<PRE><B>Question 154659</B>
When the digits of the two digit number are reversed, the new number is 9 more than the original number, and the sum of the digits of the original number is 11. What is the original number?
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Let the original number be 10t+u where t is the tens unit and u is the units digit.
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If you reverse the digits you get 10u+t
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EQUATION:
10u+t - (10t+u) = 9
t + u = 11
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Rearrange:
-9t + 9u = 9
t + u = 11
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Modify:
t -u = -1
t+u = 11
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Add to solve for t
2t = 10
t = 5
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since t+u = 11, u = 6
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The original number is 56
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Cheers,
Stan H.</PRE>