Adieu 2018

Consider the $n \times n$ matrix $A$ , where $A_{ij} \; = \; \delta_{ij} + ij \hspace{2cm} 1 \le i,j \le n$ Then we note that $A \; = \; I + \mathbf{x}\mathbf{x}^T$ where $\mathbf{x} \; = \; \left(\begin{array}{c} 1 \\ 2 \\ \vdots \\ n \end{array} \right)$ Thus we deduce that $A\mathbf{x} = \big(\Vert\mathbf{x}\Vert^2 + 1\big)\mathbf{x}$ while $A\mathbf{y} \; = \; \mathbf{y} \hspace{2cm} \mathbf{x}\cdot\mathbf{y} = 0$ If we choose an orthogonal matrix $P$ whose first column is a multiple of $\mathbf{x}$ , we deduce that $AP \; = \; P\left(\begin{array}{cccccc}\Vert\mathbf{x}\Vert^2+1 & 0 & 0 & 0 & \ldots & 0 \\ 0 & 1 & 0 & 0 & \ldots & 0 \\ 0 & 0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \ldots & 1 \end{array} \right)$ and hence $\mathrm{det}\,A \; = \; \mathrm{det}\,(P^TAP) = \Vert\mathbf{x}\Vert^2 + 1 \; = \; \tfrac16n(n+1)(2n+1) + 1$ For this problem we have $n = 2018 = 2 \times 1009$ where $1009$ is prime. It is clear that $\mathrm{det}\,A \equiv 1 \pmod{1009}$ , and we also see that $\mathrm{det}\,A \equiv 0 \pmod{2}$ . Thus we deduce that $\mathrm{det}\,A \equiv \boxed{1010} \pmod{2018}$ .

The symmetric matrix $B=[b_{ij}]$ with $b_{ij}=ij$ has rank 1, with eigenvalues 0 (with muliplicity $n-1$ ) and $\text{tr}B =\sum_{i=1}^ni^2$ . Thus the matrix $A=B+I_n$ has the eigenvalues 1 (with multiplicity $n-1$ ) and $1+\sum_{i=1}^ni^2$ so $\det A=1+\sum_{i=1}^ni^2$ . (We know that $\det A$ is the product of the eigenvalues and $\text{tr}A$ is their sum.)

In the case of $n=2018$ we find $\det A= \frac{1}{6}\times 2018\times 2019 \times 4037+1=1009 \times 673 \times 4037+1$ , which leaves a remainder of $\boxed{1010}$ .

I'm sure there is a much more elegant solution to this one. But here's my approach:

Observe that, $\det(A_n) = \text{tr}(A_n) - (n-1) = 1-n + \sum_{i=1}^{n}(1+i^2)= 1 + \sum_{i=1}^{n}i^2 = 1 + \frac{n(n+1)(2n+1)}{6}$ .

For $n = 2018$ , $\det(A_n)\text{ }\text{mod}\text{ }2018 = 1 + \frac{2018(2018+1)(2\cdot 2018+1)}{6}\text{ mod }2018 = 1 + 1009 = \boxed{1010}$ , which is the answer we seek.