riddle: camel and bananas...
we make 2 stopping points along the way: point A, and point B.
first we move as many as we can from the start to point A, which is 200u from the start. first we carry 1000, eat 200 on the way, drop 600 and return to start eating another 200. we do that again, drop 600, return with 200. then we make the final trip of 1000, eating 200 on the way and we can drop 800.
So now at point A we are holding 800+600+600=2000.
Next we travel to point B, which is 333 1/3 units from point A. we carry 1000, eat 333 1/3 on the way, drop 333 1/3 at point B and return eating 333 1/3. then we carry the second 1000, eating 333 1/3 on the way and dropping 666 2/3.
Now at point B, 200+333 1/3=533units from the start, we are holding 333 1/3 + 666 2/3 = 1000.
Now we have 1000 - 533 1/3 = 467 2/3 units left to travel, and with 1000 bananas we can do it in one trip. 1000 - 467 2/3 = 533 1/3 bananas that make it the full 1000 units uneaten.
first we move as many as we can from the start to point A, which is 200u from the start. first we carry 1000, eat 200 on the way, drop 600 and return to start eating another 200. we do that again, drop 600, return with 200. then we make the final trip of 1000, eating 200 on the way and we can drop 800.
So now at point A we are holding 800+600+600=2000.
Next we travel to point B, which is 333 1/3 units from point A. we carry 1000, eat 333 1/3 on the way, drop 333 1/3 at point B and return eating 333 1/3. then we carry the second 1000, eating 333 1/3 on the way and dropping 666 2/3.
Now at point B, 200+333 1/3=533units from the start, we are holding 333 1/3 + 666 2/3 = 1000.
Now we have 1000 - 533 1/3 = 467 2/3 units left to travel, and with 1000 bananas we can do it in one trip. 1000 - 467 2/3 = 533 1/3 bananas that make it the full 1000 units uneaten.
it doesnt sound like you can deal with fractional units, since the camel eats before he moves.
fwiw, I get 500 bananas intact when using waypoints that are at 250 and 500 units from the starting point.
fwiw, I get 500 bananas intact when using waypoints that are at 250 and 500 units from the starting point.
If you figure 3 1000 banana loads and you move each of them 1 mile at a time, each mile costs 6 bananas. At the end of 167 miles, you have 1998 bananas (2 loads). Now each mile costs 4 bananas. At the end of 250 more miles (417 total), you are down to 998 bananas. At this point, you take 998 bananas and subtract the 417, leaving 581 bananas.
I think this is close, but I see some problems.
I think this is close, but I see some problems.
The unabageler is correct. I didn't plan on partial units, but his method is correct. I'll post another riddle tomorrow. Now that you guys figured this one out, we'll try some harder ones
I hope every good camel rider has two hands to split a banana in thirds to feed 1/3 of it to the camel for the last 1/3 unit of travel.


Originally posted by pellisS2k
it doesnt sound like you can deal with fractional units, since the camel eats before he moves.
fwiw, I get 500 bananas intact when using waypoints that are at 250 and 500 units from the starting point.
it doesnt sound like you can deal with fractional units, since the camel eats before he moves.
fwiw, I get 500 bananas intact when using waypoints that are at 250 and 500 units from the starting point.



