Mongolian National Mathematical Olympiad

Question

Let a_1 < a_2 < < a_53 be positive integers such that the sum of any 27 integers is greater than the sum of the remaining 26 integers. a) Find the minimum value of a_1. b) Find all possible values of a_2, , a_53, when a_1 is at the minimum value.

Accepted Answer

Clearly, the condition on the sequence is equivalent to the condition

a_{1} + a_{2} + \dots + a_{27} > a_{28} + \dots + a_{53} . (*)

We have

a_{i + 26} \geq a_{i + 25} + 1 \geq \dots \geq a_{i} + 26

, for

1 \leq i \leq 27

. It implies that

a_{i + 26} - a_{i} \geq 26

. Thus from (*) and the above inequality, we have

a_{1} \geq 1 + i = 2 \sum 27 (a_{i + 26} - a_{i}) \geq 1 + 26 \cdot 26 = 677.

Next we show that minimum value of

a_{1}

is indeed 677. If

a_{1} = 677

, the above inequalities must be equalities. Hence

a_{i + 26} = a_{i} + 26

,

2 \leq i \leq 27

and

a_{j} = a_{j + 1} - 1 = \dots = a_{j + 26} - 26

,

2 \leq j \leq 53

.

Hence, if

a_{2} = n

then

a_{3} = n + 1, \dots, a_{53} = n + 51

and

n \geq a_{1} + 1 = 678

.

So

a_{1} = 677, a_{2} = n, a_{3} = n + 1, \dots, a_{53} = n + 51, n \geq 678

is the solution to this problem.