document.write( "Question 1162656: The positive integers 1,2,3,4,...,m are written one after another to form the integer L = 123456789101112131415 . . .. What is the smallest integer m > 2020 for which L is divisible by 9? \n" ); document.write( "

Algebra.Com's Answer #786496 by ikleyn(52781) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( " Here is another (\"elegant\") solution, which tutor @greenestamps wants to see\r
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\n" ); document.write( "\n" ); document.write( "The base for this solution is the same rule of divisibility by 9:\r
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document.write( "    The number is divisible by 9 if and only if the sum of its digits is divisible by 9.\r\n" );
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\n" ); document.write( "\n" ); document.write( "The more general form of this rule is\r
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document.write( "    For any number, the remainder of dividing by 9 is the same as the remainder of dividing by 9 the sum of its digits.\r\n" );
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\n" ); document.write( "\n" ); document.write( "For both forms, see the lesson\r
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document.write( "It follows immediately from these rules, that the remainder of division the number L(m) by 9 \r\n" );
document.write( "is equal to the remainder of divisibility by 9 the number\r\n" );
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document.write( "    1 + 2 + 3 + . . . + 9 + 10 + 11 + 12 + . . . + 99 + 100 + 101 + 102 + . . . 999 + 1000 + 1001 + 1002 + . . . + the last 4-digit number m = \r\n" );
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document.write( "    = the sum of the first \"m\" natural numbers = .\r\n" );
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document.write( "So, all we need is to check the remainder of dividing by 9 of several numbers of the form  ,\r\n" );
document.write( "starting from m = 2021, 2022, 2023, 2024 . . . \r\n" );
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document.write( "The table below (made using the function r = mod(N,9) of Excel) shows these remainders\r\n" );
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document.write( "    m                              2021    2022   2023  2024\r\n" );
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document.write( "    r =           6       3      1     0\r\n" );
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document.write( "Thus that minimal value of \"m\" the problems is asking for is  2024.    ANSWER\r\n" );
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