**Problem:**

*Some athletes start together at the same starting point for a running contest over a circular track. They keep running clockwise around forever at fixed distinct speeds. At some instant of time each
runner may be at equal distance d from the one who runs before him and the one behind him, so that the positions of the runners are evenly distributed over the track. But is there for each of them an
instant of time when he is at distance at least d from any other runner?
The conjecture that JM Wills made in 1967 says there is such an instant of time. The problem is how to prove this. This problem has been solved for up to seven runners, but the general problem remains
unsolved.
*

*In mathematical terms, we may formulate the problem for k+1 runners as follows:
Given a positive integer k and k integers d _{1} , d_{2} , ... , d_{k} without common prime divisor. Prove there is a positive real t such that for all i∈{1,2,..,k} the
product td_{i} is at distance at least 1/(k+1) from the nearest integer.*

__First approach:__

Denote the distance of any real number x from the nearest integer by ||x||.

**The case k=1**

In this case, since d_{1} has no prime divisors, d_{1} = 1. For t=1/2 we have ||td_{1}|| = ||1/2|| ≥ 1/2.

**The case k=2**

We are looking for a positive real t such that ||td_{1}|| ≥ 1/3 ∧ ||td_{2}|| ≥ 1/3.

After perhaps rearranging d_{1}, d_{2} we can proceed as follows:

Let d_{1} = r_{1}3^{i1} + r_{2}3^{i2} + ... + r_{n}3^{in} where 0 ≤ i1 < i2 < ... < in and r_{1}, r_{2}, ... , r_{n} ∈ {1,2}.

Let d_{2} = s_{1}3^{j1} + s_{2}3^{j2} + ... + s_{m}3^{jm} where i1 ≤ j1 < j2 < ... < jm and s_{1}, s_{2}, ... , s_{m} ∈ {1,2}.

Then d_{1}3^{-j1-1} = r_{1}3^{i1-j1-1} + r_{2}3^{i2-j1-1} + ... + r_{n}3^{in-j1-1} = u3^{i1-j1-1} + v where u,v are nonnegative integers
and 1 ≤ u ≤ 3^{j1-i1+1}-1 and u is not divisible by 3.

And d_{2}3^{-j1-1} = s_{1}3^{-1} + s_{2}3^{j2-j1-1} + ... + s_{m}3^{jm-j1-1}, so d_{2} is already at distance 1/3 from the
nearest integer.

We are now looking for a positive integer w such that t := w3^{-j1-1} satisfies our conditions.

If only w is not divisible by 3, this t will satisfy ||td_{2}|| = 1/3.

In order to let t satisfy ||td_{1}|| ≥ 1/3, too, we have to require furthermore that the product u*w modulo 3^{j1-i1+1} belong to the interval [3^{j1-i1}, 2*3^{j1-i1}].

But it is evident that such a w exists.

**The case k=3**

We are looking for a positive real t such that ||td_{1}|| ≥ 1/4 ∧ ||td_{2}|| ≥ 1/4 ∧ ||td_{3}|| ≥ 1/4 .

After perhaps rearranging d_{1}, d_{2}, d_{3} we can proceed as follows:

Let d_{1} = r_{11}4^{i11} + r_{12}4^{i12} + ... + r_{1n1}4^{i1n1} where
0 ≤ i_{11} < i_{12} < ... < i_{1n1} and r_{11}, r_{12}, ... , r_{1n1} ∈ {1,2,3}.

Let d_{2} = r_{21}4^{i21} + r_{22}4^{i22} + ... + r_{2n2}4^{i2n2} where
i_{11} ≤ i_{21} < i_{22} < ... < i_{2n2} and r_{21}, r_{22}, ... , r_{2n2} ∈ {1,2,3}.

Let d_{3} = r_{31}4^{i31} + r_{32}4^{i32} + ... + r_{3n3}4^{i3n3} where
i_{21} ≤ i_{31} < i_{32} < ... < i_{3n3} and r_{31}, r_{32}, ... , r_{3n3} ∈ {1,2,3}.

Then d_{1}4^{-i31-1} = r_{11}4^{i11-i31-1} + r_{12}4^{i12-i31-1} + ...
+ r_{1n1}4^{i1n1-i31-1} = u_{1}4^{i11-i31-1} + v_{1} where u_{1}, v_{1} are nonnegative integers
and 1 ≤ u_{1} ≤ 4^{ik1-i11+1}-1 and u_{1} is not divisible by 4.

And d_{2}4^{-i31-1} = r_{21}4^{i21-i31-1} + r_{22}4^{i22-i31-1} + ...
+ r_{2n2}4^{i2n2-i31-1} = u_{2}4^{i11-i31-1} + v_{2} where u_{2}, v_{2} are nonnegative integers
and 1 ≤ u_{2} ≤ 4^{i31-i11+1}-1.

And d_{3}4^{-i31-1} = r_{31}4^{i31-i31-1} + r_{32}4^{i32-i31-1} + ...
+ r_{3n3}4^{i3n3-i31-1} = u_{3}4^{-1} + v_{3} where u_{3}=r_{31} and v_{3} is a nonnegative integer.

We are now looking for a positive integer w such that t := w*4^{-i31-1} satisfies our conditions.

If only w is not divisible by 4, this t will satisfy ||td_{3}|| ≥ 1/4.

