Lời giải chi tiết:

Theo bài ra, ta có:

\(a\) chia \(8\) dư \(5\)\( \Rightarrow a - 5\,\, \vdots \,\,8\)

\(a\) chia \(10\) dư \(7\)\( \Rightarrow a - 7\,\, \vdots \,\,10\)

\(a\) chia \(15\) dư \(12\)\( \Rightarrow a - 12\,\, \vdots \,\,15\)

\(a\) chia \(20\) dư \(17\)\( \Rightarrow a - 17\,\, \vdots \,\,20\)

\( \Rightarrow \left\{ \begin{array}{l}a - 5\,\, \vdots \,\,8\\a - 7\,\, \vdots \,\,10\\a - 12\,\, \vdots \,\,15\\a - 17\,\, \vdots \,\,20\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a - 5 + 8\,\, \vdots \,\,8\\a - 7 + 10\,\, \vdots \,\,10\\a - 12 + 15\,\, \vdots \,\,15\\a - 17 + 20\,\, \vdots \,\,20\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a + 3\,\, \vdots \,\,8\\a + 3\,\, \vdots \,\,10\\a + 3\,\, \vdots \,\,15\\a + 3\,\, \vdots \,\,20\end{array} \right.\,\)

\( \Rightarrow a + 3 \in BC\left( {8;10;15;20} \right)\)

Có: \(8 = {2^3}{;^{}}10 = 2.5{;^{}}15 = 3.5{;^{}}20 = {2^2}.5\)

\(\begin{array}{l} \Rightarrow BCNN\left( {8;10;15;20} \right) = {2^3}.3.5 = 120\\ \Rightarrow a + 3 \in B\left( {120} \right) \Rightarrow a + 3\,\, \vdots \,\,120\\ \Rightarrow a + 3 + 120\,\, \vdots \,\,120 \Rightarrow a + 123\,\, \vdots \,\,120\end{array}\)

Ta lại có:  \(\left. \begin{array}{l}a \vdots 41\\123 \vdots 41\end{array} \right\} \Rightarrow a + 123\,\, \vdots \,\,41\)

Suy ra, \(\left\{ \begin{array}{l}a + 123\,\, \vdots \,\,120\\a + 123\,\, \vdots \,\,41\end{array} \right. \Rightarrow a + 123 \in BC\left( {120;\,\,41} \right)\)

\(\begin{array}{l} \Rightarrow a + 123 \in BCNN\left( {120;41} \right) = B\left( {4920} \right)\\ \Rightarrow a + 123 = 4920k\,\,\,\left( {k \in \mathbb{N}} \right)\\ \Rightarrow a = 4920k - 123\,\,\,\left( {k \in \mathbb{N}} \right)\end{array}\)

Vậy \(a = 4920k - 123\,\,\,\left( {k \in \mathbb{N}} \right)\).

Chọn D.