1) nr.consecutive sunt: a,a+1,a+2, a+3, a+4
[a+(a+1)+(a+2)+(a+3)+(a+4)]:21=9 rest 6
5a+10=21x9+6
5a=189+6-10
5a=185
a=185/5
a=37
nr.sunt:37,38,39,40,41
2) a+b+c=121
a;c=10 rest 5 rezulta a=10c+5
b:c=5 rest 4 rezulta b=5c+4
inlocuim in a+b+c=121
10c+5+5c+4+c=121
16c=121-5-4
16c=112
c=112/16
c=7
a=10c+5=10x7+5=75
a=75
b=5c+4=5x7+4=39
75+39+7=121
3)a+b+c=135
a:c=12 rest 1 rezulta a=12c+1
b:c=31 rest 2 rezulta b=31c+2
12c+1+31c+2+c=135
44c=135-1-2
44c=132
c=132/44
c=3
a=12c+1=12x3+1
a=37
b=31c+2=31x3+2=95
b=95
37+95+3=135