document.write( "Question 1012947: This confuses me, please help me out:what is the remainder when 2014^2015 is divided by 9? \n" ); document.write( "

Algebra.Com's Answer #629081 by ikleyn(52932) $\"\"$ $\"About$

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\n" ); document.write( "This confuses me, please help me out:what is the remainder when 2014^2015 is divided by 9?
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document.write( "Answer. 4.\r\n" );
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document.write( "Solution\r\n" );
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document.write( "1.\r\n" );
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document.write( "2014 is not divisible by 3.  (The divisibility by 3 rule is divisibility by 3 the sum of the digits of a number. \r\n" );
document.write( "                              The sum 2+ 0 + 1 + 4 = 7 in not divisible by 3.\r\n" );
document.write( "                              See the lesson Divisibility by 3 rule in this site).\r\n" );
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document.write( "Hence,   is not divisible by 9.\r\n" );
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document.write( "2.\r\n" );
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document.write( "The remainder of division 2014 by 9 is 7   (You can check it yourself by making long division. \r\n" );
document.write( "                                            By the way, it coincides with the number 7 above, and it is not occasional). \r\n" );
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document.write( "The remainder of division  by 9 is the same as the remainder of division  = 49 by 9 and is 4.  \r\n" );
document.write( "                                           (I refer to elementary number theory, but you can check it yourself).\r\n" );
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document.write( "The remainder of division  by 9 is the same as the remainder of division  = 343 by 9 and is 1.  \r\n" );
document.write( "                                           (Again, I refer to elementary number theory, but you can check it yourself).\r\n" );
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document.write( "After that the remainders of division  by 9 will repeat again and again in the loop 7, 4, 1, 7, 4, 1, . . . \r\n" );
document.write( "                                           periodically/cyclically with the cycle length 3.\r\n" );
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document.write( "So, the sequence of remainders of divisions  by 9 is \r\n" );
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document.write( "7, 4, 1, 7, 4, 1, . . . for n = 1, 2, 3, 4, 5, 6 . . . , (1)\r\n" );
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document.write( "i.e. is cyclical with the cycle length 3.\r\n" );
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document.write( "Thus when you get , the remainder will be WHAT?\r\n" );
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document.write( "Divide 2015 by 3 and you will get the remainder 2.\r\n" );
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document.write( "So, the number of interest is the second number in the sequence of remainders (1), i.e. 4. \r\n" );
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