KVPY 2015 6

Let $\space p(x) = \displaystyle \sum_{k=0}^m a_kx^k$ , then we have:

$\begin{aligned} p(x) & = a_m x^m + a_{m-1}x^{m-1} +a_{m-2}x^{m-2} +... + a_2x^2 + a_1x + a_0 \\ p'(x) & = m a_m x^{m-1} + (m-1)a_{m-1}x^{m-2} +(m-2)a_{m-2}x^{m-3} +... + 3a_3x^2 + 2a_2x + a_1 \\ \Rightarrow p(x) - p'(x) & = a_mx^m + (a_{m-1}-m a_m) x^{m-1} + (a_{m-2}-(m-1)a_{m-1})x^{m-2} \\ & \quad +(a_{m-3}-(m-2)a_{m-2})x^{m-3} + ... + (a_2-3a_3)x^2 + (a_1-2a_2)x + (a_0-a_1) \\ & = x^n \end{aligned}$

Equating the coefficients, we have:

$\begin{aligned} m & = n \\ a_m & = a_n = 1 \\ a_{n-1} - na_n & = 0 \quad \Rightarrow a_{n-1} = n \\ a_{n-2} & = (n-1)a_{n-1} = n(n-1) \\ a_{n-3} & = (n-2)a_{n-2} = n(n-1)(n-2) \\ ... & \quad ... \\ a_3 & = 4a_4 = n(n-1)(n-2)...(6)(5)(4) \\ a_2 & = 3a_3 = n(n-1)(n-2)...(5)(4)(3) \\ a_1 & = 2a_2 = n(n-1)(n-2)...(4)(3)(2) = n! \\ a_0 & = a_1 = n! \end{aligned}$

We note that $\space p(0) = a_0 = \boxed{n!}$

Let $D$ be the differential operator $\dfrac d{dx}$
Then, $(1-D)p=x^n$
$p=\dfrac 1{1-D}(x^n)$
$=(1+D+D^2+...)(x^n)$
$=x^n+nx^{n-1}+...+n!$
Terms other than constant term vanishes when $x=0$ is plugged in.
Answer is $n!$

My approach was calculus based. I took p(x)= y ; p'(x) = dy/dx. The given equation then reduces to a first order differential equation.

Follow standard solution approach thereafter. Do note that since p(0) = p'(0) , we infer that the constant term of the polynomial is zero.

Objective approach to accelerate calculations --> Note that putting n = 1 would be futile since one or more options matches for it. Put n = 2 and proceed to find the answer. You shall obtain 2, which is equal to 2!. Hence answer will be n!