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?
Well, this can be solved by dividing 12 balls in 4 grps of 3 each .
ReplyDeletethen 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.
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.
ReplyDelete@vaibhav..
ReplyDeleteincorrect.. your method will work only if the defective ball is heavier.. but we dont know if it is heavier or lighter
@ vaibhav..
ReplyDeleteexactly 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!:)