Find two numbers $ a$ and $b $ such that the digits of b are the same digits as a in another order and the number $a − b$ has all the digits equal to $1$?
What I thought:
A first natural serial thought where each digit of the corresponding $ a $ number in the $ b $ number is one more, but when we have to put the largest $ a $ digit in $ b $ we realize that subtraction is not possible to give $ 1. $. So another idea is to use the next digit in your favor. For example, when we subtract $ 80 - 79 = 1 $, because if we weigh the subtraction of each house individually, we realize that $ 0 $ uses a dozen of $ 8 $ to subtract it. Thinking this way and doing some examples to get a better view we get the desired $ a $ and $ b $ numbers. Let $ a = $ $987654320$, $ b =$ $ 876543209 $. It is easy to see that $ a $ and $ b $ have the same digits and they are not in the same order. It is also evident after subtracting that reasoning makes sense, since $ a - b = 111111111 $. I don't know if I could clearly show my thinking.