Don't guess the function

$f(x)$ is $2x^{5} + x^{3} + 3x + 1$ .

Consider a random polynomial $g(x)$ of order n with whole number coefficients, given by $a_{n}$ $x^{n}$ + $a_{n-1}$ $x^{n-1}$ + .... + $a_{0}$ .

Obviously $g(1)$ will be $a_{n}$ + $a_{n-1}$ + .... + $a_{0}$ . Since all the a's are whole numbers, g(1)+1 will be greater than all a's, and lets call it $A$ .

Now lets consider $g(A)$ , which will be $a_{n}$ $A^{n}$ + $a_{n-1}$ $A^{n-1}$ + .... + $a_{0}$ . If we change its base from 10 to $A$ , then we get the number $a_{n}$ $a_{n-1}$ ... $a_{0}$ , which will be the coefficients written in order.

Here $A$ is 8, and after changing its base from 10 to 8, we get 201031 . Thus $f(x)$ is $2x^{5} + x^{3} + 3x + 1$ .

Since we have to enter only the nonzero coefficients in order, the answer is 2131 .