Found this intriguing question in the book *Mathematical Circles (Russian Experience) by Dmitri Fomin, Sergey Genkin and Ilia Itenberg*. I have not done this before, and do not know, therefore, how easy or difficult this is.

There are 101 coins, and only one of them differs from the other (genuine) ones by weight. We have to determine whether this counterfeit coin is heavier or lighter than a genuine coin. How can we do this using two weighings?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy weighing!

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Assuming we don’t need to identify the particular coin that is lighter or heavier.

Place 50 coins each on either side of the balance. If the scale does not balance, then for the second weighing, split the heavier side into two piles of 25 on to the balance. If the scale now balances, the fake coin was lighter; if it does not balance, the fake coin was heavier.

If the scale balances after the first weighing, then the 101st coin is fake. replace one of the coins from the balance and place this 101st coin in its place. This will now tell us whether the fake coin was lighter or heavier.

You weigh 2 set of 33 coins (random). Two scenario

1. They weigh the same – You take 35 coins from the 2 sets (They have no counterfeit). You also take remaining 35 coins. Compare the weights. If remaining 35 coins are heavier, counterfeit is heavier, else it is lighter

2. They weigh different – You take the lighter one and you weigh it with other 33 coins. If it is again lighter (It has counterfeit coin which is lighter). If it is equal (the set discarded has a counterfeit coin which is heavier). It can never be heavier (will imply, 3 set of 33 coins weigh differently).

Alok – i think we can split the weights into 34, 34, and 33. Balance 34 and 34. If they are equal, we know the bad lot and second weighing will reveal , depending on the tilt. If they are unequal, assume that one is bad. Now replace 33 of this lot with the remaining 33 good ones. If we have removed the bad one in this exhange, things will balance out. However there is one caveat. What if the last one was bad. So we do things differently. We put 33 good ones one one side and exchange the 34th with one from the other side. If they are not balanced, the balance will tilt in the opposite direction from the first try and we will know if we had a heavy or a lighter coin.

Divide the set of 101 coins in 3 sets:

Set A of 33 coins, Set B of 33 coins and Set C of the remaining 35 coins

Weight Set A and Set B. Without loss of generality, two cases are possible:

a) Set A == Set B

b) Set A > Set B

In the first case, we know that the counterfeit coin is in Set C, and Set A and B are all genuine coins. Using the coins of A and B, we create a 35 coin set which we know is genuine. Just weighing this against C tells us whether the counterfeit coin is heavier or lighter.

In the second case, we know that Set C is all genuine coins. Remove 2 coins from Set C. And weigh Set C against Set A, if Set A == Set C, Set B has counterfeit coin and its lighter. If Set A > Set C, Set A has counterfeit coin and is heavier.

Hence, in two weighings we can figure out whether the counterfeit coin was heavier or lighter. Cheers!

Alok,

Nice one. Here’s how I would do it. Weigh one coin and other 100 separately. If 100*weight of one coin is greater than weight of the other 100 coins, then the fake coin is heavier and if less, then fake coin is lighter.

Prakhar

Alok, I think I goi it wrong. A slightly different approach – split into 33,34, and 34. Weigh 34 vs 34 and determine if 33 is bad if both weigh the same. Then weigh 33 against 33 of the good sets and determine if it is heavy or light coin. Next, if the 34 does not balance against each other, remember the way the two sides moved. Now we can split one 34 into 17 each and balance them. If they balance out, the other 34 is the bad one with light or heavy based on the previous movement. If they do not, one 17 will go up against the other and we will know that one of them is bad. But based on the way the balance had responded in the 34 vs 34 testing, we can tell if the bad set of 17 had the lighter coin or the heavier coin.

50 against 50.

If equal left out coin is odd and can be weighed against any coin.

If unequal, take any heap and 25 against 25.

If equal the other heap has the odd coin and if the other heap was heavier in the fist weighing, odd one is heavier and vice versa.

If unequal, odd one is in this heap and the odd one is heavier if this heap was heavier on the first weighing and vice versa…

Absolutely correct!