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

We want the fractional part of $$\dfrac{2^{403}}{15}$$. By definition, if we divide $$2^{403}$$ by $$15$$ we can write

$$2^{403}=15q+k$$

where $$q$$ is the quotient and $$k$$ is the remainder satisfying $$0\le k<15$$. Then

$$\dfrac{2^{403}}{15}=q+\dfrac{k}{15},$$

so the fractional part is exactly $$\dfrac{k}{15}$$. Our task is therefore to find the remainder $$k$$ when $$2^{403}$$ is divided by $$15$$; symbolically, we must evaluate

$$k\equiv 2^{403}\pmod{15}.$$

To do this efficiently we recall Euler’s theorem. First we note that $$\gcd(2,15)=1$$, so Euler’s theorem applies. The theorem states:

$$\text{If }\gcd(a,n)=1,\text{ then }a^{\phi(n)}\equiv1\pmod n.$$

Here $$n=15$$, so we compute Euler’s totient function $$\phi(15)$$. Since $$15=3\cdot5$$ with both factors prime, we have

$$\phi(15)=15\left(1-\dfrac13\right)\left(1-\dfrac15\right)=15\cdot\dfrac23\cdot\dfrac45=8.$$

Therefore, by Euler’s theorem,

$$2^{8}\equiv1\pmod{15}.$$

Now we use this congruence to break the large exponent $$403$$ into manageable parts. Write the exponent in the form

$$403=8\cdot50+3.$$

So,

$$2^{403}=2^{8\cdot50+3}=2^{8\cdot50}\cdot2^{3}=\left(2^{8}\right)^{50}\cdot2^{3}.$$

Since $$2^{8}\equiv1\pmod{15},$$ raising it to any power keeps it congruent to $$1$$:

$$\left(2^{8}\right)^{50}\equiv1^{50}\equiv1\pmod{15}.$$

Hence,

$$2^{403}\equiv1\cdot2^{3}\pmod{15}.$$

We still need $$2^{3}$$, and that is straightforward:

$$2^{3}=8.$$

So we finally have

$$2^{403}\equiv8\pmod{15}.$$

Comparing this with our earlier definition $$2^{403}=15q+k,$$ we see the remainder is

$$k=8.$$

Because $$0\le k<15,$$ this value is valid. The fractional part of $$\dfrac{2^{403}}{15}$$ is therefore $$\dfrac{8}{15},$$ and the asked integer is $$k=8$$.

Hence, the correct answer is Option C.