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If I have:
\begin{align*} N &\equiv 1 &\pmod{2}\\ N &\equiv 2 &\pmod{3}\\ N &\equiv 3 &\pmod{4}\\ &\vdots\\ N &\equiv n - 1 &\pmod{n} \end{align*} How could I solve for $N$? Is there any property relates to this problem?

Update
Base on Moron hint, we have:
\begin{align*} N + 1 &\equiv 0 &\pmod{2}\\ N + 1 &\equiv 0 &\pmod{3}\\ N + 1 &\equiv 0 &\pmod{4}\\ \vdots\\ N + 1 &\equiv 0 &\pmod{n} \end{align*} Hence, $N + 1 \equiv 0 \pmod{\mathrm{lcm}(2\cdot 3\cdots n}$

$\therefore N + 1 = lcm(2.3.4...n) * k \text{ for some integers k } $ $\implies N = lcm(2.3.4...n) * k - 1$

Does it look right?

Thanks,
Chan

  • 0
    @Chan: Yes, the result is correct now.2011-02-28

1 Answers 1

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Hint: Consider the possible values for $N+1$.

  • 0
    I got it ;) So simple. Many thanks.2011-02-28