# Not just a single number, but a kind of number

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

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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