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The sum of two numbers is $2000$ and thier LCM is $21879.$ Find the numbers.

My attempt: Let two numbers be $x, 2000-x.$ Product of two numbers is equal to the product of their lcm and hcf. So, $x(2000-x)=21879*hcf.$

Now we have two variables and one equation. So I am stuck.

But the book simply considers $x(2000-x)=21879,$ thereby $x=1989,11.$

My question is - Is it implied anywhere in the question that hcf is $1$ or is the question incomplete?

More impotantly, can we have any other pair of numbers whose sum is $2000$ and lcm is $21879 (hcf <> 1)?$

Willie Wong
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aarbee
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  • http://answers.yahoo.com/question/index?qid=20080702042039AAbPfBX – lab bhattacharjee Oct 19 '13 at 13:38
  • I am feeling a little embarassed because even after reading the detailed answer on that site, I am not confident enough to answer similar questions. I don't know what's wrong with my understanding. – aarbee Oct 19 '13 at 13:55

1 Answers1

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Note that $21879$ and $2000$ are coprime. So this means that $x$ and $2000-x$ are coprime (simply because their gcd must divide $x+(2000-x)=2000$).

Pablo Rotondo
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  • I wish I could have a detailed answer. I am still not able to see clearly and confidently. – aarbee Oct 19 '13 at 13:53
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    Let $d={\text{gcd}}(x,2000-x)$, we saw that $d|2000$, but also, since $x$ divides ${\text{lcm}}(x,2000-x) = 21879$, we have that $d | 21879$ too. But $21879$ and $2000$ are coprime, hence $d = 1$ since $d$ divides both of them.

    Now remember that we have ${\text{lcm}}(a,b) = \frac{a \cdot b}{{\text{gcd}}(a,b)}$.

     – Pablo Rotondo Oct 19 '13 at 13:56
  • oh, that's beautiful! really thanks! So gcd is not just common factor for a and b. it's also common factor for a*b and a+b. Now that I am writing it, it's appearing too obvious to me. I am sure I won't stuck on such thing in future now. thanks. – aarbee Oct 19 '13 at 14:01