3. Given that $x$ and $y$ are real numbers greater than 10, the characteristic of $\lg x$ is $a$, and the mantissa is $b$; the characteristic of $\lg y$ is $c$, and the mantissa is $d$, and $|1-a|+\sqrt{c-4}=1$, $b+d=1$, then $x y=$ $\qquad$
3. $10^{7}$

Solution: Both leading numbers $a, c$ are positive integers, $|1-a|$ is a non-negative integer, and from the given information, we deduce that $\sqrt{c-4}$ is also a non-negative integer. Therefore, we have
( I ) $\left\{\begin{array}{l}|1-a|=1 \\ \sqrt{c-4}=0\end{array}\right.$ or ( II) $\left\{\begin{array}{l}|1-a|=0 \\ \sqrt{c-4}=1 .\end{array}\right.$
Solving ( I ) yields $\left\{\begin{array}{l}a=2 \\ c=4\end{array}\right.$ or $\left\{\begin{array}{l}a=0 \\ c=4\end{array}\right.$ (discard).
Solving ( II ) yields $\left\{\begin{array}{l}a=1 \\ c=5\end{array}\right.$.
Thus, $a+c=6, \lg (x y)=\lg x+\lg y=a+b+c+d=7$,
Therefore, $x y=10^{7}$.