document.write( "Question 744576: Find the remainder when $\"1%5E2013+%2B+2%5E2013+%2B+3%5E2013+%2B+ellipsis+%2B+2012%5E2013%29\"$ is divided by 2013 .
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Algebra.Com's Answer #852568 by ikleyn(53547) $\"\"$ $\"About$

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document.write( "Let's group the addends in pairs\r\n" );
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document.write( "     + ,\r\n" );
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document.write( "     + ,\r\n" );
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document.write( "     + ,\r\n" );
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document.write( "     . . . . . . . . . . . . \r\n" );
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document.write( "     + .\r\n" );
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document.write( "Thus, all the addends in the long sum are grouped in pairs this way.\r\n" );
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document.write( "Now use the theorem (= the statement) that for all integers 'a' and 'b' and odd positive integer 'n'\r\n" );
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document.write( "the sum   +   is a multiple of (a+b), so the sum   +   is divisible by  (a+b)  with zero remainder.  \r\n" );
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document.write( "It follows from well known polynomial decomposition \r\n" );
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document.write( "     +  = ,\r\n" );
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document.write( "which works for all odd integer 'n'.\r\n" );
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document.write( "Now apply this theorem to each pair formed above.\r\n" );
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document.write( "Since for each pair the sum of integers, that are the bases, is 2013,  each and every pair is a multiple of 2013.\r\n" );
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document.write( "Hence, the entire sum of these pairs is a multiple of 2013.\r\n" );
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document.write( "It implies that long sum   +  +  + . . . +  is divisible by 2013,\r\n" );
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document.write( "i.e. gives zero remainder when is divided by 2013.\r\n" );
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\n" ); document.write( "\n" ); document.write( "At this point, the problem is solved completely.\r
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