# With One Weighing

Here's problem 31 (p. 9) from the wonderful collection Mathematical Circles (Russian Experience).

Of 101 coins, 50 are counterfeit. The weight of one genuine coin is an unknown integer, while all the counterfeit coins have the same weight that differs from the weight of a genuine coin by 1 gram. Peter has a two-pan pointer scale which shows the difference in weights between the objects placed in each pan. Peter chooses one coin and wants to determine in one weighing whether itís genuine or a fake. Can he do this?

### References

- D. Fomin, S. Genkin, I. Itenberg, Mathematical Circles (Russian Experience), AMS, 1996

|Contact| |Front page| |Contents| |Arithmetic|

Copyright © 1996-2018 Alexander Bogomolny

Of 101 coins, 50 are counterfeit. The weight of one genuine coin is an unknown integer, while all the counterfeit coins have the same weight that defers from the weight of a genuine coin by 1 gram. Peter has a two-pan pointer scale which shows the difference in weights between the objects placed in each pan. Peter chooses one coin and wants to determine in one weighing whether itís genuine or a fake. Can he do this?

### Solution 1

After Peter picks up a coin, he proceeds to separate the remaining $100$ coins into two groups of $50$ which he places on the two pans. The difference in weights will be an integer: odd or even. If the number is odd, Peter holds a counterfeit coin. Otherwise, the chosen coin is genuine. Why?

Let $L$ be the number of counterfeit coins in the left pan, $R$ their number in the right pan. We know that $L+R$ is either $49$ or $50.$ In the former case, Peter holds a counterfeit coin. The difference in weight between the two pans is $L-R$ which is odd or even simultaneously with the sum $L+R$.

It is also to imagine an experiment under which all the counterfeit coins have been placed on one of the pans. Starting with that, we swap a counterfeit coin on one pan with a genuine coin from the other pan. If $C$ is the initial number of counterfeit coins (which is either $49$ or $50)$, then after one swap, the pans will contain $C-1$ and $1$ counterfeit coins, respectively. The difference in weight will be $C-3$ - the same parity as $C.$ After the second swap, the numbers will be $C-2$ and $2,$ and so on.

### Solution 2

Let $g$ and $c$ be the weights of a genuine and a counterfeit coin, respectively. For this solution we assume that the weights are both integer: $c=g\pm 1.$ Peter picks up a coin and places it on one pan; the remaining coins go onto the other. If the lone coin is counterfeit, the difference in weights between the two pans will be:

$(49c+51g)-c=48(g\pm 1)+51g=99g\pm 48.$

In case Peter chose a genuine coin, the difference will be:

$(50c+50g)-g=50(g\pm 1)+49g=99g\pm 50.$

In the former case, the observed difference in weights is divisible by $3,$ in the latter case, it is not.

### Acknowledgment

The second solution has been communicated to me by Professor Anany Levitin who also brought the problem to my attention.

### Weighing Coins, Balls, What Not ...

- The Oddball Problem, B. Bundy
- Weighing 12 coins, Dyson and Lyness' solution
- Weighing 12 coins, W. McWorter
- Thought Less Mathematics, D. Newman
- Weighing with counterbalances
- Odd Coin Problems, J. Wert
- Six Balls, Two Weighings
- 12 Coins in Verse
- Six Misnamed Coins, Two Weighings
- A Fake Among Eight Coins
- A Stack of Fake Coins
- Five Coins - One Good, One Bad
- With One Weighing

|Contact| |Front page| |Contents| |Arithmetic|

Copyright © 1996-2018 Alexander Bogomolny

71427894