...Y),若正整数A,B满足A#B=888,则有序对(A,B)共有多少对??

发布网友发布时间：2024-10-03 05:04

共2个回答

热心网友时间：2024-10-21 13:58

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)

热心网友时间：2024-10-21 13:52

A#B=2(2AB-A-B)=888
2AB-A-B=444
2AB-(A+B)=444
A+B=2AB-444 2AB=A+B+444
AB同奇或同偶
2AB-A=444-B
A(2B-1)=444-B
A=(444-B)/(2B-1)=[-(2B-1)/2+443.5]/(2B+1)
=-1/2+443.5/(2B+1)
=[-1+443.5/(B+1/2)]*1/2
2A=-1+4435/(10B+5)

4435=5*887
10B+5=5 无正整数解
10B+5=887 无正整数解
所以,为0对