A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$. (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q?$

$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$
1. Let \( P(a, b, c) \) denote the given assertion \( a \diamondsuit (b \diamondsuit c) = (a \diamondsuit b) \cdot c \).

2. We are given that \( a \diamondsuit a = 1 \) for all nonzero real numbers \( a \).

3. To show that \( \diamondsuit \) is injective, assume \( a \diamondsuit b = a \diamondsuit c \). We need to show that \( b = c \).

4. Consider \( P(a, a, b) \):
   \[
   a \diamondsuit (a \diamondsuit b) = (a \diamondsuit a) \cdot b
   \]
   Since \( a \diamondsuit a = 1 \), we have:
   \[
   a \diamondsuit (a \diamondsuit b) = b
   \]

5. Similarly, consider \( P(a, a, c) \):
   \[
   a \diamondsuit (a \diamondsuit c) = (a \diamondsuit a) \cdot c
   \]
   Since \( a \diamondsuit a = 1 \), we have:
   \[
   a \diamondsuit (a \diamondsuit c) = c
   \]

6. Given \( a \diamondsuit b = a \diamondsuit c \), we have:
   \[
   b = a \diamondsuit (a \diamondsuit b) = a \diamondsuit (a \diamondsuit c) = c
   \]
   Thus, \( \diamondsuit \) is injective.

7. Next, consider \( P(a, a, 1) \):
   \[
   a \diamondsuit (a \diamondsuit 1) = a \diamondsuit a
   \]
   Since \( a \diamondsuit a = 1 \), we have:
   \[
   a \diamondsuit (a \diamondsuit 1) = 1
   \]
   Therefore, \( a \diamondsuit 1 = a \).

8. Now, consider \( P(a, 1, b) \):
   \[
   a \diamondsuit (1 \diamondsuit b) = a \cdot b
   \]

9. Consider \( P(1, 1, b) \):
   \[
   1 \diamondsuit (1 \diamondsuit b) = b
   \]

10. From the above, we have:
    \[
    a \diamondsuit b = a \cdot (1 \diamondsuit b)
    \]

11. Let \( f(x) = 1 \diamondsuit x \). Then:
    \[
    a \diamondsuit b = a \cdot f(b)
    \]

12. Since \( 1 \diamondsuit (1 \diamondsuit b) = b \), we have:
    \[
    f(f(b)) = b
    \]

13. From \( a \diamondsuit a = 1 \), we have:
    \[
    a \cdot f(a) = 1 \implies f(a) = \frac{1}{a}
    \]

14. Therefore, \( a \diamondsuit b = a \cdot \frac{1}{b} = \frac{a}{b} \).

15. Now, solve the equation \( 2016 \diamondsuit (6 \diamondsuit x) = 100 \):
    \[
    2016 \diamondsuit \left( \frac{6}{x} \right) = \frac{2016}{\frac{6}{x}} = 2016 \cdot \frac{x}{6} = 336x
    \]
    \[
    336x = 100 \implies x = \frac{100}{336} = \frac{25}{84}
    \]

16. The solution \( x = \frac{25}{84} \) can be written as \( \frac{p}{q} \) where \( p = 25 \) and \( q = 84 \). Since 25 and 84 are relatively prime, we have \( p + q = 25 + 84 = 109 \).

The final answer is \( \boxed{109} \)