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2014-07-17T09:57:07+05:30

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1) Note that working mod 10 yields the sum of the digits.
(10^27 + 2) = 1^27 + 2 = 3 (mod 10).
==> k = (10^27 + 2)/3 = 1 (mod 10).

So, the sum of the digits of k equals 1.
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2) Working mod 11,
9^1990 = (-2)^1990
............= 2^1990
............= (2^10)^199
............= 1^199, since 2^(11-1) = 1 (mod 11) by Fermat
............= 1.

So, the remainder equals 1.
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3) It appears to be the greatest common factor of the two numbers...

3 || 5 = 1
7 || 14 = 7
3 || 15 = 15 <-- Hmm... this should be 3?
15 || 20 = 5
17 || 34 = 17

The greatest common factor of 64 and 30 is 2:

64 || 30 = 2
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