The greatest common divisor of two positive integers is $(x+5)$ and their least common multiple is $x(x+5)$, where $x$ is a positive integer. If one of the integers is 50, what is the smallest possible value of the other one?
I know that the greatest common divisor (GCD) of two numbers divides both of them, and the least common multiple (LCM) is a multiple of both of them. So, if the GCD is $(x+5)$, then both numbers must be multiples of $(x+5)$. That means that 50 is a multiple of $(x+5)$, so $(x+5)$ must be a factor of 50. The factors of 50 are 1, 2, 5, 10, 25, and 50. I can try each of these values and see if they work for $x$. If $(x+5) = 1$, then $x = -4$, which is not a positive integer, so that's not possible.
[*: If $(x+5) = 2$, then $x = -3$, which is also not positive, so that's also not possible. If $(x+5) = 5$, then $x = 0$, which is not positive either, so that's not possible. If $(x+5) = 10$, then $x = 5$, which is positive, so that's a possible value of $x$. If $(x+5) = 25$, then $x = 20$, which is also positive, so that's another possible value of $x$. If $(x+5) = 50$, then $x = 45$, which is also positive, so that's another possible value of $x$. So, I have three possible values of $x$: 5, 20, and 45. To find the smallest possible value of the other integer, I need to minimize the LCM of 50 and the other integer. I know that the LCM is $x(x+5)$, so I want to minimize $x(x+5)$. I can see that $x(x+5)$ is an increasing function for positive $x$, so it will be smallest when $x$ is smallest. Therefore, the smallest possible value of $x$ is 5. If $x = 5$, then the GCD is $(x+5) = 10$ and the LCM is $x(x+5) = 5 \times 10 = 50$. That means that the other integer must be a multiple of 10 and a factor of 50. The smallest such number is 10. So, the smallest possible value of the other integer is 10. 