Question 1127061
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<pre>
Let  

    r = {{{2^511}}} mod 7     (1)

be the remainder of division the integer number  {{{2^511}}} by 7.


Then the remainder  {{{2^512}}} mod 14  is (2r).


Therefore, to answer the problem question, it is ENOUGH to find  r = {{{2^511}}} mod 7   and then double it.


The remainders of division  {{{2^n}}}  by 7 form a cyclical sequence

    2, 4, 1,  2, 4, 1, . . .   for  n = 1, 2, 3, 4, 5, 6,  . . .    (2)

with the length of the cycle equal 3.


r = {{{2^511}}} mod 7   is 511-th term in this periodical/cycling sequence.  


Since  511 = 510+1 = 170*3 + 1,   511-th term in this periodical/cycling sequence is the first term of the basic cycle.


Hence,  r = {{{2^511}}} mod 7  is equal to 2.


Therefore, the value of (2r), which is the problem question, is 4.
</pre>

<U>Answer</U>.  &nbsp;&nbsp;The remainder when &nbsp;2 raised to &nbsp;512 &nbsp;is divided by &nbsp;14  &nbsp;is  &nbsp;4.



Solved.