3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set $A=\left[\frac{7}{8}\right]+\left[\frac{7^{2}}{8}\right]+\cdots+\left[\frac{7^{2016}}{8}\right]$, then the remainder when $A$ is divided by 50 is . $\qquad$
Solve
$7^{n} \equiv(-1)^{n}(\bmod 8)$, so $\left[\frac{7^{k}}{8}\right]=\left\{\begin{array}{l}\frac{7^{k}+1}{8}-1, k \text { is odd, } \\ \frac{7^{k}-1}{8}, k \text { is even. }\end{array}\right.$
Then $A=\frac{7+7^{2}+\cdots+7^{2016}}{8}-1008=\frac{7\left(7^{2016}-1\right)}{48}-1008=\frac{7\left(49^{1008}-1\right)}{48}-1008$ $=\frac{7\left(49^{504}+1\right)\left(49^{252}+1\right)\left(49^{126}+1\right)\left(49^{63}+1\right)\left(49^{63}-1\right)}{48}-1008$,
Since $48\left|49^{63}-1,50\right| 49^{63}+1$, then $50 \left\lvert\, \frac{7\left(49^{1008}-1\right)}{48}\right.$.
Therefore, $A \equiv-1008 \equiv-8 \equiv 42(\bmod 50)$, which means the remainder of $A$ divided by 50 is 42.