What Is The Most Efficient

First separate the 9 ball into 3 groups of 3 balls. Compare the weight of any two group of balls. If one of the groups is heavier than the other, the heavier group contains the odd ball. However, if the masses of the two groups are equal, the unweighed group contains the odd ball. Ignore the other two groups and focus to the group with the odd ball. Compare the weight of any two balls. If one is heavier that is the odd ball. However, if the masses are equal, the last ball is the odd ball.

All in all, minimum of 2 weighings needed to find the odd ball.

This can be solved in 2 moves.

Split the 9 balls into 3 sets of 3

Weigh 2 of those sets, if one weighs more than the other, it has the heavier ball. If it doesn't, then the set you didn't weigh has the heavier ball

Once you know which set of 3 has the heavier ball, compare the weights of 2 of the balls in that set

If one weighs more then it is the heavier ball. If they weigh the same then the ball you didn't weigh is the heavier ball

Hi. the first mistake i made was i took the combination to be 4 4 1 and then i solved it .. got the answer as 3 which was not on the answer list and so turned to another combination 3 3 3 and then solved it .. obviously if the weights are balanced then the fault is in the remaining set .. else the set towards which the balance tilts is the faulty set.. then you have 3 balls left.. take 1 each on either side of the balance .. if the weights are balanced then the remaining ball is the faulty one .. else the side to which the balance tilts contains the faulty ball

Split the balls into three groups of three.

Put all three balls of the first group on one side of the balance, all three balls of the second group on the other side. If the two sets of balls do not balance, the odd ball is in the heavier group of three. If they do balance, the odd ball is in the unweighed group of three. Either way, we know which group of three balls contains the odd one.

Now put the first ball from the odd group on one side of the balance, and the second ball from the odd group on the other side of the balance. If the balls do not balance, the heavier one is the odd ball. If the scales do balance, the odd ball is the unweighed one.

Hence only two weighings are required.

divide the 9 balls into 3 group . if the weight of two group is level then the ball is in the last group . In this way find the group where the ball belong . from that group weight two balls. if its balanced then the 3rd ball is heavier ball .

take a group of 3-3 balls.put each group on both sides of the taraju.the heavier group contain the odd ball.if the weight of both groups are same then take the left three balls put 1-1 on both sides of tarazu,the heavier ball is the odd ball.if weight of both ball is same then the left ball is the odd ball.so 2 wieghings needed to find the odd ball

Two. Split into three groups of three, three, and two. weigh the two groups of three against each other. If equal, weigh the group of two to find the heavier. If one group of three is heavier pick two of the three and compare them to find the heaviest.

Let's split it into three groups of three. If we measure we can have two possible outcome. If one of the groups is greater than the other, then that group has the odd one. If they are the same, the remaining group contains the other. Now we have 3 marbles. We can use the same logic so we have 2.

Split the spheres into groups of three, three spheres per group. Compare the mass of two groups; either 1. one of the two groups are lighter or 2. they are equal which means the third group is lighter. Isolate the lighter group of three spheres. Perform the same test, taking two spheres and weighing them; either 1. one of the two spheres weighed is lighter or 2. the two spheres weighed are equal which means the third sphere is lighter. 2 times!

First, it is easy to see that one cannot find the heavy ball with one weighing

However, we can with 2. Here is the method we use: Separate the balls into 3 groups, with 3 balls each. Put two of the 3 ball groups on to the scale. If they come out to be equal, then the third group is the one with the heavier ball. Otherwise the one that is heavier on the scale is. Then we have three balls. Repeat the process with except with an individual ball, and we can find the heavy ball. This took us 2 steps, and this is the answer

You divide them into three groups of three, and place the first group on the left side and the second group on the right side. If they are even, then the odd ball is in the third group. Then you pick two balls from the third group and compare them. If they are even, the the odd ball is the third one, else the odd ball is the heavier one. Similarly, if any of the first two groups were heavier, the the odd one is in that group. Then you repeat the second process.

First of all we divide the 9 balls in 3 "teams" of 3 balls, A B C. First weighing: We put A in the first pan, B in the second pan and C away. If the pans are equal, the heavier ball is in C, otherwise is in the team shown by the balance (A or B). Second weighing: We select the "heavier team" from the first weighing and we put 1 ball in the first pan, 1 ball in the second and 1 ball away. If the pans are equal, the heavier ball is the last one, otherwise is in the lowered pan.

Split the $9$ balls into sets of $3$ each. Take any two sets of $3$ and weigh them. If they are equal, take the third, set, take any two balls from them. Weigh them. If they are equal, the third ball is the answer and if they aren't they the heavier ball is our answer.

If the two initial sets of $3$ aren't equal, then we do the same thing we did above with the set that is heavier.