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,B)共有多少对??

显示全部楼层 · 2012-3-12 18:51:47

A#B= 2(2*A*B - A - B) = 8882*A*B - A - B = 444(2B - 1)A = 444 + BA = (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 = 1B = 1、A = 445②2B - 1 = 7B = 4、A = 64③2B - 1 = 127B = 64、A = 4④2B - 1 = 889B = 445、A = 1综上(A,B)有4对(...

千问 · 2012-3-12 18:51:47

A#B=2(2AB-A-B)=8882AB-A-B=4442AB-(A+B)=444A+B=2AB-444 2AB=A+B+444AB同奇或同偶2AB-A=444-BA(2B-1)=444-BA=(444-B)/(2B-1)=[-(2B-1)/2+443.5]/(2B+1) =-1/2+443.5/(2B+1) =[-...