document.write( "Question 1194539: Hi
\n" ); document.write( "In a 3 digit number the units digit is 3 more than the tens digit and the sum of the digits is 11. If the units and hundreds digits are interchanged the number increased by 99.
\n" ); document.write( "What is original number.
\n" ); document.write( "Thanks
\n" ); document.write( "

Algebra.Com's Answer #826769 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "When the digits of a 3-digit number are reversed, the difference between the new number and the original number is 99 times the difference between the units digit and hundreds digit.

\n" ); document.write( "In this problem, the number is increased by 99 when the digits are reversed; that means in the original number the hundreds digit is 1 less than the units digit.

\n" ); document.write( "Then, given that the units digit is 3 more than the tens digit, we have

\n" ); document.write( "x = tens digit
\n" ); document.write( "x+3 = units digit (3 more than the tens digit)
\n" ); document.write( "x+2 = hundreds digit (1 less than the units digit)

\n" ); document.write( "The sum of the digits is 11:

\n" ); document.write( "x+x+3+x+2 = 11
\n" ); document.write( "3x+5=11
\n" ); document.write( "3x=6
\n" ); document.write( "x=2

\n" ); document.write( "The tens digit is x=2; the units digit is x+3=5; the hundreds digit is x+2=4.

\n" ); document.write( "ANSWER: 425

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