Question 612620
Two ways to do it:


Let the number be 10a+b, where a and b are digits. The "reversed" number is 10b+a, and 


*[tex \LARGE (10a+b) + (10b+a) = 88 \Rightarrow 11(a+b) = 88 \Rightarrow a+b = 8]


Therefore 17,26,35,44,53,62,71 are the only possible integers. 35 and 53 are the only ones that differ by two, so the answer is {35,53}.


The other way to do it is let the number be 10a + (a+2); switching the digits gives 10(a+2) + a. Do the same method as shown above.