Diophantine reciprocals I
In the following equation x, y, and n are positive integers.
$$frac{1}{x}+frac{1}{y}=frac{1}{n}$$
For n = 4 there are exactly three distinct solutions:
$$frac{1}{5}+frac{1}{20}=frac{1}{4}$$ $$frac{1}{6}+frac{1}{12}=frac{1}{4}$$ $$frac{1}{8}+frac{1}{8}=frac{1}{4}$$
What is the least value of n for which the number of distinct solutions exceeds one-thousand?
NOTE: This problem is an easier version of Problem 110; it is strongly advised that you solve this one first.
丢番图倒数I
在如下方程中,x、y、n均为正整数。
$$frac{1}{x}+frac{1}{y}=frac{1}{n}$$
对于n = 4,上述方程恰好有3个不同的解:
$$frac{1}{5}+frac{1}{20}=frac{1}{4}$$ $$frac{1}{6}+frac{1}{12}=frac{1}{4}$$ $$frac{1}{8}+frac{1}{8}=frac{1}{4}$$
使得不同的解的数目超过1000的最小n值是多少?
注意:这个问题是第110题的简单版本;强烈推荐先解决这一题。
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