设这三个自然数为 a ,a + 1,a + 2
设 a = 11k k>=1
则 13 | 11k + 1 => 13 | 6(11k+1) => 13 | (65k+k+6)
=> 13 | k + 6
设 k = 13t -6 t>=1
则有 17 | a + 2 => 17 | 11(13t-6) + 2 =>17 | 7t + 4
=> 17 | 5(7t + 4) => 17 | 34t + t + 20
=> 17 | t + 20
设t = 17p - 20 p>=2
所以 a = 11(13(17p - 20)- 6) p >=2