The sequence of all positive plaindromes are written in ascending order $1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, .....$ Then

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The rank of 2140412 is 3139

The rank of 1999991 is 2998

The $1998^{t h}$ positive palindrome is 1001001

The $2018^{t h}$ positive palindrome is 1019101

answer is A, C, D.

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Detailed Solution

1 digit palindromes $\to 9$
2 digit palindromes $\to 9 \times 1$
3 digit palindromes $\to 9 \times 10$
4 digit palindromes $\to 9 \times 10$
5 digit palindromes $\to 9 \times 10^{2}$
6 digit palindromes $\to 9 \times 10^{2}$
$1 \underline{0} \underline{0} - - - - - - \to 10$
$1 0 1 - - - - - - \to 10$
$∴$ Number at $2 (9 + 90 + 900 + 10) = 2018,$ rank is 1019101
$1 - - - - - - - - - - \to 10^{3}$
$20 - - - - - - - - - - \to 10^{2}$
$210 - - - - - - - - - - \to 10$
$211 \to 10$
$212 \to 10$
$213 \to 10$
$2140412 \to 10$
Rank of 2140412 = $2 (9 + 90 + 900) + 1000 + 100 + 40 + 1$
$= 1998 + 1000 + 100 + 40 + 1 = 3139$