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$f(x) - 10$ is a 10th degree polynomial that has roots $1, 2, 3, \ldots, 10$ . Since the roots are symmetric over the line $x = 5.5$ , the whole polynomial is also symmetric over the same line. (Proof below.) Thus $f(11) = f(5.5 + 5.5) = f(5.5 - 5.5) = f(0) = 10! = \boxed{3628800}$ , where the equality $f(5.5 + x) = f(5.5 - x)$ is because the symmetry.

Proof of this lemma:

Lemma : If $n$ is an even integer, and $f(x)$ is an $n$ th degree polynomial that has roots $a_1, a_2, a_3, \ldots, a_n$ that are symmetric over the line $x = b$ , then the whole polynomial is symmetric over the line $x = b$ .

Proof : It suffices to prove it for $b = 0$ , since we can consider the polynomial $f(x+b)$ with roots $a_1-b, a_2-b, \ldots, a_n-b$ that are now symmetric over the line $x = 0$ . Without loss of generality assume that $a_1 < a_2 < a_3 < \ldots < a_n$ . Then $a_1 = -a_n, a_2 = -a_{n-1}, \ldots$ .

We can factor $f(x) = c(x - a_1)(x - a_2)\ldots(x - a_n) = c(x + a_n)(x - a_n)(x + a_{n-1})(x - a_{n-1}) \ldots = c(x^2 - a_n^2)(x^2 - a_{n-1}^2) \ldots$ . Now we can clearly see that $f(x)$ is even, since the variable $x$ always appears as $x^2$ , and thus $f(-x) = f(x)$ for all $x$ . An even polynomial is just a polynomial that is symmetric over the line $x = 0$ , proving the claim.

$f(x)-10= a(x-1)(x-2)(x-3)(x-4)(x-5)(...)$

$f(x)= a(x-1)(x-2)(x-3)(x-4)(x-5)(...)+10$

Since $f(x)$ has constant term of 10!, $a*10!+10 = 10!$

Then, we have $f(11) = a(11-1)(11-2)...+10$

$f(11) = a*10!+10$

Since $a*10!+10 = 10!$ ,

$f(11) = 10! = \boxed {3628800}$

See the title, find the most unique number, try it ._.

as the polynomial is f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)...(x-10) =>10![putting x=11]