Question 908907

Let's call the first digit "{{{D}}}". Then the second digit must be {{{7-D}}}, so that the two digits add up to {{{7}}}. 

 The value of the original number is ten times the tens digit, plus the units digit (for example, 48 is 4*10+8). 

So, the original number is {{{10D + (7-D)}}}. 

When you reverse the digits, you get {{{10(7-D) + D}}}. If this is {{{9}}} greater, then: 
 {{{10(7-D) + D = 9 + 10D+(7-D) }}}

 Solving this for {{{D}}} gives us:
 
 {{{10(7-D) + D = 9 + 10D+(7-D) }}}
 {{{70 - 10D + D = 9 + 10D + 7 - D }}}
 {{{70 - 9D = 9 + 9D + 7}}} 
 {{{70 = 9 + 18D + 7 }}}
 {{{61 = 18D + 7}}} 
 {{{54 = 18D }}}
 {{{D = 3}}} 

 This means the other digit has to be {{{4}}}. So the number is {{{34}}}.