document.write( "Question 582293: I am more than 100 but less than 200. I am divisible by 5. The sum and product of my digits are divisible by 5. My three digits are odd. Which number am I? \n" ); document.write( "
Algebra.Com's Answer #372030 by Edwin McCravy(20065)\"\" \"About 
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document.write( "The first digit has to be 1 since it's more than 100 but less than 200.\r\n" );
document.write( "So it's\r\n" );
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document.write( "1 _ _\r\n" );
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document.write( "Since it's divisible by 5 the last digit is either 0 or 5.\r\n" );
document.write( "Since the three digits are odd, the last digit can only be 5.\r\n" );
document.write( "So it's\r\n" );
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document.write( "1 _ 5\r\n" );
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document.write( "Now we are certain that the product of the digits will be a multiple\r\n" );
document.write( "of 5.\r\n" );
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document.write( "Since the sum of those two digits is 6 and the sum of the digits \r\n" );
document.write( "is a multiple of 5, that multiple of 5 could only be 10 or 15,\r\n" );
document.write( "which would require the middle digit to be 4 or 9.  But since\r\n" );
document.write( "the digits are all odd, it can't be 4, so it has to be 9.  So the \r\n" );
document.write( "number is\r\n" );
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document.write( "1 9 5\r\n" );
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document.write( "Edwin
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