Hint 3 for the 12 billiard balls maths prize (follows hint 2)

How few possibilities must you have narrowed the choice down to before the *second* weighing?

The second weighing can at best divide the possibilities by three.

So, if you have three possibilities left after the second weighing, you can have a maximum of 3×3=9 after the first weighing and before the second weighing.

The first weighing must reduce the possibilities from 24 to 9 (or fewer: in fact it’s eight, because every option left open after every possible first weighing contains an even number of possibilities).

The second must reduce them from 8 to three.

The third must reduce them from 3 to one.

What general rule does that give you to follow in *all* the weighings?

Click here for hint 4

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