document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1162656>Question 1162656</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #786492 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=786492\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "(1) A number is divisible by 9 if the sum of its digits is divisible by 9.<br><BR>\n" );
document.write( "(2) We can ignore the 0's in the number, since they add nothing to the sum of the digits.<br><BR>\n" );
document.write( "(3) The number 123456789 is divisible by 9, because the sum of the digits is 45, which is divisible by 9.<br><BR>\n" );
document.write( "So the number L is divisible by 9 through m=9.<br><BR>\n" );
document.write( "(4) In the number 101112...979899, the digits 1 through 9 are each used the same number of times, so the sum of the digits is a multiple of the sum of the digits 1 through 9, which is divisible by 9.<br><BR>\n" );
document.write( "So the number 101112...979899 is divisible by 9; and so the number L through m=99 is divisible by 9.<br><BR>\n" );
document.write( "(5) The same argument holds for the number 100101102...997998999 -- each digit 1 through 9 is used the same number of times, so the number 100101102...997998999 is divisible by 9.<br><BR>\n" );
document.write( "So the number L through m=999 is divisible by 9.<br><BR>\n" );
document.write( "(6) For the number 100010011002...199719981999, we again have the same digits as in the number 123456789101112...997998999, plus we have 1000 1's.  Those 1000 1's have a sum of 1000, which is 1 more than a multiple of 9.<br><BR>\n" );
document.write( "So the number L through m=1999 has remainder 1 when divided by 9.<br><BR>\n" );
document.write( "From here, it looks as if we need to go one more number m at a time in order to find the first one larger than 2020 for which the integer L is divisible by 9.<br><BR>\n" );
document.write( "(perhaps another tutor will respond showing an easier way to get from here to the answer -- or perhaps to solve the whole problem more easily)<br><BR>\n" );
document.write( "We start knowing that through m=1999 the integer L has  remainder of 1 when divided by 9 -- i.e., L = 1 mod 9.<br><BR>\n" );
document.write( "We go from there, remembering that we are looking for the first m greater than 2020 for which L is divisible by 9....<br><PRE STYLE=\"white-space: pre-wrap; white-space: -moz-pre-wrap; white-space: -pre-wrap; white-space: -o-pre-wrap; word-wrap: break-word;\">\r\n" );
document.write( "   m added\r\n" );
document.write( "      sum   L mod 9\r\n" );
document.write( "  -------------------\r\n" );
document.write( "  2000  2   3\r\n" );
document.write( "  2001  3   6\r\n" );
document.write( "  2002  4   10 = 1\r\n" );
document.write( "  2003  5   6\r\n" );
document.write( "  2004  6   12 = 3\r\n" );
document.write( "  2005  7   10 = 1\r\n" );
document.write( "  2006  8   9 = 0\r\n" );
document.write( "  2007  9   9 = 0\r\n" );
document.write( "  2008 10  10 = 1\r\n" );
document.write( "  2009 11  12 = 3\r\n" );
document.write( "  2010  3   6\r\n" );
document.write( "  2011  4  10 = 1\r\n" );
document.write( "  2012  5   6\r\n" );
document.write( "  2013  6  12 = 3\r\n" );
document.write( "  2014  7  10 = 1\r\n" );
document.write( "  2015  8   9 = 0\r\n" );
document.write( "  2016  9   9 = 0\r\n" );
document.write( "  2017 10  10 = 1\r\n" );
document.write( "  2018 11  12 = 3\r\n" );
document.write( "  2019 12  15 = 6\r\n" );
document.write( "  2020  4  10 = 1\r\n" );
document.write( "  2021  5   6\r\n" );
document.write( "  2022  6  12 = 3\r\n" );
document.write( "  2023  7  10 = 1\r\n" );
document.write( "  2024  8   9 = 0</PRE><br><BR>\n" );
document.write( "ANSWER: m=2024 is the smallest integer greater than 2020 for which L is divisible by 9.<br><BR>\n" );
document.write( "Note that, from the pattern of L mod 9 for m from 2000 to 2020, we could have predicted the answer; but since the period of the pattern is relatively short it was just as easy to continue with the table.<br><BR>\n" );
document.write( "Cool problem....<br><BR>\n" );
document.write( "I will be looking to see if another tutor has an elegant way of finding the answer with less effort.<br><BR>\n" );
document.write( "-----------------------------------------------------------<br><BR>\n" );
document.write( "Thanks to tutor @ikleyn for showing a much easier path to the answer to the problem....<br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );