Front bit

The middle part and bits. For example, 12345.

The first place is 1.

The value of the middle part 234 is 2340. The number of digits is 5, so 12345 can be expressed as 1* 104+2340+5. Now, the number we are looking for is represented by N = A* 10n+X+B, where a.

B and n are unknowns. Exchange the first number for the last number.

The new number will become 2N = b *10n+x+a. It is 2n, because the new number is twice the original number. B must be greater than a .. n = 2n-n n = (b *10n+x+a)-(a *10n+x+b) n = (b-a) *10n-(b-a) n = (. 9s) So the possibility of n is only: 9999 ... (x2 =1999 ... 98)1999 ... 98 (x2 = 3999 ... 96) 2999 ... 97.3999 ... 96 (x2 = 7999. 5999 ... 94 (x2 =11999 ... 88) 6999 ... 93 (x2 =13999 ... 86) 7999 ... 92 (x2 =1599. So there are several problems that have not been solved.

The first number =a and the last number = B 2 (A *10000+B) = B *10000+A20000a+B =10000b+A19999a = 9999.