一个正整数被5、7、9、11整除
设这个正整数为X,由已知得:
X ≡ 1 (mod 5 ) ①
X ≡ 2 (mod 7 ) ②
X ≡ 3 (mod 9 ) ③
X ≡ 4 (mod 11) ④
由①得,设X=5K + 1 ⑤
把⑤代入②中得:5K+1 ≡ 2 (mod 7)
解得:K ≡ 3 (mod 7 ) ,设K=7m + 3 代入⑤中得:X=35m+16 ⑥
把⑥代入③中得:35m+16 ≡3 (mod 9 )
解得:m ≡ 4 (mod 9 ) ,设m = 9n+4 代入⑥中得:X=315n ...全部
设这个正整数为X,由已知得:
X ≡ 1 (mod 5 ) ①
X ≡ 2 (mod 7 ) ②
X ≡ 3 (mod 9 ) ③
X ≡ 4 (mod 11) ④
由①得,设X=5K + 1 ⑤
把⑤代入②中得:5K+1 ≡ 2 (mod 7)
解得:K ≡ 3 (mod 7 ) ,设K=7m + 3 代入⑤中得:X=35m+16 ⑥
把⑥代入③中得:35m+16 ≡3 (mod 9 )
解得:m ≡ 4 (mod 9 ) ,设m = 9n+4 代入⑥中得:X=315n + 156 ⑦
把⑦代入④中得:315n+156 ≡4 (mod 11 ) ,解得:n ≡ 5 (mod 11 )
设n = 11p +5 代入 ⑦中得:X =3465P + 1731
所以 P = 0 时 X取最小值为 :1731
。
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