With $$81$$ sweets wrappers. Peter can exchange for $$\frac{81}{3}=27$$ more sweets With the additional $$27$$ sweets wrappers, Peter can exchange for $$\frac{27}{3}=9$$ more sweets.  with the~~additional $$9$$ sweets wrappes. Peter can exchange for $$\frac{9}{3}=3$$ more sweets.  with the additional $$3$$ sweets wrappers. Peter can exchange for $$\frac{3}{3}=1$$ more sweet.  Therefore, the largest possible number of sweets Peter could have  eaten is $$81+27+9+3+1=121$$. 
In a candy shop, the shopkeeper allows the exchange of $$3$$ sweet wrappers for $$1$$ new sweet. Peter has $$81$$ sweets at first. What is the largest possible number of sweets that Peter could have eaten? 
A) $$119$$ 
B) $$120$$ 
C) $$121$$ 
D) $$122$$ 
E) Non of the above 