Junior 2011/7

[tanmoy]

Consider a sequence of four integers so that the GCD of first three is the same as the LCM of the last three.How many such sequences exist so that the sum of the numbers is 2011.

[Mod edit: Topic locked. Please use search option and do not post duplicate topics. This problem was previously posted here: viewtopic.php?f=13&t=684 ]

[asif e elahi]

If $a,b,c,d$ are the integers then prove that $b=c=gcd$ of $a,b,c=lcm$ of $b,c,d$.

[tanmoy]

Please give the full solution

[asif e elahi]

Let gcd of $a,b,c$=lcm of $b,c,d=x$.So $x$ divides $b,c$ and $b,c$ divides $x$.So $b=c=x$.Again $d$ divides $x=b$.Let $b=dk$.Again $b=dk$ divides $a$.Let $a=dmk$.So $a+b+c+d=dmk+2dk+d=d(mk+2k+1)=2011$.$2011$ is a prime.So $d=1$ or 2011.If$ d=2011$,then $mk+2k+1=1$,it has no positive integer solution.
So $d=1$ and $mk+2k+1=2011$
or $k(m+2)=2010=2.3.5.67$
It has $2^{4}-2=14$ solutions.
So there are $14$ such sequences.

[Labib]

Since it is a BdMO problem, please do check twice if the problem has already been posted before.
Please also remember that all the problems of BdMO since 2011 have already been posted in the forum.
Plus all the higher secondary problems of BdMO since 2007 have been posted before.

This particular problem, as you can guess now, has already been posted before - here. Please post the solutions there. It will help all.

The 2011 BdMO problem archive can be found here.