Question

如果1/A+1/B+1/C=1/2005,其中ABC为正整数,而且AB为不相同的四位数,C为五位数,求满足条件的ABC.

Answer 1

显然 1/4010 + 1/4010 = 1/2005

不妨令A = 4010,B > A,求使得1/B + 1/C = 1/4010 = 1/A成立的B、C.

可令：B = A + NA/M = (M + N)A/M = 4010(M + N)/M

显然M是A即4010的因数.

则有：

1/ [4010(M + N)/M] + 1/C = 1/4010

1/C = 1/4010 - M/4010(M+N) = N / 4010(M + N) = 1/ [4010(M + N)/N]

C = 4010(M + N)/N

N是4010(M + N)的因数

按B是4位数、C是5位数等条件,可知M/N> 10000/4010 = 1.49

则根据4010 = 10*401,可得如

M = 10,N = 1

M = 10,N = 2

M = 10,N = 4

M = 10,N = 5

M = 10,N = 10 (不符M/N范围)

……

对应的成立式子有：

1/A + 1/B + 1/C = 1/2005

1/4010 + 1/4411 + 1/44110 = 1/2005

1/4010 + 1/4812 + 1/24060 = 1/2005

1/4010 + 1/5614 + 1/14035 = 1/2005

1/4010 + 1/6015 + 1/12030 = 1/2005

当A > 4010时应有解,讨论将更复杂