Discussion
The last two digits of \(3^{123}\) are :
Answer: C
The last two digits are 27.
Work modulo 100. \(3^{10} = 59049 \equiv 49\), so \(3^{20} \equiv 49^{2} = 2401 \equiv 1\). Since \(123 = 6 \cdot 20 + 3\),
\[ 3^{123} = (3^{20})^{6} \cdot 3^{3} \equiv 1 \cdot 27 = 27 \]
The other endings belong to other exponents: \(3^{3} \equiv 27\), \(3^{7} \equiv 87\), \(3^{11} \equiv 47\), \(3^{19} \equiv 67\) (mod 100).
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