A two-digit number has digit sum $11$. Adding $27$ reverses its digits. Original number is

A two-digit number has digit sum $11$. Adding $27$ reverses its digits. Original number is
  1. $47$
  2. $74$
  3. $56$
  4. $65$

Solution

$10a+b+27 = 10b+a \Rightarrow b-a=3$. With $a+b=11$: $a=4, b=7$.

Asked in: IMO

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