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Let x be a 4-digit number abcd. Let x* be a smallest 4-digit number using a, b, c, d also. If x-x*=999, then how many possible 4-digit number x are there?

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there are 24 different numbers. Assume abcd is the lowest 4 digit number. That implies a is less than or equal to b, which <= c, which is <=d. so lowest digit is in thousands place, next lowest in hundreds ......  Now the lowest, we're calling abcd, added with 999 must use all the same digits as the original. abcd + 999 has interesting properties. The ones place of the sum must always be one less than the original digit and there must be a 1 carried over into the next column. for example, 68 + 9 = 77. Ones place is one less and there is a carry over of 1 to the next column. Now in our case, that carry over of 1 is added to digit c and added to 9, so 1+c+9 is the same as c + 10. That means you add 0 to c and carry over the 1 to the next column. So now the hundreds column becomes 1+b+9, which is b+10, so add 0 to b, carry the 1 to the thousands column. so the thousands column of the answer must be one more than the original digit.
4678
+999
is 5677
Notice ones column went down by one, middle two columns added 0 and thousands column went up one.
 
So, if I have to reuse the original digits, abcd + 999 must be dbca. Middle digits stay the same but first and last must change.
So the digit a must be d-1 from ones place. So a=1 and d=2 or a=2,d=3 or a=3,d=4 or a=4, d=5 or a=5,d=6 or a=6,d=7 or a=7,d=8 or a=8,d=9. 
So, for example, 1492+999 = 2491.  But 1492 is not the smallest possible number using those digits. Remember a<=b<=c<=d insures abcd is smallest possible number.
But, for example, 1_ _ 2 cannot fulfill this requirement unless we have repeated digits. so for these values of a and d there are three possibilities, 1112, 1122, and 1222. The other 7 values of a and d have the same problem and only have 3 possibilities each. So FINALLY, a and d have 8 possible values and once you pick those, there are three values of c and d for each case making a total of 24 numbers.