Gonitzoggoβ
This is a proof based problem added for learning purposes and does not accept submissions.
You can view the solution by clicking on the solution tab.
You can view the solution by clicking on the solution tab.
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Consider sets $A_0,A_1,\dots A_6$.
Let $A_i=\{x : x\in \mathbb{N},\space 1\leq x\leq 2023, x\equiv i \text{(mod 7)}\}$
Each set has $\frac{2023}{7}=289$ elements.
Form pairs $(x,y)$ such that
$x\in A_1, y\in A_6$
$x\in A_2, y\in A_5$
$x\in A_3, y\in A_4$
As each set has $289$ elements, we can form $289$ pairs from each pair of sets. So, we can form $3\cdot 289=867$ pairs.