Maths Olympiad Prep

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Combinatorics Difficulty 5.7 AIME, harder Prove it Bulgaria

Problem:
Find the largest possible number AA having the following property: if the numbers 1,2,,10001,2, \ldots, 1000 are ordered in arbitrary way then there exist 50 consecutive numbers with sum not less than AA.

Solution

Solution:
Let x1,x2,,x1000x_{1}, x_{2}, \ldots, x_{1000} be an arbitrary rearrangement of the numbers 1,2,,10001,2, \ldots, 1000. Set

S1=x1+x2++x50,,S20=x951+x952++x1000 S_{1} = x_{1} + x_{2} + \cdots + x_{50}, \ldots, S_{20} = x_{951} + x_{952} + \cdots + x_{1000}
Since S1++S20=500500S_{1} + \cdots + S_{20} = 500500, we have Si50050020=25025S_{i} \geq \frac{500500}{20} = 25025 for at least one index ii.

On the other hand, if a number BB has the required property then we have B25025B \leq 25025. To see this consider the rearrangement
1000,1,999,2,,501,500 1000, 1, 999, 2, \ldots, 501, 500
and take arbitrarily fifty consecutive numbers in it. If the first number is greater than 500500, then the sum of these fifty numbers is 2502525025, otherwise it is 2500025000. Hence A=25025A = 25025.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.