Mystery polynomial

Note that, because $f(1)=7$ , either $f(x)=7x^n$ for some $n$ , or all of the coefficients are less than $7$ . We can immediately check that the first option is impossible — if $f(x)=7x^n$ , then $f(7)=7^{n+1}$ , and we see that $f(7)=7597$ is not a power of $7$ . Thus, all of the coefficients are less than $7$ .

Writing $f(7)$ in base $7$ : $f(7)=7597=31102_7$ In other words: $f(7)=7597=3(7^4)+1(7^3)+1(7^2)+0(7^1)+2(7^0)$ We see that $f(x)=3x^4+x^3+x^2+2$ satisfies the conditions. (It is easy to check that $f(1)=7$ .) Since all of the coefficients are less than $7$ , and every number has only one base $7$ representation, we see this is the only solution. Thus: $\boxed{f(x)=3x^4+x^3+x^2+2}$ And thus, $f(10)=31102$ .

I liked the solution by Akiva Weinberger. My solution is here:

We know that the general polynomial is

$f(x) = { a }_{ n }{ x }^{ n } + { a }_{ n - 1 }{ x }^{ n - 1 } + ... + { a }_{ 1 }x + { a }_{ 0 }$

We are given that

$f(1) = { a }_{ n } + { a }_{ n - 1 } +...+ { a }_{ 1 } + { a }_{ 0 } = 7$

and that

$f(7) = { a }_{ n }7^{ n } + { a }_{ n - 1 }7^{ n - 1 } + ... + 7{ a }_{ 1 } + { a }_{ 0 } = 7597$ .

Now we have to look for a polynomial such that ${ 7 }^{ n } \le 7597$ and this is only possible with $n$ at most $4$ .

Thus, we have

$f(7) = 2401{ a } + 343b + 49c + 7d + e = 7597$ .

Comparing and applying simple calculations yield

$a = 3$ ; $b = 1$ ; $c = 1$ ; $d = 0$ ; $e = 2$ satisfying

$f(1) = 3 + 1 + 1 + 0 + 2 = 7$ .

Hence, $f(x) = 3{ x }^{ 4 } + { x }^{ 3 } + { x }^{ 2 } + 2$ .

Giving $f(10)= \boxed{31102}$ , the desired result.