Ted's favorite number is equal to \[1\cdot\binom{2007}1+2\cdot\binom{2007}2+3\cdot\binom{2007}3+\cdots+2007\cdot\binom{2007}{2007}.\] Find the remainder when Ted's favorite number is divided by $25$.

$\begin{array}{@{\hspace{-1em}}l@{\hspace{14em}}l@{\hspace{14em}}l} 

\textbf{(A) }0&\textbf{(B) }1&\textbf{(C) }2\\\\ 

\textbf{(D) }3&\textbf{(E) }4&\textbf{(F) }5\\\\

\textbf{(G) }6&\textbf{(H) }7&\textbf{(I) }8\\\\

\textbf{(J) }9&\textbf{(K) }10&\textbf{(L) }11\\\\

\textbf{(M) }12&\textbf{(N) }13&\textbf{(O) }14\\\\

\textbf{(P) }15&\textbf{(Q) }16&\textbf{(R) }17\\\\

\textbf{(S) }18&\textbf{(T) }19&\textbf{(U) }20\\\\

\textbf{(V) }21&\textbf{(W) }22 & \textbf{(X) }23\\\\

\textbf{(Y) }24 \end{array}$
1. Let \( T \) be Ted's favorite number. The given expression for \( T \) is:
   \[
   T = 1 \cdot \binom{2007}{1} + 2 \cdot \binom{2007}{2} + 3 \cdot \binom{2007}{3} + \cdots + 2007 \cdot \binom{2007}{2007}
   \]

2. We can add \( 0 \cdot \binom{2007}{0} \) to the sum without changing its value:
   \[
   T = 0 \cdot \binom{2007}{0} + 1 \cdot \binom{2007}{1} + 2 \cdot \binom{2007}{2} + \cdots + 2007 \cdot \binom{2007}{2007}
   \]

3. Consider the sum \( S \) where the coefficients are reversed:
   \[
   S = 2007 \cdot \binom{2007}{0} + 2006 \cdot \binom{2007}{1} + 2005 \cdot \binom{2007}{2} + \cdots + 1 \cdot \binom{2007}{2006} + 0 \cdot \binom{2007}{2007}
   \]

4. Adding \( T \) and \( S \) together, we get:
   \[
   T + S = (0 + 2007) \cdot \binom{2007}{0} + (1 + 2006) \cdot \binom{2007}{1} + (2 + 2005) \cdot \binom{2007}{2} + \cdots + (2007 + 0) \cdot \binom{2007}{2007}
   \]
   \[
   T + S = 2007 \left( \binom{2007}{0} + \binom{2007}{1} + \binom{2007}{2} + \cdots + \binom{2007}{2007} \right)
   \]

5. Using the binomial theorem, we know that:
   \[
   \sum_{k=0}^{2007} \binom{2007}{k} = 2^{2007}
   \]

6. Therefore:
   \[
   T + S = 2007 \cdot 2^{2007}
   \]

7. Since \( S \) is just a rearrangement of the terms in \( T \), we have \( T = S \). Thus:
   \[
   2T = 2007 \cdot 2^{2007}
   \]
   \[
   T = 2007 \cdot 2^{2006}
   \]

8. To find \( T \mod 25 \), we use Euler's theorem. Euler's totient function \( \phi(25) = 20 \), so:
   \[
   2^{20} \equiv 1 \pmod{25}
   \]

9. We can write:
   \[
   2^{2006} = 2^{20 \cdot 100 + 6} = (2^{20})^{100} \cdot 2^6 \equiv 1^{100} \cdot 2^6 \equiv 2^6 \pmod{25}
   \]

10. Calculating \( 2^6 \):
    \[
    2^6 = 64 \equiv 14 \pmod{25}
    \]

11. Also, we need \( 2007 \mod 25 \):
    \[
    2007 \equiv 7 \pmod{25}
    \]

12. Therefore:
    \[
    T \equiv 7 \cdot 14 \equiv 98 \equiv 23 \pmod{25}
    \]

The final answer is \( \boxed{23} \).