Find the number of solutions to the equation: 

$6\{x\}^3 + \{x\}^2 + \{x\} + 2x = 2018. $

With {x} we denote the fractional part of the number x. 
1. **Understanding the problem**: We need to find the number of solutions to the equation \(6\{x\}^3 + \{x\}^2 + \{x\} + 2x = 2018\), where \(\{x\}\) denotes the fractional part of \(x\). Recall that \(0 \leq \{x\} < 1\).

2. **Isolate the integer part**: Let \(x = [x] + \{x\}\), where \([x]\) is the integer part of \(x\) and \(\{x\}\) is the fractional part. Substituting this into the equation, we get:
   \[
   6\{x\}^3 + \{x\}^2 + \{x\} + 2([x] + \{x\}) = 2018
   \]
   Simplifying, we have:
   \[
   6\{x\}^3 + \{x\}^2 + 3\{x\} + 2[x] = 2018
   \]

3. **Bounding the equation**: Since \(0 \leq \{x\} < 1\), the term \(6\{x\}^3 + \{x\}^2 + 3\{x\}\) ranges from 0 to 10. Therefore:
   \[
   2[x] \leq 2018 < 10 + 2[x]
   \]
   Solving these inequalities for \([x]\), we get:
   \[
   2[x] \leq 2018 \implies [x] \leq 1009
   \]
   \[
   2018 < 10 + 2[x] \implies 1004 < [x]
   \]
   Combining these, we find:
   \[
   1005 \leq [x] \leq 1009
   \]

4. **Rewriting the equation**: For each integer value of \([x]\) in the range \(1005 \leq [x] \leq 1009\), we can rewrite the equation as:
   \[
   6\{x\}^3 + \{x\}^2 + 3\{x\} = 2018 - 2[x]
   \]
   Let \(t = \{x\}\). Then the equation becomes:
   \[
   6t^3 + t^2 + 3t = 2018 - 2[x]
   \]

5. **Analyzing the function**: Consider the function \(f(t) = 6t^3 + t^2 + 3t\). We need to determine if \(f(t)\) can take the values \(2018 - 2[x]\) for \(t \in [0, 1)\). Note that:
   \[
   f(0) = 0 \quad \text{and} \quad f(1) = 10
   \]
   Since \(f(t)\) is a continuous and strictly increasing function on \([0, 1)\) (as \(f'(t) = 18t^2 + 2t + 3 > 0\) for all \(t \in [0, 1)\)), it takes all values in the interval \([0, 10)\) exactly once.

6. **Checking the values**: For each \([x]\) in \([1005, 1006, 1007, 1008, 1009]\), we compute:
   \[
   2018 - 2[1005] = 8, \quad 2018 - 2[1006] = 6, \quad 2018 - 2[1007] = 4, \quad 2018 - 2[1008] = 2, \quad 2018 - 2[1009] = 0
   \]
   Each of these values lies in the interval \([0, 10)\), and since \(f(t)\) takes each value in this interval exactly once, there is exactly one \(t \in [0, 1)\) for each value.

7. **Conclusion**: Therefore, there are 5 solutions corresponding to \([x] = 1005, 1006, 1007, 1008, 1009\).

The final answer is \(\boxed{5}\)