Problem of the Week
Problem D and Solution
Again and Again and Again

Problem

When the fraction \(\frac{1}{70\,000\,000}\) is written as a decimal, which digit occurs in the 2023^rd place after the decimal point?

Solution

Notice that \(\frac{1}{70\,000\,000} = \frac{1}{10\,000\,000}\times \frac{1}{7} = 0.000\,000\,1\times \frac{1}{7}\).

Also, note that \(\frac{1}{7} = 0.\overline{142857}\). That is, when \(\frac{1}{7}\) is written as a decimal, the digits after the decimal point occur in repeating blocks of the \(6\) digits \(142857\).

Therefore, \(\frac{1}{70\,000\,000} = 0.000\,000\,1\times \frac{1}{7} = 0.000\,000\,1\times 0.\overline{142\,857} = 0.000\,000\,0\overline{14\,285\,7}\).

That is, when \(\frac{1}{70\,000\,000}\) is written as a decimal, the digits after the decimal point will be seven zeros followed by repeating blocks of the six digits \(142857\).

We see the decimal representation of \(\frac{1}{70\,000\,000}\) has the same repetition as that for \(\frac{1}{7}\), but the pattern is shifted over \(7\) places. Since \(2023-7=2016\), the 2023^rd digit after the decimal point when \(\frac{1}{70\,000\,000}\) is written as a decimal is the same as the 2016^th digit after the decimal point when \(\frac{1}{7}\) is written as a decimal.

Since \(\frac{2016}{6} = 336\), then the 2016^th digit after the decimal point occurs after exactly \(336\) repeating blocks of the \(6\) digits \(142857\). Therefore, the 2016^th digit is the last digit in the repeating block, which is \(7\).

The 2023^rd digit after the decimal point in the decimal representation of \(\frac{1}{70\,000\,000}\) is the same as the 2016^th digit after the decimal point in the decimal representation of \(\frac{1}{7}\), and is therefore \(7\).