Mysterious polynomial

Let the polynomial be $f(x) = a_{n}x^n + a_{n - 1}x^{n - 1} + ... + a_{1}x + a_0$ . Note that $f(1)$ is the sum of coefficients of the polynomial, so all coefficients are strictly smaller $f(1) = 12$ . Now, note that $f(12)$ gives the coefficients of the polynomial in base $12$ ! So, since $f(12) = 2080$ , converting $2080$ to base $12$ gives $1254$ , but note that since $a_3 = 1, a_2 = 2, a_1 = 5, a_0 = 4$ are all less than $10$ . Thus, $f(10)$ gives $\overline{a_3a_2a_1a_0} = 1254$ and $1254 \equiv \boxed{254} \pmod{1000}$

First note that $12^3<2080<12^4$ and since $f(x)$ has positive integral coefficients, we deduce that $\deg (f)<4$ .

If the degree is $1$ , then $f(x)=ax+b$ for $a,b\in\mathbb{N}$ but then $a\le 11$ and so $2080$ is not attainable.

If the degree is $2$ then $f(x)=ax^2+bx+c$ for $a,b,c\in\mathbb{N}$ , but then $a\le 10$ and so $2080$ is still not attainable.

We're left with $3$ and $2080$ can be attainable when $\deg (f)=3$ . We have $f(x)=ax^3+bx^2+cx+d$ for $a,b,c,d\in\mathbb{N}$ . Thus we get two equations:

$f(1)=a+b+c+d=12 ~~~\text{and}~~~ f(12)=1728a+144b+12c+d=2080$

Subtracting the first from the second, we get:

$1727a+143b+11c=2068 ~~~ \Longrightarrow ~ 157a+13b+c=188$

Since $a,b,c,d$ are positive integers, clearly $a=1$ . Since $b\le 9$ , we have $(b,c)=(2,5)$ . Substituting back into the previous equation, we have $d=4$ . And so:

$f(x)=x^3+2x^2+5x+4$

Giving our desired answer:

$f(10) = 1\fbox{254}$

I did a pretty tedious (but foolproof) method that doesn't require finding out the degree:

Notice that $f(12)-f(1) = 2068$ eliminates the constant term $a_{0}$ , and every term is of the form $a_{k}(12^{k}-1) = 11a_{k}(12^{k-1}+12^{k-2}+\ldots+12+1)$

Dividing 2068 by 11 gives 188. Since all the terms have $(12^{k-1}+12^{k-2}+\ldots+12+1)$ as a factor, $188 \equiv a_{n}+a_{n-1}+\ldots+a_{1} \equiv 8 \pmod{12}$

Hence, $a_{0} = 12-(a_{n}+a_{n-1}+\ldots+a_{1}) = 4$ .

Using the same method twice (but being careful to subtract $(a_{k}x^{k}+a_{k-1}x^{k-1}+\ldots+a_{1}x+a_{0})$ and divide by $x^{k}$ for each $f(x)$ ), we get $\boxed{a_{0} = 4, a_{1} = 5, a_{2} = 2}$ . The terms $a_{k}x^{k}$ when $x = 10$ and $k \geq 3$ are $\equiv 0 \pmod{1000}$ and so we only need to know the aforementioned coefficients.

Thus, $f(10) \equiv 100a_{2}+10a_{1}+a_{0} \equiv \boxed{254} \pmod{1000}$

Let $f(x)=a_n \cdot x^{n}+ ...+a_1 \cdot x +a_0$ . Since $deg(f) >0$ , we know $1 \leq a_i \leq 11$ For all $i$ . Now $f(12)=a_0 =2080=4$ mod $12$ , so $a_0=4$ . Similarly $\frac {f(12)-4}{12} = \frac {2076}{12} =173=5$ mod $12$ , so $a_1=5$ , and $\frac {f(12)-(5 \cdot 12 +4)}{12^2}=\frac{2016}{144}=14=2$ mod $12$ , so $a_2=2$ . Therefore, $f(10)=100a_2+10a_1+a_0$ mod $1000 =\boxed{254}$ mod $1000$ .

Side note: Doing one more coefficient gives $a_3=1$ , and since $a_3+a_2+a_1+a_0=1+2+5+4=12$ , we must have no more coefficients, so $f(x)=x^3+2x^2+5x+4$ .

Let’s start by finding the degree of the polynomial . As, $f(1) = 12$ , the sum of the coefficients is $12$ .Since each of them are positive , all the coefficients are less than $12$ . We have $f(12) = 2080$ and $12^3 = 1728$ .Then, without doubt the polynomial is a degree $3$ polynomial (otherwise one of the coefficients have to be negative or greater than $12$ ) .So, $a\cdot12^3 + b \cdot 12^2 + c \cdot 12 + d = 2080$ .

This implies that $a = 1$ and by similar argument $b = 2$ . Thus we are left with $12c + d = 64$ and $c+d = 9$ . Solving these two equations, we get $c = 5$ and $d = 4$ . Finally, we are done with coefficients .So, $f(10) = 1000 +200 + 50 + 4 = 1254$ ,hence the answer is $\boxed{254}$ .