= Solution
<Fermat's little theorem> states that for a prime $p$ and integer $a$,
$$
a^p\equiv a\pmod p;
$$
if $p\nmid a$, equivalently $a^{p-1}\equiv1\pmod p$. For the latter form, multiplication by $a$ permutes the nonzero <residue classes> modulo $p$. Multiplying the congruences $a,2a,\ldots,(p-1)a$ and cancelling $(p-1)!$ gives $a^{p-1}\equiv1$. The first form also covers $p\mid a$.
Since $803=50\cdot16+3$, the theorem gives
$$
3^{803}\equiv3^3=27\equiv\boxed{10}\pmod{17}.
$$
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