Note that you're allowed to put more than one ball in each pan.

Suppose you are given eight balls and you know that one ball is slightly heavier than all the others, which are of equal weight. You're also given a balancing scale with which you can compare the weight of balls by putting some in one pan and some in the other. What's the minimum number of weighings you need to establish which ball is the heavier one?

## Comments

## Its two. But you could have 9

Its two. But you could have 9 balls. Divide the balls in 3 groups of 3 balls. You compare the weight 2 of the groups. Now you known the group where the heavy ball is. You pick two balls from this group and compare them, you now know wich ball is heavier.

Harder. You have 9 balls and one has a different weight, with three weighings you can get which ball is the different ball.

## Math

With 3 weighings you can find one ball out of a total of 12 balls that are all of the same weight except one, and in addition determining whether the odd one is lighter or heavier than the other 11.

Similarly with 4 weighings you can solve the problem with 38 balls, 5 weighings for 118 balls, etc. More generally one can deduce a formula for the number f(n) of balls for n weighings, where n is any positive integer.

## Common solution

floor(log3(numberOfBalls-1))+1:

up to 3^1 (2..3) balls : 1 measure

up to 3^2 (4..9) balls : 2 measures (start with up to 3x3 groups)

up to 3^3 (10..27) balls : 3 measures (start with up to 3x9 groups)

up to 3^x balls : x measures (start with up to 3x3^(x-1)) groups

## potential solution

I think 3 weightings at most, possibly 2.

1. 4 balls on the left-hand side of the scale, 4 on the right, to determine which group of 4 has the heavier ball.

2. Of those 4 which we know has the heavier ball, divide 2/2 between right/left sides of the scale to find which has the heavier.

3. Repeat with 1/1 to find which is heaviest.

I'm hoping there's a more clever way where in step 2 you could put 1 on the left and 2 on the right to finagle the heavier ball out of 3 segments (1 left, 2 right, 1 not weighed), but I can't think of it.

## Loopholes!

Minimum number of weighings...

It's one. Weigh one ball on each side. If you're trying to do this in the smallest number possible, we'll, you weigh the two, one is heavier and you're done. It's only got a 25% chance of working but it's the still the minimum.

## Then the answer could be zero

Then the answer could be zero

Just pick a ball, but you have only12.5% chances to get right.

## Weighing balls

Should be 2. Divide into groups of 3/3/2. Weigh 3 against 3. If equal, take the 2 and weigh one against the other to find heavier ball (2 weighings total). If not equal, pick the heavier group of 3 - weigh 1 against 1 and leave the other ball to one side. If equal, heavy ball is 1 not on scales. If not equal, you have found the heavier ball (2 weighings total)

## Alright...8 balls....1 scale.

Alright...8 balls....1 scale. 8 plus 1....47!!!

I think the answer is one. But I'm skeptical because it seemed to easy of an answer.

## 8 balls - 1 scale ... 47!

What planet are you from?

## Answer

This is an oft used puzzle where the N of balls changes or you're asked to find the lighter or heavier ball.

In this case, the "easy" answer is 3 weighings. You're guaranteed to find it with 8 balls. 4v4. Then 2v2. Then 1v1.

But you can actually do it in 2 weighings. You do 3v3 and leave 2 balls unweighed.

If side a is heavier, weigh 2 of the balls. If the scale balances the third ball is heavier. If it doesn't balance the heavier side contains the heavier ball.

Same if side b is heavier.

If the scale balances then just weigh the two remaining balls against each other.

## It's a divide and conquer one

It's a divide and conquer one. I liked the one here below, with a little twist.

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You have 9 balls. 8 of them weigh the same, but one of them weighs heavier than the other 8.

How can we find the heaviest weighted ball in no more than 2 operations?

## Weighing balls

I want it to be three, winnowing down from four balls on each side, to two, to one.

But there is no requirement to weigh all the balls at once, so if you only weigh two on each side, you have a 50% chance of being able to find the heaviest ball in only two weighings and 50% chance of finding it in three.

So by further application, you have a one in four chance of finding the heaviest ball with only one weighing, and a maximum of four, by placing only one ball on each side at a time. Thus the minimum turns out to be one.

## Weighing balls

My solution is two weighings.

1. 3 balls in each pan.

2. If the first weighing balances then weigh the remaing two. If they don't balance then weigh any two from the heavier pan. If they balance then it is the other ball that is the heaviest.

## 3 v3 with 2 aside.

3 v3 with 2 aside.

This will confirm which group the odd ball is in. Once confirmed balance 2 more from that group, which will show directly or indirectly which the odd ball out is.

So 2 Weighings

## The minimum number of

The minimum number of weighings in two. In the first one, three balls are put on both sides of the balancing scale. If one side is heavier, two of the three balls are weight (one one each side). If none of these balls is heavier the ball of the three that was not weight is the heavier one. If in the first weigh both sides have equal weight the two remaining balls are weight and the heavier one is found.

## I think 3 is the correct

I think 3 is the correct answer.

## Weighing Balls

Can be done in two:

Four balls in each pan will show which group has the heavy ball. Split this group between the two pans (two in each). Note which pan (left or right) is low and thus contains the heavy ball. Take one ball from the left pan with your left hand and one from the right with your right hand.

If the scales are even then the heavy ball is in the hand where the ball was taken from the lower pan. Otherwise...

## Error

You actually weighed 3 times. 4v4, 2v2, 1v1

## Math

This continues on the comment I submitted yesterday on the ball weighing problem.

Using a balance n times, one can find out one ball out of f(n) balls that are all of the same weight except one, and also determine whether it is heavier or lighter. The formula is

f(n) = (3^n - 1)/2 - n + 2

We have f(3) = 12, f(4) = 38, f(5) = 118, f(6) = 360, etc.

The solution for the n problem is to leave f(n -1) balls untouched, divide the rest into 2 piles and put one pile to each side of the balance.

If the balance balances, then the odd ball is in the untouched f(n - 1) balls and the problem reduces to the one of f(n-1) balls.

If the balance does not balance, then move 3^(n-1) balls out from one side, move the same number of balls from the other side to replace the moved-out ones, and move the same number from the untouched pile to replace these ones. Now one of three things can happen:

1. If the scale now balances, then the odd ball is among the 3^(n-1) ones and we know whether it is heavier or lighter. Then we can find it out in n-2 tries (see below).

2. If the scale tips to change, then the oddball is among the 3^(n-1) moved across the scale and we can again find it out in n-2 tries.

3. If the scale remains the same and does not tip, then the oddball is among the unmoved ones. Then we repeat the above process by further moving f(n-2) balls.

The solution relies on the fact that if one knows the oddball is heavier or lighter, then one can find it in 1 try for 3 balls, in 2 tries for 3^2=9 balls, etc. and in n-2 tries in 3^(n-1) balls as asserted in the above.

Thus for 12 balls in 3 tries we place 4 on each side to begin. If scale

balances, then compare 3 of the remaining 4 with 3 of the known good ones. If scale unbalances, then move 3 across the scale and replace the moved 3 by 3 of the known good ones. Watch the scale and the rest is easy.

For 38 balls, place 13 on each side, etc.

## Solution

It is two for sure. You just have to divide the given balls into groups of three, three and two balls. Try it out, it you will find out.

## 3 at the most

3 at the most

1) but 4 on each side and one out , if they are equal , then which out will be the grater

2) if not , we take the greater group side and but 2 ball on each side

3) then takes the greater side and but each one on each side and decided