Which one among the following digits can never be at the units place i

Which one among the following digits can never be at the units place in $(273)^n$, where $n$ is a positive integer?

1
3
5
7
This question was previously asked in
UPSC CISF-AC-EXE – 2024
The correct answer is 5.
The units digit of a number raised to a power is determined solely by the units digit of the base number and the power. In this case, the base number is 273, and its units digit is 3. So, the units digit of $(273)^n$ is the same as the units digit of $3^n$, where n is a positive integer.
Let’s examine the pattern of the units digits of powers of 3:
$3^1$: Units digit is 3.
$3^2$: Units digit of $3 \times 3 = 9$.
$3^3$: Units digit of $9 \times 3 = 27$, which is 7.
$3^4$: Units digit of $7 \times 3 = 21$, which is 1.
$3^5$: Units digit of $1 \times 3 = 3$. (The pattern repeats)
$3^6$: Units digit of $3 \times 3 = 9$.
The cycle of the units digits of powers of 3 is 3, 9, 7, 1. This cycle has a length of 4.
For any positive integer n, the units digit of $3^n$ will be one of these four digits: 1, 3, 7, or 9.
Let’s check the given options:
A) 1: Possible (e.g., for $n=4$)
B) 3: Possible (e.g., for $n=1$)
C) 5: Not in the cycle {3, 9, 7, 1}
D) 7: Possible (e.g., for $n=3$)
Therefore, the digit 5 can never be at the units place in $(273)^n$ for any positive integer n.
The units digits of powers of any single digit follow a repeating pattern (a cycle). For example, powers of 2 have units digits 2, 4, 8, 6, 2, 4, 8, 6, … (cycle length 4). Powers of 7 have units digits 7, 9, 3, 1, 7, 9, 3, 1, … (cycle length 4). Powers of 0, 1, 5, 6 have a cycle length of 1 (always 0, 1, 5, 6 respectively). Powers of 4 and 9 have a cycle length of 2 (4, 6, 4, 6, … and 9, 1, 9, 1, …). The units digit of a number like 273 raised to a power is determined only by the units digit of 273, which is 3.
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