document.write( "Question 977731: I am 4-digit number divisible by 3.My tens digit is three times as great as my thousands digit, and sum of my digit is 15 .
\n" ); document.write( "If you reverse my digits, I am divisible by 2 as well as by 6. What am I? \n" ); document.write( "
Algebra.Com's Answer #599298 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "I disagree a little with Rich.  I think there are 8 solutions.\r\n" );
document.write( "Here's how I did it.\r\n" );
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\n" ); document.write( "I am 4-digit number
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document.write( "Let the 4-digit number be \"ABCD\" where \r\n" );
document.write( "A = the thousands digit, B = the hundreds digit, C = the tens digit, and D = the\r\n" );
document.write( "one's or units digit. \r\n" );
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\n" ); document.write( "If you reverse my digits, I am divisible by 2.
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document.write( "That says DCBA is even. So A is even, either 2,4,6, or 8\r\n" );
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\n" ); document.write( "My tens digit is three times as great as my thousands digit.
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document.write( "So C = 3A\r\n" );
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document.write( "The only even digit that A can be of these (2,4,6,or 8) that we can multiply\r\n" );
document.write( "by 3 to get another digit C is 2.  So A = 2 and C=6\r\n" );
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document.write( "Therefore the number has the form:\r\n" );
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document.write( "2B6D\r\n" );
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\n" ); document.write( "and sum of my digit is 15.
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document.write( "2+B+6+D = 15, so\r\n" );
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document.write( "B+D = 7\r\n" );
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document.write( "For that we can have any of these: \r\n" );
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document.write( "0+7 = 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1 = 7+0 = 7\r\n" );
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document.write( "So we have eight solutions:\r\n" );
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document.write( "1.  2067\r\n" );
document.write( "2.  2166\r\n" );
document.write( "3.  2265\r\n" );
document.write( "4.  2364\r\n" );
document.write( "5.  2463\r\n" );
document.write( "6.  2562\r\n" );
document.write( "7.  2661\r\n" );
document.write( "8.  2760\r\n" );
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document.write( "Let's see if any of the other clues that we did not use\r\n" );
document.write( "narrow the list down:\r\n" );
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document.write( "

\n" ); document.write( "divisible by 3
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\n" ); document.write( "A number is divisible by 3 if the sum of its digits is divisible by 3. So
\n" ); document.write( "\"divisible by 3\" is redundant (unnecessary) since the sum of the digits is 15
\n" ); document.write( "which tells us it is divisible by 3.
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\n" ); document.write( "If you reverse my digits, I am divisible by 2 as well as by 6.
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document.write( "We used the \"divisible by 2\" part above. However the \"as well as by 6\" is\r\n" );
document.write( "redundant (unnecessary) because the sum of the digits is not changed by\r\n" );
document.write( "reversing the digits.  We already know that ABCD is divisible by 3, so since its\r\n" );
document.write( "reverse is both even and divisible by 3, so it is automatically divisible by 6.\r\n" );
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document.write( "So all the other information given was unnecessary.  Therefore there are 8\r\n" );
document.write( "solutions.\r\n" );
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document.write( "Edwin
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