Estimate the number of primes among the first thousand primes divide some term of the sequence
    \[2^0+1,2^1+1,2^2+1,2^3+1,\ldots.\]
    An estimate of $E$ earns $2^{1-0.02|A-E|}$ points, where $A$ is the actual answer.


[i]2021 CCA Math Bonanza Lightning Round #5.4[/i]
To solve this problem, we need to estimate the number of primes among the first thousand primes that divide some term of the sequence \(2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, \ldots\).

1. **Understanding the Sequence**:
   The sequence given is \(2^n + 1\) for \(n = 0, 1, 2, 3, \ldots\). The terms of this sequence are:
   \[
   2^0 + 1 = 2, \quad 2^1 + 1 = 3, \quad 2^2 + 1 = 5, \quad 2^3 + 1 = 9, \quad 2^4 + 1 = 17, \quad \ldots
   \]

2. **Identifying Fermat Numbers**:
   Fermat numbers are of the form \(2^{2^k} + 1\). The first few Fermat numbers are:
   \[
   F_0 = 2^{2^0} + 1 = 3, \quad F_1 = 2^{2^1} + 1 = 5, \quad F_2 = 2^{2^2} + 1 = 17, \quad F_3 = 2^{2^3} + 1 = 257, \quad F_4 = 2^{2^4} + 1 = 65537
   \]
   These are the only known Fermat primes, as it is known that \(F_n\) for \(n \geq 5\) are composite.

3. **Prime Divisors of the Sequence**:
   We need to find the primes among the first thousand primes that divide any term of the sequence \(2^n + 1\). Note that if a prime \(p\) divides \(2^n + 1\), then \(2^n \equiv -1 \pmod{p}\), which implies \(2^{2n} \equiv 1 \pmod{p}\). Therefore, the order of 2 modulo \(p\) must be \(2n\), and \(2n\) must be a divisor of \(p-1\).

4. **Counting the Primes**:
   We need to count the number of primes among the first thousand primes that divide any term of the sequence. The first thousand primes are well-known, and we can check which of these primes divide any of the terms \(2^n + 1\).

   - For \(2^0 + 1 = 2\), the prime is 2.
   - For \(2^1 + 1 = 3\), the prime is 3.
   - For \(2^2 + 1 = 5\), the prime is 5.
   - For \(2^3 + 1 = 9\), no prime divides 9.
   - For \(2^4 + 1 = 17\), the prime is 17.
   - For \(2^5 + 1 = 33\), the prime is 3.
   - For \(2^6 + 1 = 65\), the primes are 5 and 13.
   - For \(2^7 + 1 = 129\), the primes are 3 and 43.
   - For \(2^8 + 1 = 257\), the prime is 257.
   - For \(2^9 + 1 = 513\), the primes are 3 and 171.
   - For \(2^{10} + 1 = 1025\), the primes are 5 and 41.
   - For \(2^{11} + 1 = 2049\), the primes are 3 and 683.
   - For \(2^{12} + 1 = 4097\), the prime is 17.
   - For \(2^{13} + 1 = 8193\), the primes are 3 and 2731.
   - For \(2^{14} + 1 = 16385\), the primes are 5 and 3277.
   - For \(2^{15} + 1 = 32769\), the primes are 3 and 10923.
   - For \(2^{16} + 1 = 65537\), the prime is 65537.

   From this, we see that the primes 2, 3, 5, 13, 17, 41, 43, 257, 683, and 65537 divide some term of the sequence.

5. **Conclusion**:
   The number of distinct primes among the first thousand primes that divide some term of the sequence is 10.

The final answer is \(\boxed{10}\).