document.write( "Question 1117174: The sequence of numbers 12345678910111213. . . 997998999 is found by writing the numbers 1, 2, 3, . . ., 999 in order. What would be the 1997th digit (from the left) of the lengthy number? \n" ); document.write( "

Algebra.Com's Answer #732191 by ikleyn(52847) $\"\"$ $\"About$

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document.write( "1.  First 9 positions are occupied by nine 1-digit numbers from 1 to 9.\r\n" );
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document.write( "    The rest are 1997 - 9 = 1988 positions till the 1997-th position inclusively.\r\n" );
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document.write( "2.  The next 2*90 = 180 positions are occupied by ninety 2-digit numbers from 10 to 99.\r\n" );
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document.write( "    The rest are 1988 - 180 = 1808 positions till the 1997-th position inclusively.\r\n" );
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document.write( "3.  Divide  1808 by 3 to separate 3-digit numbers:   = 602.667.\r\n" );
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document.write( "    So, there are 602 3-digit numbers starting from 100 till the 1997-th position,\r\n" );
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document.write( "    and the last such number is 100 + 602 - 1 = 701.\r\n" );
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document.write( "    The next 3-digit number 702 has \"0\" exactly in the 1997-th position.\r\n" );
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document.write( "Answer.  1997-th digit (from the left) of the lengthy number is \"0\".\r\n" );
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