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Let the number be (ab)_{10}=10a+b
Number obtained by interchanging digits is (ba)_{10}=10b+a

Take the difference and set it equal to 54

Notice that a=9 gives b=3, then the possible maximum sum of digits is 9+3=12.

So sum of the digits = \boxed{14} is never possible as 14>{12}.

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why did u take 9+3 instead of 8+2
good question, whats the maximum value of "a" ?
possible maximum sum means here u take digits 0 to 9 among these digits 9 is the big one n maximum am i right what is the logic here could u pls help me out of this qn
thats right, you can never get a sum greater than 12 as the maximum possible sum is 12