1) fie numerele a,b,c
a+b+c=269
a-n=24 ; b-n=81 ; c-n=31
a=24+n ;b=81+n ; c=31+n
a+b+c=24+n+81+n+31+n=136+3n => 136+3n=269 =>3n=133 => n=133/3
2) fie a,b,cele doua numere
a-b=87
a=b*5+15
inlocuim in prima relatie => 5b+15-b=87
4b=87-15=72 => b=72/4=18 => a=18*5+15=105 => a doua varianta
3) n=numarul
n/5=n-500
500=n-n/5
500=4/5*n => n=500*5/4=625
4) 2*[14+(5+a):6]:8=4
14+(5+a):6=16
(5+a):6=2
5+a=12
a=7