The equation you provided is a modular arithmetic problem:

$1^{17} + 2^{17} + 3^{17} + \dots + 49^{17} \equiv x \ (\text{mod}\ 17)$

We are required to find the value of $x$ in the set $\mathbb{Z}_{17}$ (which means $x$ is an integer between 0 and 16).

Solution:

1. Euler's Theorem Application: For any integer $a$ that is not divisible by 17, $a^{16} \equiv 1 \ (\text{mod}\ 17)$. Thus: $a^{17} \equiv a \ (\text{mod}\ 17)$

2. Calculation: Since $17$ is a prime number, the theorem applies to all numbers except multiples of 17. In this sum, none of the integers 1 to 49 are multiples of 17.

   Therefore: $1^{17} + 2^{17} + 3^{17} + \dots + 49^{17} \equiv 1 + 2 + 3 + \dots + 49 \ (\text{mod}\ 17)$

3. Sum of the First 49 Numbers: The sum of the first $n$ integers is given by: $\text{Sum} = \frac{n(n+1)}{2}$ For $n = 49$: $\text{Sum} = \frac{49 \times 50}{2} = 1225$

4. Modular Reduction: Now, find $1225 \mod 17$. $1225 \div 17 \approx 72 \quad \text{(quotient)}, \quad 1225 - (72 \times 17) = 1225 - 1224 = 1$ Thus: $1225 \equiv 1 \ (\text{mod}\ 17)$

So, $x = 1$.

Final Answer:

$x = 1$

Would you like further details or explanations on any step? Here are some questions related to the problem:

1. What is Euler's theorem, and how is it used in modular arithmetic?
2. How can we generalize this approach to any prime modulus?
3. What would the solution be if the problem asked for mod 19 instead of mod 17?
4. Can you explain the process of summing an arithmetic series in more detail?
5. How would the problem change if the exponent were different, say 18 instead of 17?

Tip: Modular arithmetic problems often become simpler when using properties like Euler's theorem or Fermat's little theorem.