Treasure Seekers On The Mission

13 caves are arranged in a circle at the temple of doom. One of these caves has the treasure of gems and wealth. Each day the treasure keepers can move the treasure to an adjacent cave or can keep it in the same cave. The treasure keepers allowed to move only in 1 direction only i.e. right to the current position. Every day two treasure seekers visit the place and have enough time to enter any two caves of their choice.

How do the treasure seekers ensure that they find the treasure in minimum possible days? 

This should be your advice!

Source 

Tip For Treasure Seekers

What was the mission?

One of the treasure seeker should start moving clockwise & other should anti clockwise.

One starting from Cave C1 & other from Cave C13 make sure that treasures are not in those.
 
Now if we assume it was in C2 on day 1 & keeper moving it in clockwise then on Day 6 it would be in C7. At 7th day, seeker 1 should go to C8 & seeker 2 should go to C7. If keepers had kept it in same cave after Day 6, then seeker 2 would find it or if they had moved it to C8 then seeker 1 would find it for sure.

So we require minimum 7 days to make absolutely sure that seekers find the treasure. And if keeper starts from any other position it would require less number of days. For example, starting from C3 at Day1 , it would be in C8 on the 6th day & seeker by itself.

The Color Of The Last Ball?

You have 20 Blue balls and 10 Red balls in a bag. You put your hand in the bag and take off two at a time. If they’re of the same color, you add a Blue ball to the bag. If they’re of different colors, you add a Red ball to the bag. What will be the color of the last ball left in the bag?

Note: Assume you have a big supply of Blue and Red balls for this purpose. When you take the two balls out, you don’t put them back in, so the number of balls in the bag keeps decreasing.

Once you tackle that, what if there are 20 blue balls and 11 red balls to start with?

This should be the color of last ball! 

Source 

That's The Color Of Last Ball

What was the challenge? 

There could be 3 possible combinations that we could get on each removal.

1. One is Red & other is Blue. 

In this case, we are taking off Blue & Red but adding Red back though from other source. Effectively we are taking off only Blue keeping number of Red balls same.

2. Both are Red.

We are taking off 2 Red balls but adding 1 Blue.

3. Both are Blue.

Again we are taking off both Blue balls but adding 1 from other source. Effectively, we are keeping number of Blue intact in such cases.

What we observe from this is that the Red is always taken off in pair. And if it is taken off in single then other Red takes it's place as seen in case 1 above. On the other hand, Blue is added or taken off in single.


Since there are even number of Red balls i.e.10 which only can be taken off in pair, there won't be any single Red balls at the end. Hence, last ball must be Blue.

 
Sub question's Answer :

For odd number of Red balls i.e. 11 here, if taken off in pair then the last ball would be the Red always. Hence, the last ball must be Red in the case.