Large! Larger!! Largest!!!

Let the number be $\overline{abcd}$ . Then on reversing the digits, we get, $\overline{dcba}$ . Since both of them are four digit numbers, we must have $9 \geq a,d \geq 1 ; 9 \geq b,c \geq 0$ .

Now, $\overline{abcd} - \overline{dcba} = 1000a + 100b + 10c + d - ( 1000d + 100c + 10b + a).$

$= 999a + 90b - 90c - 999d$

To maximize this, we must maximize the positive terms and minimize the negative terms.

Therefore, $a = b = 9, c=0, d =1$

and $9901 - 1099 = \boxed{8802}$

Let the 4 digit number be ABCD. Reversing it, we get DCBA.

Last digit can't be zero as on reversing we get a 4 digit number again.

First, we maximize the first digit of the required number. So, A has to be 9 and D has to be 1.

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Then, we maximize the second digit of the required number. So, B has to be 9 and C has to be 0.

So, ABCD = 9901

Thus, the difference is 8802.

Let $a,b,c,d$ be the digits of the 4 digit number of the form $1000a+100b+10c+d$

If the digits are reversed the number will be $1000d+100c+10b+a$

The difference of the two numbers is $999a+90b-90c-999d$ =9{111(a-b) + 10(b-c))

So the difference of the two four digit numbers depends on $(a-d)$ and $(b-c)$ .

The digit $d$ cannot be $0$ because that will make the first digit of the second number $0$

So to get the largest difference the digits should be $a=b=9$ , $d=1$ , $c=0$

When you substitute the values the difference will be $8802$

Finally I moved to Level 4 in Algebra by Solving this Problem. :) Solution: I tried larger num as number. So, Last number should not be 0. Because if it is, By reversing it becomes 3 digit. So, next possibility is 1. 3rd num can be 0. Then remaining 2 number I tried larger number i.e., 9. So, it is 9901. Reversing it, 1099. Diff 8802. Which is Max diff.