ID  Problem  Prize  Status 

1 
Let n_1 and n_2 be prime numbers such that n_1 \lt n_2; let n be an integer greater than 1. We want to prove that there exist two multiples of n_1, which we will denote by n_1 h and n_1 k with h and k positive, such that:
This statement is part of a larger problem, which has been tackled using the dashed line theory (see Study of existence of complementary space pairs based on second order dashed lines) and it has been proved except for one point left open, highlighted within the cited page: the aim is to prove precisely this open point, thus completing the proof. This problem is part of one of our proof strategies, the one based on dashes. Completing the proof would be the first step to prove analogous properties, where the integers involved are more than two. 
40€  No solution received 
2  Let n_1, n_2 and n_3 be integers such that 1 \lt n_1 \lt n_2 \lt n_3. It is required to find a function f: \mathbb{N}^{\star} \Rightarrow \mathbb{N} such that, for each x \gt 0:
f(x)  \frac{(n_3)^2}{2} \leq \mathrm{t\_space}_{(n_1, n_2, n_3)}(x) \leq f(x) + \frac{(n_3)^2}{2} \tag{1}
where \mathrm{t\_space}_{(n_1, n_2, n_3)}(x) indicates the xth integer not divisible by neither n_1, nor n_2, nor n_3. As a starting point, the article Calculation of \mathrm{t\_space} for dashed lines of arbitrary order can be considered. The solution to this problem would solve the last open point of the cited article, with \delta = \frac{(n_3)^2}{2}.
This choice of \delta is motivated by the fact that, if (n_1, n_2, n_3) = (2, 3, 5), then \frac{(n_3)^2}{2} = \frac{25}{2} is smaller than half the width of validity interval, which in this case ranges from 5 + 1 = 6 to 7^2  1 = 48 included (so half the width is (48  6)/2 = 21). This ensures that if f(x) belongs to the validity interval, \mathrm{t\_space}_{(n_1, n_2, n_3)}(x) must belong to it, because the difference in absolute value f(x)  \mathrm{t\_space}_{(n_1, n_2, n_3)}(x) is less than or equal to \frac{(n_3)^2}{2}, which itself is less than half the width of the interval (in the worst case, f(x) can be in the middle of the interval and \mathrm{t\_space}_{(n_1, n_2, n_3)}(x) may be near one of the bounds, but will always be within the interval). In this way it could therefore be proved that there are spaces in the validity interval of the dashed line T_3 (same definition as before) based solely on the f(x) function. Clearly, if we limit ourselves to the T_3 dashed line alone, the verification of the existence of spaces in the validity interval can be performed directly, but we hope that the solutions we will receive are potentially extendable to all dashed lines T_k.

100€  No solution received 
3  Let n_1, n_2, \ldots, n_k be integers such that 1 \lt n_1 \lt n_2 \lt \ldots \lt n_k. Consider the set
S := \left\{x > 0 \mid \begin{aligned}&x \textrm{ it is not divisible by any of the integers } n_1, n_2, ..., n_k;\\& x + 1 \textrm{ is divisible by } n_1 \end{aligned}\right\}
It is required to prove that this set can be expressed as S = \{f(x)\ \ g(x) \in \{n_2 a_2 + n_3 a_3 + ... + n_k a_k\ \ a_2 \in I_2, a_3 \in I_3, ..., a_k \in I_k\}\}
where:
It is advisable to start from the proof of the Proposition LC.5 (Spaces of a dashed line of the third order that precede a dash of the first row). Proving a theorem of this type would be essential to make progress in one of our proof strategies, the one based on dashes. 
200€  No solution received 
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