Problem 1 We will prove this by induction. Note that base case is satisfied, as I have one gold piece worth 1 at the end of day n. As the inductive hypothesis, we may assume that I have pieces of total value n at the end of day n. Note that I turn each silver piece (value 1) on day n into a gold piece (value ) and each gold piece (value ) into a silver piece and a gold piece (total value +1.) Note also that +1=, as can be verified by direct computation. Hence, after day n+1, my treasure is worth times as much as it was worth on day n. By the inductive hypothesis, this shows the treasure is worth n+1 dollars after day n+1, and we are done. Problem 2 Define the two tokens to be in critical position if the squares they are on share a corner. We assign to each square S coordinates (x,y), where x is the number of …show more content…
First, we observe that either ai and i have the same parity for all i or ai and i have opposite parity for all i. This follows from the fact that a 2n+1-flip preserves the parity of every term in the sequence and a 2n-flip reverses the parity of every term of the sequence. Hence, any finite sequence x1x2...xm that contains two consecutive terms of the same parity cannot occur anywhere in the infinite sequence. I claim that all other finite sequences can occur at the beginning of the infinite sequence. To prove that this is so, I will describe a recursive algorithm for constructing a sequence of n-flips that takes takes 1,2,3,... to a sequence beginning with a given finite sequence x1x2...xm with terms alternating in parity. Let Flips(x1x2...xm) be the output of this algorithm. It is easy to construct Flips(x1); a single x1-flip does the job. Now, in order to construct Flips(x1x2...xm), where m>1, we first perform Flips(x2...xm). Then, let i be the index of x1 in the sequence. Note that since x2 has index 1 and x1 and x2 have opposite parity, i is even. Define k=i2. Now, perform a k-flip. Note that x1 is now in position k+1