IMO 2010 Shortlist N11

Category: Number Theory

Problem

. Same as Problem N1, but the constant is replaced by . (Canada) Answer for Problem N1. n 39. Solution for Problem N1. Suppose that for some n there exist the desired numbers; we may assume that s1 s2 sn. Surely s1 ¡ 1 since otherwise 1 1 s1 0. So we have 2 ¨s1 ¨s2 1 ¨¨sn pn 1q, hence si ©i 1 for each i 1,...,n. Therefore 1 1 s1 1 1 s2 ... 1 1 sn © 1 1 1 1 ... 1 1 n 1 1 2 n n 1 1 n 1 , which implies n 1 © 2010 670 ¡39, so n ©39. Now we are left to show that n 39 fits. Consider the set t2,3,...,33,35,36,...,40,67u which contains exactly 39 numbers. We have 2 32 34 39 66 1 34 66 17 51 , p1q hence for n 39 there exists a desired example. Comment. One can show that the example p1q is unique. Answer for Problem N11. n 48. Solution for Problem N11. Suppose that for some n there exist the desired numbers. In the same way we obtain that si © i 1. Moreover, since the denominator of the fraction 7 is divisible by 67, some of si’s should be divisible by 67, so sn © si © 67. This means that © 1 2 n 1 n 1 1 66 67n ,65 which implies n © 2010 66 42 67 330 ¡47, so n ©48. Now we are left to show that n 48 fits. Consider the set t2,3,...,33,36,37,...,50,67u which contains exactly 48 numbers. We have 2 32 35 49 66 1 35 66 7 42 , hence for n 48 there exists a desired example. Comment 1. In this version of the problem, the estimate needs one more step, hence it is a bit harder. On the other hand, the example in this version is not unique. Another example is 2 46 66 329 1 66 329 7 67 5 42 . Comment 2. N11 was the Proposer’s formulation of the problem. We propose N1 according to the number of current IMO.66