document.write( "Question 1008650: The sum of the digits of a three-digit number is 11. The tens digit is three times the Hundreds digit and twice the units digit. Find the original number.\r
\n" ); document.write( "\n" ); document.write( "Help! I don't really know on how to do this.. ._.
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Algebra.Com's Answer #624237 by maxitee(11)\"\" \"About 
You can put this solution on YOUR website!
Let's take the digits in the number to be A B and C.
\n" ); document.write( " From the question,
\n" ); document.write( " A + B + C = 11\r
\n" ); document.write( "\n" ); document.write( "A=> Hundreds,
\n" ); document.write( "B=> Tens
\n" ); document.write( "C= Units.\r
\n" ); document.write( "\n" ); document.write( " B = 3A ( tens digit is three times the hundreds digit.. As in the question)
\n" ); document.write( " B = 2C (tens digit is twice the unit digit.. As in the question)\r
\n" ); document.write( "\n" ); document.write( " This therefore means that 3A = 2C = B.
\n" ); document.write( "To relate all elements together, we have.
\n" ); document.write( "C = 3A/2
\n" ); document.write( "A= 2C/3
\n" ); document.write( "B= 3A or B = 2C.\r
\n" ); document.write( "\n" ); document.write( " From the first equation, (A+B+C=11)
\n" ); document.write( "If we substitute the values of B and C in terms of A, we have the following;\r
\n" ); document.write( "\n" ); document.write( "\"A+%2B+3A+%2B+3A%2F2+=+11\"\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "\"4A%2B3A%2F2+=+11\"\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "\"11A%2F2+=11\"\r
\n" ); document.write( "\n" ); document.write( "\"11A+=+22\"\r
\n" ); document.write( "\n" ); document.write( "\"A+=+2\"\r
\n" ); document.write( "\n" ); document.write( "If A = 2, Then B = 6, Making C = 3.\r
\n" ); document.write( "\n" ); document.write( " The original number is 263\r
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\n" ); document.write( "\n" ); document.write( "(Engr. Terry, Nigeria)
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