(1982)=a~(1984)+b~(1984)+c~(1984) 4.证(k≥1)
1 Proof (1·2·3...1984)~(1/1984)<1/1984 sum from k=1 to 1984 k=1/1984·(1984(1+1984))/2=1985/2, 1984!<1985~(1984)·2~(-1984) 2 Solution 1985 can be divisible by five. 1984~(1984)=(1985-1)~(1984)=1985~(1984)-C_(1984)~1•1985~(1983)+C_(1984)~2•1985~(198)~2 +...-C_(1984)~(1983)•1985+1* The remainder obtained by dividing 1984~(1984) by 5 is 1. 3 Prove that the question is given, get l~2=a~2+b~2+c~2 and l>a l>b,l>c. ∴l~(1984)=l~2, l~(1982)=(a~2+b~2+c~2)l~(1982)=a~2l~(1982)+b~2•l~ (1982)+c~(2·1982)≥a~2·a~(1982)+b~2b~(1982)+c~2·c~(1982)=a~(1984)+b~(1984 )+c~(1984) 4. Certificate (k≥1)