二进制的颠倒数

王守恩

二进制数。正整数A×颠倒数(A)=正整数B×颠倒数(B)。譬如:

8609375=2375(100101000111)×(111000101001)3625=2755(101011000011)×(110000110101)3125,
10171175=2527(100111011111)×(111110111001)4025=3059(101111110011)×(110011111101)3325,
30832795=4555(1000111001011)×(1101001110001)6769=4835(1001011100011)×(1100011101001)6377,
17218750=4750(1001010001110)×(1110001010010)7250=5510(1010110000110)×(1100001101010)6250,x
35034175=4775(1001010100111)×(1110010101001)7337=5539(1010110100011)×(1100010110101)6325,
38354311=4847(1001011101111)×(1111011101001)7913=5371(1010011111011)×(1101111100101)7141,
39795975=4959(1001101011111)×(1011111010011)6099=6525(1100101111101)×(1111101011001)8025,
20342350=5054(1001110111110)×(1011111100110)6118=6650(1100111111010)×(1111101110010)8050,x
41227615=5819(1011010111011)×(1101110101101)7085=5995(1011101101011)×(1101011011101)6877,
51672925=6575(1100110101111)×(1111010110011)7859=6775(1101001110111)×(1110111001011)7627,

得到一串数——2375, 2527, 4555, 4750x, 4775, 4847, 4959, 5054x, 5819, 6575, ......——OEIS没有这串数。

3组解的。2371610879733375,
=36861615(10001100100111011010101111)×(11110101011011100100110001)64338225,
=36915375(10001100110100100010101111)×(11110101000100101100110001)64244529,
=45955875(10101111010011101100100011)×(11000100110111001011110101)51606261,

4组解？5组解？6组解？还会有吗？！

王守恩

接1楼——就是这串数——2375, 2527, 4555, 4775, 4847, 4959, 5819, 6575, 9035, 9115, 9135, 9179, 9287, 9359, 9575, 9583, 9823, 11039, 11451, 13135, 17675, ...——没有偶数, 至少2组解。

目的很明确: 拿这二进制的颠倒数与十进制的颠倒数比较,会不会来点启发,十进制的颠倒数太难了！