[...]... 1 for all n ∈ N In particular, f (2007) ≤ 2008 Now we present a family of examples showing that all values from 1 to 2008 can be realized Let fj (n) = max{1, n + j − 2007} for j = 1, 2, , 2007; f2008 (n) = n, 2007 n, n + 1, 2007 n We show that these functions satisfy the condition (1) and clearly fj (2007) = j To check the condition (1) for the function fj (j ≤ 2007) , note first that fj is nondecreasing... Otherwise, 2007 m+n, hence 2007 m or 2007 n In the former case we have f2008 (m) = m, while in the latter one f2008 f2008 (n) = f2008 (n) = n, providing f2008 (m) + f2008 f2008 (n) − 1 ≤ (m + n + 1) − 1 = f2008 (m + n) 11 Comment The examples above are not unique The values 1, 2, , 2008 can be realized in several ways Here we present other two constructions for j ≤ 2007, without proof:  1, n < 2007, ... Otherwise, fj (m) + fj fj (n) − 1 ≤ (m + j − 2007) + n = (m + n) + j − 2007 = fj (m + n) In the case j = 2008, clearly n + 1 ≥ f2008 (n) ≥ n for all n ∈ N; moreover, n + 1 ≥ f2008 f2008 (n) as well Actually, the latter is trivial if f2008 (n) = n; otherwise, f2008 (n) = n + 1, which implies 2007 n + 1 and hence n + 1 = f2008 (n + 1) = f2008 f2008 (n) So, if 2007 m + n, then f2008 (m + n) = m + n + 1... realized in several ways Here we present other two constructions for j ≤ 2007, without proof:  1, n < 2007,  jn hj (n) = max 1, gj (n) = j, n = 2007,  2007  n, n > 2007; Also the example for j = 2008 can be generalized In particular, choosing a divisor d > 1 of 2007, one can set n, d n, f2008,d (n) = n + 1, d n 12 A3 Let n be a positive integer, and let x and y be positive real numbers such that xn...10 A2 Consider those functions f : N → N which satisfy the condition f (m + n) ≥ f (m) + f f (n) − 1 (1) for all m, n ∈ N Find all possible values of f (2007) (N denotes the set of all positive integers.) (Bulgaria) Answer 1, 2, , 2008 Solution Suppose that a function f : N → N satisfies (1) gers m > n, by (1) we have For arbitrary positive inte- f (m)... proof 30 C3 Find all positive integers n, for which the numbers in the set S = {1, 2, , n} can be colored red and blue, with the following condition being satisfied: the set S × S × S contains exactly 2007 ordered triples (x, y, z) such that (i) x, y, z are of the same color and (ii) x + y + z is divisible by n (Netherlands) Answer n = 69 and n = 84 Solution Suppose that the numbers 1, 2, , n are... b)2 − 3rb = r 2 − rb + b2 , as claimed So, to find all values of n for which the desired coloring is possible, we have to find all n, for which there exists a decomposition n = r + b with r 2 − rb + b2 = 2007 Therefore, 9 r 2 − rb + b2 = (r + b)2 − 3rb From this it consequently follows that 3 r + b, 3 rb, and then 3 r, 3 b Set r = 3s, b = 3c We can assume that s ≥ c We have s2 − sc + c2 = 223 Furthermore, . present other two constructions for j ≤ 2007, without proof: g j (n) =      1, n < 2007, j, n = 2007, n, n > 2007; h j (n) = max  1,  jn 2007  . Also the example for j = 2008 can. f (2007) ≤ 2008. Now we present a family of examples showing that all values from 1 to 2008 can be realized. Let f j (n) = max{1, n + j 2007} for j = 1, 2, . . . , 2007; f 2008 (n) =  n, 2007. = (m + 1) + (n + 1) − 1 ≥ f 2008 (m) + f 2008  f 2008 (n)  − 1. Otherwise, 2007    m+n, hence 2007    m or 2007    n. In the former case we have f 2008 (m) = m, while in the latter