Maths Olympiad Prep

Track / Stage 6 / 375 of 400 #1375 of 1964

Problem 1375

National Olympiad, first round
Combinatorics Difficulty 6.6 Find the answer

For any set SS of integers, let f(S)f(S) denote the number of integers kk with 0≤k<20190 \le k < 2019 such that there exist s1,s2∈Ss_1, s_2 \in S satisfying s1−s2=ks_1 - s_2 = k. For any positive integer mm, let xmx_m be the minimum possible value of f(S1)+⋯+f(Sm)f(S_1) + \dots + f(S_m) where S1,…,SmS_1, \dots, S_m are nonempty sets partitioning the positive integers. Let MM be the minimum of x1,x2,…x_1, x_2, \dots, and let NN be the number of positive integers mm such that xm=Mx_m = M. Compute 100M+N100M + N.

Proposed by Ankan Bhattacharya

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

1. **Understanding the function f(S) f(S) **:
- For a set S S of integers, f(S) f(S) counts the number of integers k k in the range 0≤k<2019 0 \le k < 2019 such that there exist s1,s2∈S s_1, s_2 \in S with s1−s2=k s_1 - s_2 = k .
- This means f(S) f(S) essentially counts the distinct differences s1−s2 s_1 - s_2 modulo 2019.

2. Partitioning the positive integers:
- We need to partition the positive integers into m m nonempty sets S1,S2,…,Sm S_1, S_2, \ldots, S_m and minimize the sum f(S1)+f(S2)+⋯+f(Sm) f(S_1) + f(S_2) + \cdots + f(S_m) .

3. **Lower bound of xm x_m **:
- Consider the numbers 1,2,…,2019 1, 2, \ldots, 2019 . If we partition these into m m sets, each set Si S_i will have ai a_i elements.
- The function f(Si) f(S_i) will be at least ai a_i because the differences s1−s2 s_1 - s_2 for s1,s2∈Si s_1, s_2 \in S_i will cover at least ai a_i distinct values.
- Therefore, f(S1)+f(S2)+⋯+f(Sm)≥a1+a2+⋯+am=2019 f(S_1) + f(S_2) + \cdots + f(S_m) \geq a_1 + a_2 + \cdots + a_m = 2019 .

4. Achieving the lower bound:
- To achieve f(S1)+f(S2)+⋯+f(Sm)=2019 f(S_1) + f(S_2) + \cdots + f(S_m) = 2019 , each set Si S_i must cover exactly ai a_i distinct differences.
- This can be achieved if each Si S_i is an arithmetic sequence with a common difference that divides 2019.

5. **Possible values of ri r_i **:
- The divisors of 2019 are 1, 3, 673, and 2019.
- If Si S_i is an arithmetic sequence with common difference ri r_i , then ri r_i must be one of these divisors.

6. Covering system:
- We need to find sets Si S_i such that the sum of their reciprocals equals 1: ∑1ri=1 \sum \frac{1}{r_i} = 1 .
- Additionally, ∑⌈2019ri⌉=2019 \sum \lceil \frac{2019}{r_i} \rceil = 2019 .

7. Possible partitions:
- ri=1 r_i = 1 : This gives one set covering all integers.
- ri=3 r_i = 3 : We need three sets, each covering every third integer.
- ri=673 r_i = 673 : We need 673 sets, each covering every 673rd integer.
- ri=2019 r_i = 2019 : We need 2019 sets, each covering every 2019th integer.

8. Counting the number of solutions:
- The solutions are of the form 2019−672n 2019 - 672n (for n≤3 n \leq 3 ), 2019−2n 2019 - 2n (for n≤673 n \leq 673 ), and 1.
- This gives 676 solutions.

The final answer is 100M+N=100×2019+676=201900+676=202576 \boxed{ 100M + N = 100 \times 2019 + 676 = 201900 + 676 = 202576 } .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.