11. (10 points) A four-digit number $\overline{\mathrm{abcd}}$ and its reverse $\overline{\mathrm{dcba}}$ are both multiples of 65. Find this number.

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11. (10 points) A four-digit number $\overline{\mathrm{abcd}}$ and its reverse $\overline{\mathrm{dcba}}$ are both multiples of 65. Find this number.
【Analysis】Since a four-digit number $\overline{\mathrm{abcd}}$ and its reverse number $\overline{\mathrm{dcba}}$ are both multiples of 65, we have $1000 a+100 b+10 c+d \equiv 1000 d+100 c+10 b+a \equiv 0$ (mod65). By reasoning, we can solve the problem.

【Solution】Solution: $1000 a+100 b+10 c+d \equiv 1000 d+100 c+10 b+a \equiv 0 (\bmod 65)$.
$1001(a+d)+110(b+c) \equiv 0(\bmod 65)$.
$7 \times 11 \times 13(a+d)+2 \times 5 \times 11(b+c) \equiv 0(\bmod 65)$.
$65=13 \times 5$.
So $a+d=5,10,15 ; b+c=0,13$.
On the other hand, $1000 a+100 b+10 c+d-(1000 d+100 c+10 b+a)$
$=999(a-d)+90(b-c) \equiv 0(\bmod 65)$.
So $a-d=0,5$ or $d-a=5$.
Since $a+d$ and $a-d$ have the same parity, $a=d=5$.
If $b+c=13$, then:
$1000 a+100 b+10 c+d$
$=5000+100 b+10(13-b)+5=5005+90 b+130=65 \times 79+65 b+25 b \equiv 25 b \equiv 0(\bmod 65)$.
So $25 b \equiv 0$ (mod13), $b=0, c=13$, which is impossible ( $c$ is a digit).
Therefore, $b+c=b=c=0,1000 a+100 b+10 c+d=5005$.
Thus, the answer is: 5005.