b = a + 2c
abc : 112 = a rest 59
abc = 112×a + 59
100×a + 10×b + c = 112×a +59
10×b + c = 12×a + 59
10×(a+2c) + c = 12×a + 59
10×a + 21×c = 12×a + 59
2×a +59 = 21×c
pentru a = 2
4 + 59 = 21×c
63 = 21×c
c = 63 : 21
c = 3
b = 2+3×2 = 8
==> abc = 283
verificare:
283 : 112 = 2 rest 59