document.write( "Question 1208054: When expanded as a decimal, the fraction \"1%2F977\" has a repetend (the repeating part of the decimal) that begins right after the decimal point, and is 976 digits long. If the last three digits of the repetend are ABC, compute the digits A, B, and C.
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Algebra.Com's Answer #846198 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "To get some insight as to what is going on here, I took\r\n" );
document.write( "the long division out a few places, without skipping any\r\n" );
document.write( "steps because of 0's, as we usually do in long division:\r\n" );
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document.write( "      0.0 0 1 0 2\r\n" );
document.write( "9 7 7)1.0 0 0 0 0\r\n" );
document.write( "      0\r\n" );
document.write( "      1 0\r\n" );
document.write( "        0\r\n" );
document.write( "      1 0 0\r\n" );
document.write( "          0\r\n" );
document.write( "      1 0 0 0\r\n" );
document.write( "        9 7 7\r\n" );
document.write( "          2 3 0\r\n" );
document.write( "              0\r\n" );
document.write( "          2 3 0 0\r\n" );
document.write( "          1 9 5 4\r\n" );
document.write( "            3 4 6\r\n" );
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document.write( "The very first remainder above is 1. The decimal will start repeating \r\n" );
document.write( "when the remainder is 1 again, and the subtractions we perform to get a\r\n" );
document.write( "remainder are always from a remainder that is annexed on the right with \r\n" );
document.write( "a 0.  So to get a 1 remainder, we must be subtracting a digit multiple \r\n" );
document.write( "of 977 that ends in a 9.\r\n" );
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document.write( "The only digit multiple of 977 that ends with a 9 is 7x977 or \r\n" );
document.write( "6839, so 1 more, or 6840 must have been the number we would be \r\n" );
document.write( "subtracting 6839 from, which means the remainder just before the\r\n" );
document.write( "remainder of 1 was 684. So the last digit before the digits started\r\n" );
document.write( "repeating must have been a 7.\r\n" );
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document.write( "To have gotten the remainder 684 from a number that ended with 0, we\r\n" );
document.write( "must have subtracted a digit multiple of 977 that ended with a 6.\r\n" );
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document.write( "The only digit multiple of 977 that ends with a 6 is 8x977 or \r\n" );
document.write( "7816, so 684 more, or 8500 must have been the number we would be \r\n" );
document.write( "subtracting 6839 from, which means the remainder just before the\r\n" );
document.write( "remainder of 684 was 850. And the digit before the 7 must have\r\n" );
document.write( "been an 8.\r\n" );
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document.write( "To have gotten the remainder 850 from a number that ended with 0, we\r\n" );
document.write( "must have subtracted a digit multiple of 977 that ended with a 0,\r\n" );
document.write( "which could only have been 0x977 or 0.\r\n" );
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document.write( "So the last three digits before we got a 1 remainder must have been 087 \r\n" );
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document.write( "That's the answer 087.  A=0, B=8, C=7\r\n" );
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document.write( "Edwin

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