Gcd(b,15)=c. So c|15 and c|b. Gcd (a,c)=5. So 5|c and 5|a. Gcd(a,20)=b. So b|20 and b|a So, 5|c|15. C= 5 or 15. If c=5, 5|b|20. • b=5, a=5k for every odd k • b=10, a=10k for every odd k • b=20, a=20k for every positive integers k If c=15, 15|b|20, no solution So there are only 3 general solution: (a,b,c)=(5k,5,5) for every odd k, (10k,10,5) for every odd k, (20k,20,5) for every positive integers k

Using $a=5a', b=5b', c=5c' $ . Solving we conclude that $gcd(a',4)=b'$, so $b'$ only can be ${1,2,4}$ we divide in cases. So the answers will be

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