1989 AIME 第 9 题

先试着解答 1989 AIME 第 9 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1989 AIME 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

9.

19601960 年代,三位美国数学家证明存在正整数 nn 使 1335+1105+845+275=n5,133^5+110^5+84^5+27^5=n^5\text{,}从而推翻了欧拉的一个猜想。求 nn 的值。
One of Euler’s conjectures was disproved in the 19601960s by three American mathematicians when they showed there was a positive integer nn such that 1335+1105+845+275=n5.133^5+110^5+84^5+27^5=n^5. Find the value of n.n.
答案:144
知识点:估算指数整数运算
小提示:
估算五次方根,以缩小 nn 的整数取值范围
Estimate the fifth root to narrow the possible integer values of nn
大提示:
通过反复平方和乘法计算各个五次方,再将它们的和与附近的候选值比较
Evaluate the fifth powers by repeated squaring and multiplication, then compare their sum with the nearby candidate
解答:
直接进行整数运算,得到 1335=41,615,795,893,1105=16,105,100,000,845=4,182,119,424,275=14,348,907。\begin{aligned}133^5&=41{,}615{,}795{,}893,\\110^5&=16{,}105{,}100{,}000,\\84^5&=4{,}182{,}119{,}424,\\27^5&=14{,}348{,}907\end{aligned}\text{。}它们的和为 61,917,364,22461{,}917{,}364{,}224。反复相乘也可得 1445=61,917,364,224144^5=61{,}917{,}364{,}224,所以正整数 nn 为 144144。
Direct integer arithmetic gives 1335=41,615,795,893,1105=16,105,100,000,845=4,182,119,424,275=14,348,907.\begin{aligned}133^5&=41{,}615{,}795{,}893,\\110^5&=16{,}105{,}100{,}000,\\84^5&=4{,}182{,}119{,}424,\\27^5&=14{,}348{,}907.\end{aligned} Their sum is 61,917,364,224.61{,}917{,}364{,}224. Repeated multiplication also gives 1445=61,917,364,224,144^5=61{,}917{,}364{,}224, so the positive integer nn is 144.144.
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