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.

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TopNewestIf \(19^n \) 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 \( 100 \).

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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.

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