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Problem 40

1. Problem

Champernowne's Constant

An irrational decimal fraction is created by concatenating the positive integers:

\[0.12345678910\color{red}{1}112131415161718192021...\]

It can be seen that the 12th digit of the fractional part is 1.

If dn represents the nth digit of the fractional part, find the value of the following expression.

\[d_1 \cdot d_{10} \cdot d_{100} \cdot d_{1000} \cdot d_{10000} \cdot d_{100000} \cdot d_{1000000}\]

钱珀瑙恩数

将所有正整数连接起来构造的一个十进制无理数如下所示:

\[0.12345678910\color{red}{1}112131415161718192021...\]

可以看出小数点后第 12 位数字是 1。

如果 dn 表示上述无理数小数点后的第 n 位数字,求下式的值:

\[d_1 \cdot d_{10} \cdot d_{100} \cdot d_{1000} \cdot d_{10000} \cdot d_{100000} \cdot d_{1000000}\]

2. Solution

易知从 1 到 9 有 9 个数,从 10 到 99 有 90 个数,从 100 到 999 有 900 个数,从 1000 到 9999 有 9000 个数,从 10000 到 99999 有 90000 个数。

这一题不需要通过编程语言来解决,我们很容易找到数字与位置的对应关系:(下面的下标表示从左到右取数字的位置,从 0 开始)

  • \(A_n = n\), \(1 \le n \le 9\)
  • \(A_n = \{10 + \lfloor\frac{n-10}{2}\rfloor\}_{[(n - 10)\%2]}\), \(10 \le n \le 189\)
  • \(A_n = \{100 + \lfloor\frac{n-190}{3}\rfloor\}_{[(n - 190)\%3]}\), \(190 \le n \le 2889\)
  • \(A_n = \{1000 + \lfloor\frac{n-2890}{4}\rfloor\}_{[(n - 2890)\%4]}\), \(2890 \le n \le 38889\)
  • \(A_n = \{10000 + \lfloor\frac{n-38890}{5}\rfloor\}_{[(n - 38890\%5]}\), \(38890 \le n \le 488889\)
  • \(A_n = \{100000 + \lfloor\frac{n-488890}{6}\rfloor\}_{[(n - 488890)\%6]}\), \(488890 \le n \le 5888889\)

根据上面的公式有:

  • \(d_1 = 1\), \(d_{10} = 1\), \(d_{100} = 5\)
  • \(d_{1000} = 3\), \(d_{10000} = 7\), \(d_{100000} = 2\)
  • \(d_{1000000} = 1\)

结果为 \(1 \cdot 1 \cdot 5 \cdot 3 \cdot 7 \cdot 2 \cdot 1 = 210\)