157/225=1/(a+1/(b+1/(c+1/(d+1/e)))),abcde为非零自然数,a+b+c+d+e=?

Question

157/225=1/(a+1/(b+1/(c+1/(d+1/e)))),abcde为非零自然数,a+b+c+d+e=?

Answer 1

已知：
1.157/225=1/(a+1/(b+1/(c+1/(d+1/e)))),
2.abcde为非零自然数
则：
一.将等式两边取倒数,得如下等式：
225/157=a+1/(b+1/(c+1/(d+1/e)))
二.因为a是非零自然数,则a≥1且a是整数.
同时 1＜225/157＜2,并且abcde均为非零自然数,故1/(b+1/(c+1/(d+1/e)))＜1
由此得如下结果：
a = 1,68/157 = 1/(b+1/(c+1/(d+1/e)))
重复上面两步可得出abcde的值分别是1、2、3、4、5,于是得a+b+c+d+e=15