Iranian Combinatorics Olympiad 2021 - Advanced level - Problem 1 - Frogs on the stone

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Anindya Biswas
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Iranian Combinatorics Olympiad 2021 - Advanced level - Problem 1 - Frogs on the stone

Unread post by Anindya Biswas » Thu Aug 26, 2021 2:55 am

In the lake, there are $23$ stones arranged along a circle. There are $22$ frogs numbered $1,2,\dots,22$ (each number appears once). Initially, each frog randomly sits on a stone (several frogs might sit on the same stone). Every minute, all frogs jump at the same time as follows: the frog number $i$ jumps $i$ stones forward in the clockwise direction. (In particular, the frog number $22$ jumps $1$ stone in the counter-clockwise direction.) Prove that at some point, at least $6$ stones will be empty.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Re: Iranian Combinatorics Olympiad 2021 - Advanced level - Problem 1 - Frogs on the stone

Unread post by Mehrab4226 » Sat Sep 04, 2021 7:56 pm

Let, $k_i$ be the stone on which ith frog was initially sitting on. Let's say all move M times. Then each frog will naturally go from $k_i \to k_i +M$(Mod 23). We can work on (Mod 23)

Now if 2 or more frogs sit on the same stone then we will call one(any) of them 'domain' and the rest 'refugees'. (I know I am bad at making names). It will be sufficient if we prove that at some point there will be at least 5 refugees.

Now, the value of M(the number of moves) is between (1-23) as after 23 moves all will again come at the same initial position. So we only need to think about (1-23) values of M.

Claim-1: Each pair of frogs will have a value M(1-23) such that after M moves that pair will be on the same stone.
To make our claim true the congruence must hold in (mod 23) of course.
$k_i+iM \equiv k_j +jM$
Or,$k_i - k_j \equiv (j-i)M$
The second form holds as $(k_i - k_j) \neq 0$ and is in the complete residue class of $23$ and so is $j-i$. So we can find a value M for which the congruence holds. And M will also be in (1-23). Note this holds only because 23 is prime. $\square$

Now, there are $\binom{22}{2}=231$ pairs of frogs. And each pair will have a value M. Since there are 23 different possible values of M(or at least the ones we need to look at, by PHP we get, $\left\lceil \frac{231}{23} \right\rceil = 11$. So 11 pairs must have the same value of M.

Claim-2: There will be at least 5 refugees at some point
Let's say after $M_1$ moves [Let $M_1$ be the value of the 11 pairs] 11 of the pairs will sit on the same stone. If there were less than 5 refugees at that time we cannot find 11 pairs who sit on the same stone. A contradiction. So there must be at least 5 refugees at that time.$\square$

So there will be at least 6 stones empty at some point. $\square$
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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