# 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 2023rd 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 2023rd digit after the decimal point when $$\frac{1}{70\,000\,000}$$ is written as a decimal is the same as the 2016th digit after the decimal point when $$\frac{1}{7}$$ is written as a decimal.

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

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