A three-digit number 'xyz' is such that the number equals x!+y!+z!.Find the difference of the number formed by reversing its digits and the original.
Interesting problem
145 does the trick.
$\displaystyle 1!=1$
$\displaystyle 4!=24$
$\displaystyle 5!=120$
$\displaystyle 1!+4!+5!=145$
if you just look 7! is more that 3 digits, so none of the numbers can be 7.
$\displaystyle 6!=720$ which means the number would have to have at least a 7 in the first place, which means it is too big by above.
This narrows your choices to 1,2,3,4,5.
If you use all 4 or lower its not enough to get up into the 3 digits.
So you know there has to be a 5 in there somewhere. From there I just played around with it.