Let $p$ be a real number and $c\neq 0$ such that \[c-0.1<x^p\left(\dfrac{1-(1+x)^{10}}{1+(1+x)^{10}}\right)<c+0.1\] for all (positive) real numbers $x$ with $0<x<10^{-100}$.  (The exact value $10^{-100}$ is not important.  You could replace it with any "sufficiently small number".)

Find the ordered pair $(p,c)$.
1. **Determine the sign of \( p \):**
   - If \( p \) were nonnegative, as \( x \to 0 \), \( x^p \) would approach 0. This would make the entire expression approach 0, which contradicts the given condition that \( c \neq 0 \). Therefore, \( p \) must be negative.
   - Let \( p = -k \) where \( k > 0 \).

2. **Rewrite the expression:**
   - The given inequality is:
     \[
     c - 0.1 < x^p \left( \frac{1 - (1 + x)^{10}}{1 + (1 + x)^{10}} \right) < c + 0.1
     \]
   - Substituting \( p = -k \), we get:
     \[
     c - 0.1 < x^{-k} \left( \frac{1 - (1 + x)^{10}}{1 + (1 + x)^{10}} \right) < c + 0.1
     \]

3. **Analyze the limit as \( x \to 0 \):**
   - We need to find:
     \[
     \lim_{x \to 0} x^{-k} \left( \frac{1 - (1 + x)^{10}}{1 + (1 + x)^{10}} \right)
     \]
   - Let \( L \) be this limit. For \( L \) to exist and be finite, the expression inside the limit must balance the \( x^{-k} \) term.

4. **Apply L'Hôpital's Rule:**
   - The expression inside the limit is of the form \( \frac{0}{0} \) as \( x \to 0 \), so we apply L'Hôpital's Rule:
     \[
     \lim_{x \to 0} \frac{1 - (1 + x)^{10}}{x^k (1 + (1 + x)^{10})}
     \]
   - Differentiate the numerator and the denominator:
     \[
     \text{Numerator: } \frac{d}{dx} \left( 1 - (1 + x)^{10} \right) = -10(1 + x)^9
     \]
     \[
     \text{Denominator: } \frac{d}{dx} \left( x^k (1 + (1 + x)^{10}) \right) = k x^{k-1} (1 + (1 + x)^{10}) + x^k \cdot 10(1 + x)^9
     \]
   - Applying L'Hôpital's Rule:
     \[
     L = \lim_{x \to 0} \frac{-10(1 + x)^9}{k x^{k-1} (1 + (1 + x)^{10}) + x^k \cdot 10(1 + x)^9}
     \]

5. **Simplify the limit:**
   - As \( x \to 0 \), the term \( k x^{k-1} (1 + (1 + x)^{10}) \) dominates over \( x^k \cdot 10(1 + x)^9 \) because \( k x^{k-1} \) grows faster than \( x^k \):
     \[
     L = \lim_{x \to 0} \frac{-10(1 + x)^9}{k x^{k-1} (1 + (1 + x)^{10})}
     \]
   - Simplify further:
     \[
     L = \lim_{x \to 0} \frac{-10(1 + x)^9}{k x^{k-1} (1 + 1)} = \lim_{x \to 0} \frac{-10(1 + x)^9}{2k x^{k-1}}
     \]
   - For the limit to be finite, \( k \) must be 1:
     \[
     L = \lim_{x \to 0} \frac{-10(1 + x)^9}{2x^0} = \frac{-10}{2} = -5
     \]

6. **Determine \( c \):**
   - Since \( c \) is an integer and the limit \( L \) is -5, we conclude that \( c = -5 \).

7. **Determine \( p \):**
   - Since \( k = 1 \), \( p = -k = -1 \).

Thus, the ordered pair \((p, c)\) is \(\boxed{(-1, -5)}\).