Question: You have to make 125 packets of sugar with first one weighing 1 kg, second 2 kg, third 3 kg etc ...and 125th one weighing 125kg.You can only use one pan of the common balance for measurement for weighing sugar, the other pan had to be used for weights i.e. weights should be used for each weighing.
It has come into notice that moving weights into and out of the pan of the balance takes time and this time depends on the number on the number of weights that are moved. For example - If we need to measure 4 kg using weights 1 and 3 only, it will take twice as much time needed to measure 1 kg. Lets say we want to make sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 kg, with 1 unit of time, we place 3 kg along with 1 kg and measure 4kg with again 1 unit of time, and finally we move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4kg using weights 1 and 3 only.

Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.

Solution: NOTE : N represents a normal ball

Name the 12 balls as A1,A2,A3,A4,B1,B2,B3,B4 C1,C2,C3 and C4.We will weigh A's on one side and B's on the other side.
A1 A2 A3 A4 <---> B1 B2 B3 B4
If Both weighed same then the odd ball is among C's.
We have to find the odd ball among 4 balls using 2 weightings.For that we balance

C1 C2 with C3 N.

Now if C1 C2 equals C3 N then C4 is the odd one.

Now suppose C1 C2 is heavier than C3 N then we compare
C1 C3 to N N
Now if C1 C3 equals to N N the C2 is the odd ball.
If C1 C3 is lighter than N N, then C3 is the odd ball.
If C1 C3 is heavier than N N then C1 is the odd ball.


Coming to the situation where the first weighing resulted in unequal balance.Lets assume A's are heavier than B's.
ie .

A1 A2 A3 A4 > B1 B2 B3 B4

Now we compare

A1 A2 B1 B2 to N N N B4.


if A1 A2 B1 B2 > N N N B4

The odd ball is among A1 A2 or B4 and we also know that A1 A2 > B4 N.To find out the odd ball among A1 A2 and B4 we do compare

A1 B4 to N N .

if A1 B4 > N N then
A1 is the odd ball
if A1 B4 < b4 =" N" b2 =" N"> N B3 which we have already shown how to solve.


Solution for 13 ball problem is easy if we could solve 12 ball problem.For that lets assume we have an extra ball C5.Now if A's and B's dont weigh same we have all the steps already calculated.

Now if A's and B's weigh same then the odd one is among C's.

Now we compare

C1 C2 to C3 N.

If C1 C2 = C3 N then we can find out the odd ball from C4 and C5 by comparing it with any normal ball.
Else if C1 C2 > C3 N then we compare

C1 C3 to N N.
If C1 C3 > N N then C1 is the odd ball
If C1 C3 = N N then C2 is the odd ball
If C1 C3 < N N then C3 is the odd ball