Question

sa se determine perechile de numere naturale a b care verifica relatia
412:4+{19+2×[216-5×(a×b+1)]}×10=2013

Answer 1

412:4+{19+2×[216-5×(a×b+1)]}×10=2013
103+{19+2×[216-5×(a×b+1)]}×10=2013
{19+2×[216-5×(a×b+1)]}×10=2013-103
{19+2×[216-5×(a×b+1)]}×10=1910
19+2×[216-5×(a×b+1)]=191
2×[216-5×(a×b+1)]=191-19
2×[216-5×(a×b+1)]=172
216-5×(a×b+1)=172:2
216-5×(a×b+1)=86
5×(a×b+1)=216-86
5×(a×b+1)=130
(a×b+1)=130:5
axb+1 = 26
axb=26-1
axb=25
perechi de solutii: {1,25}{5,5}{25,1}