An algebra problem by Fatin Farhan

Algebra Level 3

x ( x + 1 ) ( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) ( x + 6 ) ( x + 7 ) ( x + 8 ) x(x+1)(x+2)(x+3)(x+4)(x+5)(x+6)(x+7)(x+8) = a x 9 + b x 8 + c x 7 + d x 6 + e x 5 + f x 4 + g x 3 + h x 2 + i x =ax^9+bx^8+cx^7+dx^6+ex^5+fx^4+gx^3+hx^2+ix . What is the last 3 3 digits of ( a + b + c + d + e + f + g + h + i ) (a+b+c+d+e+f+g+h+i) ?


The answer is 880.

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3 solutions

Putting x = 1 x = 1

a ( 1 ) 9 + b ( 1 ) 8 + c ( 1 ) 7 + d ( 1 ) 6 + e ( 1 ) 5 + f ( 1 ) 4 + g ( 1 ) 3 + h ( 1 ) 2 + i ( 1 ) a(1)^9 + b(1)^8 + c(1)^7+ d(1)^6 + e(1)^5 + f(1)^4 + g(1)^3 + h(1)^2 + i(1)

= 1 ( 1 + 1 ) ( 1 + 2 ) ( 1 + 3 ) ( 1 + 4 ) ( 1 + 5 ) ( 1 + 6 ) ( 1 + 7 ) ( 1 + 8 ) = 1(1 + 1)(1 + 2)(1 + 3)(1 + 4)(1 + 5)(1 + 6)(1+ 7)(1 + 8)

a + b + c + d + e + f + g + h + i = 9 ! \Rightarrow a + b + c + d + e + f + g + h + i = 9!

a + b + c + d + e + f + g + h + i = 362880 \Rightarrow a + b + c + d + e + f + g + h + i = 362880

Therefore Answer = 880 \boxed{880}

Christian Daang
Dec 29, 2014

Set x = 1 , the equation will be just 9! = a+b+c+d+e+f+g+h+i = 362880 So, the last 3 digit is 880. :)

Sergio Rodriguez
Jan 6, 2015

usando el teorema del binomio la suma de los coeficientes da 362880

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