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How is this fun lol

:D right nw this problem looks neat but nasty but once u see the solution u'll smile :)

Ok we will see about that

sure :D

How is it that you got so many Medals and we ain't done nothing yet

Are we gonna work it out

working cases so far

theres no solution for all different

nope there r 3 sets of solutions :)

7,3,3
1,0,0

oh wait1,0,0 not possible

7,3,3

and its permutations?
3,7,3
3,3,7

no this question is pure algebra no permutations :)

|dw:1439830819873:dw|

well i got that solution from algebra xD

the other solutions have negative numbers in them

oh my bad i see l

okay lemme look for solutions in all different case too then

ah i tried somethjing similar with +1

but got lazy and said i hope theres just no solution for all different

:D

the start which i gave is jst a partial start :) u need to find a way to get through

if it helps, there are only 3 solutions
|dw:1439831929718:dw|

yes @ganeshie8 these r the solutions :)

yes oldrin

I wasn't asking for confirmation, I'm suggesting for an actual approach

oh ok

|dw:1439833355443:dw|

yes, those follow readily from the thing I wrote above

ok what now?

Maybe considerations mod l,m,n

no just algebra its a highschool question

modular arithmetic is certainly not beyond competition high school math

xD im just saying for this question its pretty much all algebra

yes this question is all algebra :D

Oops , it should read \(2(1+l+m+n)=2(4+p+q+r)\) so$$pqr=2(4+p+q+r)$$

I got lost on the 3rd number o-o

neat trick

with the bounding

can u show me how u got to this equation again
pqr=2(p+q+r)+4

too much clutter to read through up there

^ lol gg