Let "excess" be the amount beyond the amount required to return to base. If the group travels together for a distance of 1/3, then at that point half of the group can transfer half of their remaining supplies and completely replenish the packs of the other half, while still retaining enough to return to base. Starting at that point, the remainder of the group, each of which has excess of 2/3, can travel 1/3 of the excess distance = 2/9, at which point half of the remainder transfer enough to fill the packs of the remaining half (a quarter of the original) and return home. So, we have that the remainder of the group (after consecutive halvings) can travel 1/3 + 2/9 + 4/27 + ... -> 1. I'm not sure that that counts as non-computational, though. -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Victor Miller Sent: Tuesday, July 03, 2012 10:00 AM To: math-fun Subject: [EXTERNAL] [math-fun] Travelers through the dessert After listening to Car Talk this weekend, I was inspired to create the following question: We have an inexhaustible supply of *travelers* each of whom can carry up to 1 day's worth of rations (we measure the distance in days of travel time). One of them, designated the pilgrim, wants to set off across the dessert from an oasis, aided by a cohort of n other travelers -- the helpers. Each of his helpers can travel a certain distance, transfer some of his rations to the remaining travelers (pilgrim plus remaining helpers) and turn back to the oasis (and thus needs to keep enough rations to allow him to get there). After the last helper leaves, the pilgrim keeps going. When he runs out of rations a magical genie appears and creates a new oasis for him. Let L[n] denote the longest distance (in days) that the pilgrim can travel using n helpers. It's clear that L[n] <= 2 since the last helper to leave can't be more than 1 day's distance away from the oasis. It seems intuitively clear that as n ---> infinity that L[n] ----> 2. Indeed one can work out a recursion for L[n] and solve for it explicitly. However, is there a simple, non-computational proof of this fact which doesn't involve solving for L[n]? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun