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Let's analyze the problem in detail.

We are asked to compute \( 50^{256} \mod 37 \) and check which of the provided statements are correct.

### Approach:
1. **Simplify the base modulo 37**:
   First, reduce \( 50 \mod 37 \):
   \[
   50 \equiv 50 - 37 = 13 \pmod{37}
   \]
   So, \( 50^{256} \mod 37 = 13^{256} \mod 37 \).

2. **Use Euler's Theorem**:
   Euler’s theorem states that if \( a \) and \( n \) are coprime (i.e., \( \gcd(a, n) = 1 \)), then \( a^{\phi(n)} \equiv 1 \pmod{n} \), where \( \phi(n) \) is Euler’s totient function. Since \( \gcd(13, 37) = 1 \) and \( 37 \) is a prime number, \( \phi(37) = 37 - 1 = 36 \). Therefore:
   \[
   13^{36} \equiv 1 \pmod{37}
   \]

3. **Reduce the exponent modulo 36**:
   To simplify \( 13^{256} \mod 37 \), reduce the exponent \( 256 \mod 36 \):
   \[
   256 \div 36 = 7 \text{ remainder } 4
   \]
   So, \( 256 \equiv 4 \pmod{36} \), which implies:
   \[
   13^{256} \equiv 13^4 \pmod{37}
   \]

4. **Calculate \( 13^4 \mod 37 \)**:
   Now calculate \( 13^4 \mod 37 \):
   \[
   13^2 = 169 \equiv 22 \pmod{37}
   \]
   \[
   13^4 = (13^2)^2 = 22^2 = 484 \equiv 34 \pmod{37}
   \]

Thus, \( 50^{256} \equiv 34 \pmod{37} \).

### Conclusion:
The correct answer is \( 50^{256} \equiv 34 \pmod{37} \), which matches the third statement. Therefore, the third option is the correct one.

Let me know if you'd like any further clarification or details.

Here are five related questions:
1. How do we use Euler's theorem to simplify large exponentiation in modular arithmetic?
2. What is Euler's totient function and how is it computed for prime numbers?
3. How do you compute \( a^b \mod n \) efficiently using exponentiation by squaring?
4. What are the conditions for applying Euler's theorem in modular arithmetic problems?
5. Can the Chinese Remainder Theorem be used in similar modular exponentiation problems?

**Tip**: Always try to reduce the base and exponent in modular arithmetic to keep calculations manageable.

Learn how to compute 50^256 mod 37 using modular arithmetic and Euler's Theorem. This solution covers exponentiation, modular reduction, and the application of Euler's Totient function. Suitable for advanced high school and college students.

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