Let $n$ $(n>1)$ be one of the factors of $20152015.$

Find the least $n$ -digit number factor of $20152015.$

The answer is 10001.

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$\color{#D61F06}{20152015} = (\color{#D61F06}{2015}\times\color{#20A900}{10000}) + \color{#D61F06}{2015}$ $\color{#D61F06}{20152015} = \color{#D61F06}{2015}(\color{#20A900}{10000} + 1)$ $\color{#D61F06}{20152015} = \color{#D61F06}{2015}\times \color{#3D99F6}{10001}$

We can conclude that $\color{#3D99F6}{10001}$ is a factor of $\color{#D61F06}{20152015}$

Knowing $n = 5$ (see proof below), we just have to proof that $\color{#20A900}{10000}$ is not a factor of this number because it is only $5$ -digit number that is less than $\color{#3D99F6}{10001}$

$\color{#20A900}{10000}$ has $10$ as a factor but $\color{#D61F06}{20152015}$ cannot have $10$ as a factor because $\color{#D61F06}{20152015}$ doesn't end with $0$

Thus, the number is $\boxed{\color{#3D99F6}{10001}}$

Proof that $n$ must be equal to $5$Note that $1 \leq n \leq 8$ because prime factor of a number cannot go beyond the number, which has length of $8.$

$n$

cannotbe $2$ because the last digit isnoteven, this also deduce $4, 6, 8.$$n$

cannotbe $3$ because 3cannotdivide $\color{#D61F06}{20152015}$ bydivisibility by 3 rule$2 + 0 + 1 + 5 + 2 + 0 + 1 + 5 = 16$ which cannot be divided by $3$$n$

canbe $5$ because the last digit is $5.$$n$

cannotbe $7$ because 3 iscannotdivide $\color{#D61F06}{20152015}$ bydivisibility by 7 rule$2015201-(2\times5) = 2015191 \rightarrow 201519-(2\times1) = 201517 \rightarrow20151-(2\times7) = 20137 \rightarrow 2013-(2\times7) = 1999 \rightarrow 199-(2\times9) = 181 \rightarrow 18-(2\times1) = 16$ which cannot be divided by $7$Thus, $n$ must equal to $5.$