abc , ab si restu am inteles numere scrise in baza 10
(abc-ab)/90+(bca-bc)/90 +(cab-ca)/90
tinand cont ca abc-ab = 10ab+c-ab=9ab+c si analog, avem;
(9ab+c+9bc +a+9ca+b)/90
[9 (ab+bc+ca) + (a+b+c)]/90
[9 (11a+11b+11c) +(a+b+c)]/90
[99(a+b+c) + a+b+c]/90
100( a+b+c)/90
10 (a+b+c)/9 patrat perfect pt ca radicalul trebuie sa fie rational
√(10(a+b+c)/√9=[√(10(a+b+c)]/3 rational
√(10(a+b+c)) natural
a+b+c=10
0<a<b<c<9
daca a=1, b+c=9 deci b=2 si c=7 sau b=3 sic=6, b=4, c=5
daca a=2, b+c=8, b=3, c=5; b=4 , c= 4 nu convine, pt ca b=c
cdac a=3. b+c=7 b=3 , c=4 nu convine , pt c a=b;nu mai exista solutii a<b<c, pt a≥3
Deci ( a;b;c)∈{ (1;2;7);(1;3;6) ; (1;4;5); (2;3;5)}
