a:b=3 rest 7 => b>7
a=3b+7
3a:b=c rest 6
3a=b•c+6; il inlocuim pe a
3(3b+7)=b•c+6
9b+21=b•c+6
21-6=b•c-9b
15=b(c-9); b>7
=> b=15 si c-9=1; c=10
=> a=3•b+7=3•15+7=52
a=52=>3a=3•52=156
sau: 3a=b•c+6=15•10+6=156
R: a=52 si b=15
verificare:
52:15=3 rest 7 (A)
156:15=10 rest 6 (A)
