Olympiad Maths Prep

Track / Stage 8 / 33 of 180 #1733 of 2000

Problem 1733

IMO Shortlist mid-range; USAMO P2/P5
Number theory Difficulty 8.1 Prove it

It is given nn a natural number and a circle with circumference nn. On the circle, in clockwise direction, numbers 0,1,2,n10,1,2,\dots n-1 are written, in this order and in the same distance to each other. Every number is colored red or blue, and there exists a non-zero number of numbers of each color. It is known that there exists a set S{0,1,2,n1},S2S\subsetneq \{0,1,2,\dots n-1\}, |S|\geq 2, for wich it holds: if (x,y),x<y(x,y), x<y is a circle sector whose endpoints are of distinct colors, whose distance yxy-x is in SS, then yy is in SS.

Prove that there is a divisor dd of nn different from 11 and nn for wich holds: if (x,y),x<y(x,y),x<y are different points of distinct colors, such that their distance is divisible by dd, then both x,yx,y are divisible by dd.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

1. Initial Setup and Definitions:
- Let n n be a natural number and consider a circle with circumference n n .
- On the circle, in a clockwise direction, numbers 0,1,2,,n1 0, 1, 2, \ldots, n-1 are written, each at equal distances.
- Each number is colored either red or blue, with at least one number of each color.
- There exists a set S{0,1,2,,n1} S \subsetneq \{0, 1, 2, \ldots, n-1\} with S2 |S| \geq 2 such that if (x,y) (x, y) is a circle sector with endpoints of distinct colors and yxS y - x \in S , then yS y \in S .

2. Claim:
- We claim that there exists a divisor d d of n n different from 1 1 and n n such that if (x,y) (x, y) are different points of distinct colors and their distance is divisible by d d , then both x x and y y are divisible by d d .

3. Basic Additive Number Theory:
- For any set of integers {s1,s2,,sS} \{s_1, s_2, \ldots, s_{|S|}\} such that gcd(s1,s2,,sS,n)=d \gcd(s_1, s_2, \ldots, s_{|S|}, n) = d , we can find coefficients c1,c2,,cS c_1, c_2, \ldots, c_{|S|} such that:
i=1Scisid(modn) \sum_{i=1}^{|S|} c_i s_i \equiv d \pmod{n}
- This implies that for d1 d \neq 1 , x x and x+d x + d must be the same color for every dx d \neq x .

4. Induction Proof:
- Base case: For S=1 |S| = 1 , this is just the Euclidean algorithm.
- Inductive step: Assume it is true for k=S k = |S| . This means we can find coefficients c1,,cS c_1, \ldots, c_{|S|} such that:
i=1ScisidS(modn) \sum_{i=1}^{|S|} c_i s_i \equiv d_S \pmod{n}
where dS d_S is the gcd of S={s1,,sS} S = \{s_1, \ldots, s_{|S|}\} .
- For k=S+1 k = |S| + 1 , let the new set S={s1,,sS}{t} S' = \{s_1, \ldots, s_{|S|}\} \cup \{t\} . The gcd of the new set is (dS,t)=dS (d_S, t) = d_{S'} . By the Euclidean algorithm, we can find coefficients l l and m m such that:
ldS+mt=dS ld_S + mt = d_{S'}
- Setting ci=lci c_i' = l \cdot c_i and cS+1=m c_{|S|+1}' = m , we verify that the claim holds true.

5. Extremal Chaining:
- Consider the remaining numbers (attacking numbers) in the set E={0,1,,n1}S E = \{0, 1, \ldots, n-1\} - S .
- We show that E E can be divided into two sets EA E_A and EB E_B which behave similarly to A A and B B .

6. Chaining Down:
- If eEA e \in E_A , then eb e - b (for bB b \in B ) also belongs to EA E_A , and their sizes are equal.
- This is proven by considering the minimal representation of e e and showing that it can be chained down to s1 s_1 .

7. Set Theory Argument:
- If aA a \in A and bB b \in B , then a+b a + b must not be an attacking number, or else it would lead to a contradiction.
- This implies that all members of E E must be of unison color.

8. Final Argument:
- Suppose there exists some dn d \mid n with dn d \neq n such that S+d=S S + d = S .
- The minimal d d' which satisfies this equation will work for the problem.
- If not, then all numbers from {0,1,,n1}{0} \{0, 1, \ldots, n-1\} - \{0\} have the same color, contradicting the problem statement.

9. Conclusion:
- We have shown that there exists a divisor d d of n n different from 1 1 and n n such that if (x,y) (x, y) are different points of distinct colors and their distance is divisible by d d , then both x x and y y are divisible by d d .

\blacksquare

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.