By E. J. Billington, S. Oates-Williams, A. P. Street

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**Additional info for conf. Brisbane, 1981**

**Example text**

Subtracting these two equations gives f (2f (f (x))) − f (2x) = f (2f (f (x) + f (f (x)))) − f (2f (x + f (x))). 53 If f (f (x)) > x, the left side of this equation is negative, so f (f (x) + f (f (x)) > f (x + f (x)) and f (x) + f (f (x)) < x + f (x), a contradiction. A similar contradiction occurs if f (f (x)) < x. Thus f (f (x)) = x as desired. ) 4. Let A be a matrix of zeroes and ones which is symmetric (Aij = Aji for all i, j) such that Aii = 1 for all i. Show that there exists a subset of the rows whose sum is a vector all of whose components are odd.

We are given 1997 distinct positive integers, any 10 of which have the same least common multiple. Find the maximum possible number of pairwise coprime numbers among them. Solution: The maximum number of pairwise coprime numbers in this set is 9. First, suppose there were 10 pairwise coprime numbers n1 , n2 , . . , n10 . Then the least common multiple of any 10 members of this set is lcm(n1 , n2 , . . , n10 ) = n1 n2 · · · n10 . In particular, for any other N in this set, lcm(N, n2 , · · · , n10 ) = n1 n2 · · · n10 is divisible by n1 ; as n1 is relatively prime to nj for 2 ≤ j ≤ 10, n1 divides N .

Solution: All congruences are taken modulo 2. First, changing the order in which we choose the vertices does not affect the end coloring. Also, choosing a vertex twice has no net effect on the coloring. Then choosing one set of vertices has the same effect as choosing its “complement”: the latter procedure is equivalent to choosing the first set, then choosing all the vertices. ) 34 Label the vertices 1, . . , 2n + 1. Let ai be the number of blue segments at each vertex, bi be the number of times the vertex is chosen, and B be the sum of all bi .