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    Can any1 help me with following q's?

    1) Find x, y, z belonging to Z (intergers) such that 56x+63y+72z=1 [Hint: First compute (56,63) and express it in the form (56,63) = 56s +63t with s,t belonging to Z (intergers)]. Here (56,63) means the greatest common divisor of 56 and 63.

    I found that (56,63) = 7 and that -56 +63 = 7 so s=-1 and t = 1 but im not sure how to use that to solve the problem......

    2)Let n belong to N (natural numbers: 1, 2, 3.....) and a1, a2, a3 , ..., ak belong to Z. Show that if (n,ai) = 1 (i.e n and ai are coprime for all i), then (n,a1a2a3....ak) = 1

    i know that (n, ai) =1 means that there exists intergers xi, yi such that (n,ai) = n.xi + ai.yi.....

    For (1), do what quakeeem says; it's much nicer.

    For (2), I'd think of it in terms of divisibility. (n,a_1a_2a_3....a_k) = 1 means n and a_1a_2...a_k are coprime. Prove the contrapositive; suppose a_1a_2...a_k and n are not coprime. Then for some prime p, p divides a_1a_2...a_k and n. Can (n,a_i) = 1 be true now for all i?

    For (1), what is (7,72)? Calculate THIS, and then do the same work as before, but substitute in 7 = 63 - 56 at the end.

    For (2), what Hathlan says is correct, there's a big theorem that does that one for you. If you want to do it more colloquially though, if (n,ai)=1 for all i, then n shares no prime factors with ai, and therefore also doesn't share any prime factors with a1a2.....ak, so therefore they are coprime.
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