SOLUTION: what is the three digit number that satisfies the following condition.the tens digit is greater than the ones digit, the sum of the digits is 9 and if digits are reversed and resul

Let the number be ABC and its reverse be CBA,
Then the subtraction is:

 ABC
-CBA
 ---
 198

A must be bigger than C so that the top number will be bigger than the bottom
number.  So we will need to borrow 1 from the B at the start of the subtraction.

That will reduce the B in the top middle digit to 1 less than the B just below
it.

Therefore we will also need to borrow 1 from the A as well.

If we had not needed to borrow 1 from the A, then A-C would have been 1 to
get the 1 in 198.  But since we must borrow 1 from A, A-C must be 2
instead, so that borrowing 1 from A would give the 1 in 198.

A-C cannot be as high as 6-4 because all three digits must have sum 9,
and 6 and 4 alone have sum 10.  So we try lower digits.

We try A-C = 5-3, making the number 513 since the digits must have sum 9.
But that won't do because the tens digit is less than the ones digit.

We try A-C = 4-2. Then the number would be 432 since the digits must have sum 9.
That's one solution because the tens digit is greater than the ones digit, and
432-234=198.

We try A-C = 3-1. Then the number would be 351 since the digits must have sum 9. 
That's another solution because the tens digit is greater than the ones digit,
and 351-153 = 198.

So there are TWO solutions, 432 and 351

 432          351
-234         -153
----         ----
 198          198

Edwin