To solve this puzzle follow the methodology and talk the story out.If the soldier goes alone he will eat one day's worth of food and drink one day's worth of water the first day.He will have three days worth left.If two people go both carrying four days worth of food and water they would have eight days worth of food between them.They could walk one day and each consume one day's worth and have six days worth left walk a second day and have consumed two days worth(for a total of four days worth).That would leave four days worth of food and water.Then the second man would have to turn around and back and he would need to take days worth of food with him.That would leave two days worth of food and water for the soldier which is not enough.If three men start out with twelve days worth of food and water by the end of the first day there is nine days worth left.One man could take one day's worth and travel back to the beginning point.Four days worth of food and water (three consumed and one for the return trip)would be used and eight would be left.The next day the two men consume two more allotments of food and water(8-2=6)and one man takes two of the remaining six and returns to the starting point.That leaves four days worth of food and water for the soldier to complete the hike.