Source: A small book called Problematical Recreations shown to me by Bill Cleveland. Used as Ken's POTD 3/7/94.
1.Leave 6 adjacent numbers of the face of a clock intact and rearrange the other 6 in such a way that the sum of every pair of adjacent numbers is prime. two solutions: 1 2 3 4 7 10 9 8 5 6 11 12 1 2 3 4 9 10 7 6 5 8 11 12 2.Leave as many numbers as possible in their correct positions on the face of a clock and rearrange the rest such that the sum of every pair of adjacent numbers is prime. move 4 (actually done with 2 swaps): 11 2 3 8 5 6 7 4 9 10 1 12 The standard clock has 3 locations (4 5)(7 8)(10 11) where there is a composite sum. At least one of each of these pairs must be moved. The 4 choices are to cycle either the even or odd values - none of which give all primes. So 4 moves must be the minimum. 3.Repeat the previous problem, such that the sum of every adjacent pair is composite (not prime). Similar to analysis for #2, if 5 were the minimum number of moves, then 6,9 and one from each of (11 12)(1 2)(3 4). The 4 ways to select that also include one from each pair (12,1) and (2,3) are (11 1 3)(12 1 3)(12 2 3)(12 2 4), and none produce a solution. For 6 moves, assume that 1-3,6,9,12 are to move. Only 1 goes in the 6 position and 2 goes in the 9 position. There are 4 ways to permute the remaining, one of which leads to: 6 9 12 4 5 1 7 8 2 10 11 3