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\[ 2^{1}-2^{2}+2^{3}-2^{4}+2^{5}-2^{6}+2^{7} -\ldots +2^{2013}\]

Let \(a\) be the value of the expression above. Find the last two digits of \(a.\)

Given that in the 8-digit number \(\overline{\mathrm{ABCDEFGH}}\),

(i) \(\overline{\mathrm{ABCD}}=\overline{\mathrm{EFGH}}\)

(ii) The numbers \(\overline{\mathrm{ABCDEFGH}}\) and \(\overline{\mathrm{ABCD0EFGH}}\) are both divisible by \(11\).

Let the sum of all possible values of \(\overline{\mathrm{ABCDEFGH}}\) be \(N\).

Find the digit sum of \(N\).

**Details and assumptions**:-

- \(\overline{\mathrm{ABC}}\) means the number in decimal representation with digits \(A,B,C\) i.e. \(\overline{\mathrm{ABC}} = \mathrm{100A+10B+C}\)
- The letters \(A\) to \(H\) do
**not**necessarily stand for**distinct**digits. - In the second number, the digit \(0\) is added in the middle of the 8-digit number, which makes it a 9-digit number.
- Digit sum is sum of all digits in decimal representation, digit sum of \(12023\) is \(1+2+0+2+3=8\)
- 00123 is not a 5-digit number.

Consider the sequence \( 50 + n^2 \) for positive integer \(n\):

\[51, 54, 59, 66, 75, \ldots\]

If we take the greatest common divisor of 2 consecutive terms, we obtain

\[3, 1, 1, 3, \ldots\]

What is the sum of all distinct elements in the second series?

\[ \large \frac1a+\frac1b+\frac1c= \frac1{42} \]

Let \(a\leq b \leq c\) be positive integers that satisfy the equation above. Find the maximum possible value of \(c\).

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