1, assuming that one digit is selected, one.
Then the inversion can be expressed as 100A, which is the so-called "seconds".
The difference between them is 99A, which is the so-called "third number".
2. Suppose you choose a two-digit number,10b+a.
Then the reverse can be expressed as 100a+ 10b, which is the so-called "seconds".
The difference between them is 99A, which is the so-called "third number".
3. Suppose you choose a three-digit number,100c+10b+a.
Then the inversion can be expressed as 100a+ 10b+c, which is the so-called "seconds".
The difference between them is | 99a-99c |, that is, 99 | a-c |, which is the so-called "third number".
As can be seen from the above three points, no matter what the initial number is, the third number is always the product of 99 and a one-digit number (for the convenience of discussion later, we assume that this one-digit number is n). This unit number can be 0, because when three digits are selected, when a = c, the "third digit" is 0.
N = 0, the third number = 99× 0 = 0, the fourth number = 0, and the final result = 0.
N = 1, the third number = 99× 1 = 99, the fourth number = 89 1, and the final result = 88209.
N = 2, the third number = 99× 2 = 198, the fourth number = 693, and the final result = 1372 14.
N = 3, the third number = 99× 3 = 297, the fourth number = 495, and the final result = 1470 15.
N = 4, the third number = 99× 4 = 396, the fourth number = 297, and the final result = 1 176 12.
N = 5, the third number = 99× 5 = 495, the fourth number = 99, and the final result = 49005.
N = 6, the third number = 99× 6 = 594, the fourth number = 99, and the final result = 58806.
N = 7, the third number = 99× 7 = 693, the fourth number = 297, and the final result = 20582 1.
N = 8, the third number = 99× 8 = 792, the fourth number = 495, and the final result = 392040.
N = 9, the third number = 99× 9 = 89 1, the fourth number = 693, and the final result = 6 17463.
It can be seen that with the different value of n, the "third number" is also different, and the final result is also different, so there is no duplication. As long as you remember the correspondence between the third number and the final result, you can judge the third number and calculate the fourth number according to the result.
or
Set unknown variables
Let one hundred digits be a, ten digits be b and one digit be C (A, b).
Then the number in the first step is100a+10b+c.
Then the second number is100c+10b+a.
Step 3: 1) If A > C (C > A Class A is not discussed), the third number is 100(A-C)+C-A=99(A-C).
Note that A-C here is nothing more than several possibilities, 1, 2, 3, 4, 5, 6, 7, 8, (9,0 is impossible, because C is not 0).
When you multiply it, you will find that the tenth digit of every third digit is 9.
Then the third number is 099,198,297,396,495,594,693,792,891990.
Take step four, step five. Actually, there are only numbers 099*990, 198*89 1, 297*792, 396*693, 495*594.
It is easy to find that the units are actually 0, 8, 4, 8, 0.
And their highest position is 9 1, 2, 2, 2.