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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?

 Feb 22, 2018
 #1
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+1

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

 Feb 22, 2018
 #2
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x is not 5.....

tertre  Feb 22, 2018
 #3
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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)

 Feb 23, 2018

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