When the least common multiple of two positive integers is divided by their greatest common divisor, the result is 33. If one integer is 45, what is the smallest possible value of the other integer?

notsmart2.0 Mar 16, 2020

#2**+1 **

The least common multiple will be 3*3*5*x

The highest common factor is some factor of 45

so

\(\frac{3*3*5* \boxed{x} }{\text{some factor of 45}}=\frac{33}{1}\\ \frac{3*3*5*\boxed{11}}{\boxed{3*5}}=\frac{3*11}{1}\)

So the least common multiple is 3*3*5*11

The highest common factor is 3*5

so they are both multiples of 15 and ours must be a multiple of 11 as well ... 3*5*11=165

So I think the two numbers are 45 and **165**

Melody Mar 16, 2020