Olympiad Maths Prep

Library / /15 of 15

, 2010

Algebra Difficulty 6.6 National olympiad Prove it Ukraine

Consider an arbitrary arrangement of the brackets. Numbers 20102010 and 20092009 always have signs '+' and '-', respectively, that's why the sum reaches the maximum value when the other terms will have '+' sign.

In the expression
201020092010200920102009201020092010 numbers \frac{2010 - 2009 - 2010 - 2009 - 2010 - 2009 - \dots - 2010 - 2009}{2010 \text{ numbers}}
there are somehow placed brackets and the value is calculated. Find the maximum value that can be reached. Justify your answer.

*Notice.* Left bracket can be placed only before a number and right - only after. For example, expressions 2010(20092010)-2010(-2009 - 2010) and (20102009)2010-(2010 - 2009-)2010 are incorrect.

Solution

2010(2009201020092010200920102009)2010 numbers=4035077. \frac{2010 - (2009 - 2010 - 2009 - 2010 - 2009 - \dots - 2010 - 2009)}{2010 \text{ numbers}} = 4035077.

Solution. Consider an arbitrary arrangement of the brackets. The first numbers 20102010 and 20092009 always have signs '+' and '-', respectively, that's why the sum reaches the maximum value when other terms will have '+' signs. We can achieve this as it was made in the answer. Calculating maximum value:
2010(2009201020092010200920102009)2010 numbers==20102009+2010+2009+2010+2009++2010+20092008 numbers==1+(2010+2009)1004=4035077. \begin{align*} & \underbrace{2010 - (2009 - 2010 - 2009 - 2010 - 2009 - \dots - 2010 - 2009)}_{2010 \text{ numbers}} = \\ &= 2010 - 2009 + \underbrace{2010 + 2009 + 2010 + 2009 + \dots + 2010 + 2009}_{2008 \text{ numbers}} = \\ &= 1 + (2010 + 2009) \cdot 1004 = 4035077. \end{align*}

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.