### Brain Teasers

# Odd Ball

You are given 14 balls, one of which is marked as a reference ball. 12 of the other balls are the same weight as the reference ball but the other ball may be slightly heavier, slightly lighter or the same weight as the reference ball. Using a balance scale how can you determine which is the odd ball, if any, with just three weighings?

### Answer

Split the 14 balls into three groups: one of 4 unknown balls, one of 5 unknown balls and one of 4 unknown plus the reference ball.1st balance: Balance the two groups of 5 balls.

a) If the balance is not equal, designate the heavy side balls H and the light side balls L. The group of 4 unbalanced balls must be good balls.

If the 5 unknown is heavier (5 H > 4 L + ref): take the ref ball, one H, and two L's, and place to one side.

2nd balance: Balance the remaining balls like this: 2H+L against 2H+L.

a) If the balance is equal, then one of the two L's placed to one side is a light ball or the H placed to one side is a heavy ball.

3rd balance: Balance the two L's against each other. If unequal, the lighter side is a light ball. If equal, the H is a heavy ball.

b) If the 2nd balance is not equal, then either the 2H has the heavy ball or the opposing L is light.

3rd balance: Balance the two H balls from the heavier side against each other. If equal, then L is a light ball. If not, the heavier H is a heavy ball.

If the 5 unknown is lighter (5 L > 4 H + ref): take the ref ball, one L, and two H's, and place to one side.

2nd balance: Balance the remaining balls like this: 2L+H against 2L+H.

a) If the balance is equal, then one of the two H's placed to one side is a heavy ball or the L placed to one side is a light ball.

3rd balance: Balance the two H's against each other. If unequal, the heavier side is a heavy ball. If equal, the L is a light ball.

b) If the 2nd balance is not equal, then either the 2L has the light ball or the opposing H is heavy.

3rd balance: Balance the two L balls from the lighter side against each other. If equal, then the H is a heavy ball. If not, the lighter L is a light ball.

b) If the first balance is equal, then the odd ball could be in the group of 4 unknown balls. The other 10 must all be good balls.

2nd balance: Place one of the 4 unknown balls to one side and balance the remaining three against three good reference balls from the 10 good balls.

a) If the balance is equal, then the ball placed to one side could be the odd ball.

3rd balance: Balance this ball against a reference to determine if it's good or not.

b) If the 3 unknown are heavier then: 3rd balance: balance two of the three balls against each other. If equal, the unbalanced ball is heavy. If not equal, the heavy side has a heavy ball.

c) If the 3 unknown are lighter then:

3rd balance: balance two of the three balls against each other. If equal, the unbalanced ball is lighter. If not, equal the lighter side has a lighter ball.

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## Comments

omg just looking at the explanation made my head hurt.

Good teaser...even though the answer "rolled" right past me!

Knew it had something to do with splitting them into groups, but not quite sure how. Stumped me.

Knew it had something to do with splitting them into groups, but not quite sure how. Stumped me.

The groups are 4,4, and 5, so there can't be 2 groups of 5 to weigh against each other

............. ................

that was so stupid boring and dumb,i mean its not fun when its complicating,i mean the freekin answer is complicating

There ARE two groups of 5 balls.

On is 5 unknown balls. Tne other is 4 unknown plus the reference (4+1 =5)

On is 5 unknown balls. Tne other is 4 unknown plus the reference (4+1 =5)

This is solvable without the reference ball. 1st weighing: 4 vs 4. Result A: they balance. Weigh 3 of the remaining 5 against 3 of the 8 we now know are normal. If it is unbalanced, we now know if the strange ball is heavy or light, and its in the 3, so weigh two of them against each other to find the one. If they balance, its in the two we haven't weighed, weigh one of them against one of the normal ones. Result B: One side of the 4v4 is heavy. So, we have 5N (the normal, unweighed ones), 4H from the heavy side, and 4L from the light side. Weigh 2H+1L vs 1H+1L+1N. If the first side goes heavy, its in the 2H, or the 1L on the other side. Weigh the 2H against each other to see which. If the first side goes light, its the 1L on the first side or the 1H on the second side, weigh one of them vs 1N to decide. If the scale balances, its among the 1H and 2L we didn't weigh. Weight the 2L against each other to see which.

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