Given a balance scale that is sure to break after X weighings, find an equation for the largest number of coins N, from which you can determine a fake coin that has the wrong weight if
A: You know whether the fake is lighter or heavier
B: You do not know whether the fake is lighter or heavier
(Assume only one of the N coins is fake)
(In reply to
re: solution to B by Sanjay)
4 coins in 2 weighings is definitely possible. You don't need a tree with 8 terminal branches, because you don't necessarily have to know if the fake coin is heavier or lighter.
Let's call the coins A,B,C,D.
(1) weigh A - B
case I: they're even. In which case
(2) weigh A - C.
If C is heavier or lighter, C is the fake.
If they're even, D is the fake.
case II: A is heavier than B (or lighter, doesn't matter).
(2) weigh A - C.
If they're even, B is the fake.
If not, A is the fake. QED
13 coins: divide them into groups A (4 coins) B (4 coins) and C (5 coins).
(1) weigh all A coins - all B coins
case (1)I: they're even, therefore all A and B are 'true'.
(2) weigh 3 true coins - C1, C2, C3
case (2)I: they're not even - reduces to 3 coins, one weighing, but you KNOW if it's lighter or heavier (part A of riddle).
case (2)II: they're even - the fake is C4 or C5. Weigh C4 against a true coin.
case (1)II: A coins heavier than B coins (wlg):
(2) weigh A1, A2, B1 - A3, A4, B2
If they're even - B3 or B4 is the fake, so (3) weigh B3 - true coin.
If left is heavier: A1 or A2 is heavy or B2 is light. So, (3) Weigh A1, B2 - 2 true coins. Left heavier - it's A1; right heavier - it's B2; even - it's A2. Similarly for right side heavier.
40 coins is a bit long. But it starts with 13 coins v 13, and keeping 14 on the side.
re(2): solution to B
