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查看: 1817|回复: 2
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（2x4x...x2004 - 1x3x...x2003）mod 2005
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证明
2 x 4 x 6 x 8 x....2002 x 2004 - 1 x 3 x 5 x 7 x...x 2001 x 2003
可以被 2005 除，无余数！ |
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发表于 28-4-2005 11:07 PM
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2 x 4 x 6 x 8 x....2002 x 2004 - 1 x 3 x 5 x 7 x...x 2001 x 2003
=2(1x2x3x4x...1002)-(1x3x5x7x...x2001x2003)
=(1x3x5x7..x1001)(4x8x12x16x...x2004-1003x1005...x2001x2003)
(1x3x5x7...x1001) 中有 5x401=2005 |
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发表于 28-4-2005 11:14 PM
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不好意思，有点错误
不过解发一样
2 x 4 x 6 x 8 x....2002 x 2004 - 1 x 3 x 5 x 7 x...x 2001 x 2003
=(2x1)x(2x2)x(2x3)x...(2x1002)-1x3x5x7x2001x2004
都有 5x401=2005 |
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