3. b)
2019:(n+4)=9 rest n+3
2019=9•(n+4)+n+3)
2019=9n+36+n+3
2019=10n+39
10n=2019-39
n=1980:10
n=198; n+4=202; n+3=201
impartitorul=202
verificare: 2019:202=9 rest 201
2019=9•202+201 (A)
4.
a+b+c+d=2019
b=8+2a
c=8+a+b=8+a+(8+2a);
c=16+3a
d=c-b-8=16+3a-8-2a-8
d=a
=> a+b+c+d=a+(8+2a)+16+3a+a
7a+24=2019
7a=2019-24
a=1995:7
a=285
b=8+2•285=8+570
b=578
c=8+(285+578)=8+863
c=871
d=285; sau d=871-578-8=285
285+578+871+285=2019
