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I have a trouble understanding p.7 of the following article: http://www.edb.gov.hk/attachment/en/curriculum-development/kla/ma/IMO/Nov20155-4online.pdf

which says the folllowing:

By the same reasoning as before (the chinese remainder theorem), we know that there exists an integer $a$ such that $n=m_1m_2\cdots m_9+a, 2m_1m_2\cdots m_9 +a, 3m_1m_2\cdots m_9 +a, \cdots$ all make $n+1, n+2, \cdots ,n+9$ composite.

I didn't understand why such $a$ exist. Why the same argument still holds for an "infinite system of congruence" as above? Thanks for help.

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  • $\begingroup$ Doesn't look like an infinite system to me. $\endgroup$
    – Dustan Levenstein
    Dec 27, 2015 at 16:40
  • $\begingroup$ It looks like a typo tbh. They tried to expand the meaning of a system of congruences. $\endgroup$
    – Dustan Levenstein
    Dec 27, 2015 at 16:48

1 Answer 1

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We can make $a+i \equiv 0 \pmod{m_i}$, ($ 1 \le i\le 9$) since it is given in the paper that $m_i$ are pairwise coprime.

Then, we have $n+i=m_1m_2 \cdots m_9 + (a+i) \equiv 0 \pmod{m_i}$, and by letting $a$ large enough we can avoid the bad cases where $n+i = m_i = \text{some prime}$.

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