[理工] [離散] congruence
關於解法的問題
Solve the congruence 2x=7(mod 17)
黃子嘉的書的解法:
解法1:
1=2*(-8)+17*1
=>1=2*(-8+17k)+17*(1-2k) , k屬於Z
=>7=2*7(-8+17k)+17*7(1-2k)
所以2x=7(mod 17)的解為: x = 7(-8+17k) , k屬於Z
但是這不是完整的答案吧...
應該是
解法2:
1=2*(-8)+17*1
=>7 = 2*7*(-8)+17*7*1
=>x = (-56)mod(17) = 12 (mod 17)
=>x = 12+17k, k屬於Z
這個似乎才是完整的答案?
像x=29是答案之一,但解法1的答案裡沒有29,很明顯解法1不完整
我看黃子嘉解清大98離散的congruence就是用解法2
跟他自己的參考書中的方法不同...他這不是在自打嘴巴嗎...?
到底要用哪種 = =?
請幫小弟解惑...
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※ 編輯: dacvidania 來自: 114.34.133.127 (02/09 09:07)
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