Define a sequence of positive integers $\{a_n\}$ through the recursive formula:
$a_{n+1}=a_n+b_n(n\ge 1)$,where $b_n$ is obtained by rearranging the digits of $a_n$ (in decimal representation) in reverse order (for example,if $a_1=250$,then $b_1=52,a_2=302$,and so on). Can $a_7$ be a prime?
1. **Understanding the sequence and the recursive formula:**
   The sequence $\{a_n\}$ is defined by the recursive formula:
   \[
   a_{n+1} = a_n + b_n \quad \text{for} \quad n \ge 1
   \]
   where $b_n$ is obtained by reversing the digits of $a_n$.

2. **Analyzing the sequence modulo 11:**
   We need to consider the properties of numbers under modulo 11. Specifically, we use the fact that reversing the digits of a number does not change its value modulo 11. This is because the alternating sum of the digits of a number is invariant under digit reversal.

3. **Case 1: $a_k$ has an odd number of digits:**
   If $a_k$ has $2m+1$ digits, then:
   \[
   a_k \equiv b_k \pmod{11}
   \]
   This is because the alternating sum of the digits of $a_k$ and $b_k$ will be the same.

4. **Case 2: $a_k$ has an even number of digits:**
   If $a_k$ has $2m$ digits, then:
   \[
   a_k + b_k \equiv 0 \pmod{11}
   \]
   This is because the alternating sum of the digits of $a_k$ and $b_k$ will be negatives of each other, summing to zero modulo 11.

5. **Implications for $a_7$:**
   We need to show that $a_7$ cannot be a prime number. To do this, we need to establish two things:
   - $a_7 > 11$
   - At least one of $a_1, a_2, \ldots, a_6$ has an even number of digits.

6. **Proving $a_7 > 11$:**
   Since $a_1$ is a positive integer, and each $a_{n+1}$ is obtained by adding a positive integer $b_n$ to $a_n$, the sequence $\{a_n\}$ is strictly increasing. Therefore, $a_7$ will be greater than $a_1$. Given that $a_1$ is a positive integer, it is clear that $a_7 > 11$.

7. **Proving that one of $a_1, a_2, \ldots, a_6$ has an even number of digits:**
   Since the sequence $\{a_n\}$ is strictly increasing, the number of digits in $a_n$ can only increase or stay the same as $n$ increases. If $a_1$ has an odd number of digits, then at some point, one of the $a_n$ must have an even number of digits because the number of digits will eventually increase as the sequence progresses.

8. **Conclusion:**
   Since one of $a_1, a_2, \ldots, a_6$ has an even number of digits, we have:
   \[
   a_k + b_k \equiv 0 \pmod{11} \quad \text{for some} \quad k \le 6
   \]
   This implies that $a_7 \equiv 0 \pmod{11}$, meaning $a_7$ is divisible by 11 and hence cannot be a prime number.

\[
\boxed{a_7 \text{ cannot be a prime number}}
\]