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# Nine teens

The last three digits of $$19^{n}$$is $$001$$. Define the $\color{BLUE}{\huge{second}}$ smallest solution for $$n$$, or else to prove that it is impossible.

Note by Bryan Lee Shi Yang
2 years ago

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If $$19^n$$ ends in $$001$$, then $$19^n - 1$$ ends in $$000$$,i.e, $$19^n - 1$$ is divisible by $$1000$$. Or simply, $$19^n \equiv 1 \pmod{1000}$$. By Euler's theorem, $$19^{\phi(1000)} \equiv 1 \pmod{1000}$$ gives us an upper bound of $$\phi(1000) = 400$$.

We can do better though. We see that $$19^2 \equiv 3^2 \equiv 1 \pmod{8}$$. Also, $$19^{\phi(125)} \equiv 19^{100} \equiv 1 \pmod{125}$$. So we get a smaller solution $$100$$.

At this point, I can't find a better solution. Wolfram Alpha gives the smallest solution as $$n = 50k$$ where $$k$$ is an integer, so the answer to your question is $$100$$. · 2 years ago