It sufficies to prove that the exists $n\in\mathbb{Z}^{+}$ such that $1993^{n}\equiv 1\pmod{10000}$. Of course $1993\bot 10000$ and $\phi(10000)= 4000$, so $1993^{4000}\equiv 1\pmod{10000}$ and $4000$ is the desired $n$ which ends the proof.

i know this is a very easy question but the other ones aren't easy like this.

Rzeszut wrote: ...... such that $1993^{n}\equiv 1\pmod{1000}$. you may like to edit out a typo - should be 10000 instead of 1000.

Gyan wrote: Rzeszut wrote: ...... such that $1993^{n}\equiv 1\pmod{1000}$. you may like to edit out a typo - should be 10000 instead of 1000. You're right . I corrected it.

But if $1994 \times 1993^{n}\cong 1994 (mod 10000)$, but gcd(1994,10000) is not equal to 1. Sorry if this sounds stupid

D.P.L wrote: But if $1994 \times 1993^{n}\cong 1994 (mod 10000)$, but gcd(1994,10000) is not equal to 1. Sorry if this sounds stupid That means $1994 \cdot 1993^{n}\equiv 1994 \pmod{10000}\not\Rightarrow 1993^{n}\equiv 1 \pmod{10000}$ but we still have $1993^{n}\equiv 1 \pmod{10000}\Rightarrow 1994 \cdot 1993^{n}\equiv 1994 \pmod{10000}$.