Denote the number by 10<em>a</em> + <em>b</em>, where both <em>a</em> and <em>b</em> are integers picked from {1, 2, 3, …, 9}.
"the difference of whose digits is 3" ==> |<em>a</em> - <em>b</em>| = 3
"4 times the number is equal to 7 times the number obtained by reversing the digits" ==> 4 (10<em>a</em> + <em>b</em>) = 7 (10<em>b</em> + <em>a</em>)
Simplifying the second equation, you get
40<em>a</em> + 4<em>b</em> = 70<em>b</em> + 7<em>a</em>
33<em>a</em> - 66<em>b</em> = 0
<em>a</em> - 2<em>b</em> = 0
Assume <em>a</em> > <em>b</em>, in which case we would have |<em>a</em> - <em>b</em>| = <em>a</em> - <em>b</em>, so
<em>a</em> - <em>b</em> = 3
Eliminate <em>a</em> to solve for <em>b</em>, then for <em>a</em> :
(<em>a</em> - <em>b</em>) - (<em>a</em> - 2<em>b</em>) = 3 - 0
<em>b</em> = 3 ==> <em>a</em> = 6
Then the original number is 63.
If instead we had assumed <em>a</em> < <em>b</em>, we would have had |<em>a</em> - <em>b</em>| = <em>b</em> - <em>a</em> = 3. Then
2 (<em>b</em> - <em>a</em>) + (<em>a</em> - 2<em>b</em>) = 2×3 + 0
-<em>a</em> = 6
But neither <em>a</em> nor <em>b</em> can be negative, so this case is moot.
