76. Mountain Exploration. Professor Walkinhom, a participant in the expedition, was given the task to survey the mountain from all sides at a certain height. He has to walk 100 miles around the mountain alone. The professor is capable of walking 20 miles a day, but he can only carry enough food for two days. For convenience, each daily ration is packed in a sealed box. Every day, the professor walked his 20 miles and used up a daily ration. What is the shortest time in which he could circumnavigate the mountain?

This puzzle can rightly be considered one of the most intriguing among those we have considered so far. Professor Walkinhom will need considerable ingenuity. The idea for the problem was proposed by G. F. Heath.
76. Walkingholme deposits 5 rations at the 90-mile mark (see figure) and returns to base (5 days). Then he leaves 1 ration at the 85-mile mark and returns to the 90-mile mark (1 day). He leaves 1 ration at the 80-mile mark and returns to the 90-mile mark again (1 day). He transports 1 ration to the 80-mile mark, returns to the 85-mile mark, picks up the remaining 1 ration there, and transports it to the 80-mile mark (1 day). He "deposits" 1 ration at the 70-mile mark and returns to the 80-mile mark (1 day), then returns to base (1 day). Thus, there is 1 ration left at both the 70-mile and 90-mile marks. Walkingholme transports 1 ration to the 5-mile mark and returns to base (1 day). If he needs to travel 20 miles, he can do so by reaching the 10-mile mark and returning to base.

![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-200.jpg?height=508&width=468&top_left_y=614&top_left_x=797)

He transports 4 rations to the 10-mile mark and returns to base (4 days). He leaves 1 ration at the 10-mile mark and returns to the 5-mile mark, picks up the 1 ration left there, and transports it to the 10-mile mark (1 day). He transports 2 rations to the 20-mile mark and returns to the 10-mile mark (2 days). He transports 1 ration to the 25-mile mark and returns to the 20-mile mark (1 day). He leaves 1 ration at the 30-mile mark, returns to the 25-mile mark, picks up the 1 ration left there, and transports it to the 30-mile mark (1 day). He travels to the 70-mile mark (2 days). He travels back to base (1.5 days). In total, 23.5 days.

Attempts were made to reduce this time, but all of them were based on tricks that were, in one way or another, prohibited. For example, Walkingholme was "forced" to leave not a whole daily ration but only part of it, to make a forced march, or to eat a daily ration before leaving a certain mark so that he could carry two daily rations. In the latter case, Walkingholme would actually be carrying three rations: one in his stomach and two on his back!

If the professor's route had been through a desert, the shortest time would have been 86 days, and he should have proceeded as follows.

Deposit 42 rations at the 10-mile mark from base, return to base (42 days). Carry 1 ration to the 15-mile mark, return to the first depot at the 10-mile mark from base (1 day). Leave 20 rations at the 20-mile mark from base and return to the depot at the 10-mile mark from base (20 days). Carry 1 ration to the 20-mile mark from base and return to the point 15 miles from base, pick up the 1 ration left there, and transport it to the 20-mile mark (1 day). Transport 10 rations to the point 30 miles from base and return to the 20-mile mark (10 days). Carry 1 ration to the 35-mile mark and return to the 30-mile mark (1 day). Carry 4 rations to the 40-mile mark and return to the 30-mile mark (4 days). Carry 1 ration to the 40-mile mark and return to the 35-mile mark. Pick up the 1 ration there and transport it to the 40-mile mark (1 day). Carry 2 rations to the point 50 miles from base and return to the 40-mile mark (2 days). Carry 1 ration to the 55-mile mark and return to the 50-mile mark (1 day). Transport 1 ration to the 60-mile mark and return to the 55-mile mark. Pick up the 1 ration there and transport it to the 60-mile mark (1 day). From there, make the final leg of the journey to the end point (2 days). In total, 86 days.