I+II+III+IV=140
I=IV+15
II,III,IV=nr consecutive
=> IV=I-15; III=I-16; II=I-17
I+I-17+I-16+I-15=140
4I-48=140
I=(140+48):4=188:4
I=47
II=47-17=30
III=47-16=31; sau III=II+1=31
IV=47-15=32; sau IV=III+1=32
R: I=47; II=30; III=31; IV=32
47+30+31+32=140 (A)
