#
Problem
of the Week

Problem
E and Solution

Not
Again

## Problem

When \(\frac{1}{50^{2023}}\) is
written as a decimal, it terminates.

What is the last non-zero digit in the decimal representation of
\(\frac{1}{50^{2023}}\)?

## Solution

Our first instinct might be to use our calculator to get an idea of
how the last digit behaves for the first few powers of \(\frac{1}{50}\). This might work for a
little while, however most calculators let us down too quickly.

Notice that \(\frac{1}{50^{2023}} =
\left(\frac{1}{50}\right)^{2023}=\left(\frac{1}{100} \times
2\right)^{2023} = (0.01\times 2)^{2023} = (0.01)^{2023}\times
2^{2023}\).

The last non-zero digit in the decimal representation of \(\frac{1}{50^{2023}}\) will therefore be the
last non-zero digit in the decimal representation of \((0.01)^{2023}\) multiplied by the last
digit of \(2^{2023}\).

Since the last non-zero digit in the decimal representation of \((0.01)^{2023}\) is \(1\), the last non-zero digit in the decimal
representation of \(\frac{1}{50^{2023}}\) will therefore be the
last digit of \(2^{2023}\).

We now examine the last digit of various powers of \(2\): $$\begin{align}
2^1=2 && 2^2=4 && 2^3=8 && 2^4=16\\
2^5=32 && 2^6=64 && 2^7=128 && 2^8=256
\end{align}$$ Notice that the last digit repeats every four
powers of \(2\). It is \(2\), then \(4\), then \(8\), then \(6\). This pattern continues, and we can
verify that \(2^9\) ends with a \(2\), \(2^{10}\) ends with a \(4\), \(2^{11}\) ends with an \(8\), \(2^{12}\) ends with a \(6\), and so on. We will leave it up to the
solver to explain *why* this pattern continues.

Now we need to determine the number of complete cycles of this
pattern before we get to \(2^{2023}\).
Since \(\frac{2023}{4} =
505\frac{3}{4}\), it follows that there are there are \(505\) complete cycles. Since \(505\times 4 = 2020\), this means \(2^{2020}\) ends with a \(6\), \(2^{2021}\) ends with a \(2\), \(2^{2022}\) ends with a \(4\), and \(2^{2023}\) ends with an \(8\).

Since \(2^{2023}\) ends with an
\(8\), \(\frac{1}{50^{2023}}\) also ends with an
\(8\).