3. Given that $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio 2, and $\left\{b_{n}\right\}$ is an arithmetic sequence with the first term 2 and common difference 5. The numbers that appear in both sequences are arranged in ascending order to form the sequence $\left\{x_{n}\right\}$. Then $x_{100}=$ $\qquad$
3. $2^{397}$.

It is easy to see that the numbers in the sequence $\left\{a_{n}\right\}$ are integer powers of 2, and the numbers in the sequence $\left\{b_{n}\right\}$ leave a remainder of 2 when divided by 5.

Notice that the remainders when $2^{0}, 2^{1}, 2^{2}, 2^{3}, 2^{4}, 2^{5}, 2^{6}, 2^{7}, \cdots$ are divided by 5 are $1,2,4,3,1,2,4,3, \cdots$, and in the sequence $\left\{a_{n}\right\}$, only numbers of the form $2^{4 k-3}$ leave a remainder of 2 when divided by 5.
Therefore, numbers of the form $2^{4 k-3}$ appear in $\left\{b_{n}\right\}$.
Thus, $x_{n}=2^{4 n-3}$.
In particular, $x_{1 \infty}=2^{397}$.