Divisible polynomial

Assuming we want a nonzero polynomial...

Since $f(0)$ must be a multiple of $6$ , the constant coefficient of $f(x)$ must be zero. Since $f(1)$ must be a multiple of $6$ , $f(x)$ must have a multiple of $6$ nonzero coefficients. It is easy to check that $f(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x$ satisfies the divisibility requirements, since $f(0), f(1), f(2), f(3), f(4)$ and $f(5)$ are all multiples of $6$ . It is also clear that $f(x)$ has the smallest possible degree of all polynomials that satisfy the divisibility requirement, and hence is the polynomial with the smallest value of $f(2)$ . Thus makes the answer $\boxed{126}$ .

$f(x)$ is always a multiple of 6, therefore it is always even and always a multiple of 3.

If the constant term were 1, then $f(any-even-number)$ would be odd. Therefore the constant term is $0$ .

Now consider the number of coefficients which are 1. These precede the terms which I shall call Terms, upper-case T.

Consider $f(n)$ where $n$ is odd. All the Terms are odd, therefore there must be an even number of Terms.

Consider $f(m)$ where $m)$ is $1(mod 3)$ . All powers of $m$ are also $1(mod 3)$ . The sum of the Terms must be a multiple of 3, therefore the number of Terms is a multiple of 3.

Therefore the number of Terms is a multiple of 6.

As stated in the question, f(2) encodes the coefficients when written in binary. As a shorter binary number is always less than a longer, we need to find the lowest order of polynomial to have 6 Terms. (order = the largest power in the polynomial,or the largest Term).

It can be seen that $x^6+x^5+x^4+x^3+x^2+x^1$ is the lowest order of polynomial that could possibly meet the criteria and generates $f(2)=\boxed{126}$

All other valid polynomials must include a Term $> x^6$ and therefore generate $f(2)>128$