The greatest common divisor of two integers is (x+3) and their least common multiple is x(x+3), where x is a positive integer. If one of the integers is 40, what is the smallest possible value of the other one?
The divisors of 40 are
1 2 4 5 8 10 20 and 40
But x is a positive integer.. and since x + 3 divides 40......the possible values for x are
1 2 5 7 17 37
So when
x = 1 LCM = 4 ( impossible....the LCM is at least 40)
x = 2 LCM = 10
x = 5 LCM = 40
So x = 5 and the other integer is 8