Solución
Tenemos que
N = 10 a + b y N' = 10 b + a
0  a  9 0  b  9
Entonces,
- 1) N' = 2N - 1
10 b + a = 20 a + 2 b - 1 19 a = 8 b + 1
Tenemos que 19 a = 3 a = 1 (mod 8) a = 3, b = 7 N = 37, N' = 73
- 2) N' = 2N + 1
10 b + a = 20 a + 2 b + 1 19 a = 8 b - 1
Tenemos que 19 a = 3 a = -1 = 7 (mod 8) a = 5, b > 9, imposible
- 3) N' = 3N - 1
10 b + a = 30 a + 3 b - 1 29 a = 7 b + 1
Tenemos que 29 a = a = 1 (mod 7) a = 1, b = 4 N = 14, N'= 41 ó a = 8 b > 9, imposible.
- 4) N' = 3N + 1
10 b + a = 30 a + 3 b + 1 29 a = 7 b - 1
Tenemos que 29 a = a = -1 = 6 (mod 7) a = 6, b > 9, imposible
Saludos
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