2025 的特殊含义

据野史记载, 2025 是一个年份(美丽的废话)。

我们都知道, 2025=45 \times 45=45^2

当然, 2025 也可以分解成 3^4 \times 5^2 ,分解过程如下(当然,这是一个可以口算出来的东西,不过我们可以利用上述结论来推算出来):

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,45^2=2025
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5 \times 9)^2=2025
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5^2 \times 9^2=2025
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9^2 \times 5^2=2025
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3^2)^2 \times 5^2=2025
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3^{2 \times 2} \times 5^2=2025
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3^4 \times 5^2=2025

可得 3^4 \times 5^2=2025 。是的,我们学过数论的都知道, 2025 的因数个数为:
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4+1) \times (2+1)=5 \times 3=15

而又因为 2025 的组成 35 是最小的两个奇质数,所以 2025 为最小的拥有 15 个因数的奇数。

拓展:有 15 个因数的最小数为 144 ,可以分解为 2^4 \times 3^2144 的因数分别是 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1、2、3、4、6、8、9、12、16、18、24、36、48、72、144。‌

接下来,我们的目光来到 n^3 上, 2025 虽然好像跟三次方没一点关系,不过他的确为 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 的和。

这就是 2025 的特殊含义,还有什么含义可以下面留言我来补充。

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2025 还是拥有 15 个因数的最小奇数

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但是你并没有说他是最小的!

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image

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e呵呵眼瞎

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把它倒过来再除以10,向下取整

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520.2?

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我来补充一下:

1+2+3+……+63=2025

2025 分成 2025 (20+25)^2=2025

2025 还是个勾股数: 27^2+36^2=45^2=2025

45^2=2025=1012+1013,45^2+1012^2=1013^2

还有一个更冷门的:如果你仔细算一算你会发现 2025 等于 99 乘法表中元素的总和

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@2345安全卫士 写了这么多,给个解决方案呗(逃

如果不够多我还能写

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最后一个可以用 2025=1^3+2^3+...+9^3 来推出来

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是的

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求最小有 n 个因数的数可以在 O(n) 的时间复杂度内解决 。
首先用线性筛筛出 <n 的所有质数。接着枚举 n 的因数(忽略 1 )。
接下来考虑 \text{dp}
定义 \text{dp} 数组与 \text{maxPrimeIdx} 数组(为了大家理解意思我故意写这么长的)。
其中 \text{maxPrimeIdx} 数组储存的是质数的下标。
然后看状态转移方程:

dp_i=\min\{dp_j\times (p_{maxPrimeIdx_{j+1}})^{(\dfrac{d_i}{d_j}-1)}\} \,\,\,\,\,\,\,\,\, (d_j\mid d_i)\\ maxPrimeIdx_i=maxPrimeIdx_j+1

枚举防式跟完全背包差不多。

我课上想出来的,不知道对不对。

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另外:
2520110 的最小公倍数,与 2025 只是改变了数字顺序

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对,这个我知道,等到2520年说吧[doge]

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对,这个我知道,等我的第 5 代出生了,再说吧(

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我上网搜了一下,似乎求最小有 n 个因数的数都是 \text{DFS} 做的,没搜到 \text{DP}

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也是搜到了

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发现 luogu 上有这道题,还是蓝题(P1128)

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2025继续加油(超过ZZY!)

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