题目

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.

提问时间：2021-04-01

答案

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