An algebra problem by Fatin Farhan

Algebra Level 3

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

3 solutions

Siddhartha Srivastava
Feb 9, 2014

Putting $x = 1$

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

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

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

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

Therefore Answer = $\boxed{880}$

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