An interesting puzzle at the Futility Closet page today. It first appeared in 1967 in Eureka the Cambridge math journal. No answer was ever published. I am guessing the answer is none (if they don’t have to all arrive together), but I am having trouble proving it. So am I right?

Charles, Donald, and Edward are triplets.

The triplets (whose abilities at walking, cycling, and donkey riding are identical) always leave home together at the last possible minute and arrive at school together on the last stroke of the bell.

They used to walk the 4 1/2 miles, and so had to set out at 8.00; then they acquired a bicycle and found that they did not have to leave home until 8.15 (Charles rode it for the first 1 1/2 miles, left it, and walked on; Donald walked 1 1/2 miles, cycled 1 1/2 miles, and walked again; Edward walked 3 miles and cycled the rest). More recently they have been given a donkey. After experiments to determine the donkey’s speed and to verify that it stood stock still when left, they found that — using the bicycle and the donkey — they did not need to leave home until 8.25. There were several schemes of changing over which they could use to do this, of course; but naturally they chose a scheme which involved the minimum number of changes. Going to school tomorrow Charles will start on foot and Edward will arrive on foot. How far will Donald walk?