Question

Afla nr.''n''' din : [(n+260:2)×3+4]×5=1985

Answer 1

[(n+260:2) x 3 +4]= 1985:5=397
[( n+130) x3+4]= 397
(n+130) x3 +4 =397
(n+130 ) x 3= 397-4= 393
n+130= 393:3= 131
n= 131-130
n=1

Answer 2

[(n+260:2) x3 +4]x5=1985
(n+130x3+4)x5=1985
(n+390+4)x5=1985
Paranteza o notam cu a :
ax5=1985
a=1985:5=397
si apoi zicem
n+390+4=397
n+394=397
n=497-494
n=3