Function can be tricky

The function $f(x)$ is of the form $f(x) = \displaystyle \sum_{k=0}^n a_kx^{b_k}$ , where $a_k \in \mathbb Z^+$ , $b_k$ not necessary integer except for $b_0 = 0$ .

From $f(0) = 2046$ , $\implies a_0 = 2046$ . From $f(1) = 2047$ and $a_k \in \mathbb Z^+$ , we can only have one coefficient and it must be equal to 1, that is $a_k = 1$ , therefore the function is $f(x) = x^b + 2046$ .

From $f(2048) = 2048$ , we have:

$\begin{aligned} 2048^b + 2046 & = 2048 \\ 2^{11b} & = 2^1 \\ \implies 11b & = 1 \\ b & = \frac{1}{11} \\ \implies f(x) & = x^\frac{1}{11} + 2046 \\ \implies f(177147) & = 177147^\frac{1}{11} + 2046 \\ & = (3^{11})^\frac{1}{11} + 2046 \\ & = \boxed{2049} \end{aligned}$

$f(x)=x^\frac{1}{11}+2046$

$f(0)=2046$ implies the constant term =2046

$f(1)=2047$ implies the coefficient =1(Since the function is with non-negative coefficient(s) only , it implies that $f(x)=x^k+2046$ , when $k$ is a constant

Hence, $f(2048)=2048^k+2046=2048$

$2048^k=2$

$k = \frac{1}{11}$

$f(177147) = 177147^\frac{1}{11} +2046 = 3+2046 = 2049$