Question

The integers from 1 to n are written in increasing order from left to right on a blackboard. David
and Goliath play the following game: starting with David, the two players alternate erasing any two
consecutive numbers and replacing them with their sum or product. Play continues until only one
number on the board remains. If it is odd, David wins, but if it is even, Goliath wins. Find the 2011th
smallest positive integer greater than 1 for which David can guarantee victory.

(copied off math tournament questionnaire)

Answer 1

If n is odd , then the last move is for Goliath . No matter what two numbers
are on the board, Goliath can combine them to make an even number.
so if n is odd then Goliath will win.

Answer 2

but lets work on even n's

Answer 3

replace the numbers by residues modulo

Answer 4

so we have \[1,0,1,...,1,0\]

Answer 5

david sums up 1 and 0 furthest to the right to 1 making the sequence 1 0 1 0 01

Answer 6

we get 4022 eventually

Answer 7

i think he can do it ... without considering Goliath's move...so yes 2011th even number is answer

Answer 8

even with goliath's good chance, david can still restore a winning game

Answer 9

yeah thats right. nice problem