# Primes classifications [closed]

1) If $p_1$$p_2$,...,$p_k$ be different primes and m = product of primes $p_1$,$p_2$,...,$p_k$ . How to prove that, when N = $N_1$ + $N_2$+...+$N_k$, where the prime factors of $N_i$ (here i is running from 1 to k) are exactly {$p_1$,$p_2$,...,$p_k$} \ {$p_i$} for each i, then (N, m) = 1.

2) For any given integer N with (N, m) = 1, use the Chinese Remainder Theorem to determine integers $M_i$ for which N is congruent to $M_i$ (mod m), where i is from 1 to k. The prime factors of $M_i$ are exactly {$p_1$,$p_2$,...,$p_k$} \ {$p_i$} and 1 < or equal to $M_i$ less than or equal to m,for each i.

I need complete discussion on both questions by proof or generalization.

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## closed as too localized by Andres Caicedo, Ｊ. Ｍ., Asaf Karagila, Zev ChonolesOct 10 '11 at 13:38

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This sounds like homework. If it is then please tag it as such and try to explain what you have tried so far. –  Brandon Carter Oct 5 '11 at 6:35
Hint for #1: Can you show that none of the primes $p_i$ divide $N$? Agree with Brandon in the sense that irrespective of whether this is homework or you are studying by yourself, you should show what you have done. That way you get better answers, and will benefit more from them. –  Jyrki Lahtonen Oct 5 '11 at 6:38
Unfortunately, when the mathematicians are failed to answer my question, they say that my question will come under HOME WORK category. Moreover, What this man edited my question. Nothing. Whichever I have given in my post, the same he/she posted. It is very very Unfortunate...and sorry If I am wrong. –  gandhi Oct 5 '11 at 6:40
@gandhi You have to realize that this question looks like it is a copy/pasted exercise from a textbook. The reputable mathematicians here can solve it no sweat. But if it is, indeed, HW, then we would do a disservice to you, your fellow classmates and a number of other people by giving a full answer. Hinting that we cannot answer your question only serves to irritate us, and is counterproductive. Well, enough of that. What can you do with the given hints and Mose Wintner's answer? Think!! –  Jyrki Lahtonen Oct 5 '11 at 6:51
@Jyrki! I am so sorry! –  gandhi Oct 5 '11 at 11:53

For 1, consider that the only prime factors of $m$ are the $p_i$. For each $N_i$, factor $p_i$ out of the other summands in $N$ and reduce it modulo $p_i$ to conclude that $p_i$ does not divide $N$ for any $i$.
For 2, use the $N_i$ from 1.