\[\large N = \overbrace{a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0}^{n+1 \ \text{digits}} \\ \large \Updownarrow \\ \large N= 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0\]

Let \(N\) be a positive integer as defined above.

A certain kind of positive integer \(\chi\) exists such that

\[\large \dfrac{N}{\chi} = \dfrac{\overbrace{a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0}^{n+1 \ \text{digits}}}{\chi} = 0. \ \overline{a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0}\]

What is \(\chi\)?

**Details and Assumptions:**

- \(N = \overbrace{a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0}^{n+1 \ \text{digits}}\) denotes that \(N\) is a positive integer that contains \(n+1\) digits in the order of digits: \(a_n, a_{n-1}, a_{n-2}, ... ,a_2, a_1, a_0\) from the leftmost digit to the rightmost digit (i.e. from the digit with highest place value \(a_n\) to the digit in the units place \(a_0\)).
- \(0. \ \overline{a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0}\) denotes that it is the decimal representation of the rational number \(\dfrac{N}{\chi}\), where ' \(\overline{\cdots}\) ' denotes the recurring digits in the decimal expansion. Here, these recurring digits will have the same order as that of order of digits of \(N\) from left to right.

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