If the fourth number is \geq 12, then the last three numbers must sum to at least $12+$ $2 \cdot 12+2^{2} \cdot 12=84>79$. This is impossible, so the fourth number must be less than 12. Then the only way we can have the required divisibilities among the first four numbers is if they are $1,2,4,8$. So the last two numbers now sum to $79-15=64$. If we call these numbers $8 a, 8 a b(a, b>1)$ then we get $a(1+b)=a+a b=8$, which forces $a=2, b=3$. So the last two numbers are 16,48.