## Nov 8, 2009

### The twelve balls problem

Sam has 12 balls. All but one are of equal weight. Sam doesn't know whether the defective ball is lighter or heavier than the normal balls. Sam also has a comparison balance (for weighing). However, Sam can use the balance only 3 times. How would he find out which is the defective ball?

1. Well, this can be solved by dividing 12 balls in 4 grps of 3 each .
then one can proceed logically and get that odd ball and can also find out whther is it heavier or lighter.

This prob is actually a submodel of a general prob of finding the odd ball frm the (3^n-3)/2.

2. we dont need to divide in 4 groups,half will also do.after 1st weighing we r left with 6,after 2nd weighing we have 3 coins left.next time weigh any 2,we get the heavier one,if both balance then the left one is the heavier one.

3. @vaibhav..
incorrect.. your method will work only if the defective ball is heavier.. but we dont know if it is heavier or lighter

4. @ vaibhav..
exactly i agree with harsha!! Ur method works bt only fr a particular case !!
the only sol is by dividing it in 4 grp of 3 each!!
Thnk abt the general solution!! A nice one!:)