coefficient problem

This is the correct way, I guess. I too did the same way. But still is there a good expansion?

Please explain your method more clearly.

I did the same way.

i did the same way great method to a great problem

https://brilliant.org/problems/find-the-areaonly-your-logic-can-help-you/?group=3UHxOzwinQpA&ref_id=384997

PLEASE TRY TO DO THIS AWSOME PROBLEM TOO..post a solution if you get......................i am waiting for an awesome solution that i made while creating this problem

Brilliant Problem And solution!

I did the same way, but missed 1, 2, 5.
$\color{#D61F06}{Let~F ( k, ~m, ~n )=(x^k-k)(x^m-m)(x^n-n)~~~and~~~F ( 0,0,r)=F(r),~ ~and ~~F(0,q,r)=F(q,r)\\ \therefore~~ F (1, . . . ,12 )~=(x-1)(x^2-2)(x^3-3)....(x^{11}-11)(x^{12}-12)\\ =x^{70+8} - ax^{77} \pm bx^{76} \pm cx^{75} \pm dx^{74} \pm ex^{73} \pm fx^{72} \pm gx^{71}\pm {\Large ?}x^{70} \pm ...........................\\ \therefore~ \dfrac{F(1, . . .12)} {F ( p,q,r)}\\ ={\Large ?}x^{70} \pm .........where~p+q+r = - 8.\\ Various~ p, q, r~ ~for~~ ~ p+q+r = - 8 ~~(integer ~composition~of~~-8~possible~here):-\\ (-8)~~~~||~~~(-1,-7),~~~~(-2,-6),~~~~(-3,-5), ~~~~||~~~(-1,-2,-5),~~~~(-1,-3,-4).\\ \therefore~~{\Large ?}= -8 +(-1*-7)+(-2*-6)+(-3*-5)+(-1*-2*-5)+(-1*-3*-4)=\Huge~~~~ \color{#D61F06}{4}. } \\$

Working with brackets of the type ( aX^p+b), a,b are signed integers , p is +tive integer..
We brake up the brackets into two groups.
First group where the product of brackets give highest power of x we desire.
Let P1 be the product of coefficient of each x term.
Second group has the rest of brackets, if any .
Product of constants of the second group be P2. If there is no second group, P2=1.
P1 * P2 is the desired coefficient. If all X term has 1 as coefficient, P1=1.
Brake up of the brackets can be in more than one path. In that case, algebraic sum of P1 * P2 from each path is the desired coefficient.
$\color{#D61F06}{ Let~us~take~an~example:-\\ (aX+p)(bX^2+q)(cX^3+r)(dX^3+s)\\ =abcdX^9+bcd*pX^8+acd*qX^7+(abc*s+abd*r+bcd*p)X^6+(bc*ps+bd*ps)X^5\\ +(ac*ps+ad*pr)X^4+(ab*rs+c*pqs)X^3+~b*prsX^2+~a*qrsX^1+~pqrs.~~~~~HOW? }\\ X^9,~only ~one~group. We~use~all~four~brackets. ~~~~ \implies~abcdX^9. \\ X^8~obtain~by~(bX^2+q)(cX^3+r)(dX^3+s)=bcdX^9 +.... ,~~and~~~(ax+p)=....+p~~\implies~~bcdpX^8. \\ X^7~obtain~just ~as ~for ~X^8. ~~by~(aX+p)(cX^3+r)(dX^3+s)=acdX^7~+ .... ~~and~~bX^2+q=....+ q~~~\implies~acdqX^7. \\ X^6~obtain~by~THREE~diffrent~paths:- \\ \ \ \ \ \ [1]~~(aX+p)(b X^2+q)(cX^3+r)=abc X^6~+....~~~and~~(dX^3+s)=....+s~~~~ \implies~~abcs X^6.\\ \ \ \ \ \ [2]~~(aX+p)(b X^2+q)(dX^3+s)=abd X^6~+....and~~(cX^3+r)=.... + r~~\implies ~~abdr X^6.\\ \ \ \ \ \ [3]~~(c X^3+r)(dX^3+s)=cd X^6+....~~~and~~(aX+p)(b X^2+q)=....+~pq~~\implies cdpq X^6.\\ \implies~~(abcs+abdr + cdpq)X^6. \\ X^5~obtain~by~TWO~diffrent~paths:- \\ \ \ \ \ \ [1]~~(b X^2+q)(cX^3+r)=bc X^5~+....~~~and~~(aX+p)(dX^3+s)=....+~ps~~\implies bcps X^5.\\ \ \ \ \ \ [2]~~(b X^2+q)(dX^3+s)=bd X^6~+....~~~and~~(aX+p)(cX^3+r)=.... +~pr~~\implies bdpr X^5.\\ \therefore ~(~~bcps~+~bdpr)X^5. \\ X^4~obtain~by~TWO~diffrent~paths. \\ {\large~First ~group},~~~1^{st}~and~3^{rd}~brackets,~~~ ~and~~~1^{st}~and~4^{th}~brackets.~~~Rest~ for~second~group.~Just~as~for ~X^5 ~above.\\ X^3~obtain~by~THREE~diffrent~paths:- \\ \ \ \ \ \ [1]~~(aX+p)(b X^2+q)=ab X^3~+....~~~and~~(cX^3+r)(dX^3+s)=.... ~+rs~~\implies abrs X^3.\\ \ \ \ \ \ [2]~~(cX^3+r)=cX^3~+.... ~~~and~~(aX+p)(b X^2+q)(dX^3+s)=.... +~pqs~~~\implies~cpqsC^3 \\ \ \ \ \ \ [3]~~(dX^3+s)=dX^3~+.... ~~~and~~(aX+p)(b X^2+q)(cX^3+r)=....+~pqr.~~~\implies~~dpqrX^3 \\ X^2~obtain~by ~ (b X^2-q)=bX^2+.... ~~~and~~~~(aX+p)(c X^3+r)(dX^3+s)=....~+prs~~\implies~ bprsX^2.\\ X~by~(aX+p)~and ~ ~~(b X^2+q)(c X^3+r)(dX^3+s), ~~So~~aqrsX.\\ Constant ~term=pqrs.\\ \text{The problem given is much simple. 1)All X coefficients are 1. }\\ \text{2)Powers of X and constants are consecutive integers. }$

Did it in the same way. Nice one