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发帖数: 23432 | | m********4
发帖数: 1837 | 2 I tried and may figure it out this kind of slow way:
Lemma 1a: If a|c, b|c, and a, b are coprime, then ab|c.
Lemma 1b: lcm(a,b)*gcd(a,b)=ab.
Proof: Let m=lcm(a,b), d=gcd(a,b). Then a|m, b|m.
=> d*(a/d)|m, d*(b/d)|m.
a/d|m/d, b/d|m/d.
Because a/d and b/d are coprime by the
definition of gcd, then applying Lemma 1 we would get
(a/d)(b/d)|m/d.
=> ab/(d^2)=x*m/d
ab=xmd, x is an integer. By the
definition of lcm, x must be 1. Otherwise, we would have xmd|m, which
implies xd|1. This contradicts with x=!1.
Lemma 2a: If ab|cb, then a|c.
Lemma 2b(actually this is an important theorem):
gcd(ac, bc)=c*gcd(a,b)
Proof. Let gcd(a,b)=d, gcd(ac, bc)=d'. We need to prove
both directions of division between cd and d'.
Direction 1. cd|d' is an easier direction.
d|a, d|b => cd|ac, dc|bc. Then dc is a division number.
By the definition of gcd, we would get dc|d'.
Direction 2. d'|cd.
d'|ac, d'|bc and c|d' by the previous result. Therefore,
applying Lemma 2a we would have d'/c|a, d'/c|b, which would imply d'/c is a
division number of both a and b. Hence by the definition of gcd, d'/c|d =>d'
|cd.
Then it's straightforward to solve your question.
lcm(ac, bc)= ac*bc/gcd(ac,bc)= abcc/(c*gcd(a,b))=abc/gcd(a,b)=c*lcm(a,b)
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