# Determinant and Divisibility

Problem 9 from the Fifth International Internet Mathematical Olympiad for students run by the Ariel Center in Samaria (Israel) does not need for its solution but the knowledge of the definition of determinants and their basic properties.

The rows of a determinant of a $3\times 3$ matrix consist of three consecutive digits of certain three-digit numbers, all of which are divisible by $17$. Prove that the determinant is also divisible by $17$.

Solution

The rows of a determinant of a $3\times 3$ matrix consist of three consecutive digits of certain three-digit numbers, all of which are divisible by $17$. Prove that the determinant is also divisible by $17$.

### Solution

Let the determinant $D$ be

$D= \left | \begin{array}{ccc} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{array} \right | .$

Applying column operations we get successively

$10^{3}D = \left | \begin{array}{ccc} 10^{2}a_{11}&10a_{12}&a_{13} \\ 10^{2}a_{21}&10a_{22}&a_{23} \\ 10^{2}a_{31}&10a_{32}&a_{33} \end{array} \right | = \left | \begin{array}{ccc} 10^{2}a_{11}+10a_{12}+a_{13} & 10a_{12}&a_{13} \\ 10^{2}a_{21}+10a_{22}+a_{23} & 10a_{22}&a_{23} \\ 10^{2}a_{31}+10a_{32}+a_{33} & 10a_{32}&a_{33} \end{array} \right | .$

All terms in the first column are divisible by $17$ and thus, by definition, the whole determinant is divisible by $17$.

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