document.write( "Question 360515: What are the final two digits of 71997?
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Algebra.Com's Answer #257247 by Edwin McCravy(20054)\"\" \"About 
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What are the final two digits of 71997?
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document.write( "70 = 1 (consider that as 01)\r\n" );
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document.write( "71 = 7 (consider that as 07)\r\n" );
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document.write( "72 = 49\r\n" );
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document.write( "73 = 343\r\n" );
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document.write( "74 = 2401\r\n" );
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document.write( "75 = 16807\r\n" );
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document.write( "76 = 117649\r\n" );
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document.write( "77 = 823543\r\n" );
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document.write( "78 = 5764801\r\n" );
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document.write( "...\r\n" );
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document.write( "So we see that the last two digits of powers of 7 are\r\n" );
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document.write( "01, 07, 49, 43, 01, 07, 49, 43, \r\n" );
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document.write( "ad the last two digits continue to repeat in groups of 4 forever. \r\n" );
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document.write( "So to find the last two digits of any positive integral power of\r\n" );
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document.write( "7, we can just divide the power by 4 and take only the remainder.\r\n" );
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document.write( "If the remainder is 0, the last two digits are 01,\r\n" );
document.write( "If the remainder is 1, the last two digits are 07,\r\n" );
document.write( "If the remainder is 2, the last two digits are 49,\r\n" );
document.write( "If the remainder is 3, the last two digits are 43,\r\n" );
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document.write( "To find 71997, we divide 1997 by 4\r\n" );
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document.write( "          499  \r\n" );
document.write( "       4)1997\r\n" );
document.write( "         16\r\n" );
document.write( "          39 \r\n" );
document.write( "          36\r\n" );
document.write( "           37 \r\n" );
document.write( "           36\r\n" );
document.write( "            1\r\n" );
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document.write( "The remainder is 1, so the last two digits of 71997 are 07.\r\n" );
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document.write( "Edwin
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