The answer is that only x=11 yields a fair game. I solved this using a recurrence. Let $f[N]$ be optimal expected amount of money that Bob has to pay Alice to find the best answer. Clearly, $f[... View answer Accepted answer 21 votes 1.Split the input into blocks of size 3. Make the following encryption : AAA : BBBBB AAB : BBBAB ABA : BBABB ABB : BABBB BAA : ABBBB BAB : BABAB BBA : ABBAB BBB : ABABB This leaves us with what to ... View answer Accepted answer 14 votes 2048 Proof Each move changes the number of black numbers remaining by 0 or 2, and thus cannot change the parity. Thus initial states with an even number of black numbers can never be brought to the ... View answer Accepted answer 13 votes Explanation View answer Accepted answer 12 votes Does an algorithm exist? Yes. Consider every valid state of the Rubik's cube. It can be brought to the solved state in 20 moves or less. For each state, apply the sequence of moves followed by its ... View answer Accepted answer 12 votes View answer 12 votes Here is a permutation that works upto n=12: 1 6 11 4 5 10 15 8 9 14 19 12 13 18 23 16 17 22 27 20 21 26 31 24 25 30 35 28 29 34 39 32 33 38 43 36 37 42 47 40 41 46 51 44 45 50 3 48 49 2 7 52 ... View answer Accepted answer 10 votes Reason View answer Accepted answer 7 votes (a) and (d) are correct, while the rest are false. Reason This is because the digit sum of 2N (or 5N) is just the sum of digit sums of 2*d_i (or 5*d_i) where d_i are the digits of N. This is not ... View answer Accepted answer 6 votes View answer Accepted answer 5 votes View answer 5 votes The ship is days old. Explanation: View answer Accepted answer 5 votes Since there were exactly 32 empty spaces, I tried to push the ability of my computer and enumerate the best possible move starting from all$2^{32}\$ states. For each state, the code finds the maximum ...

Solution Explanation The positive and negative clues are supposed to mean:

The highest number of coins that Stanlio can guarantee for himself is: Proof: Pseudocode initialise best[][][] with 0 for i=1 to 100 best [i]=-inf for j=1 to 9 best [i][j]=i ...

2. 3. 4. 5. 6A. 6D. 7. 8. 9. 10. 11. 12. 13. 14. 16. 17A. 17D. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36A. 36D. 37. 38. 39. 40. ...

Assuming that multiple consecutive moves of the same block are counted as one, here is a solution:

(See 9) ??? ??? (See 3) ??? (See 12) ??? (See 21) Thanks to rand al'thor for (7) and Christian Rau for (15)

Reasoning: