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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, 7 months 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, 7 months ago

19= 20 - 1 . when we do any even power of (20-1) it will end up with last digit 1 succeeded by 3 zeroes if n is a multiple of 50 (binomial expansion) non multiple may give 1 or 2 zeroes before 1 but multiple of 50 gives 3 zeroes . so ans is 100.

- 2 years, 4 months ago