Greatest common divisor

Number Theory Level pending

Let a , b , a n d c a, b, and \ c be positive integers such that gcd ( a , b c ) = 1 \gcd(a,bc)=1 .

What can I say about gcd ( a , b ) a n d gcd ( a , c ) \gcd(a,b) \ and \gcd(a,c) = ?

g c d ( a , c ) = 1 gcd(a,c)=1 only g c d ( a , b ) = 1 gcd(a,b)=1 only g c d ( a , b ) = 1 a n d g c d ( a , c ) gcd(a,b)=1 \ and \ gcd(a,c) = 1 1 none of the above

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2 solutions

Hana Wehbi
May 29, 2016

Since g c d ( a , b c ) = 1 gcd(a,bc)=1 , then there is no prime that divides both a a n d b c a \ and\ bc .

This means that there is no prime that divides both a a and b b or both a a and c c .

So, if there is no prime that divides both a a and b b or both a a and c c \implies gcd(a,c)=1 and gcd(a,b)=1

展豪 張
May 29, 2016

gcd ( a , b ) , gcd ( a , c ) gcd ( a , b c ) \gcd(a,b),\gcd(a,c)|\gcd(a,bc)
gcd ( a , b ) = gcd ( a , c ) = 1 \therefore\gcd(a,b)=\gcd(a,c)=1

Since, we don't know what are the values of a , b , c a, b, c . It is a risk to approach it that way? We need "divides" in our solution, true.

Hana Wehbi - 5 years ago

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Sorry I don't understand what you say. Can you elaborate it further?

展豪 張 - 5 years ago

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