###
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.

## No comments:

Post a Comment