Seems Easy?
Consider the set of numbers . We delete two of them say and and in their place we put only one number . After performing the operation 2011 times what is the number that is left over.
Note- It's an old NMTC Junior level problem
The answer is 2012.
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We see that the product P = ( a + 1 ) ( b + 1 ) . . . ( n + 1 ) is invariant, where a , b , . . . , n are members of the set A .
Proof:- Let the numbers be a , b , c , . . . n . Then initially, P = ( a + 1 ) ( b + 1 ) ( c + 1 ) . . . ( n + 1 ) .. After replacing a , b with a b + a + b , the product now becomes P = ( a b + a + b + 1 ) ( c + 1 ) . . . ( n + 1 ) = ( a + 1 ) ( b + 1 ) ( c + 1 ) . . . ( n + 1 ) . Therefore P remains invariant.
Now initially P = ( 1 + 1 ) ( 1 / 2 + 1 ) ( 1 / 3 + 1 ) . . . ( 1 / 2 0 1 2 ) = 2 ( 3 / 2 ) ( 4 / 3 ) . . . ( 2 0 1 3 / 2 0 1 2 ) = 2 0 1 3 .
Let the number left be n . Then P = n + 1 . Therefore, n = 2 0 1 2 .