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The greatest common divisor of two integers is (x+8) and their least common multiple is x(x+8), where x is a positive integer. If one of the integers is 40, what is the smallest possible value of the other one?

 Dec 3, 2020
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x + 8 = 40


x =32


x(x + 8) =32(32 + 8) =1280 - the 2nd smallest number.

 

LCM(1280, 40) =1280

GCD(1280, 40) = 40

 Dec 3, 2020

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