25. Bernardo chose a three-digit positive integer $N$, and wrote its base-5 and base-6 numbers on the blackboard. Later, Leroy only saw the two numbers written by Bernardo, and mistakenly took them as decimal positive integers. He added them to get the positive integer $S$. For example, if $N=749$, the numbers Bernardo wrote are 10444 and 3245, and Leroy's sum is $S=$ 13689. Then there are ( ) ways to choose $N$ such that the last two digits of the number $S$ are the same as the last two digits of $2N$.
(A) 5
(B) 10
(C) 15
(D) 20
(E) 25
25. E.
$$
\begin{array}{l}
\text { Let } N \equiv 6 c + a \left(\bmod 6^{2}\right), \\
N \equiv 5 d + b \left(\bmod 5^{2}\right),
\end{array}
$$

where, $a, c \in (0,1, \cdots, 5), b, d \in \{0,1, \cdots, 4\}$.
$$
\begin{aligned}
& \text { From } N \equiv b(\bmod 5) \\
& \Rightarrow 2 N \equiv 2 b(\bmod 10) \\
\Rightarrow & a + b \equiv 2 b(\bmod 10) \\
\Rightarrow & a = b .
\end{aligned}
$$

Let $N_{0} = N - a \equiv 10 e \left(\bmod 10^{2}\right)$.
$$
\begin{array}{l}
\text { From } N_{0} \equiv 5 d \left(\bmod 5^{2}\right) \\
\Rightarrow 5 d \equiv 10 e \left(\bmod 5^{2}\right) \\
\Rightarrow d \equiv 2 e(\bmod 5) .
\end{array}
$$

From the given conditions,
$$
\begin{array}{l}
c + d \equiv 2 e(\bmod 10) \\
\Rightarrow c \equiv 0(\bmod 5) \\
\Rightarrow c + d = 2 e \text { or } 2 e - 10 .
\end{array}
$$

If $c = 0$, then
$$
\begin{aligned}
(d, e) = & (0,0), (0,5), (2,1), (2,6), \\
& (4,2), (4,7) ;
\end{aligned}
$$

If $c = 5$, then
$$
(d, e) = (1,3), (1,8), (3,4), (3,9) .
$$

In summary, $(c, d)$ has 5 choices, and $a = b$ has 5 choices.
By the Chinese Remainder Theorem, the equations (1) and (2) determine 25 solutions for $N$ in the interval $\left[99 + 1, 99 + 6^{2} \times 5^{2}\right]$.