If the fractional part of the number $\dfrac{2^{403}}{15}$ is $\dfrac{k}{15}$, then $k$ is equal to

Question

If the fractional part of the number $\dfrac{2^{403}}{15}$ is $\dfrac{k}{15}$, then $k$ is equal to

My attempt is as follows:-

$$\dfrac{2^{13}\cdot 2^{390}}{15}$$ $$2^{13}\cdot\dfrac{\displaystyle{390\choose 0}+\displaystyle{390\choose 1}+\displaystyle{390\choose 2}+\displaystyle{390\choose 3}+\cdots\cdots+\displaystyle{390\choose 390}}{15}$$

As $390$ is divisible by $15$, so fractional part would come from $\displaystyle{390\choose 0},\displaystyle{390\choose 390}$, hence fractional part would be fractional part of $\dfrac{2^{13}\cdot2}{15}$

$$\dfrac{2^{14}}{15}=\dfrac{16384}{15}=1092+\dfrac{4}{15}$$

Hence fractional part should be $\dfrac{4}{15}$ and $k$ should be $4$ but answer is $8$. What am I missing here. I tried to find my mistake but didn't get any breakthrough.

Answer 1

Note that $2^4 = 16 \equiv 1 \pmod{15}$. So:

$$ 2^{403} = 2^{400} \cdot 2^3 = (2^4)^{100} \cdot 2^3 \equiv 1^{100} \cdot 8 \equiv 8 \pmod{15} $$