#
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 \(7\).