7)a:7=12rest r si r>2
a=7x12+r deci a=84+r
Avem regula rest<impartitor rezulta ca r este3,4,5,6.
a=84+3=87 ; a=84+4=88 ; a=84+5=89 ; a=84+6=90
Raspuns: a este 87,88,89,90.
**[(a-2)x3+26]:25=8
(a-2)x3+26=25x8=200
(a-2)x3=200-26=174
a-2=174:3=58 deci a=60
""3+3x[3+3x(3+3x3)]=(3xa+3)x3+3
(3-urile ingroate se reduc deci le barezi)
3+3x(3+3x3)=3xa+3(din nou ce e ingroat se reduce)
3+3x3=a deci a=12
