How many natural number $\displaystyle n$ with $\displaystyle n\leq2009$ which satisfies $\displaystyle lcm(n,2009)+gcd(n,2009)=n+2009$ ?
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Originally Posted by GTK X Hunter this is........ uhmmm... 5, isn't it? How many natural number $\displaystyle n$ with $\displaystyle n\leq2009$ which satisfies $\displaystyle lcm(n,2009)+gcd(n,2009)=n+2009$ ? May be, but won't you need to explain why? CB
Originally Posted by GTK X Hunter How many natural number $\displaystyle n$ with $\displaystyle n\leq2009$ which satisfies $\displaystyle lcm(n,2009)+gcd(n,2009)=n+2009$ ? Since more errors occur at the start or end points, my guess is that you missed that 1 works as well as 2009.
no i didn't miss 1 nor 2009... the 5 natural numbers satisfying that equation are 1,7,41,287,2009 ryt???!
Originally Posted by GTK X Hunter no i didn't miss 1 nor 2009... the 5 natural numbers satisfying that equation are 1,7,41,287,2009 ryt???! Well those do, but do you not have to prove that there are no others? CB
Originally Posted by CaptainBlack Well those do, but do you not have to prove that there are no others? CB Hi CB I might be able to attempt on that Let gcd(n,2009) = x lcm(n,2009) = n2009/x Substitute and solve the quadratic, it comes in a nice form. And, this infact can be generalized. Thanks
Originally Posted by GTK X Hunter no i didn't miss 1 nor 2009... the 5 natural numbers satisfying that equation are 1,7,41,287,2009 ryt???! did u miss 49?
Originally Posted by aman_cc did u miss 49? ooouuuppsss... sorry... yeap, i miss 49, good So there are 6 natural numbers including 49 satisfying the equation ryt??
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