8. Given positive real numbers $x, y (x>y)$ satisfy $x y=490$, and $(\lg x-\lg 7)(\lg y-\lg 7)=-\frac{143}{4}$, then in decimal notation, the integer part of $x$ has $\qquad$ digits.
(Zhang Jia provided the problem)
8. 8 .

From $x y=490(x>y>0)$, we get $\lg x+\lg y=2 \lg 7+1$, which means
$$
(\lg x-\lg y)+(\lg y-\lg 7)=1 .
$$

Also, $(\lg x-\lg 7)(\lg y-\lg 7)=-\frac{143}{4}$, so $\lg x-7, \lg y-7$ are the two roots of the quadratic equation $t^{2}-t-\frac{143}{4}=0$ in terms of $t$. It is easy to see that $\lg x-\lg 7$ is the larger root, solving which gives
$$
\lg x-\lg 7=\frac{13}{2},
$$

which means $\lg x=\lg 7+6.5$. Since $\sqrt{10}<7<10$, we have $0.5<\lg 7<1$. Therefore, $7<\lg x<7.5 \Rightarrow 10^{7}<x<10^{8}$. Thus, $x$ has 8 digits in its integer part.