fie a primul numar
atunci a+2=b/3 (fiind impare consecutive sunt distantate cu cate 2)
si
b/3+2=c/4
deci
b=3a+6
si
c=4b/3+8=(4/3)*(3a+6)+8=4a+8+8=4a+16
atunci a+b+c=190 devine
a+3a+6+4a+16=190
8a+22=190
8a=168
a=21
b=3a+6=63+6=69 si intr-adevar 69:3=23
c=4a+16=84+16=100 si intr-adevar 100:4=25
cam grea grea grea