There are 109 soldiers in a camp. Every night three of them go on watch patrol. Prove that it can be arranged so that after a while, every pair of soldiers has shared watch exactly three times.

Lindsey Ibarra

Lindsey Ibarra

Answered question

2022-09-23

There are 109 soldiers in a camp. Every night three of them go on watch patrol. Prove that it can be arranged so that after a while, every pair of soldiers has shared watch exactly three times.

Answer & Explanation

Kinowagenqe

Kinowagenqe

Beginner2022-09-24Added 4 answers

Suppose n = 2 m + 1 and consider the triples ( i , i + j , i + 2 j ) where i = 1 , 2 , 3 , , ( 2 m + 1 ) and j = 1 , 2 , 3 , , m .. We can construct a ( 2 m + 1 ) × m array of triples with ( i , i + j , i + 2 j ) in the ith row and jth column. We work modulo 2 m + 1..
For example here's a solution for n = 9.
( 1 , 2 , 3 ) ( 1 , 3 , 5 ) ( 1 , 4 , 7 ) ( 1 , 5 , 9 ) ( 2 , 3 , 4 ) ( 2 , 4 , 6 ) ( 2 , 5 , 8 ) ( 2 , 6 , 1 ) ( 3 , 4 , 5 ) ( 3 , 5 , 7 ) ( 3 , 6 , 9 ) ( 3 , 7 , 2 ) ( 4 , 5 , 6 ) ( 4 , 6 , 8 ) ( 4 , 7 , 1 ) ( 4 , 8 , 3 ) ( 5 , 6 , 7 ) ( 5 , 7 , 9 ) ( 5 , 8 , 2 ) ( 5 , 9 , 4 ) ( 6 , 7 , 8 ) ( 6 , 8 , 1 ) ( 6 , 9 , 3 ) ( 6 , 1 , 5 ) ( 7 , 8 , 9 ) ( 7 , 9 , 2 ) ( 7 , 1 , 4 ) ( 7 , 2 , 6 ) ( 8 , 9 , 1 ) ( 8 , 1 , 3 ) ( 8 , 2 , 5 ) ( 8 , 3 , 7 ) ( 9 , 1 , 2 ) ( 9 , 2 , 4 ) ( 9 , 3 , 6 ) ( 9 , 4 , 8 )
The number of triples = m ( 2 m + 1 ) = ( 2 m + 1 m ) = the number of distinct pairs of integers in S = { 1 , 2 , 3 , , 2 m + 1 } and all triples are unique, where the order of the elements is taken into account. (For suppose ( x , y , z ) is in our array, then x = i , y = i + j and z = i + 2 j for some i , j and thus if x < y this triple is in the xth row and ( y x )th column, and if x > y it is in the xth row and ( y x + 2 m + 1 )th column. So its position is uniquely determined by its elements.
Moreover, if ( x , y , z ) = ( i , i + j , i + 2 j ) is in the array then none of ( x , z , y ) , ( z , y , x ) and ( y , x , z ) can be in the array, for if one of the positions of the elements is fixed the latter three triples cannot obey the construction rule, that its elements increase by j , since we know that x, y and z increase by j .
Now consider a pair of integers ( a , b ) ,, which appears in a given triple, ( a , b , c ). There cannot exist another triple in the array ( a , b , d ) ,, say, with c d since c is uniquely determined by a and b .
The pair ( a , b ) cannot appear in more than three triples, in whatever order (we've already noted that we cannot have both ( a , b , c ) and ( b , a , c ) hence the three and not six), since the remaining element is uniquely determined by a and b . However, it cannot appear in less than three triples since the number of triples equals the number of pairs of distinct integers in S , so this would mean that another pair must appear in more than three triples, a contradiction.
Thus we have a solution for any odd n 3..
So a solution for n = 11 is:
( 1 , 2 , 3 ) ( 1 , 3 , 5 ) ( 1 , 4 , 7 ) ( 1 , 5 , 9 ) ( 1 , 6 , 11 ) ( 2 , 3 , 4 ) ( 2 , 4 , 6 ) ( 2 , 5 , 8 ) ( 2 , 6 , 10 ) ( 2 , 7 , 1 ) ( 3 , 4 , 5 ) ( 3 , 5 , 7 ) ( 3 , 6 , 9 ) ( 3 , 7 , 11 ) ( 3 , 8 , 2 ) ( 4 , 5 , 6 ) ( 4 , 6 , 8 ) ( 4 , 7 , 10 ) ( 4 , 8 , 1 ) ( 4 , 9 , 3 ) ( 5 , 6 , 7 ) ( 5 , 7 , 9 ) ( 5 , 8 , 11 ) ( 5 , 9 , 2 ) ( 5 , 10 , 4 ) ( 6 , 7 , 8 ) ( 6 , 8 , 10 ) ( 6 , 9 , 1 ) ( 6 , 10 , 3 ) ( 6 , 11 , 5 ) ( 7 , 8 , 9 ) ( 7 , 9 , 11 ) ( 7 , 10 , 2 ) ( 7 , 11 , 4 ) ( 7 , 1 , 6 ) ( 8 , 9 , 10 ) ( 8 , 10 , 1 ) ( 8 , 11 , 3 ) ( 8 , 1 , 5 ) ( 8 , 2 , 7 ) ( 9 , 10 , 11 ) ( 9 , 11 , 2 ) ( 9 , 1 , 4 ) ( 9 , 2 , 6 ) ( 9 , 3 , 8 ) ( 10 , 11 , 1 ) ( 10 , 1 , 3 ) ( 10 , 2 , 5 ) ( 10 , 3 , 7 ) ( 10 , 4 , 9 ) ( 11 , 1 , 2 ) ( 11 , 2 , 4 ) ( 11 , 3 , 6 ) ( 11 , 4 , 8 ) ( 11 , 5 , 10 )

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?