document.write( "Question 370041: i am a 3 digit number divisble by 7 but not 2 the sum of my digits is 4
\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #263712 by Edwin McCravy(20056) $\"\"$ $\"About$

You can put this solution on YOUR website!

Since I am not divisible by 2, I am an odd integer.\r\n" );
document.write( "\r\n" );
document.write( "Odd integers all end with 1, 3, 5, 7, 9. \r\n" );
document.write( "\r\n" );
document.write( "Since the sum of my 3 digits is only 4, my last digit can only be 1 or 3.\r\n" );
document.write( "\r\n" );
document.write( "If my last digit were 3, then that would leave only 1 for my first two digits\r\n" );
document.write( "to have as a sum.  Then I could be only 103, but 103 is not a multiple of 7.\r\n" );
document.write( "  \r\n" );
document.write( "Therefore my last digit must be 1.\r\n" );
document.write( "\r\n" );
document.write( "Since my first two digits must have sum 3, they can only be 12, 21 or 30.\r\n" );
document.write( "\r\n" );
document.write( "So I can only be 121, 211, or 301.  Of these only 301 is divisible by\r\n" );
document.write( "7.  So I am 301.\r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( "

\n" );