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I have 3 intervals :

$I = \{1, 4, 7, 10, 13, 16, 19, ...\}$ defined by $1 + 3k$, $k \in \mathbb{N}$

$J = \{1, 5, 9, 13, 17, 21, 25, ...\}$ defined by $1 + 4k$, $k \in \mathbb{N}$

$K = \{1, 6, 11, 16, 21, 26, 31, ...\}$ defined by $1 + 5k$, $k \in \mathbb{N}$

And I have to find numbers common for the 3 intervals. So, I made an algorithm to find them, which is working fine. But I wonder if it is possible to find these numbers by mathematical way ?

I tried to draw these intervals as straight line and to find intersection points of these lines, but, of course, it's not working and I don't have any other idea.

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  • $\begingroup$ Just a short hint: Think about remainders and modular arithmetic. $3$, $4$ and $5$ are pairwise coprime, meaning that no two of them share a prime factor. Observe that $3 \cdot 4 = 12$, and $I \cap J = \{ 1, 13, 25, \ldots \}$. $\endgroup$
    – m_l
    Jul 29, 2013 at 7:31
  • $\begingroup$ You can actually find some numbers in the intersection easily and make a guess. Then, go check the Chinese Remainder Theorem at en.wikipedia.org/wiki/Chinese_remainder_theorem $\endgroup$ Jul 29, 2013 at 7:33
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    $\begingroup$ The math behind this is the Chinese Remainder Theorem $\endgroup$ Jul 29, 2013 at 7:34
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    $\begingroup$ Those aren't intervals. $\endgroup$ Jul 29, 2013 at 7:36
  • $\begingroup$ Thanks for the Chinese remainder theorem. It helps me a lot. $\endgroup$ Jul 29, 2013 at 7:48

1 Answer 1

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All the numbers $1 + 60k$, $k \in \mathbb{N}$

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  • $\begingroup$ It's false because 61 is all intervals. $\endgroup$ Jul 29, 2013 at 7:45
  • $\begingroup$ It should be $1+60k$ ($k\in \mathbb{N}$). And $60 = 3\times 4 \times 5$. $\endgroup$
    – user37238
    Jul 29, 2013 at 7:55
  • $\begingroup$ I am sorry I thought 1+2K was there to $\endgroup$ Jul 29, 2013 at 8:28

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