A problem on polynomial Let $p(x)$ be polynomial of degree $10$. It is given that $p(0)=1$ and $p(x)=x$ for $x=1$ to $10$. What is the value of $p(11)$. 
I am able to show that it is certainly greater than or equal to $10$, but unable to find the exact value.  
 A: well, you may easily factor $p(x)-x$, and from this you should not have many troubles to find the result. (Remember that the value in 0 is used to find the value of the multiplicative constant, after factorizing $p(x)-x$)
A: Hint: given the degree, these values specify uniquely the polynomial $p$. If you don't see why (and want to use "heavy machinery"), look at Lagrange interpolating polynomials; otherwise, just consider the fact that any degree-$d$ polynomial has at most $d$ roots, and that you can easily build an explicit degree-$10$ polynomial $q$ such that $\Delta=p-q$ (polynomial with degree at most $10$) is zero on $0,1,\dots,10$. Then prove with the above remark that $\Delta\equiv 0$, and conclude that $p=q$.
Then compute $q(11)$.
A: Consider the polynomial $q(x):=p(x)-x$. This will likewise be of degree $10$ (why?), and it can readily be seen that $1,2,...,10$ are roots of this polynomial. They are in fact the only roots of the polynomial, and each has multiplicity $1$ (why?). Hence, $$q(x)=\alpha(x-1)(x-2)\cdots(x-10)$$ for some nonzero constant $\alpha$. Use the fact that $q(0)=1$ to find $\alpha$. Finding $q(11)$ will then be straightforward, and from there we find $p(11)$.
