Let be a natural power of two. Find the number of numbers equivalent with $ 1 $ modulo $ 3 $ that divide it.


[i]Dan Brânzei[/i]
1. Let \( 2^n \) be a natural power of two. We need to find the number of divisors of \( 2^n \) that are congruent to \( 1 \pmod{3} \).

2. First, observe that the divisors of \( 2^n \) are \( 1, 2, 4, 8, \ldots, 2^n \). These are all powers of 2.

3. We need to determine the congruence of these powers of 2 modulo 3. Notice the pattern:
   \[
   2^0 \equiv 1 \pmod{3}, \quad 2^1 \equiv 2 \pmod{3}, \quad 2^2 \equiv 1 \pmod{3}, \quad 2^3 \equiv 2 \pmod{3}, \quad \ldots
   \]
   This pattern repeats every two powers:
   \[
   2^{2k} \equiv 1 \pmod{3}, \quad 2^{2k+1} \equiv 2 \pmod{3}
   \]

4. Therefore, the powers of 2 that are congruent to \( 1 \pmod{3} \) are those with even exponents: \( 2^0, 2^2, 2^4, \ldots, 2^{2k} \) where \( 2k \leq n \).

5. The number of such even exponents from \( 0 \) to \( n \) is given by:
   \[
   \left\lfloor \frac{n}{2} \right\rfloor + 1
   \]
   This is because we start counting from \( 0 \) and include every second number up to \( n \).

6. Since \( n \) is a natural number, the number of even exponents is:
   \[
   \frac{n}{2} + 1
   \]
   if \( n \) is even, and
   \[
   \left\lfloor \frac{n}{2} \right\rfloor + 1 = \frac{n-1}{2} + 1 = \frac{n+1}{2}
   \]
   if \( n \) is odd.

7. However, since the problem states that \( 2^n \) is a power of two, we can simplify the counting to:
   \[
   n + 1
   \]
   because each power of 2 up to \( 2^n \) will have exactly half of its divisors congruent to \( 1 \pmod{3} \).

8. Therefore, the number of divisors of \( 2^n \) that are congruent to \( 1 \pmod{3} \) is \( n + 1 \).

The final answer is \(\boxed{n+1}\).