If $P(x)$ be a polynomial of degree $2015$ with leading coefficient $1$ such that $P(0)=2014, P(1)=2013, P(2)=2012,...,P(2014)=0$ and $P(2015)=n!-a$, where $n$ and $a$ are natural numbers, find $(n+a)$.


Obviously the constant is $2014$. The sum of the coefficients is $2013$. And it can be observed that $P(x+1)-P(x)=1$

But then what?

  • $\begingroup$ Can it be observed that $P(x+1)-P(x)=1$ for all $x$? If that is true, then we have $P(x)=2014-x$, which contradicts the leading coefficient statement. Try interpolation formulas, those will find the polynomial that matches those points. $\endgroup$ – Simply Beautiful Art Nov 13 '16 at 14:40
  • $\begingroup$ Hmm. We haven't done interpolation formulae in class yet, so I assume this can be done without that? $\endgroup$ – Shashank Holla Nov 13 '16 at 14:42
  • $\begingroup$ This question is a tiny bit flawed: if $(n,a)$ is a solution, then so is $(n+k, (n+k)!-n!+a)$ for all $k \in \mathbb N$. So perhaps it should ask for the smallest possible value of $n+a$. $\endgroup$ – TonyK Nov 13 '16 at 15:10
  • $\begingroup$ @TonyK, It's given that the degree is 2015, so there has to be exactly one solution. If the degree weren't mentioned, we would have a minimum value $\endgroup$ – Dhanvi Sreenivasan Nov 13 '16 at 15:20
  • $\begingroup$ @Dhanvi: You have missed my point. In this case we can take e.g. $n=2025,a=2025!-2015!+1$, and we still have $P(2015)=n!-a$. $\endgroup$ – TonyK Nov 13 '16 at 15:55

We define a new function $g(x) = p(x) + x - 2014$.

Now, we know degree of $p(x)$ is 2015. Hence, degree of $g(x)$ is also 2015

Now, $g(x)$ has roots $0,1,2,3...2014$. Hence, we write $g(x)$ as

$$g(x) = x(x-1)(x-2)....(x-2014)$$ $$\implies p(x) +x - 2014 = x(x-1)(x-2)...(x-2014) $$ $$\implies p(x) = x(x-1)(x-2)....(x-2014) -x + 2014$$

Hence, $p(2015) = 2015! - 2015 + 2014 = 2015! -1 \implies n+a = 2016$

  • 1
    $\begingroup$ Brilliant! Thanks! $\endgroup$ – Shashank Holla Nov 13 '16 at 14:43
  • $\begingroup$ This might seem a bit weird, but what made you get that idea? $\endgroup$ – Shashank Holla Nov 13 '16 at 14:44
  • 1
    $\begingroup$ I was a JEE aspirant at one time :P.. I don't really have a concrete reason as to how I got it $\endgroup$ – Dhanvi Sreenivasan Nov 13 '16 at 14:46
  • 2
    $\begingroup$ @Shashank It is obvious what $p(x)$ was for $x\in\{0,1,2,\dots,2014\}$, and since it was a polynomial, he found another polynomial $2014-x$ that matched those points so that his $g(x)=0$ at all $x\in\{0,1,2,\dots,2014\}$, and since the first coefficient was $1$, he factored it simply and solve for $p(x)$. $\endgroup$ – Simply Beautiful Art Nov 13 '16 at 14:47
  • 1
    $\begingroup$ I had exactly the same idea, but @Dhanvi was too quick for me :-( $\endgroup$ – TonyK Nov 13 '16 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.