+884 The greatest common divisor of two 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?
GCD{a, 50} = x + 5
LCM{a, 50} =x(x + 5)
One simple solution is:
a = 10, and x = 5, since:
GCD{10, 50} =10, and x + 5 = 10, therefore x=5
LCM{10, 50} =50, and x(x + 5) = 50 =x^2 + 5x, therefore x = 5
+118608 Sorry Tetre but I agree with guest. x=5
but the question asks what is the smallest value of the other interger is and that is the pronumeral 'a'
The smallest possible value of a is 10
This is how I did it.
The lowest common multiple of 2 numbers a and b is \(\frac{ab}{HCF(a,b)}\)
SO we have
\(GCD(a, 50) = x + 5\\ LCM(a, 50) =x(x + 5)=\frac{50a}{x+5}\\ x(x + 5)=\frac{50a}{x+5}\\ x(x+5)^2=50a\\ a=\frac{x(x+5)^2}{50}\qquad \text{where a and b } \in Z \ge 1\\ \)
I want the smallest positive integral value of a and I can see that as x isncreases a will increase
So I want the smallest possible value of x
Straight off I can see that x must be a multiple of 5
Try x=5 a=5*100/50 = 10 which is an integer
so the smallest value of the second number is 10
(Just as our guest already found)
