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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.
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.