Question

对任意两个正整数X.Y,定义一个运算#,为X#Y=2(2XY-X-Y),若正整数A,B满足A#B=888,则有序对(A,

Answer 1

A#B

= 2(2*A*B - A - B) = 888

2*A*B - A - B = 444

(2B - 1)A = 444 + B

A = (444 + B)/(2B - 1) = 1/2 * (2B - 1 + 889)*(2B - 1) = 1/2 [1 + 889/(2B - 1)]

即推得2B - 1 整除889,且除得的商是奇数.

889=1×7×127

则

①2B - 1 = 1

B = 1、A = 445

②2B - 1 = 7

B = 4、A = 64

③2B - 1 = 127

B = 64、A = 4

④2B - 1 = 889

B = 445、A = 1

综上(A,B)有4对(445,1)；(4,64)；(64,4)；(1,445)