Solution b4a004b0-7f69-49fa-baa6-627585022842

Points: 2 pts

Lemma 1: We can arrange the values in a grid and simplify.

Let $S=\sum_{i=1}^{2025}\lfloor\sqrt{i}\rfloor$.

\GridStep{\RawGrid}{ \textbf{Original terms}\par\smallskip
Fill a $45\times45$ grid in successive \mbox{L-shaped} layers.
The dots represent omitted rows and columns.
}

\GridArrow

\GridStep{\IntegerGrid}{ \textbf{Evaluate the floors}\par\smallskip
The first column becomes $1,2,3,\ldots,45$.
}

\GridArrow

\GridStep{\RearrangedGrid}{ \textbf{Shift the first column}\par\smallskip
Move each entry down one row and insert $0$ at the top.
The entry $45$ is left over.
}

Lemma 2: We can sum over all values.

For $1\leq k\leq44$, the layer of $k$'s in the rearranged grid contains
$(k+1)^2-k^2=2k+1$ cells. Hence
\[
S-45
=1(3)+2(5)+\cdots+44(89)
=\sum_{k=1}^{44}k(2k+1).
\]
Using the summation identities, we find
\[
\begin{aligned}
S-45
&=2\sum_{k=1}^{44}k^2+\sum_{k=1}^{44}k\\[2mm]
&=2\cdot\frac{44\cdot45\cdot89}{6}
  +\frac{44\cdot45}{2}\\[2mm]
&=58\,740+990\equiv0\pmod{10}.
\end{aligned}
\]
Therefore,
\[
S\equiv45\equiv5\pmod{10}.
\]

Final answer:
5

Notation definitions (retained for meaning):
\newcommand{\RawGrid}{\includegraphics{grid_raw.pdf}}
\newcommand{\IntegerGrid}{\includegraphics{grid_integer.pdf}}
\newcommand{\RearrangedGrid}{\includegraphics{grid_rearranged.pdf}}
\newcommand{\GridStep}[2]{  \par\noindent
  \begin{minipage}[c]{7.5cm}
    #1
  \end{minipage}\hfill
  \begin{minipage}[c]{5.5cm}
    \raggedright
    #2
  \end{minipage}
  \par
}
\newcommand{\GridArrow}{  \par\vspace{1mm}\noindent
  \makebox[6.25cm][c]{$\Downarrow$}  \par\vspace{1mm}
}