document.write( "Question 962626: In a 3 digit number unit digit is one more than the hundred digit are in a ratio 1:2:3 .if the diff of original number and the number obtained by reversing the digit is 594 find the number \n" ); document.write( "
Algebra.Com's Answer #588168 by ankor@dixie-net.com(22740)\"\" \"About 
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let a = the hundreds digit
\n" ); document.write( "let b = the tens
\n" ); document.write( "let c = the units
\n" ); document.write( "In a 3 digit number, unit digit is more than the hundred digit,
\n" ); document.write( "c > a
\n" ); document.write( ":
\n" ); document.write( " are in a ratio 1:2:3; therefore:
\n" ); document.write( "b = 2a
\n" ); document.write( "c = 3a
\n" ); document.write( ":
\n" ); document.write( "if the diff of original number and the number obtained by reversing the digit is 594,
\n" ); document.write( "Reverse number - original number 594
\n" ); document.write( "(100c + 10b + a) - (100a + 10b + c) = 594
\n" ); document.write( "100c + 10b + a - 100a - 10b - c = 594
\n" ); document.write( "Combine like terms
\n" ); document.write( "99c - 99a = 594
\n" ); document.write( "simplify, divide by 99
\n" ); document.write( "c - a = 6
\n" ); document.write( "we know c = 3a
\n" ); document.write( "3a - a = 6
\n" ); document.write( "2a = 6
\n" ); document.write( "a = 3
\n" ); document.write( "then
\n" ); document.write( "b = 2(3) = 6
\n" ); document.write( "c = 3(30 = 9
\n" ); document.write( ":
\n" ); document.write( " find the number: 369 is the original number
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "See if that checks out
\n" ); document.write( "963 - 369 = 594 \n" ); document.write( "
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