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If the three digit numbers: $x17, 3y6, 12z$ where $x,y,z$ are integers from $0-9$ are divisible by a fixed constant $k,$ then the determinant

$$\left|\begin{matrix} x & 3 & 1 \\ 7 & 6 & z \\ 1 & y & 2\end{matrix}\right|$$

must be divisible by?

mfl
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Adarsh
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1 Answers1

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Adding to the second row $100$ hundred times the first one and $10$ times the third one we get:

$$\left|\begin{matrix} x & 3 & 1 \\ 7 & 6 & z \\ 1 & y & 2\end{matrix}\right|=\left|\begin{matrix}x & 3 & 1 \\ 100x+10+7 & 300+10y+6 & 100+20+z \\ 1 & y & 2\end{matrix}\right|.$$

Now the elements of the second row are all multiple of $k$ by assumption. Since all the elements in the determinant are integer numbers the determinant must be a multiple of $k.$

mfl
  • 29,399
  • Okay, got that. Thank you. Want to know how did you get to the approach of solving this problem? What was your aim? – Adarsh Jun 21 '14 at 12:39