In order to let t satisfy ||td_{1}|| ≥ 1/4 and ||td_{2}|| ≥ 1/4 too, we have to require furthermore that the products u_{1}*w and u_{2}*w modulo
4^{i31-i11+1} belong to the interval [(4^{i31-i11}, 3*4^{i31-i11}].

It seems evident that such a w exists.

**The general case**

We are looking for a positive real t such that ||td_{1}|| ≥ 1/(k+1) ∧ ||td_{2}|| ≥ 1/(k+1) ∧ .... ∧ ||td_{k}|| ≥ 1/(k+1) .

After perhaps rearranging d_{1}, d_{2}, ... , d_{k} we can proceed as follows:

Let d_{1} = r_{11}(k+1)^{i11} + r_{12}(k+1)^{i12} + ... + r_{1n1}(k+1)^{i1n1} where
0 ≤ i_{11} < i_{12} < ... < i_{1n1} and r_{11}, r_{12}, ... , r_{1n1} ∈ {1,2,..k}.

Let d_{2} = r_{21}(k+1)^{i21} + r_{22}(k+1)^{i22} + ... + r_{2n2}(k+1)^{i2n2} where
i_{11} ≤ i_{21} < i_{22} < ... < i_{2n2} and r_{21}, r_{22}, ... , r_{2n2} ∈ {1,2,..k}.

...

Let d_{k} = r_{k1}(k+1)^{ik1} + r_{k2}(k+1)^{ik2} + ... + r_{knk}(k+1)^{iknk} where
i_{(k-1)1} ≤ i_{k1} < i_{k2} < ... < i_{knk} and r_{k1}, r_{k2}, ... , r_{knk} ∈ {1,2,..k}.

Then d_{1}(k+1)^{-ik1-1} = r_{11}(k+1)^{i11-ik1-1} + r_{12}(k+1)^{i12-ik1-1} + ...
+ r_{1n1}(k+1)^{i1n1-ik1-1} = u_{1}(k+1)^{i11-ik1-1} + v_{1} where u_{1}, v_{1} are nonnegative integers
and 1 ≤ u_{1} ≤ (k+1)^{ik1-i11+1}-1 and u_{1} is not divisible by (k+1).

And d_{2}(k+1)^{-ik1-1} = r_{21}(k+1)^{i21-ik1-1} + r_{22}(k+1)^{i22-ik1-1} + ...
+ r_{2n2}(k+1)^{i2n2-ik1-1} = u_{2}(k+1)^{i11-ik1-1} + v_{2} where u_{2}, v_{2} are nonnegative integers
and 1 ≤ u_{2} ≤ (k+1)^{ik1-i11+1}-1.

...

And d_{k}(k+1)^{-ik1-1} = r_{k1}(k+1)^{ik1-ik1-1} + r_{k2}(k+1)^{ik2-ik1-1} + ...
+ r_{knk}(k+1)^{iknk-ik1-1} = u_{k}(k+1)^{-1} + v_{k} where u_{k}=r_{k1} and v_{k} is a nonnegative integer.

We are now looking for a positive integer w such that t := w*(k+1)^{-ik1-1} satisfies our conditions.

If only w is not divisible by k+1, this t will satisfy ||td_{k}|| ≥ 1/(k+1).

In order to let t satisfy ||td_{i}|| ≥ 1/(k+1) for i=1,2,..,k-1, too, we have to require furthermore that the product u_{i}*w modulo (k+1)^{ik1-i11+1}
belong to the interval [(k+1)^{ik1-i11}, k*(k+1)^{ik1-i11}].

We have to prove that such a w exists.

**A computer program**

We may use the following Pascal program to search for an approximation of the minimal t if k is at most 100 and the k given integers are within standard pascal range:

program unsolved;

var i,j,k:integer; t:real; d:array[1..100] of integer; noggoed:boolean;

function norm(x:real):real;

begin

norm:=abs(x-round(x))

end;

begin

writeln('k?'); readln(k);

writeln('Geef ',k:3,' positieve integers');

for j:=1 to k do begin writeln('d[',j:3,']?'); readln(d[j]) end;

t:=0;

repeat

t:=t+0.00001;

noggoed:=true; i:=0;

while ((i < k) and noggoed) do

begin i:=i+1; if norm(t*d[i]) < (1/(k+1)) then noggoed:=false end;

if noggoed then

begin

writeln(t:13:5); for j:=1 to k do writeln(norm(t*d[j]):13:10);

writeln; readln

end

until noggoed

end.

**Problem:**

*Take an arbitrary positive integer n. If n is even, change it into n/2. If n is odd, change it into 3n+1. Show that if we repeat this procedure, we get 1 in the end.*

__First approach:__

If n is even, then the probability that n/2 is even is 1/2 again. We count n ---> n/2 as one step.

If n is odd, then 3n+1 is even and the probability that (3n+1)/2 is even is 1/2 again. We count n ---> 3n+1 ---> (3n+1)/2 as one step.

If we neglect the +1 in (3n+1)/2 then after k steps we expect to have n*3^{(Σ(j=0 to k) j*(k over j))/2k}/2^{k} =
n*3^{k/2}/2^{k} = n*(√3/2)^{k}, which tends to 0 if k ---> ∞ unless we get 1 at some point.

So we should get 1 in the end.