11. (5 points) If the six-digit number $\overline{201 a b 7}$ is divisible by 11 and 13, then the two-digit number $\overline{a b}=$

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【Solution】Solution: According to the divisibility rule of 11, the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is a multiple of 11.
$7+a+0-(2+1-b)=a+4-b=0$ or 11.
If $a+4-b=11, a-b=7$, only $8-1=9-2=7$, the six-digit numbers 201817, 201927 are not multiples of 13.

If $a+4-b=0, a+4=b$, only $0+4=4,1+4=5,2+4=6,3+4=7,4+4=8,5+4=9$ apply.

The six-digit numbers 201047, 201157, 201267, 201377, 201487, 207597 formed only have 201487 as a multiple of 13.

Therefore: $\overline{\mathrm{ab}}=48$.
The answer is: 48.