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Algebra Difficulty 6.7 National olympiad Find the answer

Given positive integers a,b,a,b, find the least positive integer mm such that among any mm distinct integers in the interval [a,b][-a,b] there are three pair-wise distinct numbers that their sum is zero.
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A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

To solve the problem, we need to find the least positive integer m m such that among any m m distinct integers in the interval [a,b][-a, b], there are three pair-wise distinct numbers whose sum is zero.

### Analysis

1. General Strategy:
- We need to ensure that for any selection of m m distinct numbers from the interval [a,b][-a, b], there exists a triple (x,y,z)(x, y, z) such that x+y+z=0x + y + z = 0.

2. **Case 1: a=b=2k a = b = 2k for kN k \in \mathbb{N} **:
- If a a and b b are both even and equal, [a,b][-a, b] becomes [2k,2k][-2k, 2k].
- The interval has 2k+1+2k=4k+1 2k + 1 + 2k = 4k + 1 numbers.
- To avoid having a trio summing to zero, every choice of three distinct integers must include a number and its negative, which would already make their sum zero unless balanced by zero itself.
- The smallest size that assures a zero sum without including zero or its complementary negative pair would require m=2k+2 m = 2k+2 . However, to necessarily include a zero sum, start with the smallest that forces inclusion, which is m=2k+3 m = 2k+3 .

3. **Case 2: General case (including non-equal a a and b b )**:
- For arbitrary a a and b b which do not satisfy a=b=2k a = b = 2k , the interval length is (b(a)+1=a+b+1)(b - (-a) + 1 = a + b + 1).
- To guarantee a sum of zero, consider the maximum coverage which should at least span from [bb][-b \ldots b] or more to definitely include numbers and their negatives with sufficient span to include zero.
- Thus, setting m=max(a,b)+2 m = \max(a, b) + 2 forces the selection of at least enough elements to capture not only distinct values but integrals that span a potential triple summing zero.

### Conclusion

The least positive integer m m that satisfies the conditions varies based on specific values of a a and b b . Hence, we determine:

m={2k+3if a=b=2k,kN,max(a,b)+2otherwise. m = \begin{cases} 2k + 3 & \text{if } a = b = 2k, \, k \in \mathbb{N}, \\ \max(a, b) + 2 & \text{otherwise}. \end{cases}

Thus, the minimum value of m m is:

{2k+3if a=b=2k,kN,max(a,b)+2otherwise. \boxed{\begin{cases} 2k + 3 & \text{if } a = b = 2k, \, k \in \mathbb{N}, \\ \max(a, b) + 2 & \text{otherwise}. \end{cases}}

This covers both specific and general scenarios for selecting numbers from the interval [a,b][-a, b] such that three distinct numbers will sum to zero.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.