## Le Monde puzzle [#928] A combinatorics Le Monde mathematical puzzle:

How many distinct integers between 0 and 16 can one pick so that all positive differences are distinct?

If k is the number of distinct integers, the number of positive differences is

1+2+…+(k-1) = k(k-1)/2,

which cannot exceed 16, because it is a subset of {1,2,…,16}, meaning k cannot exceed 6 if all differences are distinct. From there, picking k integers at random makes it easy to check for the condition:

```k=6
N=16
x=sort(sample(0:N,k-1))
y=outer(x[-1],x[-k],"-")
while (max(duplicated(y[!upper.tri(y)]))==1){
x=sort(sample(0:N,k-1))
y=outer(x[-1],x[-k],"-")}
```

which quickly returns for k=5

```> x
 0 1 7 12 15
```

as a solution. And is still running for k=6, meaning there is apparently no solution for k=6. (An exhaustive search shows there is indeed no solution for k=6 and N=16, while there are several for k=6 and N=17.) Now, reading the puzzle solution of Le Monde today, on September 09, I discovered that the authors proposed a sequence of length 7, (0,1,2,4,5,7,11,16), which does not work since 1-0=2-1… and proved that 8 is an impossible value by quite a convoluted argument. Did I misread again?!

In the earlier version of the R code posted today, I used

`...y[lower.tri]...`

which does not include the diagonal, instead of the proper

`...y[!upper.tri(y)]..`

a mistake that led to a wrong solution for k=6, as pointed out by Stephan.

### 2 Responses to “Le Monde puzzle [#928]”

1. stephankolassasap Says:

Even in your solution, you have 5-1 = 9-5 and 12-9 = 15-12…

• xi'an Says:

This seems to be the puzzle of the wrong answers…! I forgot the diagonal in the duplicate check. Thanks!

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