BCDEF / F = ABCDE
(BCDE*10 + F)/F = ABCDE
BCDE*10/F + 1 = ABCDE
显然BCDE*10能被F整除.
令BCDE0 = X*F,则有：
X*F + F = F * (10000*A + X*F/10)
X + 1 = 10000A + X*F/10
10X + 10 = 100000A + XF
(10 - F)X = 100000A - 10
X = (100000A - 10) / (10 - F)
或写成X = 10(10000A - 1) / (10 - F)
一、
当A ≥2时,F必然须小于5
分别代入F = 2、3、4,只得一组F = 3,A = 2
此时X = 28570
有BCDE = 8571,等式
85713 = 3 * 28571
二、
当A = 1时,分别代入F = 2、3、……9,有两组解,解略.得等式：
66664 = 4 * 16666
99995 = 5 * 19999