notam prima zi cu "a", a doua zi cu "b" iar a treia zi cu "c"
a=3*b+1
c=(b/2)-1
a+b+c=1809⇒3b+1+b+(b/2)-1=1809⇒(aducem la acelasi numitor)⇒6b+2+2b+b-2=3618⇒9b=3618⇒b=402
Solutie:
a=3b+1=3*402+1=1206+1=1207, prima zi
b=402, a doua zi
c=(b/2)-1=(402/2)-1=201-1=200, a treia zi