Discussion
What is the last digit in \(3^{55}\) ?
Answer: D
The last digit of \(3^{55}\) is 7.
Last digits of powers of 3 repeat with period 4:
\[ \begin{aligned} &3^{1} = 3,\ 3^{2} = 9,\ 3^{3} = 27, \\ &3^{4} = 81,\ 3^{5} = 243, \ldots \end{aligned} \]
so the cycle of last digits is 3, 9, 7, 1. Since \(55 = 4 \times 13 + 3\), \(3^{55}\) has the same last digit as \(3^{3}\), which is 7. In congruence form, \(3^{4} \equiv 1 \pmod{10}\), so \(3^{55} = (3^{4})^{13} \cdot 3^{3} \equiv 27 \equiv 7\).
The digit 1 would need an exponent divisible by 4 and 3 an exponent of the form \(4k + 1\). The digit 5 never occurs, because no power of 3 is divisible by 5.
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