abc divizibil cu => c apartine {0,5}
Ma = abc + bca + cab = 111(a+b+c)/3
Ma = aab + bba = 111(a+b)/2
222(a+b+c) = 333(a+b)
222a + 222b + 222c = 333a + 333b
222c = 111a + 111b
222c = 111(a+b)
2c = a+b
Cazul c = 0
a = 0
b = 0
Nu are sens
Cazul c=5
Si avem mai multe "subcazuri" :
a=1, b = 9
a = 2, b = 8
a = 3, b = 7
a = 4, b = 6
a = 5, b = 5
a = 6, b = 4
a = 7, b = 3
a = 8, b = 2
a = 9, b = 1
Deci numerele alea sunt
195, 285, 375, 465, 555, 645, 735, 825, 915.
