Question 783453
Let the digits be
{{{n}}}, {{{n+1}}} and {{{n+2}}}
The largest possible number is
the one with the largest digit ({{{n+2}}}) in the hundreds place,
the smaller digit ({{{n}}}) in the ones place,
and the intermediate digit ({{{n+1}}}) in the tens place.
Its value is
{{{(n+2)*100+(n+1)*10+n=n*100+2*100+n*10+1*10+n=100n+200+10n+10+n=111n+210}}}
To get the smallest possible number we reverse the order,
placing {{{n}}} in the hundreds place,
{{{n+1}}} in the tens place, and {{{n+2}}} in the ones place.
The value of that samllest number is
{{{n*100+(n+1)*10+(n+2)=100n+n*10+1*10+n+2=100n+10n+10+n+2=111n+12}}}
The difference is
{{{111n+210-(111n+12)=111n+210-111n-12=210-12=highlight(198)}}}
 
EXAMPLE:
With the digits 5, 6, and 7,
the largest number we can make is 765;
the smallest is 567,
and the difference is
765 - 567 = 198