If gcd(a,b)=d\gcd(a, b) = dgcd(a,b)=d, then which statement is ALWAYS true about gcd(2a+3b,5a−2b)\gcd(2a + 3b, 5a - 2b)gcd(2a+3b,5a−2b)?
gcd(2a+3b,5a−2b)=d\gcd(2a + 3b, 5a - 2b) = dgcd(2a+3b,5a−2b)=d
gcd(2a+3b,5a−2b)=2d\gcd(2a + 3b, 5a - 2b) = 2dgcd(2a+3b,5a−2b)=2d
gcd(2a+3b,5a−2b)\gcd(2a + 3b, 5a - 2b)gcd(2a+3b,5a−2b) is a multiple of ddd
Cannot be determined without knowing aaa and bbb