Track / Stage 6 / 108 of 400 #1108 of 2000

Problem 1108

National olympiad, first round

Algebra Difficulty 6.2 Prove it 7th JBMO · JBMO

Problem:

Natural numbers $1, 2, 3, \ldots, 2003$ are written in an arbitrary sequence $a_{1}, a_{2}, a_{3}, \ldots, a_{2003}$ . Let $b_{1} = 1 a_{1}$ , $b_{2} = 2 a_{2}$ , $b_{3} = 3 a_{3}$ , $\ldots$ , $b_{2003} = 2003 a_{2003}$ , and $B$ be the maximum of the numbers $b_{1}, b_{2}, b_{3}, \ldots, b_{2003}$ .

a) If $a_{1} = 2003, a_{2} = 2002, a_{3} = 2001, \ldots, a_{2002} = 2, a_{2003} = 1$ , find the value of $B$ .

b) Prove that $B \geq 1002^{2}$ .

Problem:

Numerele $1, 2, 3, \ldots, 2003$ sunt scrise într-un şir $a_{1}, a_{2}, a_{3}, \ldots, a_{2003}$ . Fie $b_{1} = 1 a_{1}$ , $b_{2} = 2 a_{2}$ , $b_{3} = 3 a_{3}$ , $\ldots$ , $b_{2003} = 2003 a_{2003}$ şi $B$ maximul numerelor $b_{1}, b_{2}, b_{3}, \ldots, b_{2003}$ .

a) Dacă $a_{1} = 2003, a_{2} = 2002, a_{3} = 2001, \ldots, a_{2003} = 1$ , găsiți valoarea lui $B$ .

b) Demonstrați că $B \geq 1002^{2}$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution:

a) Using the inequality between the arithmetical and geometrical mean, we obtain that $b_{n} = n (2004 - n) \leq \left( \frac{n + (2004 - n)}{2} \right)^{2} = 1002^{2}$ for $n = 1, 2, 3, \ldots, 2003$ . The equality holds if and only if $n = 2004 - n$ , i.e. $n = 1002$ . Therefore, $B = b_{1002} = 1002 \times (2004 - 1002) = 1002^{2}$ .

b) Let $a_{1}, a_{2}, a_{3}, \ldots, a_{2003}$ be an arbitrary order of the numbers $1, 2, 3, \ldots, 2003$ . First, we will show that numbers $1002, 1003, 1004, \ldots, 2003$ cannot occupy the places numbered $1, 2, 3, \ldots, 1001$ only. Indeed, we have $(2003 - 1002) + 1 = 1002$ numbers and $1002$ places. This means that at least one of the numbers $1002, 1003, 1004, \ldots, 2003$ , say $a_{m}$ , lies on a place which number $m$ is greater than $1001$ . Therefore, $B \geq m a \geq 1002 \times 1002 = 1002^{2}$ .

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