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We have learnt how to find the unit digits of base ending with the numbers \(0\), \(1\), \(4\), \(5\), \(6\) and \(9\).
We shall learn how to find the unit digits of the base ending with the numbers \(2\), \(3\), \(7\) and \(8\).
Let us observe the pattern for the base ending with the number \(2\).
Pattern for the number \(2\):
\(2^1 = 2\)
\(2^2 = 2 \times 2 = 4\)
\(2^3 = 2 \times 2 \times 2 = 8\)
\(2^4 = 2 \times 2 \times 2 \times 2 = 16\)
\(2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32\)
Here, in the above pattern, we can see that \(2^5\) and \(2^1\) has the same unit digit \(2\). Therefore, after every \(4^{th}\) multiplication, the unit digit will be \(2\).
Pattern for the number \(3\):
\(3^1 = 3\)
\(3^2 = 3 \times 3 = 9\)
\(3^3 = 3 \times 3 \times 3 = 27\)
\(3^4 = 3 \times 3 \times 3 \times 3 = 81\)
\(3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243\)
We can see that \(3^5\) and \(3^1\) has the same unit digit \(3\). Therefore, after every \(4^{th}\) multiplication, the unit digit will be \(3\).
Pattern for the number \(7\):
\(7^1 = 7\)
\(7^2 = 7 \times 7 = 49\)
\(7^3 = 7 \times 7 \times 7 = 343\)
\(7^4 = 7 \times 7 \times 7 \times 7 = 2401\)
\(7^5 = 7 \times 7 \times 7 \times 7 \times 7 = 16807\)
Thus, in \(7^5\) and \(7^1\) has the same unit digit \(7\). Therefore, after every \(4^{th}\) multiplication, the unit digit will be \(7\).
Pattern for the number \(8\):
\(8^1 = 8\)
\(8^2 = 8 \times 8 = 64\)
\(8^3 = 8 \times 8 \times 8 = 512\)
\(8^4 = 8 \times 8 \times 8 \times 8 = 4096\)
\(8^5 = 8 \times 8 \times 8 \times 8 \times 8 = 32768\)
Thus, in \(8^5\) and \(8^1\) has the same unit digit \(8\). Therefore, after every \(4^{th}\) multiplication, the unit digit will be \(8\).
Example:
1. Find the unit digit of the number \(2^{294}\).
Solution:
Let us divide the power \(294\) by \(4\). We have the remainder as \(2\).
That is, in the pattern, \(2^2 = 2 \times 2 = 4\).
Hence, the unit digit is \(4\).
2. Find the unit digit of the number \(3^{596}\).
Solution:
Let us divide the power \(596\) by \(4\). We have the remainder as \(0\).
Since, after every \(4^{th}\) multiplication, the unit digit will be \(3\). Thus, when the remainder is \(0\), our answer should be \(4^{th}\) power.
That is, \(3^4 = 3 \times 3 \times 3 \times 3 = 81\)
Hence, the unit digit is \(1\).
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