# I think i have somewhat of an idea, but im not sure can someone help?

Counterfeit Coin 1: In a group of five ‘gold’ coins one of them is counterfiet. and thus weights more than the rest but looks exactly like the others. Using a pan balance, what is the smallest number of balancings needed to identify the fake cold coin?

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Minimum one balancing is required as the fake coin has one weight and the other coins another weight. If you are lucky only a single weighing will enable you to choose the fake one: otherwise two, three or even four weighings may be needed.

So you have HNNNN, where you KNOW that it weighs more than the rest.

You have exactly 2 logical choices for your first move: Weigh 1 coin or weigh 2 coins. Let’s see what happens with each:

FIRST WEIGH ONE COIN:

If it’s unbalanced, you pick the heavier one, and you’re done in 1.

If it’s balanced, you have 3 coins left. On your second try, do another 1 vs 1 weighing, and if that’s still balanced, you’re done because it’s the leftover coin.

MAX number of weighings: 2

FIRST WEIGH TWO COINS:

If it’s balanced, you know the fifth coin is the counterfeit.

If it’s unbalanced, take the two coins from the heavier side, and weigh them individuall to see which is heavier.

MAX number of weighings: 2

Either method means you need at LEAST two weighings.

Put 2 coins in each pan. one is left out.

If the pan balances then it’s the coin not on the scale.

If the pan tilts to one side then take the coins on the heavy side and put each

on either pan. Whichever way it tils is theheavier coin.

So, you could get lucky and it could have been the original coin left out of the weighing or at

most you need 2 weighings

Two balancings.

Weight two vs two:either one side is heavier and it is one of the two on the heavier side or the two sides balance and the coin not weighted is the counterfeit

Then weigh one coin on each side of the balance from the first case to find the heaviest.

Simple!

Bear in mind the rule that questions will provide you with the most information if the answer has a 50-50 chance of being right.

If you have a number of coins, you have to split them in 2 equal groups and weigh them both. So, when you ask, "Which group is the fake coin in?" there is a 50% chance of it being in one group, and a 50% chance of it being in the other group. When you weigh the groups, the group containing the fake coin will be heavier. Our criteria laid down by information theory is met.

However, in your example, you have 5 coins, which won’t divide evenly into 2 groups.

So, what you do is put 2 coins in one pan, and 2 coins in the other. If they both balance, then the fake coin is the one remaining.

If they don’t balance, the fake coin must be in the heavier pan. Remove the lighter pair, and put one of the heavier pair in one pan, and the other of the pair in the other pan. The heavier coin will be the fake.

So you only need 2 balancings to be sure of finding the fake.